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Parabolic PDEs

We consider the heat or diffusion equation in one spatial dimension given by the PDE2

2 22 t x

u(x,t) u(x,t)u u

t x

= =

x

where u(x, t) = the temperature and t = time with x = position value. This equation is used todescribe the diffusion in a medium. Here we focus on the diffusion of heat (temperature) over time.In general, we need boundary conditions to ensure the existence of a unique solution. Suchconditions involve the initial temperature distribution and what is happening along the boundary.We consider the following heat equation.

2

a

b

a b finite interval

0

0 for all a b initial temperature distribution

a for 0 B.C. at left end of the spatial intervalb for 0 B.C. at right end of the spatial inter

t xxu u x

t

u(x, ) f(x) x

u( ,t) u (t) tu( ,t) u (t) t

=

=

= = val

One physical situation is to consider a one dimensional homogeneous rod. Then constant 2 iscalled the diffusion coefficient, representing the thermal diffusivity of the material in the rod. Theheat equation models the spread of heat from regions of higher temperature to regions of lowertemperature.

In heat transfer analysis, thermal diffusivity (symbol: , but note that the symbols , D, and k areall commonly used) is the thermal conductivity divided by the volumetric heat capacity. It has the SI

unit of m/s.

where:

k : thermal conductivity (SI units: W/(mK)) International System of Units (abbreviated SI) : density (kg/m)cp : specific heat capacity (J/(kgK))

The denominator of the thermal diffusivity expression above, cp , can be identified as thevolumetric heat capacity with the SI unit of J/(mK).

Substances with high thermal diffusivity rapidly adjust their temperature to that of theirsurroundings, because they conduct heat quickly in comparison to their volumetric heat capacity or'thermal bulk'.

From Wikipedia: http://en.wikipedia.org/wiki/Thermal_diffusivity

http://en.wikipedia.org/wiki/Thermal_diffusivityhttp://en.wikipedia.org/wiki/Thermal_diffusivity

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Go to http://www.math.duke.edu/education/ccp/materials/engin/pdeintro/pde1.htmlConsider the following as a work sheet.

Introduction to the One-Dimensional Heat EquationA Sample ProblemWe will examine solutions to a simple second-order linear partial differential equation -- the one-dimensional heat equation. The heat equation models the flow of heat in a rod that is insulatedeverywhere except at the two ends. Solutions of this equation are functions of two variables -- onespatial variable (position along the rod) and time. The "one-dimensional" in the description of thedifferential equation refers to the fact that we are considering only one spatial dimension.

Imagine a thin rod that is given an initial temperature distribution, then insulated on the sides. Theends of the rod are kept at the same fixed temperature; e.g., suppose at the start of theexperiment, both ends are immediately plunged into ice water. We are interested in how the

temperatures along the rod vary with time. Suppose that the rod has a length L (in meters), and weestablish a coordinate system along the rod as illustrated below.

Let u(x, t) represent the temperature at the point x meters along the rod at time t (in seconds). Westart with an initial temperature distribution u(x, 0) = f(x) such as the one represented by the graphin Figure 1 (with L = 2 meters).

Figure 1.

http://www.math.duke.edu/education/ccp/materials/engin/pdeintro/pde1.htmlhttp://www.math.duke.edu/education/ccp/materials/engin/pdeintro/pde1.html

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The partial differential equation

ut = a2 uxx

is used to model one-dimensional temperature evolution. We will not discuss the derivation of this

equation here. The most important features of this equation are the secondspatial derivative uxxand the firstderivative with respect to time, ut. The positive constant a2represents the thermal

diffusivityof the rod. It depends on the thermal conductivity of the material composing the rod, thedensity of the rod, and the specific heat of the rod.

