Solving Laplace Equation

8/13/2019 Solving Laplace Equation

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Laplaces Equation

Separation of variables two examples

Laplaces Equation in Polar Coordinates

Derivation of the explicit form

An example from electrostatics A surprising application of Laplaces eqn

Image analysis

This bit is NOT examined


2/42

0

yx 2

2

2

2

=

+

02 =

Laplaces Equation

In the vector calculus course, this appears as where

=

y

x

Note that the equation has no dependence on time, just on the spatial

variables x,y. This means that Laplaces Equation describes steady state

situations such as:

steady state temperature distributions

steady state stress distributions

steady state potential distributions (it is also called thepotential equation

steady state flows, for example in a cylinder, around a corner,


3/42

Steady state stress analysis problem, which satisfies Laplacesequation; that is, a stretched elastic membrane on a rectangular formerthat has prescribed out-of-plane displacements along the boundaries

y

x

w = 0

w = 0w = 0

axsinww 0 =

a

bw(x,y) is the displacement inz-direction

x

yz

0

y

w

x

w2

2

2

2

=

+

Stress analysis example: Dirichlet conditions

Boundary conditions

To solve:

axxa

wbxw

byyaw

axxw

byyw

=

=

==

0sin),(

00),(

00)0,(

00),0(

0 for,

for,

for,

for,


4/42

Solution by separation of variables

kY

Y

X

X

Y

Y

X

X

YXYX

yYxXyxw

=

=

=

+

=+

=

0

0

)()(),(

from which

and so

as usual

where kis a constant that is either equal to, >, or < 0.


5/42

Case k=0)()(),()( DCyyYBAxxX +=+=

)(),(:0

0),(0)(0

000),0(

DCyAxyxwB

yxwyYDC

DCByw

+==

==

====

withContinue

so,then,if

or

ACxyyxw

DwA

ADxxw

=

==

==

),(

0)00

00)0,(

withContinue

or(soEither

0),(0000),( ==== yxwCAACayyaw or

That is, the case k=0is not possible


6/42

Case k>0

)sincos)(sinhcosh(),(

,2

yDyCxBxAyxw

k

++=

=

thatsothatSuppose

)sincos(sinh),(0

0),(0

0)sincos(0),0(

00sinh,10cosh

yDyCxByxwA

yxwDC

yDyCAyw

+==

==

=+=

==

withContinue

thatRecall

yxBDyxwC

yxwB

xBCxw

sinsinh),(0

0),(0

0sinh0)0,(

==

=

==

withContinue

0),(00

0sinsinh0),(

==

==

yxwDB

yaBDyaw

oreitherso

Again, we find that the case k>0is not possible


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Final case k


8/42

Solution

a

yn

a

xnKyxw

n

n

sinhsin),(

1

=

=Applying the first three

boundary conditions, we

have

a

bsinh

wK 01

=We can see from this that n must take only one value, namely 1, so that

which gives: a

bn

a

xn

Ka

x

w n n

sinhsinsin 10

=

=

and the final solution to the stress distribution is

a

y

a

x

a

b

wyxw

sinhsin

sinh),( 0=

a

xsinw)b,x(w 0

=The final boundary condition is:

Check out: http://mathworld.wolfram.com/HyperbolicSine.html


9/42

abn

axnKxfw

n

n sinhsin)(1

0

==

Then

and as usual we use orthogonality formulae/HLT to find the Kn

y

x

w = 0

w = 0w = 0

)x(fww 0=

a

b

More general boundary condition


10/42

Types of boundary condition1. The value is specified at each point on

the boundary: Dirichlet conditions2. The derivative normal to the boundary is

specified at each point of the boundary:

Neumann conditions3. A mixture of type 1 and 2 conditions is specified

),( yx

),( yxn

Johann Dirichlet (1805-1859)

http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Dirichlet.html

Carl Gottfried Neumann (1832 -1925)

http://www-history.mcs.st-

andrews.ac.uk/history/Mathematicians/Neumann_Carl.html


11/42

A mixed condition problemy

x

w = 0

w = 0

xa

ww2

sin0

=

a

b

0=

x

w

0y

w

x

w2

2

2

2

=

+

To solve:

Boundary conditions

axxawbxw

byx

waxxw

byyw

ax

=

=

=

=

=

02sin),(

00

00)0,(

00),0(

0 for,

for,

for,

for,

A steady state heat transfer

problem

There is no flow of heat across

this boundary; but it does not

necessarily have a constant

temperature along the edge


12/42

Solution by separation of variables

kY

Y

X

X

Y

Y

X

X

YXYX

yYxXyxw

=

=

=

+

=+

=

0

0

)()(),(

from which

and so

as usual

where kis a constant that is either equal to, >, or < 0.


