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# Chapter 8 Partial Differential Equations - msharpmath · the Umbrella ‘pde’ for steady and...

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[100] 008 Chapter 8 Partial Differential Equations, Tutorial by www.msharpmath.com 1 [100] 008 revised on 2012.10.30 cemmath The Simple is the Best Chapter 8 Partial Differential Equations 8-1 Initial-Boundary Value Problems (Initial BVPs) 8-2 Steady Elliptic PDEs 8-3 System of Steady Elliptic PDEs 8-4 Data Extraction 8-5 Transient Elliptic PDEs 8-6 Summary of Syntax 8-7 References In this chapter, we treat the partial differential equations (PDEs) in two different ways. One is using the Umbrella bvpfor parabolic-elliptic partial differential equations (i.e., initial-boundary value problems). The other is using the Umbrella pdefor steady and transient elliptic PDEs of second order. Section 8-1 Initial-Boundary Value Problems (Initial BVPs) Systems of parabolic and elliptic partial differential equations (PDEs) in one spatial variable and time are frequently called the initial-boundary value problems (initial BVPs). Prevailing methods of solving initial BVPs are based on repeated solutions of BVPs with varying initial guesses. In this regard, initial BVPs are treated here. Syntax for Initial BVPs. The syntax for initial BVPs is bvp .@t[n=101,g=1](ta,tb) .x[n=101,g=1](xa,xb) { IC } ( DE, [ BCs] ) where the character @is introduced to denote the partial differentiation with
Transcript

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

1

[100] 008 revised on 20121030

cemmath

The Simple is the Best

Chapter 8 Partial Differential Equations

8-1 Initial-Boundary Value Problems (Initial BVPs)

8-3 System of Steady Elliptic PDEs

8-4 Data Extraction

8-5 Transient Elliptic PDEs

8-6 Summary of Syntax

8-7 References

In this chapter we treat the partial differential equations (PDEs) in two

different ways One is using the Umbrella lsquobvprsquo for parabolic-elliptic partial

differential equations (ie initial-boundary value problems) The other is using

the Umbrella lsquopdersquo for steady and transient elliptic PDEs of second order

Section 8-1 Initial-Boundary Value Problems (Initial

BVPs)

Systems of parabolic and elliptic partial differential equations (PDEs) in

one spatial variable and time are frequently called the initial-boundary value

problems (initial BVPs) Prevailing methods of solving initial BVPs are based

on repeated solutions of BVPs with varying initial guesses In this regard initial

BVPs are treated here

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

where the character lsquorsquo is introduced to denote the partial differentiation with

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

2

respect to time Basically three partial derivatives with respect to time

can be handled in Cemmath where the prime denotes the partial differentiation

with respect to a spatial variable In using time derivatives such as f the first

independent variable of Hub must be time and proper IC should be prescribed

inside the Hub field

Single PDE Let us consider a single PDE the formulation of which is

discussed in ref [1]

where IC and BCs are given as

A first trial to solve this example over a range of 0 2t is

gt initial BVP

gt bvp t[21](02) x[21](01) Hub span

u = sin(pix) Hub field (initial condition)

( Stem

-pipiu + u = 0 partial differential equation

[ u = 0 piexp(-t)+u = 0 ] boundary conditions

)

peep plot2[6](u) Spoke

And Figure 1 displays six curves corresponding to Spoke lsquoplot2[6](u)rsquo

Regardless of the time span defined in the Hub span Spoke lsquoplot2[n=6g=1](u)rsquo

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

3

for and otherwise The default values will be used if not

specified explicitly For this case time integration is primarily based on the

time grid specified as

bvp t[21](02)

which corresponds to t = [ 00 01 02 03 hellip 19 20 ] However time

nodes for plot are prescribed by lsquoplot2[6](u)rsquo which corresponds to t = [00

04 08 12 16 20 ] and six curves shown in Figure 1

Figure 1 Initial BVP

Using Spoke lsquoplotrsquo or lsquoplot3rsquo yields a three-dimensional plot Nevertheless

both lsquoplotrsquo and lsquoplot2rsquo cannot be used simultaneously The following

commands

gt initial BVP

gt bvp t[21](02) x[21](01)

u = sin(pix) (

pipiu - u = 0 [ u = 0 piexp(-t)+u = 0 ]

)

peep plot[6] ( u 2+01sinh(uu) )

illustrates the use of Spoke lsquoplotrsquo The results are presented in Figure 2 where

two surfaces corresponding to u and 01sinh(uu) are shown The surface on

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

4

the right of Figure 2 has no meaningful interpretation and is just prepared to

show the general capability of mathematical expressions for lsquoplotrsquo

Figure 2 3D plot for Initial BVP

Data Extraction The data fields in the numerical solution can be

extracted by using Spoke lsquotogorsquo For example let us consider again the PDE

discussed above

gt Data extraction for initial BVP

gt t = (02)span(4)

gt x = (01)span(5)

gt bvp [t][x] u = sin(pix) (

pipiu - u = 0

[ u = 0 piexp(-t)+u = 0 ]

) togo( X = x U = u Up = u Un = sqrt( uu+uu ) )

gt X U Up Un u

extract the fields x u u sqrt(uu+uu) and save them in corresponding

matrices as can be seen below

X =

[ 00000000 00000000 00000000 00000000 ]

[ 02500000 02500000 02500000 02500000 ]

[ 05000000 05000000 05000000 05000000 ]

[ 07500000 07500000 07500000 07500000 ]

[ 10000000 10000000 10000000 10000000 ]

U =

[ 00000000 00000000 00000000 00000000 ]

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5

[ 07071068 04088247 02419403 01492274 ]

[ 10000000 05832507 03535360 02260789 ]

[ 07071068 04323081 02865326 02032533 ]

[ 00000000 00752341 01060874 01143207 ]

Up =

[ 36568542 19112923 11221755 06826950 ]

[ 20000000 13593053 08133469 05111245 ]

[ 00000000 00361025 00794188 01036875 ]

[ -20000000 -12436432 -06154460 -02862922 ]

[ -36568542 -16129484 -08281156 -04251691 ]

Un =

[ 36568542 19112923 11221755 06826950 ]

[ 21213203 14194536 08485684 05324632 ]

[ 10000000 05843670 03623466 02487223 ]

[ 21213203 13166392 06788775 03511056 ]

[ 36568542 16147021 08348832 04402704 ]

u =

[ 00000000 06826950 ]

[ 01492274 05111245 ]

[ 02260789 01036875 ]

[ 02032533 -02862922 ]

[ 01143207 -04251691 ]

The i -th column in each matrix corresponds to the field solution at time

where for this case In addition the variable u in the

Hub field contains u = [ u u ] at the final stage of solution The actual output

for u is

u =

[ 0 068269 ]

[ 014923 051112 ]

[ 022608 010369 ]

[ 020325 -028629 ]

[ 011432 -042517 ]

where is shown in the last element of the second

column as expected from BCs In prescribing IC it is also possible to specify

derivatives by the following syntax

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6

u = [ u(x) u (x) ]

When the initial guess is given as u = u(x) derivatives of the initial guess are

evaluated internally with centered difference scheme Nevertheless it is wise to

specify the known mathematical expressions of derivatives for accuracy

improvement

Surface Plot Next example is the Black-Scholes PDE discussed in ref

[1]

where

And IC and BCs are given as

This example is solved by

gt Black-Scholes PDE

gt (rs) = (006508) k = r(s^22)

gt (ab) = (log(25)log(75)) (totf) = (05)

gt bvp t[11](totf) x[21](ab)

u = max(exp(x)-10) (

-u + u + (k-1)u - ku = 0

[ u = 0 u = (7-5exp(-kt))5 ]

)

peep plot[11](u)

And the results are shown in the left of Figure 3 Changing lsquoplotrsquo to lsquoplot2rsquo

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7

yields a 2D plot shown in the right of Figure 3

Figure 3 3D and 2D plot for initial BVP

System of PDEs An example for a system of PDEs is taken from ref [1]

The PDEs are written as

where the initial conditions for are

And the boundary conditions for are

Since the utility of Umbrella is handling equations with minimum amount of

modification it is not surprising to note that the following Cemmath commands

gt system of initial BVP

gt bvp t[5](002) x[10](01) Hub span

u = [ 1+05cos(2pix) -pisin(2pix) ]

v = [ 1-05cos(2pix) pisin(2pix) ] Hub field

(

-u + 05u + 1(1+vv) = 0 [ u = 0 u = 0 ]

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8

-v + 05v + 1(1+uu) = 0 [ v = 0 v = 0 ]

)peep

togo( U = u U1 = u V = v V1 = v KE = 05(uu+vv) )

plot[6](uv+2 4 + 05(uu+vv))

gt u v U U1 V V1 KE

produce the desired solutions The results are shown in Figure 4 and Figure 5

In particular the energy field defined as

is known to vanish as time approaches an infinity (see [ref 1]) Such a behavior

is represented in the energy field shown in Figure 5

Figure 4 Solution to a system of PDEs (u and v)

Figure 5 Solution to a system of PDEs (energy field)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

9

In addition we extracted the fields at specified time steps by lsquot[5](002)rsquo ie

t = [ 00 005 01 015 02 ]

and the values are listed as follows (only two cases are listed to save space)

U =

[ 15 12688 11677 113 1123 ]

[ 1383 12117 11398 11164 11163 ]

[ 10868 1067 10693 1082 10996 ]

[ 075 090265 098904 10428 10805 ]

[ 053015 079542 093671 10173 1068 ]

[ 053015 079542 093671 10173 1068 ]

[ 075 090265 098904 10428 10805 ]

[ 10868 1067 10693 1082 10996 ]

[ 1383 12117 11398 11164 11163 ]

[ 15 12688 11677 113 1123 ]

KE =

[ 0 0 0 0 0 ]

[ 40779 10564 025155 0059868 001424 ]

[ 9572 24796 059047 014053 0033427 ]

[ 74022 19175 045662 010867 0025849 ]

[ 11545 029908 007122 001695 00040317 ]

[ 11545 029908 007122 001695 00040317 ]

[ 74022 19175 045662 010867 0025849 ]

[ 9572 24796 059047 014053 0033427 ]

[ 40779 10564 025155 0059868 001424 ]

[ 59208e-031 40123e-051 86999e-070 30184e-087 16755e-103 ]

Note that the first column of U is exactly the same as the initial condition

Our next example is also adopted from ref [1] This problem has boundary

layers at both ends of the interval and a rapid change occurs in small time

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10

where The initial conditions for

are

And the boundary conditions for are

Due to the presence of rapid change in time and the boundary layers in space

spanning numerical grids must reflect such behavior Therefore geometrically

increasing grid spans are employed

gt boundary layers and rapid change in small time

gt tout = [0 0005 001 005 01 05 1 15 2]

gt bvpt[2114](02) x[21-11](01) u = 1 v = 0

(

-u+0024u-( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ u = 0 u = 1 ]

-v+0170v+( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ v = 0 v = 0 ]

)

relax(04) maxiter(400) peep

plot[tout](u 2+v )

It should be noted that the function

is repeated for each equation This is because our solution is based on sequential

procedure and hence iteration is imposed at each time step Another feature

worthy of note is the use of Spoke lsquoplotrsquo If a pre-defined matrix is used inside

lsquo[ ]rsquo curves are selected from the corresponding times Numerical results are

shown in Figure 6

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

11

Figure 6 Solution to a system of PDs (u and v)

Steady elliptic PDEs involving two or three independent spatial variables

are treated here For generality three spatial coordinates are considered in

explaining Umbrella lsquopdersquo Nevertheless only licensed version can handle three

spatial coordinates

Standard Form The standard form of elliptic PDEs with three spatial

coordinates is written here as

where And are all of the same

sign from elliptic nature

To designate the partial derivatives our notations are innovated

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12

Then the PDE of interest can be rewritten as

cuxxu + cuyyu^^ + cuzzu~~ + cuxu

+ cuyu^ + cuzu~ + cuu = cs

where nonlinear terms must be properly expressed as coefficients or source

terms Those terms with lsquo rsquo are participating the discretization procedure The

boundary conditions (BCs) also has a standard form

which can be rewritten as

The primary limitation of Umbrella lsquopdersquo at present is that it must be of second-

order in each coordinate

Syntax for PDE The syntax of Umbrella lsquopdersquo for steady two-

dimensional space is

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where two grid spans for x and y would be familiar to readers at this point

Therefore discussed below is the BCs in the Stem Since there exist two

independent spatial coordinates BCs must be specified in a pre-defined

sequence for consistency The BCs in the Stem are treated one-by-one for each

coordinate

[ x-left BC x-right BC y-left BC y-right BC ]