1. Explain why the units ofa2must be (length)2 / time.

Typical values for the diffusivity constant are given in the table below

Material a2

Silver 1.71Copper 1.14

Aluminum 0.86

Cast Iron 0.12

Granite 0.011

Brick 0.0038

Water 0.00144

The function u(x, t) that models heat flow should satisfy the partial differential equation. However,in addition, we expect it to satisfy two other conditions. For this example, we fix the temperature atthe two ends of the rod, i.e., we specify u(0, t) and u(L, t). In our sample problem, we will assumethat both ends are kept at 0 degrees Celsius:

u(0,t) = u(L,t) = 0 for all t > 0.

These are called boundary conditions since they are imposed on the values of the desiredfunction at the boundaries of the spatial domain.

The remaining condition represents the initial temperature distribution

u(x,0) = f(x),

where f(x) is the temperature at position x at time t = 0. All together, the model function u(x, t) thatwe seek should satisfy

ut = a2 uxx

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u(0,t) = u(L,t) = 0 for all t > 0

u(x,0) = f(x).

In this module we are not concerned with finding symbolicdescriptions of the solutions of such

problems. Rather, we will look graphically at the solutions to see

how they evolve in time; how they depend on the boundary conditions; how they depend on the initial condition.

2. Without usingthe applet dothe following.

For the initialtemperaturedistribution givenin Figure 1, sketchwhat you think thetemperaturedistribution u(x,t1)will be a shorttime after thestart, at t = t1.

What will thedistribution beafter asomewhatlongertimet = t2?

Do you think that the temperature distribution will approach a steady state (that is, not change withtime)? Explain.

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5. We may also look at the graph of the temperature function u as both x and t vary. Thissurface can be thought of as an infinite number of time snapshots stacked up one after theother. Identify on this surface each of the curves you obtained in Steps 3 and 4.

The picture below is from the website. Execute the MATLAB fileheatbkd_duke_heat_ex(0,2,0,1,20,20); to obtain a surface you can rotate and print out. On theprint out use a heavy marker to trace the curves you obtained in # 3 and 4 on the surfacegenerated in MATLAB.

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0

0.5

1

1.5

20

0.10.2

0.30.4

0.50.6

0.70.8

0.91

-2

-1

0

1

2

3

4

t-axis, k = 0.05

HEAT Equation by Backward-Difference Method

X-axis, h =0.1

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Approximating the Heat Equation Using Finite Differences

2

a

b

a b finite interval

0

0 for all a b initial temperature distributiona for 0 B.C. at left end of the spatial interval

b for 0 B.C. at right end of the spatial inter

t xxu u x

t

u(x, ) f(x) xu( ,t) u (t) t

u( ,t) u (t) t

=

= =

= val

Select a positive integer m to discretize the spatial

interval [a, b], setting the stepsizeb a

hm

= . Similarly

choose integer n to determine a discretization of the

time interval [0, T] with stepsizeT

kn

= . Hence we have

a two-dimensional space-time grid withi

j

x a ih i 0 1 m

t jk j 0 1 n

, , , ...,

, , , ...,

= + =

=

We use the finite difference approximations

i j i ji j

u x t k u x tx t

kt

( , ) ( , )u ( , )

+ which is O(k)

i j i j ixx i j 2

u x h t 2u x t u x h tx t

h

( , ) ( , ) ( ,u ( , )

+ +

j )

which is O(h2). The approximation forut is a forwarddifference and that foruxx is a centered difference.

At an interior grid point the heat equation isi jx t( , )2

i j xx i jx t x t 0tu ( , ) u ( , ) = and the finite

difference approximation scheme where we use ij ijw u is given by

i j 1 i j i 1 j i j i 1 j2

2

w w w 2w w0

k h

+ + + =

which has local truncation error O(k + h2).

