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Computational Fluid Dynamics IPPPPIIIIWWWW

Theory ofPartial Differential

Equations

Instructor: Hong G. ImUniversity of Michigan

Fall 2001


• Basic Properties of PDE

• Quasi-linear First Order Equations

- Characteristics

- Linear and Nonlinear Advection Equations

• Quasi-linear Second Order Equations

- Classification: hyperbolic, parabolic, elliptic

- Domain of Dependence/Influence

- Ill-posed Problems

Outline

Computational Fluid Dynamics IPPPPIIIIWWWW• The order of PDE is determined by the highest

derivatives• Linear if no powers or products of the unknown

functions or its partial derivatives are present.

• Quasi-linear if it is true for the partial derivatives of highest order.

Definitions

02, =+∂∂+

∂∂=

∂∂+

∂∂ xf

yf

xff

yfy

xfx

222

2

2

2

, fyfy

xfxf

yf

xff =

∂∂+

∂∂=

∂∂+

∂∂

Computational Fluid Dynamics IPPPPIIIIWWWW• The general solution of PDE are arbitrary functions

of specific functions.

Definitions

)2(;02 yxfyf

xf +==

∂∂−

∂∂ ψ

)()(

;02

22

2

2

dycxbyaxfyfC

yxfB

xfA

+++=

=∂∂+

∂∂∂+

∂∂

φψ


Quasi-linear first order partial differential equations


cyfb

xfa =

∂∂+

∂∂

),,(),,(),,(

fyxccfyxbbfyxaa

===

Consider the quasi-linear first order equation

where the coefficients are functions of x,y, and f, but not the derivatives of f:


The solution of this equations defines a single valued surface f(x,y) in three-dimensional space:

f=f(x,y)

f

x

y


dyyfdx

xfdf

∂∂+

∂∂=

An arbitrary change in f is given by:

f=f(x,y)

f

x

y

f

dxdy

df


cyfb

xfa =

∂∂+

∂∂

dyyfdx

xfdf

∂∂+

∂∂=

0),,(1,, =⋅

−

∂∂

∂∂ cba

yf

xf

0),,(1,, =⋅

−

∂∂

∂∂ dfdydx

yf

xf

⇒

⇒

The original equation and the conditions for a small change can be rewritten as:


0),,(1,, =⋅

−

∂∂

∂∂ cba

yf

xf 0),,(1,, =⋅

−

∂∂

∂∂ dfdydx

yf

xf

),,(),,( cbadsdfdydx =

;;; cdsdfb

dsdya

dsdx ===

ba

dydx =

⇒

⇒

Both

Picking the displacement in the direction of (a,b,c)

Separating the components

Characteristics


;;; cdsdfb

dsdya

dsdx ===

ba

dydx =

The three equations

Given the initial conditions:

);();();( ooo sffsyysxx ===

the equations can be integrated in time


x(s); y(s); f (s)

x

y

x(0); y(0); f (0)


0=∂∂+

∂∂

xfU

tf

;0;;1 ===dsdfU

dsdx

dsdt

;0; == dfUdtdx

Consider the linear advection equation

The characteristics are given by:

or

Which shows that the solution moves along straight characteristics without changing its value


0=∂∂+

∂∂

xfU

tf

Graphically:

Notice that these results are specific for:

ft

f

x


fxfU

tf −=

∂∂+

∂∂

Add a source

t

f

f

x

;;;1 fdsdfU

dsdx

dsdt −===

tff

fdtdfU

dtdx

−=⇒

−==

e)0(

;


or

Moving wave with decaying amplitude


0=∂∂+

∂∂

xff

tf

;0;;1 ===dsdff

dsdx

dsdt

;0; == dffdtdx

Consider a nonlinear (quasi-linear) advection equation


or

The slope of the characteristics depends on the value of f(x,t).


t

fx

t3

t2

t1

Multi-valued solution


02

2

=∂∂−

∂∂+

∂∂

xf

xff

tf ε

Why unphysical solutions?

- Because mathematical equation neglects some physical process (dissipation)

Additional condition is required to pick out the physicallyRelevant solution

Correct solution is expected from Burgers equation with

Burgers Equation

0→εEntropy Condition (to be continued)


Quasi-linear second order partial differential equations


dy

fcyxfb

xfa =

∂∂+

∂∂∂+

∂∂

2

22

2

2

a = a(x,y, f )b = b(x, y, f )c = c(x, y, f )

d = d(x, y, f )

Consider a general quasi-linear PDE

where


dy

fcyxfb

xfa =

∂∂+

∂∂∂+

∂∂

2

22

2

2

xfv

∂∂=

yfw

∂∂=

dywc

yvb

xva =

∂∂+

∂∂+

∂∂

0=∂∂−

∂∂

yv

xw

∂∂

∂=∂∂

∂xyf

yxf 22

Starting with

Define

PDE becomes

with


=

∂∂∂∂

−

+

∂∂∂∂

001ad

ywyvacab

xwxv

sAuu yx =+

dywc

yvb

xva =

∂∂+

∂∂+

∂∂ 0=

∂∂−

∂∂

yv

xw

In matrix form

=

wv

u

Are there lines in the x-y plane, along which the solution is determined by an ordinary differential equation?