Discretization Method It is important to understand how the mesh is

constructed and the discretization is performed in solving PDEs With two

independent spatial variables (say ) the grid span in each direction is

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

13

defined

Then the nodes for solution are selected to be the center of a quadrangle

enclosed by the four points

For boundaries the nodes for solution are selected to be the center of two

adjacent points It is worthy of note that the major weakness arises at four

corners of the whole domain since the solution at these corners are extrapolated

from the inside or the boundaries

Therefore for the syntax

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

the mesh grid for computation has a dimension of This

means that the column of matrix represents

Single Elliptic PDE Our first example is very simple as shown below

which can be solved by

gt PDE of 2nd-order

gt pde x[10](01) y[10](01) (

u + u^^ + 10 = 0

[ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u 0 )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

14

The results are shown in Figure 7 Note that the computational mesh

corresponds to the field 0 and is shown in the right of Figure 7

Figure 7 Solution and mesh for a single PDE

As was mentioned above based on the pre-defined mesh

computational mesh of dimension is re-generated as shown in Figure

7 This is due to the treatment of control volume for discretizing the partial

differential equation The solution shown in the left of Figure 7 bears a

symmetry in both directions as expected

Comparison with Exact Solution Another example has an exact solution

The PDE of interest is written as

We solve this example by

gt PDE with exact solution

gt pde x[10](-11) y[10](-11) (

u + u^^ = 6

[ u = 1+2yy u = 1+2yy u = xx+2 u = xx+2 ]

)

plot( u -u+xx+2yy )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

15

togo( Error = -u+xx+2yy )

gt Error

The u field is shown in Figure 8 and the error distribution is shown in Figure 9

Figure 8 A single PDE with exact solution

Figure 9 Error distribution

It is worthy of note that data at four corners are linearly extrapolated This is the

reason why there occur noticeable discrepancies at four corners in Figure 9

Conservative Form The second-order terms in some PDEs are

frequently expressed in conservative forms Thus we introduce the following

syntax for conservative forms

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

2

respect to time Basically three partial derivatives with respect to time

can be handled in Cemmath where the prime denotes the partial differentiation

with respect to a spatial variable In using time derivatives such as f the first

independent variable of Hub must be time and proper IC should be prescribed

inside the Hub field

Single PDE Let us consider a single PDE the formulation of which is

discussed in ref [1]

where IC and BCs are given as

A first trial to solve this example over a range of 0 2t is

gt initial BVP

gt bvp t[21](02) x[21](01) Hub span

u = sin(pix) Hub field (initial condition)

( Stem

-pipiu + u = 0 partial differential equation

[ u = 0 piexp(-t)+u = 0 ] boundary conditions

)

peep plot2[6](u) Spoke

And Figure 1 displays six curves corresponding to Spoke lsquoplot2[6](u)rsquo

Regardless of the time span defined in the Hub span Spoke lsquoplot2[n=6g=1](u)rsquo

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

3

for and otherwise The default values will be used if not

specified explicitly For this case time integration is primarily based on the

time grid specified as

bvp t[21](02)

which corresponds to t = [ 00 01 02 03 hellip 19 20 ] However time

nodes for plot are prescribed by lsquoplot2[6](u)rsquo which corresponds to t = [00

04 08 12 16 20 ] and six curves shown in Figure 1

Figure 1 Initial BVP

Using Spoke lsquoplotrsquo or lsquoplot3rsquo yields a three-dimensional plot Nevertheless

both lsquoplotrsquo and lsquoplot2rsquo cannot be used simultaneously The following

commands

gt initial BVP

gt bvp t[21](02) x[21](01)

u = sin(pix) (

pipiu - u = 0 [ u = 0 piexp(-t)+u = 0 ]

)

peep plot[6] ( u 2+01sinh(uu) )

illustrates the use of Spoke lsquoplotrsquo The results are presented in Figure 2 where

two surfaces corresponding to u and 01sinh(uu) are shown The surface on

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

4

the right of Figure 2 has no meaningful interpretation and is just prepared to

show the general capability of mathematical expressions for lsquoplotrsquo

Figure 2 3D plot for Initial BVP

Data Extraction The data fields in the numerical solution can be

extracted by using Spoke lsquotogorsquo For example let us consider again the PDE

discussed above

gt Data extraction for initial BVP

gt t = (02)span(4)

gt x = (01)span(5)

gt bvp [t][x] u = sin(pix) (

pipiu - u = 0

[ u = 0 piexp(-t)+u = 0 ]

) togo( X = x U = u Up = u Un = sqrt( uu+uu ) )

gt X U Up Un u

extract the fields x u u sqrt(uu+uu) and save them in corresponding

matrices as can be seen below

X =

[ 00000000 00000000 00000000 00000000 ]

[ 02500000 02500000 02500000 02500000 ]

[ 05000000 05000000 05000000 05000000 ]

[ 07500000 07500000 07500000 07500000 ]

[ 10000000 10000000 10000000 10000000 ]

U =

[ 00000000 00000000 00000000 00000000 ]

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5

[ 07071068 04088247 02419403 01492274 ]

[ 10000000 05832507 03535360 02260789 ]

[ 07071068 04323081 02865326 02032533 ]

[ 00000000 00752341 01060874 01143207 ]

Up =

[ 36568542 19112923 11221755 06826950 ]

[ 20000000 13593053 08133469 05111245 ]

[ 00000000 00361025 00794188 01036875 ]

[ -20000000 -12436432 -06154460 -02862922 ]

[ -36568542 -16129484 -08281156 -04251691 ]

Un =

[ 36568542 19112923 11221755 06826950 ]

[ 21213203 14194536 08485684 05324632 ]

[ 10000000 05843670 03623466 02487223 ]

[ 21213203 13166392 06788775 03511056 ]

[ 36568542 16147021 08348832 04402704 ]

u =

[ 00000000 06826950 ]

[ 01492274 05111245 ]

[ 02260789 01036875 ]

[ 02032533 -02862922 ]

[ 01143207 -04251691 ]

The i -th column in each matrix corresponds to the field solution at time

where for this case In addition the variable u in the

Hub field contains u = [ u u ] at the final stage of solution The actual output

for u is

u =

[ 0 068269 ]

[ 014923 051112 ]

[ 022608 010369 ]

[ 020325 -028629 ]

[ 011432 -042517 ]

where is shown in the last element of the second

column as expected from BCs In prescribing IC it is also possible to specify

derivatives by the following syntax

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

6

u = [ u(x) u (x) ]

When the initial guess is given as u = u(x) derivatives of the initial guess are

evaluated internally with centered difference scheme Nevertheless it is wise to

specify the known mathematical expressions of derivatives for accuracy

improvement

Surface Plot Next example is the Black-Scholes PDE discussed in ref

[1]

where

And IC and BCs are given as

This example is solved by

gt Black-Scholes PDE

gt (rs) = (006508) k = r(s^22)

gt (ab) = (log(25)log(75)) (totf) = (05)

gt bvp t[11](totf) x[21](ab)

u = max(exp(x)-10) (

-u + u + (k-1)u - ku = 0

[ u = 0 u = (7-5exp(-kt))5 ]

)

peep plot[11](u)

And the results are shown in the left of Figure 3 Changing lsquoplotrsquo to lsquoplot2rsquo

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

7

yields a 2D plot shown in the right of Figure 3

Figure 3 3D and 2D plot for initial BVP

System of PDEs An example for a system of PDEs is taken from ref [1]

The PDEs are written as

where the initial conditions for are

And the boundary conditions for are

Since the utility of Umbrella is handling equations with minimum amount of

modification it is not surprising to note that the following Cemmath commands

gt system of initial BVP

gt bvp t[5](002) x[10](01) Hub span

u = [ 1+05cos(2pix) -pisin(2pix) ]

v = [ 1-05cos(2pix) pisin(2pix) ] Hub field

(

-u + 05u + 1(1+vv) = 0 [ u = 0 u = 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

8

-v + 05v + 1(1+uu) = 0 [ v = 0 v = 0 ]

)peep

togo( U = u U1 = u V = v V1 = v KE = 05(uu+vv) )

plot[6](uv+2 4 + 05(uu+vv))

gt u v U U1 V V1 KE

produce the desired solutions The results are shown in Figure 4 and Figure 5

In particular the energy field defined as

is known to vanish as time approaches an infinity (see [ref 1]) Such a behavior

is represented in the energy field shown in Figure 5

Figure 4 Solution to a system of PDEs (u and v)

Figure 5 Solution to a system of PDEs (energy field)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

9

In addition we extracted the fields at specified time steps by lsquot[5](002)rsquo ie

t = [ 00 005 01 015 02 ]

and the values are listed as follows (only two cases are listed to save space)

U =

[ 15 12688 11677 113 1123 ]

[ 1383 12117 11398 11164 11163 ]

[ 10868 1067 10693 1082 10996 ]

[ 075 090265 098904 10428 10805 ]

[ 053015 079542 093671 10173 1068 ]

[ 053015 079542 093671 10173 1068 ]

[ 075 090265 098904 10428 10805 ]

[ 10868 1067 10693 1082 10996 ]

[ 1383 12117 11398 11164 11163 ]

[ 15 12688 11677 113 1123 ]

KE =

[ 0 0 0 0 0 ]

[ 40779 10564 025155 0059868 001424 ]

[ 9572 24796 059047 014053 0033427 ]

[ 74022 19175 045662 010867 0025849 ]

[ 11545 029908 007122 001695 00040317 ]

[ 11545 029908 007122 001695 00040317 ]

[ 74022 19175 045662 010867 0025849 ]

[ 9572 24796 059047 014053 0033427 ]

[ 40779 10564 025155 0059868 001424 ]

[ 59208e-031 40123e-051 86999e-070 30184e-087 16755e-103 ]

Note that the first column of U is exactly the same as the initial condition

Our next example is also adopted from ref [1] This problem has boundary

layers at both ends of the interval and a rapid change occurs in small time

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

10

where The initial conditions for

are

And the boundary conditions for are

Due to the presence of rapid change in time and the boundary layers in space

spanning numerical grids must reflect such behavior Therefore geometrically

increasing grid spans are employed

gt boundary layers and rapid change in small time

gt tout = [0 0005 001 005 01 05 1 15 2]

gt bvpt[2114](02) x[21-11](01) u = 1 v = 0

(

-u+0024u-( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ u = 0 u = 1 ]

-v+0170v+( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ v = 0 v = 0 ]

)

relax(04) maxiter(400) peep

plot[tout](u 2+v )

It should be noted that the function

is repeated for each equation This is because our solution is based on sequential

procedure and hence iteration is imposed at each time step Another feature

worthy of note is the use of Spoke lsquoplotrsquo If a pre-defined matrix is used inside

lsquo[ ]rsquo curves are selected from the corresponding times Numerical results are

shown in Figure 6

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

11

Figure 6 Solution to a system of PDs (u and v)

Steady elliptic PDEs involving two or three independent spatial variables

are treated here For generality three spatial coordinates are considered in

explaining Umbrella lsquopdersquo Nevertheless only licensed version can handle three

spatial coordinates

Standard Form The standard form of elliptic PDEs with three spatial

coordinates is written here as

where And are all of the same

sign from elliptic nature

To designate the partial derivatives our notations are innovated

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

12

Then the PDE of interest can be rewritten as

cuxxu + cuyyu^^ + cuzzu~~ + cuxu

+ cuyu^ + cuzu~ + cuu = cs

where nonlinear terms must be properly expressed as coefficients or source

terms Those terms with lsquo rsquo are participating the discretization procedure The

boundary conditions (BCs) also has a standard form

which can be rewritten as

The primary limitation of Umbrella lsquopdersquo at present is that it must be of second-

order in each coordinate

Syntax for PDE The syntax of Umbrella lsquopdersquo for steady two-

dimensional space is

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where two grid spans for x and y would be familiar to readers at this point

Therefore discussed below is the BCs in the Stem Since there exist two

independent spatial coordinates BCs must be specified in a pre-defined

sequence for consistency The BCs in the Stem are treated one-by-one for each

coordinate

[ x-left BC x-right BC y-left BC y-right BC ]

Discretization Method It is important to understand how the mesh is

constructed and the discretization is performed in solving PDEs With two

independent spatial variables (say ) the grid span in each direction is

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

13

defined

Then the nodes for solution are selected to be the center of a quadrangle

enclosed by the four points

For boundaries the nodes for solution are selected to be the center of two

adjacent points It is worthy of note that the major weakness arises at four

corners of the whole domain since the solution at these corners are extrapolated

from the inside or the boundaries

Therefore for the syntax

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

the mesh grid for computation has a dimension of This

means that the column of matrix represents

Single Elliptic PDE Our first example is very simple as shown below

which can be solved by

gt PDE of 2nd-order

gt pde x[10](01) y[10](01) (

u + u^^ + 10 = 0

[ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u 0 )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