We see that the initial condition 0u(x, ) f(x)= provides the values for i = 0, 1, 2, , m (alongthe bottom row in the figure we know the values ofu) so that when j = 1 (meaning at t = k) we

have an explicit expression for . To see this set j = 0 in the finite difference approximation

scheme and solve for ;

i 0w

i 1w

i 1w

( )2

i 1 i 0 i 10 i 0 i 1 02 2i 1 i 0 i 10 i 1 02 2 2

w w w 2w w 2 k k0 w 1 w w w

k h h h

+ +

+ = = +

+

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This provides values for the values ofu in the second row of the points in the grid. Hence in

general we can solve for values using values previously computed in the j-th row of the grid.

In general we have

i j 1w +

( )2

2i j 1 i j i 1 j i 1 j2 2

2 k kw 1 w w w

h h

+ +

= + +

The stencil for this finite difference expression is

Since we have an explicit scheme we can develop a matrix formulation to estimate the temperature

at each time step. Let

10 1 j0

2 0 2 j12 0 j

2

m 1m 10 m 1 j

w wf x

w wf xk

h

f xw w

( ) ( )

( )

( ), W , W

( )

= = = =

## #

1 2

and let A be the (m-1) (m-1) tridiagonal matrix

1 2 0 0

1 2 0 0

0

0

0 0

A

=

" "

"

% % % % #

# % % % %

# " % % %

" "

.

Then we have wherej j 1( ) ( ) ( )W AW b= + j 1

0 j 1

j 1

m j 1

w

0

0

w

( )b

=

# . The components of are values

from the boundary conditions; here j = 1, 2, , n. We call this the Forward Difference Method.

j 1( )b

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Example: Consider the parabolic equation PDE

2

2

u 1 uon 0 x 1 0 t 1

t 16 x

u 0 t u 1 t 0 boundary conditions

u x 0 2 2 x initial condition

,

( , ) ( , )

( , ) sin( )

=

= =

=

We see that 21

16 = . Let h = k = 0.25. Hence 2

2 2

k 1 0 25

16 4h 0 25

.

.

1= =

=

j 1

. We have m = 4

hence matrix A is 3 3 and . In this case the forward difference method is

given by where

0 5 0 25 0

0 25 0 5 0 25

0 0 25 0 5

. .

A . . .

. .

=

j j 1( ) ( ) ( )W AW b= + j 10

0

0

( )b =

(since the boundary conditions are zero). Thus

the algorithm for this problem is j j 1( ) ( )W AW = j = 1, 2, , 4. Here and we have the

following set of approximations.

0

2

0

2

( )W

=

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A sketch of the approximate values is shown next.

Source: Bradie, Example 10.1

MATLAB command used: w=heatfd_Bradie_ex_10_1(0,1,0,1,4,4) ;(Omitting the; at the end will display the approximations at the time steps with the boundary

values affixed to top and bottom of the columns.)

w=heatfd_Bradie_ex_10_1(0,1,0,1,10,10); w=heatfd_Bradie_ex_10_1(0,1,0,1,20,20);

00.2

0.40.6

0.81

0

0.5

1

-3

-2

-1

0

1

2

X-axis, h =0.1

HEAT Equation by Forward-Difference Method

t-axis, k = 0.1

00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

1

-3

-2

-1

0

1

2

X-axis, h =0.05

HEAT Equation by Forward-Difference Method

t-axis, k = 0.05

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The MATLAB program heatfd_Bradie_ex_10_1

% Program 8.1 Forward difference method for heat equation% Input: space interval [xl,xr], time interval [yb,yt], INPUT description% number of space steps M, number of time steps N% Output: solution w

% Example usage: w=heatfd(0,1,0,1,10,250)

%Original by Sauer; Modified by D. Hill 3/6/07, 3/10/10function w=heatfd(xl,xr,yb,yt,M,N)c=1/16; % diffusion coefficient this can change change to suit the problemh=(xr-xl)/M; k=(yt-yb)/N; m=M-1; n=N;sigma=c*k/(h*h);a=diag(1-2*sigma*ones(m,1))+diag(sigma*ones(m-1,1),1);a=a+diag(sigma*ones(m-1,1),-1); % define matrix a%a