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

0yx 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 statesituations such as:

• steady state temperature distributions

• steady state stress distributions

• 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’sequation; 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 = 0w = 0

axsinww 0π

=

a

b

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

x

yz

0yw

xw

2

2

2

2

=∂∂

+∂∂

Stress analysis example: Dirichlet conditions

Boundary conditionsTo solve:

axxa

wbxw

byyawaxxwbyyw

≤≤=

≤≤=≤≤=≤≤=

0sin),(

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

0 for ,

for , for , for ,

π

Solution by separation of variables

kYY

XX

YY

XX

YXYXyYxXyxw

=′′

−=′′

=′′

+′′

=′′+′′=

0

0)()(),(

from which

and so

as usual …

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

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

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

000),0(

DCyAxyxwByxwyYDC

DCByw

+==≡≡==

===⇒=

withContinue so , then , if

or

ACxyyxwDwA

ADxxw

==≡=

=⇒=

),(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

yDyCxBxAyxwk

ααααα

++==

that so that Suppose

)sincos(sinh),(00),(00)sincos(0),0(

00sinh,10cosh

yDyCxByxwAyxwDCyDyCAyw

ααα

αα

+=⇒=≡⇒===+⇒=

==

withContinue

that Recall

yxBDyxwCyxwB

xBCxw

αα

α

sinsinh),(00),(0

0sinh0)0,(

=⇒=≡⇒=

=⇒=

withContinue

0),(000sinsinh0),(

≡⇒===⇒=

yxwDByaBDyaw

or either so αα

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

Final case k<0

)sinhcosh)(sincos(),(

2

yDyCxBxAyxwk

ααααα

++=−=

that Suppose

)sinhcosh(sin),(000

0)sinhcosh(0),0(

yDyCxByxwAwDC

yDyCAyw

ααα

αα

+=⇒=≡⇒===+⇒=

withcontinue usual, as

yxBDyxwCwB

xBCxw

αα

α

sinhsin),(000

0sin0)0,(

=⇒=≡⇒==⇒=

withcontinue

ya

nxa

nBDyxwa

na

wDByaBDyaw

nπππαα

αα

sinhsin),(0sin

0000sinhsin0),(

=⇒=⇒=

≡⇒===⇒=

or

Solution

ayn

axnKyxw

nn

ππ sinhsin),(1∑∞

=

=Applying the first three boundary conditions, we have

absinh

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

which gives:a

bna

xnKaxw

nn

πππ sinhsinsin1

0 ∑∞

=

=

and the final solution to the stress distribution is

ay

ax

ab

wyxw πππ sinhsin

sinh),( 0=

axsinw)b,x(w 0π

=The final boundary condition is:

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

abn

axnKxfw

nn

ππ 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

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

),( yxn∂∂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 problemy

x

w = 0

w = 0

xa

ww2

sin0π

=

a

b

0=∂∂

xw

0yw

xw

2

2

2

2

=∂∂

+∂∂

To solve:

Boundary conditions

axxa

wbxw

byxw

axxwbyyw

ax

≤≤=

≤≤=∂∂

≤≤=≤≤=

=

02

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

kYY

XX

YY

XX

YXYXyYxXyxw

=′′

−=′′

=′′

+′′

=′′+′′=

0

0)()(),(

from which

and so

as usual …

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

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

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

000),0(

DCyAxyxwByxwyYDC

DCByw

+==≡≡==

===⇒=

withContinue so , then , if

or

ACxyyxwDwA

ADxxw

==≡=

=⇒=

),(0)00

00)0,(

withContinue or (so Either

0),(0000 ≡⇒==⇒=⇒=∂∂

=

yxwCAACyxw

ax

or

That is, the case k=0 is not possible

Case k>0

)sincos)(sinhcosh(),(,2

yDyCxBxAyxwk

ααααα

++==

that so that Suppose

)sincos(sinh),(00),(00)sincos(0),0(

00sinh,10cosh

yDyCxByxwAyxwDCyDyCAyw

ααα

αα

+=⇒=≡⇒===+⇒=

==

withContinue

that Recall

yxBDyxwCyxwB

xBCxw

αα

α

sinsinh),(00),(0

0sinh0)0,(

=⇒=≡⇒=

=⇒=

withContinue

0),(00

0sincosh0

≡⇒==

=⇒=∂∂

=

yxwDB

yaBDxw

ax

or either so

ααα

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

Final case k<0

)sinhcosh)(sincos(),(

2

yDyCxBxAyxwk

ααααα

++=−=

that Suppose

)sinhcosh(sin),(000

0)sinhcosh(0),0(

yDyCxByxwAwDC

yDyCAyw

ααα

αα

+=⇒=≡⇒===+⇒=

withcontinue usual, as

yxBDyxwCwB

xBCxw

αα

α

sinhsin),(000

0sin0)0,(

=⇒=≡⇒==⇒=

withcontinue

ya

nxa

nBDyxwa

na

wDB

yaBDxw

n

ax

2)12(sinh

2)12(sin),(

2)12(0cos

000

0sinhcos0

πππαα

ααα

−−=⇒

−=⇒=

≡⇒==

=⇒=∂∂

=

or

Solution y

anx

anKyxw

nn 2

)12(sinh2

)12(sin),(1

ππ −−=∑

∞

=

Applying the first three boundary conditions, we have

ba

wK

2sinh

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

which gives: ba

nxa

nKaxw

nn 2

)12(sinh2

)12(sin2

sin1

0πππ −−

=∑∞

=

and the final solution to the stress distribution is

ay

ax

ab

wyxw πππ sinhsin

sinh),( 0=

axwbxw

2sin),( 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

rθ

Radius ahalfupper on the ),,( UzrV =θ

0112

2

2

2

22

22 =

∂∂

+∂∂

+∂∂

+∂∂

=∇zVV

rrV

rrVV

θ

I could simply TELL you that Laplace’sEquation 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 Polar Coordinates

yθ

r

x

θcosrx =θsinry =

22 yxr +=

⎟⎠⎞

⎜⎝⎛= −

xytan 1θ

02

2

2

22 =

∂∂

+∂∂

=∇yu

xuu ),,( θrxx = ),r(yy θ=where

0),(),(),(

2 =∇

=

θ

θ

ruruyxu

So, Laplace’s Equation is

We next derive the explicit polar form of Laplace’s Equation in 2D

xu

xr

ru

xu

∂∂

∂∂

+∂∂

∂∂

=∂∂ θ

θ

xu

xxu

xr

ru

xxr

ru

xu

∂∂

⎟⎠⎞

⎜⎝⎛∂∂

∂∂

+∂∂

∂∂

+∂∂

⎟⎠⎞

⎜⎝⎛∂∂

∂∂

+∂∂

∂∂

=∂∂ θ

θθ

θ 2

2

2

2

2

2

Use the product rule to differentiate again

and the chain rule again to get these derivatives

xru

xr

ru

rru

x ∂∂

⎟⎠⎞

⎜⎝⎛∂∂

∂∂

+∂∂

⎟⎠⎞

⎜⎝⎛∂∂

∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

∂∂ θ

θ xru

xr

ru

∂∂

∂∂∂

+∂∂

∂∂

=θ

θ

2

2

2

xu

xru

ru

x ∂∂

⎟⎠⎞

⎜⎝⎛∂∂

∂∂

+∂∂

⎟⎠⎞

⎜⎝⎛∂∂

∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

∂∂ θ

θθθθ xu

xr

ru

∂∂

∂∂

+∂∂

∂∂∂

=θ

θθ 2

22

(*)

Recall the chain rule:

θcosrx = θsinry = 22 yxr += ⎟⎠⎞

⎜⎝⎛= −

xytan 1θ

The 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

,

,

rxy

yrxy

x

rx

yry

x

rx

yr

ry

xr

=∂∂

=∂∂

=∂∂

−=∂∂

=∂∂

=∂∂

θθ

θθin like manner ….