14

The results are shown in Figure 7 Note that the computational mesh

corresponds to the field 0 and is shown in the right of Figure 7

Figure 7 Solution and mesh for a single PDE

As was mentioned above based on the pre-defined mesh

computational mesh of dimension is re-generated as shown in Figure

7 This is due to the treatment of control volume for discretizing the partial

differential equation The solution shown in the left of Figure 7 bears a

symmetry in both directions as expected

Comparison with Exact Solution Another example has an exact solution

The PDE of interest is written as

We solve this example by

gt PDE with exact solution

gt pde x[10](-11) y[10](-11) (

u + u^^ = 6

[ u = 1+2yy u = 1+2yy u = xx+2 u = xx+2 ]

)

plot( u -u+xx+2yy )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

15

togo( Error = -u+xx+2yy )

gt Error

The u field is shown in Figure 8 and the error distribution is shown in Figure 9

Figure 8 A single PDE with exact solution

Figure 9 Error distribution

It is worthy of note that data at four corners are linearly extrapolated This is the

reason why there occur noticeable discrepancies at four corners in Figure 9

Conservative Form The second-order terms in some PDEs are

frequently expressed in conservative forms Thus we introduce the following

syntax for conservative forms

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

3

for and otherwise The default values will be used if not

specified explicitly For this case time integration is primarily based on the

time grid specified as

bvp t[21](02)

which corresponds to t = [ 00 01 02 03 hellip 19 20 ] However time

nodes for plot are prescribed by lsquoplot2[6](u)rsquo which corresponds to t = [00

04 08 12 16 20 ] and six curves shown in Figure 1

Figure 1 Initial BVP

Using Spoke lsquoplotrsquo or lsquoplot3rsquo yields a three-dimensional plot Nevertheless

both lsquoplotrsquo and lsquoplot2rsquo cannot be used simultaneously The following

commands

gt initial BVP

gt bvp t[21](02) x[21](01)

u = sin(pix) (

pipiu - u = 0 [ u = 0 piexp(-t)+u = 0 ]

)

peep plot[6] ( u 2+01sinh(uu) )

illustrates the use of Spoke lsquoplotrsquo The results are presented in Figure 2 where

two surfaces corresponding to u and 01sinh(uu) are shown The surface on

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

4

the right of Figure 2 has no meaningful interpretation and is just prepared to

show the general capability of mathematical expressions for lsquoplotrsquo

Figure 2 3D plot for Initial BVP

Data Extraction The data fields in the numerical solution can be

extracted by using Spoke lsquotogorsquo For example let us consider again the PDE

discussed above

gt Data extraction for initial BVP

gt t = (02)span(4)

gt x = (01)span(5)

gt bvp [t][x] u = sin(pix) (

pipiu - u = 0

[ u = 0 piexp(-t)+u = 0 ]

) togo( X = x U = u Up = u Un = sqrt( uu+uu ) )

gt X U Up Un u

extract the fields x u u sqrt(uu+uu) and save them in corresponding

matrices as can be seen below

X =

[ 00000000 00000000 00000000 00000000 ]

[ 02500000 02500000 02500000 02500000 ]

[ 05000000 05000000 05000000 05000000 ]

[ 07500000 07500000 07500000 07500000 ]

[ 10000000 10000000 10000000 10000000 ]

U =

[ 00000000 00000000 00000000 00000000 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

5

[ 07071068 04088247 02419403 01492274 ]

[ 10000000 05832507 03535360 02260789 ]

[ 07071068 04323081 02865326 02032533 ]

[ 00000000 00752341 01060874 01143207 ]

Up =

[ 36568542 19112923 11221755 06826950 ]

[ 20000000 13593053 08133469 05111245 ]

[ 00000000 00361025 00794188 01036875 ]

[ -20000000 -12436432 -06154460 -02862922 ]

[ -36568542 -16129484 -08281156 -04251691 ]

Un =

[ 36568542 19112923 11221755 06826950 ]

[ 21213203 14194536 08485684 05324632 ]

[ 10000000 05843670 03623466 02487223 ]

[ 21213203 13166392 06788775 03511056 ]

[ 36568542 16147021 08348832 04402704 ]

u =

[ 00000000 06826950 ]

[ 01492274 05111245 ]

[ 02260789 01036875 ]

[ 02032533 -02862922 ]

[ 01143207 -04251691 ]

The i -th column in each matrix corresponds to the field solution at time

where for this case In addition the variable u in the

Hub field contains u = [ u u ] at the final stage of solution The actual output

for u is

u =

[ 0 068269 ]

[ 014923 051112 ]

[ 022608 010369 ]

[ 020325 -028629 ]

[ 011432 -042517 ]

where is shown in the last element of the second

column as expected from BCs In prescribing IC it is also possible to specify

derivatives by the following syntax

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

6

u = [ u(x) u (x) ]

When the initial guess is given as u = u(x) derivatives of the initial guess are

evaluated internally with centered difference scheme Nevertheless it is wise to

specify the known mathematical expressions of derivatives for accuracy

improvement

Surface Plot Next example is the Black-Scholes PDE discussed in ref

[1]

where

And IC and BCs are given as

This example is solved by

gt Black-Scholes PDE

gt (rs) = (006508) k = r(s^22)

gt (ab) = (log(25)log(75)) (totf) = (05)

gt bvp t[11](totf) x[21](ab)

u = max(exp(x)-10) (

-u + u + (k-1)u - ku = 0

[ u = 0 u = (7-5exp(-kt))5 ]

)

peep plot[11](u)

And the results are shown in the left of Figure 3 Changing lsquoplotrsquo to lsquoplot2rsquo

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

7

yields a 2D plot shown in the right of Figure 3

Figure 3 3D and 2D plot for initial BVP

System of PDEs An example for a system of PDEs is taken from ref [1]

The PDEs are written as

where the initial conditions for are

And the boundary conditions for are

Since the utility of Umbrella is handling equations with minimum amount of

modification it is not surprising to note that the following Cemmath commands

gt system of initial BVP

gt bvp t[5](002) x[10](01) Hub span

u = [ 1+05cos(2pix) -pisin(2pix) ]

v = [ 1-05cos(2pix) pisin(2pix) ] Hub field

(

-u + 05u + 1(1+vv) = 0 [ u = 0 u = 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

8

-v + 05v + 1(1+uu) = 0 [ v = 0 v = 0 ]

)peep

togo( U = u U1 = u V = v V1 = v KE = 05(uu+vv) )

plot[6](uv+2 4 + 05(uu+vv))

gt u v U U1 V V1 KE

produce the desired solutions The results are shown in Figure 4 and Figure 5

In particular the energy field defined as

is known to vanish as time approaches an infinity (see [ref 1]) Such a behavior

is represented in the energy field shown in Figure 5

Figure 4 Solution to a system of PDEs (u and v)

Figure 5 Solution to a system of PDEs (energy field)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

9

In addition we extracted the fields at specified time steps by lsquot[5](002)rsquo ie

t = [ 00 005 01 015 02 ]

and the values are listed as follows (only two cases are listed to save space)

U =

[ 15 12688 11677 113 1123 ]

[ 1383 12117 11398 11164 11163 ]

[ 10868 1067 10693 1082 10996 ]

[ 075 090265 098904 10428 10805 ]

[ 053015 079542 093671 10173 1068 ]

[ 053015 079542 093671 10173 1068 ]

[ 075 090265 098904 10428 10805 ]

[ 10868 1067 10693 1082 10996 ]

[ 1383 12117 11398 11164 11163 ]

[ 15 12688 11677 113 1123 ]

KE =

[ 0 0 0 0 0 ]

[ 40779 10564 025155 0059868 001424 ]

[ 9572 24796 059047 014053 0033427 ]

[ 74022 19175 045662 010867 0025849 ]

[ 11545 029908 007122 001695 00040317 ]

[ 11545 029908 007122 001695 00040317 ]

[ 74022 19175 045662 010867 0025849 ]

[ 9572 24796 059047 014053 0033427 ]

[ 40779 10564 025155 0059868 001424 ]

[ 59208e-031 40123e-051 86999e-070 30184e-087 16755e-103 ]

Note that the first column of U is exactly the same as the initial condition

Our next example is also adopted from ref [1] This problem has boundary

layers at both ends of the interval and a rapid change occurs in small time

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

10

where The initial conditions for

are

And the boundary conditions for are

Due to the presence of rapid change in time and the boundary layers in space

spanning numerical grids must reflect such behavior Therefore geometrically

increasing grid spans are employed

gt boundary layers and rapid change in small time

gt tout = [0 0005 001 005 01 05 1 15 2]

gt bvpt[2114](02) x[21-11](01) u = 1 v = 0

(

-u+0024u-( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ u = 0 u = 1 ]

-v+0170v+( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ v = 0 v = 0 ]

)

relax(04) maxiter(400) peep

plot[tout](u 2+v )

It should be noted that the function

is repeated for each equation This is because our solution is based on sequential

procedure and hence iteration is imposed at each time step Another feature

worthy of note is the use of Spoke lsquoplotrsquo If a pre-defined matrix is used inside

lsquo[ ]rsquo curves are selected from the corresponding times Numerical results are

shown in Figure 6

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

11

Figure 6 Solution to a system of PDs (u and v)

Steady elliptic PDEs involving two or three independent spatial variables

are treated here For generality three spatial coordinates are considered in

explaining Umbrella lsquopdersquo Nevertheless only licensed version can handle three

spatial coordinates

Standard Form The standard form of elliptic PDEs with three spatial

coordinates is written here as

where And are all of the same

sign from elliptic nature

To designate the partial derivatives our notations are innovated

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

12

Then the PDE of interest can be rewritten as

cuxxu + cuyyu^^ + cuzzu~~ + cuxu

+ cuyu^ + cuzu~ + cuu = cs

where nonlinear terms must be properly expressed as coefficients or source

terms Those terms with lsquo rsquo are participating the discretization procedure The

boundary conditions (BCs) also has a standard form

which can be rewritten as

The primary limitation of Umbrella lsquopdersquo at present is that it must be of second-

order in each coordinate

Syntax for PDE The syntax of Umbrella lsquopdersquo for steady two-

dimensional space is

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where two grid spans for x and y would be familiar to readers at this point

Therefore discussed below is the BCs in the Stem Since there exist two

independent spatial coordinates BCs must be specified in a pre-defined

sequence for consistency The BCs in the Stem are treated one-by-one for each

coordinate

[ x-left BC x-right BC y-left BC y-right BC ]

Discretization Method It is important to understand how the mesh is

constructed and the discretization is performed in solving PDEs With two

independent spatial variables (say ) the grid span in each direction is

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

13

defined

Then the nodes for solution are selected to be the center of a quadrangle

enclosed by the four points

For boundaries the nodes for solution are selected to be the center of two

adjacent points It is worthy of note that the major weakness arises at four

corners of the whole domain since the solution at these corners are extrapolated

from the inside or the boundaries

Therefore for the syntax

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

the mesh grid for computation has a dimension of This

means that the column of matrix represents

Single Elliptic PDE Our first example is very simple as shown below

which can be solved by

gt PDE of 2nd-order

gt pde x[10](01) y[10](01) (

u + u^^ + 10 = 0

[ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u 0 )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

14

The results are shown in Figure 7 Note that the computational mesh

corresponds to the field 0 and is shown in the right of Figure 7

Figure 7 Solution and mesh for a single PDE

As was mentioned above based on the pre-defined mesh

computational mesh of dimension is re-generated as shown in Figure

7 This is due to the treatment of control volume for discretizing the partial

differential equation The solution shown in the left of Figure 7 bears a

symmetry in both directions as expected

Comparison with Exact Solution Another example has an exact solution

The PDE of interest is written as

We solve this example by

gt PDE with exact solution

gt pde x[10](-11) y[10](-11) (

u + u^^ = 6

[ u = 1+2yy u = 1+2yy u = xx+2 u = xx+2 ]

)

plot( u -u+xx+2yy )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

15

togo( Error = -u+xx+2yy )

gt Error

The u field is shown in Figure 8 and the error distribution is shown in Figure 9

Figure 8 A single PDE with exact solution

Figure 9 Error distribution

It is worthy of note that data at four corners are linearly extrapolated This is the

reason why there occur noticeable discrepancies at four corners in Figure 9

Conservative Form The second-order terms in some PDEs are

frequently expressed in conservative forms Thus we introduce the following

syntax for conservative forms

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

4

the right of Figure 2 has no meaningful interpretation and is just prepared to

show the general capability of mathematical expressions for lsquoplotrsquo

Figure 2 3D plot for Initial BVP

Data Extraction The data fields in the numerical solution can be

extracted by using Spoke lsquotogorsquo For example let us consider again the PDE

discussed above

gt Data extraction for initial BVP

gt t = (02)span(4)

gt x = (01)span(5)

gt bvp [t][x] u = sin(pix) (

pipiu - u = 0

[ u = 0 piexp(-t)+u = 0 ]