13/42

Case k=0)()(),()( DCyyYBAxxX +=+=

)(),(:0

0),(0)(0

000),0(

DCyAxyxwB

yxwyYDC

DCByw

+==

==

====

withContinue

so,then,if

or

ACxyyxw

DwA

ADxxw

=

==

==

),(

0)00

00)0,(

withContinue

or(soEither

0),(0000 ====

=

yxwCAACyx

w

ax

or

That is, the case k=0is not possible


14/42

Case k>0

)sincos)(sinhcosh(),(

,2

yDyCxBxAyxw

k

++=

=

thatsothatSuppose

)sincos(sinh),(0

0),(0

0)sincos(0),0(

00sinh,10cosh

yDyCxByxwA

yxwDC

yDyCAyw

+==

==

=+=

==

withContinue

thatRecall

yxBDyxwC

yxwB

xBCxw

sinsinh),(0

0),(0

0sinh0)0,(

==

=

==

withContinue

0),(00

0sincosh0

==

==

=

yxwDB

yaBDx

w

ax

oreitherso

Again, we find that the case k>0is not possible


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Final case k


16/42

Solutiony

a

nx

a

nKyxw

n

n2

)12(sinh

2

)12(sin),(

1

=

=

Applying the first three

boundary conditions, we

have

ba

wK

2sinh

01 =We can see from this that n must take only one value, namely 1, so that

which gives: ba

nx

a

nK

a

xw

nn 2

)12(sinh

2

)12(sin

2sin

10

=

=

and the final solution to the stress distribution is

a

y

a

x

a

b

wyxw

sinhsin

sinh),( 0=

a

xwbxw

2sin),( 0

=The final boundary condition is:

Check out: http://mathworld.wolfram.com/HyperbolicSine.html


17/42

PDEs in other coordinates

In the vector algebra course, we find that it

is often easier to express problems incoordinates other than (x,y), for example

in polar coordinates (r,)

Recall that in practice, for example for

finite element techniques, it is usual to use

curvilinear coordinates but we wont gothat far

We illustrate the solution of Laplaces Equation using polar

coordinates**Kreysig, Section 11.11, page 636


18/42

A problem in electrostatics

Thin strip of insulating material

0V

r

Radius a

halfupperon the

),,( UzrV =

011

2

2

2

2

22

22 =

+

+

+

=

z

VV

rr

V

rr

VV

I could simply TELL you that LaplacesEquation in cylindrical polars is:

This is a cross section of a

charged cylindrical rod.

brief time out while

I DERIVE this


19/42

2D Laplaces Equation in Polar Coordinates

y

r

x

cosrx=

sinry=

22yxr +=

=

x

ytan

1

02

2

2

22 =

+

=

y

u

x

uu ),,( rxx= ),r(yy =where

0),(

),(),(2 =

=

ru

ruyxu

So, Laplaces Equation is

We next derive the explicit polar form of Laplaces Equation in 2D


20/42

x

u

x

r

r

u

x

u

+

=

x

u

xx

u

x

r

r

u

xx

r

r

u

x

u

+

+

+

=

2

2

2

2

2

2

Use the product rule to differentiate again

and the chain rule again to get these derivatives

xr

u

x

r

r

u

rr

u

x

+

=

xr

u

x

r

r

u

+

=

2

2

2

x

u

x

ru

r

u

x

+

=

x

u

x

r

r

u

+

=

2

22

(*)

Recall the chain rule:


21/42

cosrx= sinry=22

yxr +=

=

x

ytan

1The required partial derivatives

rx

xrx

xrryxr =

=

+= 22222

ry

yr=

Similarly,

42

2

42

2

22

3

2

2

2

3

2

2

2

2,

2

,

,

r

xy

yr

xy

x

r

x

yr

y

x

r

x

y

r

r

y

x

r

=

=

=

=

=

=

in like manner .


22/42


23/42

011

0

2

2

22

2

2

2

2

2

=+

+

=

+

u

rr

u

rr

u

y

u

x

u

toequivalentis


24/42

Example of Laplace in Cylindrical Polar Coordinates (r,, z)

Consider a cylindrical capacitor

y

r

z

x

zP(r,,z)

011

2

2

2

2

22

22 =

+

+

+

=

z

VV

rr

V

rr

VV

Laplaces Equation in

cylindrical polars is:

Thin strip of insulating material

0V

r

Radius ahalfupperon the

),,( UzrV =

In the polar system, note that thesolution must repeat itself every =2

2:0),a(V

0:U),a(V

=

=

Boundary conditions

V should remain finite at r=0


25/42

There is no variation in Vin the z-direction, so

0=

z

V

0V

r

1

r

V

r

1

r

V2

2

22

2

=

+

+

Using separation of variables( ) ( )rRV=

0R

r

1R

r

1R

2

=++

2rR

rRR +=

As before, this means

constanta,krR

rRR=

+=

2

This means we can treat it as a 2D problem


26/42

The case k=0

( )

( )DrCbarV

DrCrRr

RR

ba

++=

+==+

+==

ln)(),(

ln)(0

)(0

soand

The solution has to be periodic in 2: a=0

The solution has to remain finite as r0: c=0

constanta,gbdrV ==),(


27/42

The case k


28/42

The case k>0

( )

( )nnnnnn

n

n

n

n

nn

rDrCnBnArV

rDrCrRRr

n

r

RR

nBnAn

nk

++=

+==+

+==+

=

)sincos(),(

)(0

)sincos()(0

2

2

2

2

thatSuppose

Evidently, this is periodic with period 2

To remain finite as r0 0=nD


29/42

The solution( ) ++=

n

nn

n nBnArgrV sincos),(

( )

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Solving Laplace Equation