4

2

2

2

43

2

3

2

2

2

2

2

2

2 22ryu

rxyu

rxy

ru

ry

ru

rx

ru

xu

θθθ ∂∂

+∂∂

+−

∂∂∂

+∂∂

+∂∂

=∂∂

Similarly, 4

2

2

2

43

2

3

2

2

2

2

2

2

2 22rxu

rxyu

rxy

ru

rx

ru

ry

ru

yu

θθθ ∂∂

+∂∂

−∂∂

∂+

∂∂

+∂∂

=∂∂

So Laplace’s Equation in polars is 011

2

2

22

2

2

2

2

2

=∂∂

+∂∂

+∂∂

=∂∂

+∂∂

θu

rru

rru

yu

xu

Back to Laplace’s Equation in polar coordinates

Plugging in the formula for the partials on the previous page to the formulae on the one before that we get:

011

0

2

2

22

2

2

2

2

2

=∂∂

+∂∂

+∂∂

=∂∂

+∂∂

θu

rru

rru

yu

xu

to equivalent is

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

Consider a cylindrical capacitor

yθ

r

z

x

zP(r,θ,z)

0112

2

2

2

22

22 =

∂∂

+∂∂

+∂∂

+∂∂

=∇zVV

rrV

rrVV

θ

Laplace’s Equation in cylindrical polars is:

Thin strip of insulating material

0V

rθ

Radius ahalfupper on the ),,( UzrV =θ

In the polar system, note that the solution must repeat itself every θ = 2π

πθπθθπθθθ2:0),a(V

0:U),a(V≤≤∀=≤≤∀=

Boundary conditions

V should remain finite at r=0

There is no variation in V in the z-direction, so 0=∂∂

zV

0Vr1

rV

r1

rV

2

2

22

2

=∂∂

+∂∂

+∂∂

θUsing separation of variables

( ) ( )θΘrRV =

0Rr1R

r1R 2 =′′+′+′′ ΘΘΘ

2rRrRR ′+′′

=′′

−ΘΘ

As before, this means

constant a ,krR

rRR=

′+′′=

ΘΘ′′

− 2

This means we can treat it as a 2D problem

The case k=0

( )

( )DrCbarV

DrCrRrRR

ba

++=

+=⇒=′

+′′

+=Θ⇒=Θ′′

ln)(),(

ln)(0

)(0

θθ

θθ

so and

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

The solution has to remain finite as r→0: c=0

constant a ,gbdrV ==),( θ

The case k<0

( )

( )mm

mm

mm

rDrCrRRrm

rRR

mBmAmmk

+=⇒=+′

+′′

+=Θ⇒=Θ+Θ′′

−=

−)(0

sinhcosh)(0

2

2

2

2

that Suppose

θθθ

The solution has to be periodic in θ, with period 2π. This implies that 0),(0 ≡⇒== θrVBA mm

The case k>0

( )( )n

nn

nnn

nn

nn

nn

rDrCnBnArV

rDrCrRRrn

rRR

nBnAnnk

−

−

++=

+=⇒=−′

+′′

+=Θ⇒=Θ+Θ′′

=

)sincos(),(

)(0

)sincos()(0

2

2

2

2

θθθ

θθθ

that Suppose

Evidently, this is periodic with period 2π

To remain finite as r→0 0=nD

The solution( )∑ ++=

nnn

n nBnArgrV θθθ sincos),(

( )⎩⎨⎧

≤<≤≤

==++∑ πθππθ

θθθ20

0),(sincos

UaVnBnAag

nnn

n

Notice that we have not yet applied the voltage boundary condition!! Now is the time to do so

