POTENTIAL THEORY AND ELLIPTIC PARTIAL DIFFERENTIAL EQUATIONS Study Results

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Chapter

Title Page Number

0 INTRODUCTION 2

1 PRELIMINARIES 3

2 POTENTIAL THEORY AND ELLIPTICPARTIAL DIFFERENTIAL EQUATIONS

11

3 APPLICATIONS 31

CONTENTS

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INTRODUCTION

Differential equations serve as an effective tool to study the changes

in the mathematical as well as the physical world. Partial differential

equation of second order arises frequently in mathematical physics. It is of

this reason that the study of such equations are of great value. This project

work is targeting at ”Potential theory and Elliptic partial differential

equations”. Laplace’s equation is a second order partial differential

equation which is elliptic in nature.

This project includes three chapters. The first chapter includes some

important definitions and necessary theorems which are needed for the

following chapters. Second chapter consist potential theory of Laplace’s

equation, solutions of Laplace’s equation, fundamental solution of Green’s

function for Laplace’s equation and examples. Third chapter highlights

some important applications of Laplace’s equation in mathematics as well

as in Physics such as Dirichlet problem for the upper half plane, the

velocity potential for an irrotational flow of an incompressible fluid etc.

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CHAPTER-1

PRELIMINARIES

Definition 1:1

The Laplace’s equation in two dimension is, ∇2U= ∂2u∂x2 +

∂2u∂ y2=0

That is, ∇2u=uxx+uyy=0

A solution of two dimensional Laplace’s equation is called two

dimensional harmonic function.

Definition 1:2

The m- dimensional case of Laplace’s equation is called potential

equation. It is denoted by ∆mu=uxαxα=0

That is, ux1x1+ux

2x2…u x

mxm=0

Definition 1:3

Any problem in which we are required to find a function u is called a

boundary value problem for Laplace’s equation. Let D be the interior of a

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simple, closed, smooth curve B and f be a conscious function defined on

the boundary B

The problem of finding a harmonic function u ( x , y ) in D such that it

coincides with f on the boundary B is called the first boundary value

problem or Dirichlet problem.

The problem of finding a function u ( x , y ) such that it is harmonic in D

and satisfies ∂u∂n=f( s) on B,(where ∂∂n is the directional derivative along the

outward normal) with the condition ∫B

❑

f (s )ds=0 is called the second

boundary value problem or Neumann problem.

The problem of finding a function u ( x , y ) which is harmonic in D and

satisfies the condition ∂u∂n +h ( s)u (s )=0 on B where h ( s)≥0 ,

And h ( s)≢0is called the third boundary value problem or Robin

problem.

The fourth boundary value problem involves finding a function

u ( x , y ) which is harmonic in D and satisfies the boundary conditions of

different types on different portions of the boundary B.

For example,

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u=f 1 (s )on B1 , ∂u∂n=f 2 ( s ) on B2,where B1UB2=B

Definition 1:4

Let A ux x+Bu yy+Cuy y=F (x , y ,u , ux , uy ) ……………….(1)

Where A( x , y ) , B( x , y ) and C( x , y ) are functions of x and y and⎾0 be a

curve in the x , y – plane. The problem of finding the solution u ( x , y ) of the

partial differential equation (1) in the neighbourhood of ⎾0 satisfying the

following conditions.

u=f ( s ) ……………….(2)

∂u∂n

=g (s ) ……………….(3)

On ⎾0 (s ) is called a Cauchy problem. The conditions (2) and (3) are

called the Cauchy conditions.

Example: The problem of vibrations of an infinite string is a Cauchy

problem.

Definition 1:5

If U( x , y ) and V( x , y ) are functions defined inside and on the boundary

B of the closed region D, then

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∫D

❑

[ ∂U∂x + ∂V∂ y ]dS=∫

B

❑

(Udy−Vdx)

Let

U=Ψ ∂∅∂ x

, V=Ψ ∂∅∂ y

Then

∫D

❑

(Ψ x∅ x+Ψ ∅ xx+Ψ x∅ x+Ψ ∅ yy)dS=∫B

❑

Ψ ∂∅∂n

ds…….(4)

On interchanging ϕ and Ψ and subtracting one from the other ,we get

∫D

❑

(Ψ ∇2∅−∅ ∇2Ψ )dS=∫B

❑

(Ψ ∂∅∂n

−∅ ∂Ψ∂n

)ds ……….(5)

The identities (4) and (5) are called Green,s identities

Theorem 1:1 . Fourier integral theorem

Let f( x ) be a continuous and absolutely integrable function in (−∞,∞ ).

ThenF ( f )=¿F(α ) called the Fourier transform of f( x ) is defined as

F(α ) =F ( f )= 1√2π ∫

−∞

∞

f ( x ) eiαa dx

Then

f( x )=F−1 (F )= 1√2π ∫

−∞

∞

F (α ) e−iαxdα

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Is called the inverse Fourier transform of F(α )

Example 1 : Find the Fourier transform of

f( x )=e−|x|

We have

F(α ) = 1√2 π ∫

−∞

∞

e−|x|eiαx dx

= 1√2 π (∫

−∞

0

exe iαxdx+∫0

∞

e− xe iαxdx)

= 1√2 π [ 1

1+iα+ 1

1−iα ]

=√ 2π [ 1

1+α2 ]Theorem 1:2 . Convolution Theorem

Let f( x ) and g( x ) possess Fourier transforms. Define

( f∗g ) ( x )= 1√2π ∫

−∞

∞

f ( x−ξ ) g (ξ )dξ ,

Called the convolution of the function f and g over the interval (−∞,∞ ) .

