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### Transcript of Part 5 Laplace Equation jmb/lectures/pdelecture5.pdf · PDF file Steady state stress...

• Laplace’s Equation

• Separation of variables – two examples • Laplace’s Equation in Polar Coordinates

– Derivation of the explicit form – An example from electrostatics

• A surprising application of Laplace’s eqn – Image analysis – This bit is NOT examined

• 0 yx 2 2

2

2

= ∂ ∂

+ ∂ ∂ φφ

02 =∇ φ

Laplace’s 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 Laplace’s Equation describes steady state situations such as:

• steady state potential distributions (it is also called the potential equation

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

• Steady state stress analysis problem, which satisfies Laplace’s equation; that is, a stretched elastic membrane on a rectangular former that has prescribed out-of-plane displacements along the boundaries

y

x

w = 0

w = 0 w = 0

a xsinww 0 π

=

a

b

w(x,y) is the displacement in z-direction

x

y z

0 y w

x w

2

2

2

2

= ∂ ∂

+ ∂ ∂

Stress analysis example: Dirichlet conditions

Boundary conditions To solve:

axx a

wbxw

byyaw axxw byyw

≤≤=

≤≤= ≤≤= ≤≤=

0sin),(

00),( 00)0,( 00),0(

0 for ,

for , for , for ,

π

• Solution by separation of variables

k Y Y

X X

Y Y

X X

YXYX yYxXyxw

= ′′

−= ′′

= ′′

+ ′′

=′′+′′ =

0

0 )()(),(

from which

and so

as usual …

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

• Case k=0 )()(),()( DCyyYBAxxX +=+=

)(),(:0 0),(0)(0

000),0(

DCyAxyxwB yxwyYDC

DCByw

+== ≡≡==

===⇒=

withContinue so , then , if

or

ACxyyxw DwA

= =≡=

=⇒=

),( 0)00

00)0,(

withContinue or (so Either

0),(0000),( ≡⇒==⇒=⇒= yxwCAACayyaw or

That is, the case k=0 is not possible

• Case k>0

)sincos)(sinhcosh(),( ,2

yDyCxBxAyxw k

αααα α

++= =

that so that Suppose

)sincos(sinh),(0 0),(0 0)sincos(0),0(

00sinh,10cosh

yDyCxByxwA yxwDC yDyCAyw

ααα

αα

+=⇒= ≡⇒== =+⇒=

==

withContinue

that Recall

yxBDyxwC yxwB

xBCxw

αα

α

sinsinh),(0 0),(0

0sinh0)0,(

=⇒= ≡⇒=

=⇒=

withContinue

0),(00 0sinsinh0),(

≡⇒== =⇒=

yxwDB yaBDyaw

or either so αα

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

• Final case k

• Solution

a yn

a xnKyxw

n n

ππ sinhsin),( 1 ∑ ∞

=

=Applying the first three boundary conditions, we have

a bsinh

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

which gives: a

bn a

xnK a xw

n n

πππ sinhsinsin 1

0 ∑ ∞

=

=

and the final solution to the stress distribution is

a y

a x

a b

wyxw πππ sinhsinsinh ),( 0=

a xsinw)b,x(w 0 π

=The final boundary condition is:

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

• a bn

a xnKxfw

n n

ππ sinhsin)( 1

0 ∑ ∞

=

= Then

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

y

x

w = 0

w = 0 w = 0

)x(fww 0=

a

b

More general boundary condition

• Types of boundary condition

1. The value is specified at each point on the boundary: “Dirichlet conditions”

2. The derivative normal to the boundary is specified at each point of the boundary: “Neumann conditions”

3. A mixture of type 1 and 2 conditions is specified

),( yxφ

),( yx n∂ ∂φ

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

• A mixed condition problem y

x

w = 0

w = 0

x a

ww 2

sin0 π

=

a

b

0= ∂ ∂

x w

0 y w

x w

2

2

2

2

= ∂ ∂

+ ∂ ∂

To solve:

Boundary conditions

axx a

wbxw

by x w

axxw byyw

ax

≤≤=

≤≤= ∂ ∂

≤≤= ≤≤=

=

0 2

sin),(

00

00)0,( 00),0(

0 for ,

for ,

for , for ,

π

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

• Solution by separation of variables

k Y Y

X X

Y Y

X X

YXYX yYxXyxw

= ′′

−= ′′

= ′′

+ ′′

=′′+′′ =

0

0 )()(),(

from which

and so

as usual …

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

• Case k=0 )()(),()( DCyyYBAxxX +=+=

)(),(:0 0),(0)(0

000),0(

DCyAxyxwB yxwyYDC

DCByw

+== ≡≡==

===⇒=

withContinue so , then , if

or

ACxyyxw DwA

= =≡=

=⇒=

),( 0)00

00)0,(

withContinue or (so Either

0),(0000 ≡⇒==⇒=⇒= ∂ ∂

=

yxwCAACy x w

ax

or

That is, the case k=0 is not possible

• Case k>0

)sincos)(sinhcosh(),( ,2

yDyCxBxAyxw k

αααα α

++= =

that so that Suppose

)sincos(sinh),(0 0),(0 0)sincos(0),0(

00sinh,10cosh

yDyCxByxwA yxwDC yDyCAyw

ααα

αα

+=⇒= ≡⇒== =+⇒=

==

withContinue

that Recall

yxBDyxwC yxwB

xBCxw

αα

α

sinsinh),(0 0),(0

0sinh0)0,(

=⇒= ≡⇒=

=⇒=

withContinue

0),(00

0sincosh0

≡⇒==

=⇒= ∂ ∂

=

yxwDB

yaBD x w

ax

or either so

ααα

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

• Final case k

• Solution y

a nx

a nKyxw

n n 2

)12(sinh 2

)12(sin),( 1

ππ −− =∑

=

Applying the first three boundary conditions, we have

b a

wK

2 sinh

0 1 π=We can see from this that n must take only one value, namely 1, so that

which gives: b a

nx a

nK a xw

n n 2

)12(sinh 2

)12(sin 2

sin 1

0 πππ −−

=∑ ∞

=

and the final solution to the stress distribution is

a y

a x

a b

wyxw πππ sinhsinsinh ),( 0=

a xwbxw

2 sin),( 0

π =The final boundary condition is:

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

• PDEs in other coordinates…

• In the vector algebra course, we find that it is often easier to express problems in coordinates other than (x,y), for example in polar coordinates (r,Θ)

• Recall that in practice, for example for finite element techniques, it is usual to use curvilinear coordinates … but we won’t go that far

We illustrate the solution of Laplace’s Equation using polar coordinates*

*Kreysig, Section 11.11, page 636

• A problem in electrostatics

Thin strip of insulating material

0V

r θ

Radius a halfupper on the ),,( UzrV =θ

011 2 2

2

2

22

2 2 =

∂ ∂

+ ∂ ∂

+ ∂ ∂

+ ∂ ∂

=∇ z VV

rr V

rr VV

θ

I could simply TELL you that Laplace’s Equation in cylindrical polars is:

This is a cross section of a charged cylindrical rod.

… brief time out while I DERIVE this

• 2D Laplace’s Equation in