) togo( X = x U = u Up = u Un = sqrt( uu+uu ) )

gt X U Up Un u

extract the fields x u u sqrt(uu+uu) and save them in corresponding

matrices as can be seen below

X =

[ 00000000 00000000 00000000 00000000 ]

[ 02500000 02500000 02500000 02500000 ]

[ 05000000 05000000 05000000 05000000 ]

[ 07500000 07500000 07500000 07500000 ]

[ 10000000 10000000 10000000 10000000 ]

U =

[ 00000000 00000000 00000000 00000000 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

5

[ 07071068 04088247 02419403 01492274 ]

[ 10000000 05832507 03535360 02260789 ]

[ 07071068 04323081 02865326 02032533 ]

[ 00000000 00752341 01060874 01143207 ]

Up =

[ 36568542 19112923 11221755 06826950 ]

[ 20000000 13593053 08133469 05111245 ]

[ 00000000 00361025 00794188 01036875 ]

[ -20000000 -12436432 -06154460 -02862922 ]

[ -36568542 -16129484 -08281156 -04251691 ]

Un =

[ 36568542 19112923 11221755 06826950 ]

[ 21213203 14194536 08485684 05324632 ]

[ 10000000 05843670 03623466 02487223 ]

[ 21213203 13166392 06788775 03511056 ]

[ 36568542 16147021 08348832 04402704 ]

u =

[ 00000000 06826950 ]

[ 01492274 05111245 ]

[ 02260789 01036875 ]

[ 02032533 -02862922 ]

[ 01143207 -04251691 ]

The i -th column in each matrix corresponds to the field solution at time

where for this case In addition the variable u in the

Hub field contains u = [ u u ] at the final stage of solution The actual output

for u is

u =

[ 0 068269 ]

[ 014923 051112 ]

[ 022608 010369 ]

[ 020325 -028629 ]

[ 011432 -042517 ]

where is shown in the last element of the second

column as expected from BCs In prescribing IC it is also possible to specify

derivatives by the following syntax

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

6

u = [ u(x) u (x) ]

When the initial guess is given as u = u(x) derivatives of the initial guess are

evaluated internally with centered difference scheme Nevertheless it is wise to

specify the known mathematical expressions of derivatives for accuracy

improvement

Surface Plot Next example is the Black-Scholes PDE discussed in ref

[1]

where

And IC and BCs are given as

This example is solved by

gt Black-Scholes PDE

gt (rs) = (006508) k = r(s^22)

gt (ab) = (log(25)log(75)) (totf) = (05)

gt bvp t[11](totf) x[21](ab)

u = max(exp(x)-10) (

-u + u + (k-1)u - ku = 0

[ u = 0 u = (7-5exp(-kt))5 ]

)

peep plot[11](u)

And the results are shown in the left of Figure 3 Changing lsquoplotrsquo to lsquoplot2rsquo

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

7

yields a 2D plot shown in the right of Figure 3

Figure 3 3D and 2D plot for initial BVP

System of PDEs An example for a system of PDEs is taken from ref [1]

The PDEs are written as

where the initial conditions for are

And the boundary conditions for are

Since the utility of Umbrella is handling equations with minimum amount of

modification it is not surprising to note that the following Cemmath commands

gt system of initial BVP

gt bvp t[5](002) x[10](01) Hub span

u = [ 1+05cos(2pix) -pisin(2pix) ]

v = [ 1-05cos(2pix) pisin(2pix) ] Hub field

(

-u + 05u + 1(1+vv) = 0 [ u = 0 u = 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

8

-v + 05v + 1(1+uu) = 0 [ v = 0 v = 0 ]

)peep

togo( U = u U1 = u V = v V1 = v KE = 05(uu+vv) )

plot[6](uv+2 4 + 05(uu+vv))

gt u v U U1 V V1 KE

produce the desired solutions The results are shown in Figure 4 and Figure 5

In particular the energy field defined as

is known to vanish as time approaches an infinity (see [ref 1]) Such a behavior

is represented in the energy field shown in Figure 5

Figure 4 Solution to a system of PDEs (u and v)

Figure 5 Solution to a system of PDEs (energy field)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

9

In addition we extracted the fields at specified time steps by lsquot[5](002)rsquo ie

t = [ 00 005 01 015 02 ]

and the values are listed as follows (only two cases are listed to save space)

U =

[ 15 12688 11677 113 1123 ]

[ 1383 12117 11398 11164 11163 ]

[ 10868 1067 10693 1082 10996 ]

[ 075 090265 098904 10428 10805 ]

[ 053015 079542 093671 10173 1068 ]

[ 053015 079542 093671 10173 1068 ]

[ 075 090265 098904 10428 10805 ]

[ 10868 1067 10693 1082 10996 ]

[ 1383 12117 11398 11164 11163 ]

[ 15 12688 11677 113 1123 ]

KE =

[ 0 0 0 0 0 ]

[ 40779 10564 025155 0059868 001424 ]

[ 9572 24796 059047 014053 0033427 ]

[ 74022 19175 045662 010867 0025849 ]

[ 11545 029908 007122 001695 00040317 ]

[ 11545 029908 007122 001695 00040317 ]

[ 74022 19175 045662 010867 0025849 ]

[ 9572 24796 059047 014053 0033427 ]

[ 40779 10564 025155 0059868 001424 ]

[ 59208e-031 40123e-051 86999e-070 30184e-087 16755e-103 ]

Note that the first column of U is exactly the same as the initial condition

Our next example is also adopted from ref [1] This problem has boundary

layers at both ends of the interval and a rapid change occurs in small time

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

10

where The initial conditions for

are

And the boundary conditions for are

Due to the presence of rapid change in time and the boundary layers in space

spanning numerical grids must reflect such behavior Therefore geometrically

increasing grid spans are employed

gt boundary layers and rapid change in small time

gt tout = [0 0005 001 005 01 05 1 15 2]

gt bvpt[2114](02) x[21-11](01) u = 1 v = 0

(

-u+0024u-( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ u = 0 u = 1 ]

-v+0170v+( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ v = 0 v = 0 ]

)

relax(04) maxiter(400) peep

plot[tout](u 2+v )

It should be noted that the function

is repeated for each equation This is because our solution is based on sequential

procedure and hence iteration is imposed at each time step Another feature

worthy of note is the use of Spoke lsquoplotrsquo If a pre-defined matrix is used inside

lsquo[ ]rsquo curves are selected from the corresponding times Numerical results are

shown in Figure 6

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

11

Figure 6 Solution to a system of PDs (u and v)

Steady elliptic PDEs involving two or three independent spatial variables

are treated here For generality three spatial coordinates are considered in

explaining Umbrella lsquopdersquo Nevertheless only licensed version can handle three

spatial coordinates

Standard Form The standard form of elliptic PDEs with three spatial

coordinates is written here as

where And are all of the same

sign from elliptic nature

To designate the partial derivatives our notations are innovated

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

12

Then the PDE of interest can be rewritten as

cuxxu + cuyyu^^ + cuzzu~~ + cuxu

+ cuyu^ + cuzu~ + cuu = cs

where nonlinear terms must be properly expressed as coefficients or source

terms Those terms with lsquo rsquo are participating the discretization procedure The

boundary conditions (BCs) also has a standard form

which can be rewritten as

The primary limitation of Umbrella lsquopdersquo at present is that it must be of second-

order in each coordinate

Syntax for PDE The syntax of Umbrella lsquopdersquo for steady two-

dimensional space is

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where two grid spans for x and y would be familiar to readers at this point

Therefore discussed below is the BCs in the Stem Since there exist two

independent spatial coordinates BCs must be specified in a pre-defined

sequence for consistency The BCs in the Stem are treated one-by-one for each

coordinate

[ x-left BC x-right BC y-left BC y-right BC ]

Discretization Method It is important to understand how the mesh is

constructed and the discretization is performed in solving PDEs With two

independent spatial variables (say ) the grid span in each direction is

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

13

defined

Then the nodes for solution are selected to be the center of a quadrangle

enclosed by the four points

For boundaries the nodes for solution are selected to be the center of two

adjacent points It is worthy of note that the major weakness arises at four

corners of the whole domain since the solution at these corners are extrapolated

from the inside or the boundaries

Therefore for the syntax

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

the mesh grid for computation has a dimension of This

means that the column of matrix represents

Single Elliptic PDE Our first example is very simple as shown below

which can be solved by

gt PDE of 2nd-order

gt pde x[10](01) y[10](01) (

u + u^^ + 10 = 0

[ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u 0 )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

14

The results are shown in Figure 7 Note that the computational mesh

corresponds to the field 0 and is shown in the right of Figure 7

Figure 7 Solution and mesh for a single PDE

As was mentioned above based on the pre-defined mesh

computational mesh of dimension is re-generated as shown in Figure

7 This is due to the treatment of control volume for discretizing the partial

differential equation The solution shown in the left of Figure 7 bears a

symmetry in both directions as expected

Comparison with Exact Solution Another example has an exact solution

The PDE of interest is written as

We solve this example by

gt PDE with exact solution

gt pde x[10](-11) y[10](-11) (

u + u^^ = 6

[ u = 1+2yy u = 1+2yy u = xx+2 u = xx+2 ]

)

plot( u -u+xx+2yy )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

15

togo( Error = -u+xx+2yy )

gt Error

The u field is shown in Figure 8 and the error distribution is shown in Figure 9

Figure 8 A single PDE with exact solution

Figure 9 Error distribution

It is worthy of note that data at four corners are linearly extrapolated This is the

reason why there occur noticeable discrepancies at four corners in Figure 9

Conservative Form The second-order terms in some PDEs are

frequently expressed in conservative forms Thus we introduce the following

syntax for conservative forms

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

5

[ 07071068 04088247 02419403 01492274 ]

[ 10000000 05832507 03535360 02260789 ]

[ 07071068 04323081 02865326 02032533 ]

[ 00000000 00752341 01060874 01143207 ]

Up =

[ 36568542 19112923 11221755 06826950 ]

[ 20000000 13593053 08133469 05111245 ]

[ 00000000 00361025 00794188 01036875 ]

[ -20000000 -12436432 -06154460 -02862922 ]

[ -36568542 -16129484 -08281156 -04251691 ]

Un =

[ 36568542 19112923 11221755 06826950 ]

[ 21213203 14194536 08485684 05324632 ]

[ 10000000 05843670 03623466 02487223 ]

[ 21213203 13166392 06788775 03511056 ]

[ 36568542 16147021 08348832 04402704 ]

u =

[ 00000000 06826950 ]

[ 01492274 05111245 ]

[ 02260789 01036875 ]

[ 02032533 -02862922 ]

[ 01143207 -04251691 ]

The i -th column in each matrix corresponds to the field solution at time

where for this case In addition the variable u in the

Hub field contains u = [ u u ] at the final stage of solution The actual output

for u is

u =

[ 0 068269 ]

[ 014923 051112 ]

[ 022608 010369 ]

[ 020325 -028629 ]

[ 011432 -042517 ]

where is shown in the last element of the second

column as expected from BCs In prescribing IC it is also possible to specify

derivatives by the following syntax

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

6

u = [ u(x) u (x) ]

When the initial guess is given as u = u(x) derivatives of the initial guess are

evaluated internally with centered difference scheme Nevertheless it is wise to

specify the known mathematical expressions of derivatives for accuracy

improvement

Surface Plot Next example is the Black-Scholes PDE discussed in ref

[1]

where

And IC and BCs are given as

This example is solved by

gt Black-Scholes PDE

gt (rs) = (006508) k = r(s^22)

gt (ab) = (log(25)log(75)) (totf) = (05)

gt bvp t[11](totf) x[21](ab)

u = max(exp(x)-10) (

-u + u + (k-1)u - ku = 0

[ u = 0 u = (7-5exp(-kt))5 ]

)

peep plot[11](u)

And the results are shown in the left of Figure 3 Changing lsquoplotrsquo to lsquoplot2rsquo

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

7

yields a 2D plot shown in the right of Figure 3

Figure 3 3D and 2D plot for initial BVP

System of PDEs An example for a system of PDEs is taken from ref [1]

The PDEs are written as

where the initial conditions for are

And the boundary conditions for are

Since the utility of Umbrella is handling equations with minimum amount of

modification it is not surprising to note that the following Cemmath commands

gt system of initial BVP

gt bvp t[5](002) x[10](01) Hub span

u = [ 1+05cos(2pix) -pisin(2pix) ]

v = [ 1-05cos(2pix) pisin(2pix) ] Hub field

(

-u + 05u + 1(1+vv) = 0 [ u = 0 u = 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

8

-v + 05v + 1(1+uu) = 0 [ v = 0 v = 0 ]

)peep

togo( U = u U1 = u V = v V1 = v KE = 05(uu+vv) )

plot[6](uv+2 4 + 05(uu+vv))

gt u v U U1 V V1 KE

produce the desired solutions The results are shown in Figure 4 and Figure 5

In particular the energy field defined as

is known to vanish as time approaches an infinity (see [ref 1]) Such a behavior

is represented in the energy field shown in Figure 5

Figure 4 Solution to a system of PDEs (u and v)

