Transcript

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Partial Differential Equations - Background

• Physical problems are governed by many PDEs

• Some are governed by first order PDEs

• Numerous problems are governed by second order PDEs

• A few problems are governed by fourth-order PDEs.

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Examples

2

2

x

TC

t

T:)D1(EquationConductionHeat

2

2

2

2

y

T

x

TC

t

T:)D2(EquationConductionHeat

0yx

:EquationLaplaceD22

2

2

22

)y,x(fyx

:EquationPoissonD22

2

2

22

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Examples (contd.)

2

2

2

2

x

uC

t

u:EquationWaveD1

2

2

2

2

2

2

y

u

x

uC

t

u

:)membranevibrating(EquationWaveD2

)plateVibrating(02

t

u2

h4

y

u4

2y

2x

u4

24

x

u4

D

)beamVibrating(04

x

u4

C2

t

u2

:EquationsthOrder4

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Classification of Partial Differential Equations (PDEs)

There are 6 basic classifications:

(1) Order of PDE

(2) Number of independent variables

(3) Linearity

(4) Homogeneity

(5) Types of coefficients

(6) Canonical forms for 2nd order PDEs

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

(1) Order of PDEs

The order of a PDE is the order of the highest partial derivative in the equation.

Examples:

(2nd order)

(1st order)

(3rd order)

2

2

x

utu

xu

tu

xsinx

uu

tu

3

3

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

(2) Number of Independent Variables

Examples:

(2 variables: x and t)

(3 variables: r, , and t)

2

2

x

utu

2

2

22

2 u

r

1ru

r1

r

utu

(3) Linearity

PDEs can be linear or non-linear. A PDE is linear if the dependent variable and all its derivatives appear in a linear fashion (i.e. they are not multiplied together or squared for example.

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Examples: (Linear)

(Non-linear)

(Linear)

(Non-linear)

(Non-linear)

(Linear)

(Non-linear)

tsinx

ue

t

u2

2t

2

2

0y

uy

x

u2

2

2

2

0tu

x

uu

2

2

0uyu

yxu

x 2

y2

2

2

eusinxu

x

u

xsiny

uyx

u2

x

u2

22

2

2

1yu

uxu 2

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

(4) Homogeneity

A PDE is called homogenous if after writing the terms in order, the right hand side is zero.

Examples:

(Non-homogeneous)

(Homogeneous)

(Homogenous)

)y,x(fy

u

x

u2

2

2

2

0tu

x

u2

2

ux

u

t

u2

2

2

2

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Examples

(Non-homogeneous)

(Homogeneous)

5utu

xu

5ut

)5u(x

)5u(

(5) Types of Coefficients

If the coefficients in front of each term involving the dependent variable and its derivatives are independent of the variables (dependent or independent), then that PDE is one with constant coefficients.

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Examples

(Variable coefficients)

(C constant; constant coefficients)

0y

ux

x

u2

22

2

2

0t

uC

x

u2

2

2

2

(6) Canonical forms for 2nd order PDEs (Linear)

(Standard Form)GFuyu

Exu

Dy

uC

yxu

Bx

uA

2

22

2

2

where A, B, C, D, E, F, and G are either real constants or real-valued functions of x and/or y.

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

PDE is Elliptic 0AC4B2

PDE is Hyperbolic 0AC4B2

PDE is Parabolic 0AC4B2

Parabolic PDE solution “propagates” or diffuses

Hyperbolic PDE solution propagates as a wave

Elliptic PDE equilibrium

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

This terminology of elliptic, parabolic, and hyperbolic, reflect the analogy between the standard form for the linear, 2nd order PDE and conic sections encountered in analytical geometry:

for which when one obtains the equation for an ellipse, when one obtains the equation for a parabola, and when

one gets the equation for a hyperbola.

0FEyDxCyBxyAx 22 0AC4B2

0AC4B2

0AC4B2

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Examples

(a) Here, A=1, B=0, C=2, D=E=F=G=0 B2-4AC = 0 - 4(1)(2) = -8 < 0 this equation is elliptic.

(b)

Here, A=1, B=0, C=-2, D=E=F=G=0 B2-4AC = 0 - 4(1)(-2) = 8 > 0 this equation is hyperbolic.

(c)

Here, A=1, B=0, E=-2, C=D=F=G=0 B2-4AC = 0 - 4(1)(0) = 0 this equation is parabolic.

0y

u2

x

u2

2

2

2

0y

u2

x

u2

2

2

2

0yu

2x

u2

2

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Examples

(d) Here, A=1, B=-4, C=1, D=E=F=G=0 B2-4AC = 16 - 4(1)(1) = 12 > 0 this equation is hyperbolic.

(e)

Here, A=3, B=-4, C=-5, D=E=F=G=0 B2-4AC = 16 - 4(3)(-5) = 76 > 0 this equation is hyperbolic.

(f) Here, A=3, B=-4, C=-5, D=8, E=-9, F=6, G=27exy B2-4AC = 16 - 4(3)(-5) > 0 this equation is hyperbolic.