yv

xv

xy

yv

xv

dxdv

∂∂+

∂∂=

∂∂

∂∂+

∂∂= α

∂∂=xyα

adl

yv

xwl

yw

ac

yv

ab

xvl 121 =

∂∂−

∂∂+

∂∂+

∂∂+

∂∂

adl

yw

xwl

yv

xvl 121 =

∂∂+

∂∂+

∂∂+

∂∂ αα

Is this ever equal to

yw

xw

xy

yw

xw

dxdw

∂∂+

∂∂=

∂∂

∂∂+

∂∂= α


α21 lacl =

α121 llabl =−

adl

yv

xwl

yw

ac

yv

ab

xvl 121 =

∂∂−

∂∂+

∂∂+

∂∂+

∂∂

adl

yw

xwl

yv

xvl 121 =

∂∂+

∂∂+

∂∂+

∂∂ αα

Computational Fluid Dynamics IPPPPIIIIWWWWSolution determined by ODE (Characteristics exist) only if

α

α

21

121

lacl

llabl

=

=−

0;001

2

1 =−

=

−−−

αα

αjiji lAl

ll

acab

Or, in matrix form:

Compatibility Relation


0=+

−−

ac

ab αα

( )acbba

421 2 −±=α

=

−−−

001

2

1

ll

acab

αα

The determinant is:

Solving for α

0T =− IA α


042 =− acb

042 >− acb

042 <− acb

Two real characteristics

One real characteristics

No real characteristics

( )acbba

421 2 −±=α


02

22

2

2

=∂∂−

∂∂

yfc

xf

dy

fcyxfb

xfa =

∂∂+

∂∂∂+

∂∂

2

22

2

2

Examples

Comparing with the standard form

shows that ;0;;0;1 2 =−=== dccba

041404 2222 >=⋅⋅+=− ccacb

Hyperbolic

Wave Equation


02

2

2

2

=∂∂+

∂∂

yf

xf

Examples


shows that ;0;1;0;1 ==== dcba

0411404 22 <−=⋅⋅−=− acb

Elliptic

Laplace Equation

dy

fcyxfb

xfa =

∂∂+

∂∂∂+

∂∂

2

22

2

2


2

2

yfD

xf

∂∂=

∂∂

dy

fcyxfb

xfa =

∂∂+

∂∂∂+

∂∂

2

22

2

2

Examples


shows that ;0;;0;0 =−=== dDcba

00404 22 =⋅⋅+=− Dacb

Parabolic

Heat Equation


02

2

2

2

=∂∂+

∂∂

yf

xf EllipticLaplace equation

02

22

2

2

=∂∂−

∂∂

yfc

xf

HyperbolicWave equation

2

2

yfD

xf

∂∂=

∂∂ ParabolicHeat equation

Examples


( )acbba

421 2 −±=α

xy

∂∂=α

Hyperbolic Parabolic Elliptic

042 >− acb 042 =− acb 042 <− acb


02

22

2

2

=∂∂−

∂∂

xfc

tf

First write the equation as a system of first order equations

;;xfcv

tfu

∂∂=

∂∂=

0

0

=∂∂−

∂∂

=∂∂−

∂∂

xuc

tv

xvc

tuyielding

Introduce

Wave equation

from the PDE

sincextf

txf

∂∂∂=

∂∂∂ 22


Domain of Domain of InfluenceInfluence

Domain of Domain of DependenceDependence

t

x

P

Two characteristic linescdx

dtcdx

dt 1;1 −=+=

cdxdt 1+=

cdxdt 1−=


Ill-posed problems

Computational Fluid Dynamics IPPPPIIIIWWWWIll-posed Problems

2

2

2

2

xf

tf

∂∂−=

∂∂

Consider the initial value problem:

02

2

2

2

=∂∂+

∂∂

xf

tf

This is simply Laplace’s equation

which has a solution if ∂f/∂x or f are given on the boundaries


Here, however, this equation appeared as an initial value problem, where the only boundary conditions available are at t = 0. Since this is a second order equation we will need two conditions, we may assume are that f and ∂f/∂x are specified at t = 0.

ikxetaf )(=

akdt

ad 22

2

=

Consider separation of variables

PDE becomes


)0( ;)0( ikaa

akdt

ad 22

2

=

determined by initial conditions

General solutionktkt BeAeta −+=)(

BA,

Therefore,

∞→)(ta as ∞→t

Ill-posed Problem

Computational Fluid Dynamics IPPPPIIIIWWWWIll-posed Problems: Summary

IVP ill-posedBVP well-posed

IVP well-posed BVP ill-posed

IVP well posedBVP ill-posed

N/ANo characteristics

Infinite speed ofcharacteristics

Finite speeds of characteristics

Solutions are always smooth

Solutions are always smooth

Solutions may not be smooth

Data needed over entire boundary

Data needed over part of boundary

Data needed over part of boundary

No real characteristics

One real characteristics

Two real characteristics

EllipticParabolicHyperbolic

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