Figure 5 Solution to a system of PDEs (energy field)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

9

In addition we extracted the fields at specified time steps by lsquot[5](002)rsquo ie

t = [ 00 005 01 015 02 ]

and the values are listed as follows (only two cases are listed to save space)

U =

[ 15 12688 11677 113 1123 ]

[ 1383 12117 11398 11164 11163 ]

[ 10868 1067 10693 1082 10996 ]

[ 075 090265 098904 10428 10805 ]

[ 053015 079542 093671 10173 1068 ]

[ 053015 079542 093671 10173 1068 ]

[ 075 090265 098904 10428 10805 ]

[ 10868 1067 10693 1082 10996 ]

[ 1383 12117 11398 11164 11163 ]

[ 15 12688 11677 113 1123 ]

KE =

[ 0 0 0 0 0 ]

[ 40779 10564 025155 0059868 001424 ]

[ 9572 24796 059047 014053 0033427 ]

[ 74022 19175 045662 010867 0025849 ]

[ 11545 029908 007122 001695 00040317 ]

[ 11545 029908 007122 001695 00040317 ]

[ 74022 19175 045662 010867 0025849 ]

[ 9572 24796 059047 014053 0033427 ]

[ 40779 10564 025155 0059868 001424 ]

[ 59208e-031 40123e-051 86999e-070 30184e-087 16755e-103 ]

Note that the first column of U is exactly the same as the initial condition

Our next example is also adopted from ref [1] This problem has boundary

layers at both ends of the interval and a rapid change occurs in small time

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

10

where The initial conditions for

are

And the boundary conditions for are

Due to the presence of rapid change in time and the boundary layers in space

spanning numerical grids must reflect such behavior Therefore geometrically

increasing grid spans are employed

gt boundary layers and rapid change in small time

gt tout = [0 0005 001 005 01 05 1 15 2]

gt bvpt[2114](02) x[21-11](01) u = 1 v = 0

(

-u+0024u-( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ u = 0 u = 1 ]

-v+0170v+( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ v = 0 v = 0 ]

)

relax(04) maxiter(400) peep

plot[tout](u 2+v )

It should be noted that the function

is repeated for each equation This is because our solution is based on sequential

procedure and hence iteration is imposed at each time step Another feature

worthy of note is the use of Spoke lsquoplotrsquo If a pre-defined matrix is used inside

lsquo[ ]rsquo curves are selected from the corresponding times Numerical results are

shown in Figure 6

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

11

Figure 6 Solution to a system of PDs (u and v)

Steady elliptic PDEs involving two or three independent spatial variables

are treated here For generality three spatial coordinates are considered in

explaining Umbrella lsquopdersquo Nevertheless only licensed version can handle three

spatial coordinates

Standard Form The standard form of elliptic PDEs with three spatial

coordinates is written here as

where And are all of the same

sign from elliptic nature

To designate the partial derivatives our notations are innovated

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

12

Then the PDE of interest can be rewritten as

cuxxu + cuyyu^^ + cuzzu~~ + cuxu

+ cuyu^ + cuzu~ + cuu = cs

where nonlinear terms must be properly expressed as coefficients or source

terms Those terms with lsquo rsquo are participating the discretization procedure The

boundary conditions (BCs) also has a standard form

which can be rewritten as

The primary limitation of Umbrella lsquopdersquo at present is that it must be of second-

order in each coordinate

Syntax for PDE The syntax of Umbrella lsquopdersquo for steady two-

dimensional space is

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where two grid spans for x and y would be familiar to readers at this point

Therefore discussed below is the BCs in the Stem Since there exist two

independent spatial coordinates BCs must be specified in a pre-defined

sequence for consistency The BCs in the Stem are treated one-by-one for each

coordinate

[ x-left BC x-right BC y-left BC y-right BC ]

Discretization Method It is important to understand how the mesh is

constructed and the discretization is performed in solving PDEs With two

independent spatial variables (say ) the grid span in each direction is

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

13

defined

Then the nodes for solution are selected to be the center of a quadrangle

enclosed by the four points

For boundaries the nodes for solution are selected to be the center of two

adjacent points It is worthy of note that the major weakness arises at four

corners of the whole domain since the solution at these corners are extrapolated

from the inside or the boundaries

Therefore for the syntax

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

the mesh grid for computation has a dimension of This

means that the column of matrix represents

Single Elliptic PDE Our first example is very simple as shown below

which can be solved by

gt PDE of 2nd-order

gt pde x[10](01) y[10](01) (

u + u^^ + 10 = 0

[ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u 0 )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

14

The results are shown in Figure 7 Note that the computational mesh

corresponds to the field 0 and is shown in the right of Figure 7

Figure 7 Solution and mesh for a single PDE

As was mentioned above based on the pre-defined mesh

computational mesh of dimension is re-generated as shown in Figure

7 This is due to the treatment of control volume for discretizing the partial

differential equation The solution shown in the left of Figure 7 bears a

symmetry in both directions as expected

Comparison with Exact Solution Another example has an exact solution

The PDE of interest is written as

We solve this example by

gt PDE with exact solution

gt pde x[10](-11) y[10](-11) (

u + u^^ = 6

[ u = 1+2yy u = 1+2yy u = xx+2 u = xx+2 ]

)

plot( u -u+xx+2yy )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

15

togo( Error = -u+xx+2yy )

gt Error

The u field is shown in Figure 8 and the error distribution is shown in Figure 9

Figure 8 A single PDE with exact solution

Figure 9 Error distribution

It is worthy of note that data at four corners are linearly extrapolated This is the

reason why there occur noticeable discrepancies at four corners in Figure 9

Conservative Form The second-order terms in some PDEs are

frequently expressed in conservative forms Thus we introduce the following

syntax for conservative forms

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

6

u = [ u(x) u (x) ]

When the initial guess is given as u = u(x) derivatives of the initial guess are

evaluated internally with centered difference scheme Nevertheless it is wise to

specify the known mathematical expressions of derivatives for accuracy

improvement

Surface Plot Next example is the Black-Scholes PDE discussed in ref

[1]

where

And IC and BCs are given as

This example is solved by

gt Black-Scholes PDE

gt (rs) = (006508) k = r(s^22)

gt (ab) = (log(25)log(75)) (totf) = (05)

gt bvp t[11](totf) x[21](ab)

u = max(exp(x)-10) (

-u + u + (k-1)u - ku = 0

[ u = 0 u = (7-5exp(-kt))5 ]

)

peep plot[11](u)

And the results are shown in the left of Figure 3 Changing lsquoplotrsquo to lsquoplot2rsquo

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

7

yields a 2D plot shown in the right of Figure 3

Figure 3 3D and 2D plot for initial BVP

System of PDEs An example for a system of PDEs is taken from ref [1]

The PDEs are written as

where the initial conditions for are

And the boundary conditions for are

Since the utility of Umbrella is handling equations with minimum amount of

modification it is not surprising to note that the following Cemmath commands

gt system of initial BVP

gt bvp t[5](002) x[10](01) Hub span

u = [ 1+05cos(2pix) -pisin(2pix) ]

v = [ 1-05cos(2pix) pisin(2pix) ] Hub field

(

-u + 05u + 1(1+vv) = 0 [ u = 0 u = 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

8

-v + 05v + 1(1+uu) = 0 [ v = 0 v = 0 ]

)peep

togo( U = u U1 = u V = v V1 = v KE = 05(uu+vv) )

plot[6](uv+2 4 + 05(uu+vv))

gt u v U U1 V V1 KE

produce the desired solutions The results are shown in Figure 4 and Figure 5

In particular the energy field defined as

is known to vanish as time approaches an infinity (see [ref 1]) Such a behavior

is represented in the energy field shown in Figure 5

Figure 4 Solution to a system of PDEs (u and v)

Figure 5 Solution to a system of PDEs (energy field)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

9

In addition we extracted the fields at specified time steps by lsquot[5](002)rsquo ie

t = [ 00 005 01 015 02 ]

and the values are listed as follows (only two cases are listed to save space)

U =

[ 15 12688 11677 113 1123 ]

[ 1383 12117 11398 11164 11163 ]

[ 10868 1067 10693 1082 10996 ]

[ 075 090265 098904 10428 10805 ]

[ 053015 079542 093671 10173 1068 ]

[ 053015 079542 093671 10173 1068 ]

[ 075 090265 098904 10428 10805 ]

[ 10868 1067 10693 1082 10996 ]

[ 1383 12117 11398 11164 11163 ]

[ 15 12688 11677 113 1123 ]

KE =

[ 0 0 0 0 0 ]

[ 40779 10564 025155 0059868 001424 ]

[ 9572 24796 059047 014053 0033427 ]

[ 74022 19175 045662 010867 0025849 ]

[ 11545 029908 007122 001695 00040317 ]

[ 11545 029908 007122 001695 00040317 ]

[ 74022 19175 045662 010867 0025849 ]

[ 9572 24796 059047 014053 0033427 ]

[ 40779 10564 025155 0059868 001424 ]

[ 59208e-031 40123e-051 86999e-070 30184e-087 16755e-103 ]

Note that the first column of U is exactly the same as the initial condition

Our next example is also adopted from ref [1] This problem has boundary

layers at both ends of the interval and a rapid change occurs in small time

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

10

where The initial conditions for

are

And the boundary conditions for are

Due to the presence of rapid change in time and the boundary layers in space

spanning numerical grids must reflect such behavior Therefore geometrically

increasing grid spans are employed

gt boundary layers and rapid change in small time

gt tout = [0 0005 001 005 01 05 1 15 2]

gt bvpt[2114](02) x[21-11](01) u = 1 v = 0

(

-u+0024u-( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ u = 0 u = 1 ]

-v+0170v+( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ v = 0 v = 0 ]

)

relax(04) maxiter(400) peep

plot[tout](u 2+v )

It should be noted that the function

is repeated for each equation This is because our solution is based on sequential

procedure and hence iteration is imposed at each time step Another feature

worthy of note is the use of Spoke lsquoplotrsquo If a pre-defined matrix is used inside

lsquo[ ]rsquo curves are selected from the corresponding times Numerical results are

shown in Figure 6

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

11

Figure 6 Solution to a system of PDs (u and v)

Steady elliptic PDEs involving two or three independent spatial variables

are treated here For generality three spatial coordinates are considered in

explaining Umbrella lsquopdersquo Nevertheless only licensed version can handle three

spatial coordinates

Standard Form The standard form of elliptic PDEs with three spatial

coordinates is written here as

where And are all of the same

sign from elliptic nature

To designate the partial derivatives our notations are innovated

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

12

Then the PDE of interest can be rewritten as

cuxxu + cuyyu^^ + cuzzu~~ + cuxu

+ cuyu^ + cuzu~ + cuu = cs

where nonlinear terms must be properly expressed as coefficients or source

terms Those terms with lsquo rsquo are participating the discretization procedure The

boundary conditions (BCs) also has a standard form

which can be rewritten as

The primary limitation of Umbrella lsquopdersquo at present is that it must be of second-

order in each coordinate

Syntax for PDE The syntax of Umbrella lsquopdersquo for steady two-

dimensional space is

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where two grid spans for x and y would be familiar to readers at this point

Therefore discussed below is the BCs in the Stem Since there exist two

independent spatial coordinates BCs must be specified in a pre-defined

sequence for consistency The BCs in the Stem are treated one-by-one for each

coordinate

[ x-left BC x-right BC y-left BC y-right BC ]

Discretization Method It is important to understand how the mesh is

constructed and the discretization is performed in solving PDEs With two

independent spatial variables (say ) the grid span in each direction is

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

13

defined

Then the nodes for solution are selected to be the center of a quadrangle

enclosed by the four points

For boundaries the nodes for solution are selected to be the center of two

adjacent points It is worthy of note that the major weakness arises at four

corners of the whole domain since the solution at these corners are extrapolated

from the inside or the boundaries

Therefore for the syntax

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

the mesh grid for computation has a dimension of This

means that the column of matrix represents

Single Elliptic PDE Our first example is very simple as shown below

which can be solved by

gt PDE of 2nd-order

gt pde x[10](01) y[10](01) (

u + u^^ + 10 = 0

[ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u 0 )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

14

The results are shown in Figure 7 Note that the computational mesh

corresponds to the field 0 and is shown in the right of Figure 7

Figure 7 Solution and mesh for a single PDE

As was mentioned above based on the pre-defined mesh

computational mesh of dimension is re-generated as shown in Figure

7 This is due to the treatment of control volume for discretizing the partial

differential equation The solution shown in the left of Figure 7 bears a

symmetry in both directions as expected

Comparison with Exact Solution Another example has an exact solution

The PDE of interest is written as

We solve this example by

gt PDE with exact solution

gt pde x[10](-11) y[10](-11) (

u + u^^ = 6

[ u = 1+2yy u = 1+2yy u = xx+2 u = xx+2 ]