0y

uyx

u4

x

u2

22

2

2

xy2

22

2

2

e27u6yu

9xu

8y

u5

yxu

4x

u3

0y

u5

yxu

4x

u3

2

22

2

2

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Examples

(g)

Here, A=y, B=0, C=-1, D=E=F=G=0 B2-4AC = 0 - 4(y)(-1) = 4y for y>0, this equation is hyperbolic; for y=0, this equation is parabolic; for y<0, this equation is elliptic.

0y

u

x

uy

2

2

2

2

Hyperbolic

x

y

Elliptic Parabolic

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Examples

(h)

Here, A=1, B=2x, C=1-y2, D=E=F=G=0 B2-4AC = 4x2 - 4(1)(1-y2) = 4x2+4y2-4 or x2+y2 >,=,< 0

0y

u)y1(

yxu

x2x

u2

22

2

2

2

Elliptic

Hyperbolic

x

y

Parabolic onsurface of circle

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Examples

(i)

Here, A=1, B=-y, C=0, D=E=F=G=0 B2-4AC = y2 for y=0, this equation is parabolic; for y0, this equation is hyperbolic.

0uyu

yxu

xyx

uy

x

u 2

2

2

Hyperbolic

Hyperbolic

x

y

Parabolic

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Example

(j)

Here, A=sin2x, B=sin2x, C=cos2x, D=E=F=G=0 B2-4AC = sin22x-4sin2xcos2x = 4sin2xcos2x-4sin2xcos2x = 0 this equation is parabolic everywhere.

xy

u)x(cos

yxu

)x2(sinx

u)x(sin

2

22

2

2

22

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Example

(k)

This must be converted to 2nd order form first:

and

subtracting,

Now, A=1, B=0, C=-1, D=E=F=G=0 B2-4AC=4 > 0

Hyperbolic.

0xu

tu

0xtu

t

u 2

2

2

0

x

utx

u2

22

0x

u

t

u2

2

2

2

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Example

(l)

Again, convert to 2nd order form first:

and

adding,

Again, A=1, B=0, C=-1, D=E=F=G=0 B2-4AC = 4 > 0 Hyperbolic.

0xu

tu

0xtu

t

u 2

2

2

0

x

utx

u2

22

0x

u

t

u2

2

2

2

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Wave equation (hyperbolic)

The wave equation has the form:

This equation can be factored as follows:

This implies that x+t and x-t define characteristic directions, i.e. directions along which the PDE will collapse into an ODE.

0t

u

x

u2

2

2

2

utxtxt

uxu

txt

u

x

u2

2

2

2

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Wave equation (contd.)

Let =x+t and =x-t

Similarly,

Thus, and

xxx

ttt

2tx

2tx

0u

4u22t

u

x

u 2

2

2

2

2

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Wave equation (contd.)

Thus, and

and

or, by integrating again,

0u

0u

)(*fu

)(*g

u

direction0xtheintraveling.e.i,wavetraveling

rightwardrepresents

direction0xtheintraveling.e.i,wavetraveling

leftwardrepresents

)tx(g)tx(f)(g)(fu

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Note that there are other important equations in mathematical physics, such as:

Schroedinger eq. 1-D

which is a wave equation by virtue of the imaginary constant i. Note that but for the “i” (= ), this equation would be parabolic. However, the “i” makes all the difference and this is a wave equation (hyperbolic).

)x(Vxm2t

i2

22

1

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

We now return to the case study problem for adiabatic ( ), frictionless, quasi-1D flow:

For simplicity, we have assumed that ne=0 (non-ionized flow) for now.

0Q

0uAP2u

)1(RT

xA1

2u

)1(RT

t

xP

x)Au(

A1

t)u(

0x

)uA(A1

t

22

2

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Recall that for steady flow, we found that M = 1 when dA/dx = 0 (choking or sonic condition),

where

•It turns out that this system of equations exhibits elliptic character for M<1, and hyperbolic character for M>1.•Thus, M=1, the choking point, exists to delineate the elliptic and hyperbolic regions of the flow.

RTu

M

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

• If we are interested in transient, i.e. time-dependent solutions, analytical solutions do not exist (except for drastic simplifications) and the governing equations must be solved numerically.

• If we are interested in obtaining steady state solutions, they can be obtained numerically as well (in quasi-1D flow, analytical solutions were obtained), in one of two ways:– Direct solution of steady state equations, or

– Time-marching from an arbitrary initial state to the steady state.

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Conservation equations for quasi-1D, isothermal flow

At steady state,

• An analytical solution can be obtained for this case.

RTP

ttanconsT

0xu

uxP

ttanconsmuA

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

Conservation equations for quasi-1D, isothermal flow (contd.)

• A steady state solution can also be obtained by time-marching, i.e. solving the unsteady equations:

RTP

ttanconsTxP

x)Au(

A1

t)u(

0x

)uA(A1

t2

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