∫ =π

πθθ2

0

),( UdaVIntegrating V from 0 to 2π:

Left hand side: ( )

2

2sincos2

0

Ug

gdnBnAagn

nnn

=

=++∫ ∑

so and

πθθθπ

Solving for Am and Bm

( )∑ ++=n

nnn nBnAaUaV θθθ sincos

2),(

mAaA

aAdmU

dmnBnAadmUdmaV

mm

m

mm

nnn

n

all for so and ,

,00

0cos

cos)sincos(cos2

cos),(

0

2

0

2

0

2

0

==

+=

++=

∫

∫ ∫ ∫∑

π

πθθ

θθθθθθθθθ

π

π π π

)12(2

cos2

sin

sin)sincos(sin2

sin),(

2

00

2

0

2

0

2

0

−==

+⎥⎦⎤

⎢⎣⎡=

++=

∫

∫ ∫ ∫∑

nmaBmU

aBmmUdmU

dmnBnAadmUdmaV

mm

mm

nnn

n

odd for ,

π

πθθθ

θθθθθθθθθ

ππ

π π π

We apply the orthogonality relationships:

So far, the solution is

∑ −−

+= −−

n

nn nr

anUUrV θπ

θ )12sin()12(

22

),( )12()12(

( ) θπ

θ )1n2sin(a)1n2(

rU22U,rV

1n)1n2(

)1n2(

−−

+= ∑∞

=−

−

Check for r = a, θ = π/2:

( )2

)1n2sin(a)1n2(

aU22U,rV

1n)1n2(

)1n2( ππ

θ −−

+= ∑∞

=−

−

⎥⎦⎤

⎢⎣⎡ ++++= ......

25sin

51

23sin

31

2sinU2

2U πππ

π

⎥⎦⎤

⎢⎣⎡ +−+−+= ......

71

51

311U2

2U

π

U4

U22U

=⎥⎦⎤

⎢⎣⎡+=π

π

0V

rθ

U

An application in image analysis

• We saw that the Gaussian is a solution to the heat/diffusion equation

• We have studied Laplace’s equation• The next few slides hint at the application

of what we have done so far in image analysis

• This is aimed at engaging your interest in PDEs … it is not examined

Laplace’s Equation in image analysis

Image fragment Edge map

How do we compute the edges?

inte

nsity

Signal position, x

d in

tens

ity/ d

x

Constant gradient + max at the step. Note amplified noise

d2in

tens

ity/ d

x2

Zero crossing at step. But doubly amplified noise

• Remove the noise by smoothing • Find places where the second derivative of the image is zero

Gaussian smoothing

Blurring with a Gaussian filter is one way to tame noise

Zero crossings of a second derivative, isotropic operator, after Gaussian smoothing

02 =∇ smoothIAn application of Laplace’s Equation!

Limits of isotropic Gaussian blurring

A noisy image Gaussian blurring

Gaussian is isotropic – takes no account of orientation of image features – so it gives crap edge features

0)(2 =∗∇ IGσ

As the blurring is increased, by increasing the standard deviation of theGaussian, the structure of the image is quickly lost.

Can we do better? Can we make blurring respect edges?

Anisotropic diffusion

22

||

/||11);(

or k,constant somefor ,);(

));((

2

2

kItxg

etxg

ItxgI

kI

Tt

σ

σ

∇+=

=

∇∇=∂∇

−

This is a non-linear version of Laplace’s Equation, in which the blurring is small across an edge feature (low gradient) and large along an edge.

Anisotropic blurring of the noisy image

Top right: Gaussian

Bottom left and right: different anisotropic blurrings

Example of

anisotropic diffusion –brain MRI images, which are very noisy.

Top: Gaussian blur

Middle and bottom: anistropic blur

Anisotropic blurring of the house image retaining important structures at different degrees of non-linear blurring

Two final examples:

Left – original image

Right – anistropic blurring