)

plot( u -u+xx+2yy )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

15

togo( Error = -u+xx+2yy )

gt Error

The u field is shown in Figure 8 and the error distribution is shown in Figure 9

Figure 8 A single PDE with exact solution

Figure 9 Error distribution

It is worthy of note that data at four corners are linearly extrapolated This is the

reason why there occur noticeable discrepancies at four corners in Figure 9

Conservative Form The second-order terms in some PDEs are

frequently expressed in conservative forms Thus we introduce the following

syntax for conservative forms

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

7

yields a 2D plot shown in the right of Figure 3

Figure 3 3D and 2D plot for initial BVP

System of PDEs An example for a system of PDEs is taken from ref [1]

The PDEs are written as

where the initial conditions for are

And the boundary conditions for are

Since the utility of Umbrella is handling equations with minimum amount of

modification it is not surprising to note that the following Cemmath commands

gt system of initial BVP

gt bvp t[5](002) x[10](01) Hub span

u = [ 1+05cos(2pix) -pisin(2pix) ]

v = [ 1-05cos(2pix) pisin(2pix) ] Hub field

(

-u + 05u + 1(1+vv) = 0 [ u = 0 u = 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

8

-v + 05v + 1(1+uu) = 0 [ v = 0 v = 0 ]

)peep

togo( U = u U1 = u V = v V1 = v KE = 05(uu+vv) )

plot[6](uv+2 4 + 05(uu+vv))

gt u v U U1 V V1 KE

produce the desired solutions The results are shown in Figure 4 and Figure 5

In particular the energy field defined as

is known to vanish as time approaches an infinity (see [ref 1]) Such a behavior

is represented in the energy field shown in Figure 5

Figure 4 Solution to a system of PDEs (u and v)

Figure 5 Solution to a system of PDEs (energy field)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

9

In addition we extracted the fields at specified time steps by lsquot[5](002)rsquo ie

t = [ 00 005 01 015 02 ]

and the values are listed as follows (only two cases are listed to save space)

U =

[ 15 12688 11677 113 1123 ]

[ 1383 12117 11398 11164 11163 ]

[ 10868 1067 10693 1082 10996 ]

[ 075 090265 098904 10428 10805 ]

[ 053015 079542 093671 10173 1068 ]

[ 053015 079542 093671 10173 1068 ]

[ 075 090265 098904 10428 10805 ]

[ 10868 1067 10693 1082 10996 ]

[ 1383 12117 11398 11164 11163 ]

[ 15 12688 11677 113 1123 ]

KE =

[ 0 0 0 0 0 ]

[ 40779 10564 025155 0059868 001424 ]

[ 9572 24796 059047 014053 0033427 ]

[ 74022 19175 045662 010867 0025849 ]

[ 11545 029908 007122 001695 00040317 ]

[ 11545 029908 007122 001695 00040317 ]

[ 74022 19175 045662 010867 0025849 ]

[ 9572 24796 059047 014053 0033427 ]

[ 40779 10564 025155 0059868 001424 ]

[ 59208e-031 40123e-051 86999e-070 30184e-087 16755e-103 ]

Note that the first column of U is exactly the same as the initial condition

Our next example is also adopted from ref [1] This problem has boundary

layers at both ends of the interval and a rapid change occurs in small time

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

10

where The initial conditions for

are

And the boundary conditions for are

Due to the presence of rapid change in time and the boundary layers in space

spanning numerical grids must reflect such behavior Therefore geometrically

increasing grid spans are employed

gt boundary layers and rapid change in small time

gt tout = [0 0005 001 005 01 05 1 15 2]

gt bvpt[2114](02) x[21-11](01) u = 1 v = 0

(

-u+0024u-( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ u = 0 u = 1 ]

-v+0170v+( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ v = 0 v = 0 ]

)

relax(04) maxiter(400) peep

plot[tout](u 2+v )

It should be noted that the function

is repeated for each equation This is because our solution is based on sequential

procedure and hence iteration is imposed at each time step Another feature

worthy of note is the use of Spoke lsquoplotrsquo If a pre-defined matrix is used inside

lsquo[ ]rsquo curves are selected from the corresponding times Numerical results are

shown in Figure 6

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

11

Figure 6 Solution to a system of PDs (u and v)

Steady elliptic PDEs involving two or three independent spatial variables

are treated here For generality three spatial coordinates are considered in

explaining Umbrella lsquopdersquo Nevertheless only licensed version can handle three

spatial coordinates

Standard Form The standard form of elliptic PDEs with three spatial

coordinates is written here as

where And are all of the same

sign from elliptic nature

To designate the partial derivatives our notations are innovated

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

12

Then the PDE of interest can be rewritten as

cuxxu + cuyyu^^ + cuzzu~~ + cuxu

+ cuyu^ + cuzu~ + cuu = cs

where nonlinear terms must be properly expressed as coefficients or source

terms Those terms with lsquo rsquo are participating the discretization procedure The

boundary conditions (BCs) also has a standard form

which can be rewritten as

The primary limitation of Umbrella lsquopdersquo at present is that it must be of second-

order in each coordinate

Syntax for PDE The syntax of Umbrella lsquopdersquo for steady two-

dimensional space is

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where two grid spans for x and y would be familiar to readers at this point

Therefore discussed below is the BCs in the Stem Since there exist two

independent spatial coordinates BCs must be specified in a pre-defined

sequence for consistency The BCs in the Stem are treated one-by-one for each

coordinate

[ x-left BC x-right BC y-left BC y-right BC ]

Discretization Method It is important to understand how the mesh is

constructed and the discretization is performed in solving PDEs With two

independent spatial variables (say ) the grid span in each direction is

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

13

defined

Then the nodes for solution are selected to be the center of a quadrangle

enclosed by the four points

For boundaries the nodes for solution are selected to be the center of two

adjacent points It is worthy of note that the major weakness arises at four

corners of the whole domain since the solution at these corners are extrapolated

from the inside or the boundaries

Therefore for the syntax

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

the mesh grid for computation has a dimension of This

means that the column of matrix represents

Single Elliptic PDE Our first example is very simple as shown below

which can be solved by

gt PDE of 2nd-order

gt pde x[10](01) y[10](01) (

u + u^^ + 10 = 0

[ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u 0 )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

14

The results are shown in Figure 7 Note that the computational mesh

corresponds to the field 0 and is shown in the right of Figure 7

Figure 7 Solution and mesh for a single PDE

As was mentioned above based on the pre-defined mesh

computational mesh of dimension is re-generated as shown in Figure

7 This is due to the treatment of control volume for discretizing the partial

differential equation The solution shown in the left of Figure 7 bears a

symmetry in both directions as expected

Comparison with Exact Solution Another example has an exact solution

The PDE of interest is written as

We solve this example by

gt PDE with exact solution

gt pde x[10](-11) y[10](-11) (

u + u^^ = 6

[ u = 1+2yy u = 1+2yy u = xx+2 u = xx+2 ]

)

plot( u -u+xx+2yy )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

15

togo( Error = -u+xx+2yy )

gt Error

The u field is shown in Figure 8 and the error distribution is shown in Figure 9

Figure 8 A single PDE with exact solution

Figure 9 Error distribution

It is worthy of note that data at four corners are linearly extrapolated This is the

reason why there occur noticeable discrepancies at four corners in Figure 9

Conservative Form The second-order terms in some PDEs are

frequently expressed in conservative forms Thus we introduce the following

syntax for conservative forms

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

8

-v + 05v + 1(1+uu) = 0 [ v = 0 v = 0 ]

)peep

togo( U = u U1 = u V = v V1 = v KE = 05(uu+vv) )

plot[6](uv+2 4 + 05(uu+vv))

gt u v U U1 V V1 KE

produce the desired solutions The results are shown in Figure 4 and Figure 5

In particular the energy field defined as

is known to vanish as time approaches an infinity (see [ref 1]) Such a behavior

is represented in the energy field shown in Figure 5

Figure 4 Solution to a system of PDEs (u and v)

Figure 5 Solution to a system of PDEs (energy field)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

9

In addition we extracted the fields at specified time steps by lsquot[5](002)rsquo ie

t = [ 00 005 01 015 02 ]

and the values are listed as follows (only two cases are listed to save space)

U =

[ 15 12688 11677 113 1123 ]

[ 1383 12117 11398 11164 11163 ]

[ 10868 1067 10693 1082 10996 ]

[ 075 090265 098904 10428 10805 ]

[ 053015 079542 093671 10173 1068 ]

[ 053015 079542 093671 10173 1068 ]

[ 075 090265 098904 10428 10805 ]

[ 10868 1067 10693 1082 10996 ]

[ 1383 12117 11398 11164 11163 ]

[ 15 12688 11677 113 1123 ]

KE =

[ 0 0 0 0 0 ]

[ 40779 10564 025155 0059868 001424 ]

[ 9572 24796 059047 014053 0033427 ]

[ 74022 19175 045662 010867 0025849 ]

[ 11545 029908 007122 001695 00040317 ]

[ 11545 029908 007122 001695 00040317 ]

[ 74022 19175 045662 010867 0025849 ]

[ 9572 24796 059047 014053 0033427 ]

[ 40779 10564 025155 0059868 001424 ]

[ 59208e-031 40123e-051 86999e-070 30184e-087 16755e-103 ]

Note that the first column of U is exactly the same as the initial condition

Our next example is also adopted from ref [1] This problem has boundary

layers at both ends of the interval and a rapid change occurs in small time

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

10

where The initial conditions for

are

And the boundary conditions for are

Due to the presence of rapid change in time and the boundary layers in space

spanning numerical grids must reflect such behavior Therefore geometrically

increasing grid spans are employed

gt boundary layers and rapid change in small time

gt tout = [0 0005 001 005 01 05 1 15 2]

gt bvpt[2114](02) x[21-11](01) u = 1 v = 0

(

-u+0024u-( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ u = 0 u = 1 ]

-v+0170v+( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ v = 0 v = 0 ]

)

relax(04) maxiter(400) peep

plot[tout](u 2+v )

It should be noted that the function

is repeated for each equation This is because our solution is based on sequential

procedure and hence iteration is imposed at each time step Another feature

worthy of note is the use of Spoke lsquoplotrsquo If a pre-defined matrix is used inside

lsquo[ ]rsquo curves are selected from the corresponding times Numerical results are

shown in Figure 6

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

11

Figure 6 Solution to a system of PDs (u and v)

Steady elliptic PDEs involving two or three independent spatial variables

are treated here For generality three spatial coordinates are considered in

explaining Umbrella lsquopdersquo Nevertheless only licensed version can handle three

spatial coordinates

Standard Form The standard form of elliptic PDEs with three spatial

coordinates is written here as

where And are all of the same

sign from elliptic nature

To designate the partial derivatives our notations are innovated

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

12

Then the PDE of interest can be rewritten as

cuxxu + cuyyu^^ + cuzzu~~ + cuxu

+ cuyu^ + cuzu~ + cuu = cs

where nonlinear terms must be properly expressed as coefficients or source

terms Those terms with lsquo rsquo are participating the discretization procedure The

boundary conditions (BCs) also has a standard form

which can be rewritten as

The primary limitation of Umbrella lsquopdersquo at present is that it must be of second-

order in each coordinate

Syntax for PDE The syntax of Umbrella lsquopdersquo for steady two-

dimensional space is

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where two grid spans for x and y would be familiar to readers at this point

Therefore discussed below is the BCs in the Stem Since there exist two

independent spatial coordinates BCs must be specified in a pre-defined

sequence for consistency The BCs in the Stem are treated one-by-one for each

coordinate

[ x-left BC x-right BC y-left BC y-right BC ]

Discretization Method It is important to understand how the mesh is

constructed and the discretization is performed in solving PDEs With two

independent spatial variables (say ) the grid span in each direction is

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

13

defined

Then the nodes for solution are selected to be the center of a quadrangle

enclosed by the four points

For boundaries the nodes for solution are selected to be the center of two

adjacent points It is worthy of note that the major weakness arises at four

corners of the whole domain since the solution at these corners are extrapolated

from the inside or the boundaries

Therefore for the syntax

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

the mesh grid for computation has a dimension of This

means that the column of matrix represents

Single Elliptic PDE Our first example is very simple as shown below

which can be solved by

gt PDE of 2nd-order

gt pde x[10](01) y[10](01) (

u + u^^ + 10 = 0

[ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u 0 )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

14

The results are shown in Figure 7 Note that the computational mesh

corresponds to the field 0 and is shown in the right of Figure 7

Figure 7 Solution and mesh for a single PDE

As was mentioned above based on the pre-defined mesh

computational mesh of dimension is re-generated as shown in Figure

7 This is due to the treatment of control volume for discretizing the partial

differential equation The solution shown in the left of Figure 7 bears a

symmetry in both directions as expected

Comparison with Exact Solution Another example has an exact solution

The PDE of interest is written as

We solve this example by

gt PDE with exact solution

gt pde x[10](-11) y[10](-11) (

u + u^^ = 6

[ u = 1+2yy u = 1+2yy u = xx+2 u = xx+2 ]

)

plot( u -u+xx+2yy )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

15

togo( Error = -u+xx+2yy )

gt Error

The u field is shown in Figure 8 and the error distribution is shown in Figure 9

Figure 8 A single PDE with exact solution

Figure 9 Error distribution

It is worthy of note that data at four corners are linearly extrapolated This is the

reason why there occur noticeable discrepancies at four corners in Figure 9

Conservative Form The second-order terms in some PDEs are

frequently expressed in conservative forms Thus we introduce the following

syntax for conservative forms

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

9

In addition we extracted the fields at specified time steps by lsquot[5](002)rsquo ie

t = [ 00 005 01 015 02 ]

and the values are listed as follows (only two cases are listed to save space)

U =

[ 15 12688 11677 113 1123 ]

[ 1383 12117 11398 11164 11163 ]

[ 10868 1067 10693 1082 10996 ]

[ 075 090265 098904 10428 10805 ]

[ 053015 079542 093671 10173 1068 ]

[ 053015 079542 093671 10173 1068 ]

[ 075 090265 098904 10428 10805 ]

[ 10868 1067 10693 1082 10996 ]

[ 1383 12117 11398 11164 11163 ]

[ 15 12688 11677 113 1123 ]

KE =

[ 0 0 0 0 0 ]

[ 40779 10564 025155 0059868 001424 ]

[ 9572 24796 059047 014053 0033427 ]

[ 74022 19175 045662 010867 0025849 ]

[ 11545 029908 007122 001695 00040317 ]

[ 11545 029908 007122 001695 00040317 ]

[ 74022 19175 045662 010867 0025849 ]

[ 9572 24796 059047 014053 0033427 ]

[ 40779 10564 025155 0059868 001424 ]

[ 59208e-031 40123e-051 86999e-070 30184e-087 16755e-103 ]

Note that the first column of U is exactly the same as the initial condition

Our next example is also adopted from ref [1] This problem has boundary

layers at both ends of the interval and a rapid change occurs in small time

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

10

where The initial conditions for

are

And the boundary conditions for are

Due to the presence of rapid change in time and the boundary layers in space

spanning numerical grids must reflect such behavior Therefore geometrically

increasing grid spans are employed

gt boundary layers and rapid change in small time

gt tout = [0 0005 001 005 01 05 1 15 2]

gt bvpt[2114](02) x[21-11](01) u = 1 v = 0

(

-u+0024u-( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ u = 0 u = 1 ]

-v+0170v+( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ v = 0 v = 0 ]

)

relax(04) maxiter(400) peep

plot[tout](u 2+v )

It should be noted that the function

is repeated for each equation This is because our solution is based on sequential

procedure and hence iteration is imposed at each time step Another feature

worthy of note is the use of Spoke lsquoplotrsquo If a pre-defined matrix is used inside

lsquo[ ]rsquo curves are selected from the corresponding times Numerical results are

shown in Figure 6

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

11

Figure 6 Solution to a system of PDs (u and v)

Steady elliptic PDEs involving two or three independent spatial variables

are treated here For generality three spatial coordinates are considered in

explaining Umbrella lsquopdersquo Nevertheless only licensed version can handle three

spatial coordinates

Standard Form The standard form of elliptic PDEs with three spatial

coordinates is written here as

where And are all of the same

sign from elliptic nature

To designate the partial derivatives our notations are innovated

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

12

Then the PDE of interest can be rewritten as

cuxxu + cuyyu^^ + cuzzu~~ + cuxu

+ cuyu^ + cuzu~ + cuu = cs

where nonlinear terms must be properly expressed as coefficients or source

terms Those terms with lsquo rsquo are participating the discretization procedure The

boundary conditions (BCs) also has a standard form

which can be rewritten as

The primary limitation of Umbrella lsquopdersquo at present is that it must be of second-

order in each coordinate

Syntax for PDE The syntax of Umbrella lsquopdersquo for steady two-

dimensional space is

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where two grid spans for x and y would be familiar to readers at this point

Therefore discussed below is the BCs in the Stem Since there exist two

independent spatial coordinates BCs must be specified in a pre-defined

sequence for consistency The BCs in the Stem are treated one-by-one for each

coordinate

[ x-left BC x-right BC y-left BC y-right BC ]

Discretization Method It is important to understand how the mesh is

constructed and the discretization is performed in solving PDEs With two

independent spatial variables (say ) the grid span in each direction is

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

13

defined

Then the nodes for solution are selected to be the center of a quadrangle

enclosed by the four points

For boundaries the nodes for solution are selected to be the center of two

adjacent points It is worthy of note that the major weakness arises at four

corners of the whole domain since the solution at these corners are extrapolated

from the inside or the boundaries

Therefore for the syntax

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

the mesh grid for computation has a dimension of This

means that the column of matrix represents

Single Elliptic PDE Our first example is very simple as shown below

which can be solved by

gt PDE of 2nd-order

gt pde x[10](01) y[10](01) (

u + u^^ + 10 = 0

[ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u 0 )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

14

The results are shown in Figure 7 Note that the computational mesh

corresponds to the field 0 and is shown in the right of Figure 7

Figure 7 Solution and mesh for a single PDE

As was mentioned above based on the pre-defined mesh

computational mesh of dimension is re-generated as shown in Figure

7 This is due to the treatment of control volume for discretizing the partial

differential equation The solution shown in the left of Figure 7 bears a

symmetry in both directions as expected

Comparison with Exact Solution Another example has an exact solution

The PDE of interest is written as

We solve this example by

gt PDE with exact solution

gt pde x[10](-11) y[10](-11) (

u + u^^ = 6

[ u = 1+2yy u = 1+2yy u = xx+2 u = xx+2 ]

)

plot( u -u+xx+2yy )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

15

togo( Error = -u+xx+2yy )

gt Error

The u field is shown in Figure 8 and the error distribution is shown in Figure 9

Figure 8 A single PDE with exact solution

Figure 9 Error distribution

It is worthy of note that data at four corners are linearly extrapolated This is the

reason why there occur noticeable discrepancies at four corners in Figure 9

Conservative Form The second-order terms in some PDEs are

frequently expressed in conservative forms Thus we introduce the following

syntax for conservative forms

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

10

where The initial conditions for

are

And the boundary conditions for are

Due to the presence of rapid change in time and the boundary layers in space

spanning numerical grids must reflect such behavior Therefore geometrically

increasing grid spans are employed

gt boundary layers and rapid change in small time

gt tout = [0 0005 001 005 01 05 1 15 2]

gt bvpt[2114](02) x[21-11](01) u = 1 v = 0

(

-u+0024u-( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ u = 0 u = 1 ]

-v+0170v+( exp(573(u-v))-exp(-1146(u-v)) ) = 0

[ v = 0 v = 0 ]

)

relax(04) maxiter(400) peep

plot[tout](u 2+v )

It should be noted that the function

is repeated for each equation This is because our solution is based on sequential

procedure and hence iteration is imposed at each time step Another feature

worthy of note is the use of Spoke lsquoplotrsquo If a pre-defined matrix is used inside

lsquo[ ]rsquo curves are selected from the corresponding times Numerical results are

shown in Figure 6

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

11

Figure 6 Solution to a system of PDs (u and v)

Steady elliptic PDEs involving two or three independent spatial variables

are treated here For generality three spatial coordinates are considered in

explaining Umbrella lsquopdersquo Nevertheless only licensed version can handle three

spatial coordinates

Standard Form The standard form of elliptic PDEs with three spatial

coordinates is written here as

where And are all of the same

sign from elliptic nature

To designate the partial derivatives our notations are innovated

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

12

Then the PDE of interest can be rewritten as

cuxxu + cuyyu^^ + cuzzu~~ + cuxu

+ cuyu^ + cuzu~ + cuu = cs

where nonlinear terms must be properly expressed as coefficients or source

terms Those terms with lsquo rsquo are participating the discretization procedure The

boundary conditions (BCs) also has a standard form

which can be rewritten as

The primary limitation of Umbrella lsquopdersquo at present is that it must be of second-

order in each coordinate

Syntax for PDE The syntax of Umbrella lsquopdersquo for steady two-

dimensional space is

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where two grid spans for x and y would be familiar to readers at this point

Therefore discussed below is the BCs in the Stem Since there exist two

independent spatial coordinates BCs must be specified in a pre-defined

sequence for consistency The BCs in the Stem are treated one-by-one for each

coordinate

[ x-left BC x-right BC y-left BC y-right BC ]

Discretization Method It is important to understand how the mesh is

constructed and the discretization is performed in solving PDEs With two

independent spatial variables (say ) the grid span in each direction is

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

13

defined

Then the nodes for solution are selected to be the center of a quadrangle

enclosed by the four points

For boundaries the nodes for solution are selected to be the center of two

adjacent points It is worthy of note that the major weakness arises at four

corners of the whole domain since the solution at these corners are extrapolated

from the inside or the boundaries

Therefore for the syntax

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

the mesh grid for computation has a dimension of This

means that the column of matrix represents

Single Elliptic PDE Our first example is very simple as shown below

which can be solved by

gt PDE of 2nd-order

gt pde x[10](01) y[10](01) (

u + u^^ + 10 = 0

[ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u 0 )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

14

The results are shown in Figure 7 Note that the computational mesh

corresponds to the field 0 and is shown in the right of Figure 7

Figure 7 Solution and mesh for a single PDE

As was mentioned above based on the pre-defined mesh

computational mesh of dimension is re-generated as shown in Figure

7 This is due to the treatment of control volume for discretizing the partial

differential equation The solution shown in the left of Figure 7 bears a

symmetry in both directions as expected

Comparison with Exact Solution Another example has an exact solution

The PDE of interest is written as

We solve this example by

gt PDE with exact solution

gt pde x[10](-11) y[10](-11) (

u + u^^ = 6

[ u = 1+2yy u = 1+2yy u = xx+2 u = xx+2 ]

)

plot( u -u+xx+2yy )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

15

togo( Error = -u+xx+2yy )

gt Error

The u field is shown in Figure 8 and the error distribution is shown in Figure 9

Figure 8 A single PDE with exact solution

Figure 9 Error distribution

It is worthy of note that data at four corners are linearly extrapolated This is the

reason why there occur noticeable discrepancies at four corners in Figure 9

Conservative Form The second-order terms in some PDEs are

frequently expressed in conservative forms Thus we introduce the following

syntax for conservative forms

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

11

Figure 6 Solution to a system of PDs (u and v)

Steady elliptic PDEs involving two or three independent spatial variables

are treated here For generality three spatial coordinates are considered in

explaining Umbrella lsquopdersquo Nevertheless only licensed version can handle three

spatial coordinates

Standard Form The standard form of elliptic PDEs with three spatial

coordinates is written here as

where And are all of the same

sign from elliptic nature

To designate the partial derivatives our notations are innovated

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

12

Then the PDE of interest can be rewritten as

cuxxu + cuyyu^^ + cuzzu~~ + cuxu

+ cuyu^ + cuzu~ + cuu = cs

where nonlinear terms must be properly expressed as coefficients or source

terms Those terms with lsquo rsquo are participating the discretization procedure The

boundary conditions (BCs) also has a standard form

which can be rewritten as

The primary limitation of Umbrella lsquopdersquo at present is that it must be of second-

order in each coordinate

Syntax for PDE The syntax of Umbrella lsquopdersquo for steady two-

dimensional space is

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where two grid spans for x and y would be familiar to readers at this point

Therefore discussed below is the BCs in the Stem Since there exist two

independent spatial coordinates BCs must be specified in a pre-defined

sequence for consistency The BCs in the Stem are treated one-by-one for each

coordinate

[ x-left BC x-right BC y-left BC y-right BC ]

Discretization Method It is important to understand how the mesh is

constructed and the discretization is performed in solving PDEs With two

independent spatial variables (say ) the grid span in each direction is

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

13

defined

Then the nodes for solution are selected to be the center of a quadrangle

enclosed by the four points

For boundaries the nodes for solution are selected to be the center of two

adjacent points It is worthy of note that the major weakness arises at four

corners of the whole domain since the solution at these corners are extrapolated

from the inside or the boundaries

Therefore for the syntax

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

the mesh grid for computation has a dimension of This

means that the column of matrix represents

Single Elliptic PDE Our first example is very simple as shown below

which can be solved by

gt PDE of 2nd-order

gt pde x[10](01) y[10](01) (

u + u^^ + 10 = 0

[ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u 0 )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

14

The results are shown in Figure 7 Note that the computational mesh

corresponds to the field 0 and is shown in the right of Figure 7

Figure 7 Solution and mesh for a single PDE

As was mentioned above based on the pre-defined mesh

computational mesh of dimension is re-generated as shown in Figure

7 This is due to the treatment of control volume for discretizing the partial

differential equation The solution shown in the left of Figure 7 bears a

symmetry in both directions as expected

Comparison with Exact Solution Another example has an exact solution

The PDE of interest is written as

We solve this example by

gt PDE with exact solution

gt pde x[10](-11) y[10](-11) (

u + u^^ = 6

[ u = 1+2yy u = 1+2yy u = xx+2 u = xx+2 ]

)

plot( u -u+xx+2yy )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

15

togo( Error = -u+xx+2yy )

gt Error

The u field is shown in Figure 8 and the error distribution is shown in Figure 9

Figure 8 A single PDE with exact solution

Figure 9 Error distribution

It is worthy of note that data at four corners are linearly extrapolated This is the

reason why there occur noticeable discrepancies at four corners in Figure 9

Conservative Form The second-order terms in some PDEs are

frequently expressed in conservative forms Thus we introduce the following

syntax for conservative forms

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

12

Then the PDE of interest can be rewritten as

cuxxu + cuyyu^^ + cuzzu~~ + cuxu

+ cuyu^ + cuzu~ + cuu = cs

where nonlinear terms must be properly expressed as coefficients or source

terms Those terms with lsquo rsquo are participating the discretization procedure The

boundary conditions (BCs) also has a standard form

which can be rewritten as

The primary limitation of Umbrella lsquopdersquo at present is that it must be of second-

order in each coordinate

Syntax for PDE The syntax of Umbrella lsquopdersquo for steady two-

dimensional space is

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where two grid spans for x and y would be familiar to readers at this point

Therefore discussed below is the BCs in the Stem Since there exist two

independent spatial coordinates BCs must be specified in a pre-defined

sequence for consistency The BCs in the Stem are treated one-by-one for each

coordinate

[ x-left BC x-right BC y-left BC y-right BC ]

Discretization Method It is important to understand how the mesh is

constructed and the discretization is performed in solving PDEs With two

independent spatial variables (say ) the grid span in each direction is

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

13

defined

Then the nodes for solution are selected to be the center of a quadrangle

enclosed by the four points

For boundaries the nodes for solution are selected to be the center of two

adjacent points It is worthy of note that the major weakness arises at four

corners of the whole domain since the solution at these corners are extrapolated

from the inside or the boundaries

Therefore for the syntax

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

the mesh grid for computation has a dimension of This

means that the column of matrix represents

Single Elliptic PDE Our first example is very simple as shown below

which can be solved by

gt PDE of 2nd-order

gt pde x[10](01) y[10](01) (

u + u^^ + 10 = 0

[ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u 0 )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

14

The results are shown in Figure 7 Note that the computational mesh

corresponds to the field 0 and is shown in the right of Figure 7

Figure 7 Solution and mesh for a single PDE

As was mentioned above based on the pre-defined mesh

computational mesh of dimension is re-generated as shown in Figure

7 This is due to the treatment of control volume for discretizing the partial

differential equation The solution shown in the left of Figure 7 bears a

symmetry in both directions as expected

Comparison with Exact Solution Another example has an exact solution

The PDE of interest is written as

We solve this example by

gt PDE with exact solution

gt pde x[10](-11) y[10](-11) (

u + u^^ = 6

[ u = 1+2yy u = 1+2yy u = xx+2 u = xx+2 ]

)

plot( u -u+xx+2yy )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

15

togo( Error = -u+xx+2yy )

gt Error

The u field is shown in Figure 8 and the error distribution is shown in Figure 9

Figure 8 A single PDE with exact solution

Figure 9 Error distribution

It is worthy of note that data at four corners are linearly extrapolated This is the

reason why there occur noticeable discrepancies at four corners in Figure 9

Conservative Form The second-order terms in some PDEs are

frequently expressed in conservative forms Thus we introduce the following

syntax for conservative forms

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

13

defined

Then the nodes for solution are selected to be the center of a quadrangle

enclosed by the four points

For boundaries the nodes for solution are selected to be the center of two

adjacent points It is worthy of note that the major weakness arises at four

corners of the whole domain since the solution at these corners are extrapolated

from the inside or the boundaries

Therefore for the syntax

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

the mesh grid for computation has a dimension of This

means that the column of matrix represents

Single Elliptic PDE Our first example is very simple as shown below

which can be solved by

gt PDE of 2nd-order

gt pde x[10](01) y[10](01) (

u + u^^ + 10 = 0

[ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u 0 )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

14

The results are shown in Figure 7 Note that the computational mesh

corresponds to the field 0 and is shown in the right of Figure 7

Figure 7 Solution and mesh for a single PDE

As was mentioned above based on the pre-defined mesh

computational mesh of dimension is re-generated as shown in Figure

7 This is due to the treatment of control volume for discretizing the partial

differential equation The solution shown in the left of Figure 7 bears a

symmetry in both directions as expected

Comparison with Exact Solution Another example has an exact solution

The PDE of interest is written as

We solve this example by

gt PDE with exact solution

gt pde x[10](-11) y[10](-11) (

u + u^^ = 6

[ u = 1+2yy u = 1+2yy u = xx+2 u = xx+2 ]

)

plot( u -u+xx+2yy )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

15

togo( Error = -u+xx+2yy )

gt Error

The u field is shown in Figure 8 and the error distribution is shown in Figure 9

Figure 8 A single PDE with exact solution

Figure 9 Error distribution

It is worthy of note that data at four corners are linearly extrapolated This is the

reason why there occur noticeable discrepancies at four corners in Figure 9

Conservative Form The second-order terms in some PDEs are

frequently expressed in conservative forms Thus we introduce the following

syntax for conservative forms

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

14

The results are shown in Figure 7 Note that the computational mesh

corresponds to the field 0 and is shown in the right of Figure 7

Figure 7 Solution and mesh for a single PDE

As was mentioned above based on the pre-defined mesh

computational mesh of dimension is re-generated as shown in Figure

7 This is due to the treatment of control volume for discretizing the partial

differential equation The solution shown in the left of Figure 7 bears a

symmetry in both directions as expected

Comparison with Exact Solution Another example has an exact solution

The PDE of interest is written as

We solve this example by

gt PDE with exact solution

gt pde x[10](-11) y[10](-11) (

u + u^^ = 6

[ u = 1+2yy u = 1+2yy u = xx+2 u = xx+2 ]

)

plot( u -u+xx+2yy )

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

15

togo( Error = -u+xx+2yy )

gt Error

The u field is shown in Figure 8 and the error distribution is shown in Figure 9

Figure 8 A single PDE with exact solution

Figure 9 Error distribution

It is worthy of note that data at four corners are linearly extrapolated This is the

reason why there occur noticeable discrepancies at four corners in Figure 9

Conservative Form The second-order terms in some PDEs are

frequently expressed in conservative forms Thus we introduce the following

syntax for conservative forms

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

15

togo( Error = -u+xx+2yy )

gt Error

The u field is shown in Figure 8 and the error distribution is shown in Figure 9

Figure 8 A single PDE with exact solution

Figure 9 Error distribution

It is worthy of note that data at four corners are linearly extrapolated This is the

reason why there occur noticeable discrepancies at four corners in Figure 9

Conservative Form The second-order terms in some PDEs are

frequently expressed in conservative forms Thus we introduce the following

syntax for conservative forms

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

16

These conservative forms can be treated as long as they are the highest terms

This implies that either (p)u or u should appear once but not together

An example is

where

This PDE can be solved by

gt PDE in conservative form

gt pde xi[21](-11) eta[21]( -11 ) (

(exp(5xi)+1)u + (exp(5eta)+1)u^^ + 10 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 10

Figure 10 Conservative PDE

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

17

Next example is given below

where

For this example we employ non-uniform grid spans in both directions

gt PDE with nonuniform grid

gt pde x[21](-33) y[21]( -sqrt(10-xx)4-x2 ) (

(exp(x)+1)u + (exp(y)+1)u^^ + uu + u^ + u + 2 = 0

[ u = 0 u = 0 u = 0 u = 0 ]

)

plot(u)

And the result is shown in Figure 11

Figure 11 PDE in an irregular geometry

Section 8-3 System of Steady Elliptic PDEs

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

18

System of PDEs For coupled steady elliptic PDEs we have the

following syntax

pde x[n=31g=1](ab) y[n=31g=1]( c(x)d(x) )

( DE [BCs] DE [BCs] hellip )

where each PDE must have its own dependent variable as well as DE and BCs

with

Due to the presence of nonlinear terms in both u and v it may be helpful to use

the under-relaxation for convergence The following commands

gt System of steady elliptic PDEs

gt pde x[30](01) y[30](01) (

u + u^^ + 5 - uu-vu^ = 0 [ u = 0 u = 0 u = 0 u

= 0 ]

v + v^^ - 3 - uv-vv^ = 0 [ v = 0 v = 0 v = 0 v

= 0 ]

)plot( uv-01 )

are executed to get the results shown in Figure 12

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

19

Figure 12 system of PDEs

Section 8-4 Data Extraction

Spoke lsquotogorsquo To extract field data we use Spoke lsquotogorsquo as before To

illustrates this we consider again the first example discussed above

and the following Cemmath commands

gt Data extraction

gt pde x[4](01) y[5](01) (

u + u^^ + 5 = 0 [ u = 1 u = 1 u = 1 u = 1 ]

)

plot( u )

togo( X = x Y = y U = u )

gt X Y U

Note that lsquopde x[4](01) y[5](01)rsquo results in

mesh grid as shown in Figure 13

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

20

Figure 13 Data extraction

Data extraction for this case yields the following matrices

X =

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

[ 0 016667 05 083333 1 ]

Y =

[ 0 0 0 0 0 ]

[ 0125 0125 0125 0125 0125 ]

[ 0375 0375 0375 0375 0375 ]

[ 0625 0625 0625 0625 0625 ]

[ 0875 0875 0875 0875 0875 ]

[ 1 1 1 1 1 ]

U =

[ 1 1 1 1 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 12487 13758 12487 1 ]

[ 1 11445 12063 11445 1 ]

[ 1 1 1 1 1 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

21

Section 8-5 Transient Elliptic PDEs

Syntax for PDE The syntax of Umbrella lsquopdersquo for transient two-

dimensional space is

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

where time must be the first grid span Spoke lsquotogorsquo for this case results in a

matrix of dimension

Therefore each column contains a field solution at a give time step

Single Transient PDE Our first example for a single transient PDE is

described as

which can be solved by

gt pdet[4](001)x[4](01)y[4](01) u=0 (

-u+u + u^^ + 10 = 0 [ u=0 u=0 u=0 u=0 ]

)

peep

plot[4]( u )

togo( U=u ) U

First the result from Spoke lsquotogorsquo is listed as

U =

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

22

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 020392 032831 040445 ]

[ 0 026275 045002 057668 ]

[ 0 020392 032831 040445 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 016275 024752 029412 ]

[ 0 020392 032831 040445 ]

[ 0 016275 024752 029412 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

[ 0 0 0 0 ]

As was mentioned each column consists of a solution at specified time

For example the first column corresponds to the initial

condition of As the column index increases it is easy to note that the

maximum value increases due to evolution of transient solution to the steady-

state solution Meanwhile time evolution of solution is drawn in Figure 14

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

23

Figure 14 Time evolution of solution

System of Transient PDEs An example for a system of transient

elliptic PDEs is written below

With the initial condition

And the boundary conditions

This system of transient elliptic PDEs can be solved by

gt pde t[4](0015) x[30](01) y[30](01) u=0 v=1 (

-u - uu-vu^ + u + u^^ + 6 = 0

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

24

[ u=0 u=0 u=0 u=0 ]

-v - uv-vv^ + v + v^^ - 4 = 0

[ v=1 v=1 v=1 v=1 ]

) peepplot( uv ) uv

The results are shown in Figure 15

Figure 15 Time evolution of solution u and v

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

[100] 008 Chapter 8 Partial Differential Equations Tutorial by wwwmsharpmathcom

25

Section 8-6 Summary of Syntax

Syntax for Initial BVPs The syntax for initial BVPs is

bvp t[n=101g=1](tatb) x[n=101g=1](xaxb) IC ( DE [ BCs] )

pde x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs])

pde x[n=31g=1](ab)y[n=31g=1](c(x)d(x)) ( DE [BCs] DE [BCs] )

pde t[ntime=31g=1](totf)

x[m=31g=1](ab) y[n=31g=1]( c(x)d(x) ) ( DE [BCs] )

Section 8-7 References

[1] DJ Higham and NJ Higham MATLAB(R) Program GuideBook 2ed

SIAM (2005)

[2] MATLAB(R) 7 Mathematics The MathWorks Inc (2009)

Recommended