F ( f∗g )=F ( f )F (g )

Theorem 1:3 . Maximum Principle

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Suppose that u ( x , y ) is harmonic in a bounded domain D and

continuous in

D=D∪B. Then u attain its maximum on the boundary B of D.

Theorem 1:4 . Minimum Principles

Suppose that u ( x , y ) is harmonic in a bounded domain D and

continuous on D=D∪B. Then u attains its minimum on the boundary B of

D

Theorem 1:5 . Uniqueness Theorem

The solution of the Dirichlet problem, if it exists, is unique

Proof

Suppose u1 ( x , y ) and u2 ( x , y ) are two solutions of the Dirichlet

problem. That is ∇2u1=0 in D and u1=f (s ) on B

∇2u2=0in D and u2=f (s ) on B

As u1 and u2 are harmonic in D, (u1−u2 ) is also harmonic in D.

However (u1−u2 ) =0 on B

By the maximum and minimum principle, (u1−u2 )≡0 in D.Hence the

theorem.

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Theorem 1:6 .

The necessary condition for the Neumann problem

Let u be a solution of the Neumann problem

∇2u=0in D,And ∂u∂n

=f ( s ) on B

Then

∫B

❑

f (s )ds=0

Proof

Put Ψ=1 and ϕ =u in. Then we get

∫B

❑

f (s )ds=0

Thus f( s) cannot be arbitrarily prescribed as it has to satisfy the above

condition.

Theorem 1:7 . Gauss divergence theorem

If F⃗ is a vector point function, finite and differentiable in a reagion V

of space bounded by a closed surfaces S, then the surface integral of the

normal component ofF⃗taken over S is equal to the volume integral of the

divergence of F⃗ taken over V

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That is, ∫s

❑

F⃗ n⃗ ds=∭v

❑

(∇ . F⃗ )dv

Where n is the unit vector in the direction positive (outward drawn)

normal to S.

CHAPTER 2

POTENTIAL THEORY AND ELLIPTIC

PARTIAL DIFFERENTIAL EQUATIONS

Definition 2:1

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The general linear homogeneous second order partial differential

equation in m-space variablex1 x2…….. xm is

Lu =aαβuxαxβ+bαuxα+cu=0 , α , β=1,2……m……………….(1)

Where the coefficients aαβ,bα and c are continuous functions of the

independent variables x1 , x2 ,…….. xmand aαβ. This equation is said to be

elliptic in a domain D of m-dimensional space, when the quadratic form.

Q( λ )=aαβ λα λβ ……………….(2)

Can be expressed as the sum of squares with coefficients of the same

sign, or equivalently Q( λ ) is either positive or negative definite in D. The

simplest case is that of the Laplaces equation or potential equation.

∆mu=uxαxα=0 ……………….(3)

Definition 2:2

A function u ( x ) is called harmonic in D, if u ( x )∈C°

In D+∂ D∈C2 in D and ∆mu=0 in D

General Solution of the potential equation

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In the case of two or three variables, the general solution of the potential

equation can be easily obtained. For m=2(x1=x , x2= y ), this is the real or

imaginary part of any analytic function of the complex variable +iy . For

m=3 (x1=x , x2= y , x3=z ), consider an arbitrary function p(w ,t ) analytic in the

complex variable w for fixed real t. Then, for arbitrary values of t, both the

real and imaginary parts of the function.

u=p (Z+ ix cost+ iy sin t , t ) ……………….(4)

Of the real variables x , y , z are solutions of the equation ∆u=0. Further

solutions may now be obtained by superposition.

u=∫a

b

p ( z+ix cos t+iy sin t , t )dt……………….(5)

If u ( x , y ) is a solution of Laplace’s equation in a domain D of the ( x , y )

plane, the function

υ ( x , y )=u¿ ……………….(6)

Also satisfies the potential equation and is regular in the domain D1

obtained from D by inversion with respect to the unit circle.

In general in m-dimensions, if u (x1 , x2……. xm ) satisfies the potential

equation in a bounded domain D, then.

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υ= 1rm−2 u [ x1

r2 ,x2

r2 ,……xm

r2 ] , r2=xα xα ……………….(7)

Also satisfies the potential equation and is regular in the region D1

obtained from D by inversion with respect to the m-dimensional unit

sphere. Therefore, except for the factor r2−m, the harmonic character of a

function is invariant under inversions with respect to spheres.

Examples 2:1

Dirichlet problem for a circle in the x , y plane.

Let the circle C be given by |ξ|=R, where ξ=x+iy. The problem is to find

u ( x , y ) such that

Δ2u=uxx+uyy=0 ……………….(8)

Where u=f (θ ) on C

Where θ is the angular coordinate on C;

That is ,ξ=Re iθ on C

0W

AR2/W

B

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Fig 2.1.wandR2

w are inverse points with respect to C:|Z|=R

We shall present here a constructional proof for the existence of the

solution.

That is, we shall derive an expression for the solution. Let F (ξ ) be an

analytic function in the region enclosed by C, such that the real part of F(ξ )

on |ξ| =f(θ ). Let w be a complex number in the region. The inverse point of

w with respect to C is R2/w, which lies out side C. Here w denotes the

complex conjugate of w. According to the Cauchy integral formula.

F (w )= 12πi∫c

❑ F (ξ )ξ−w

dξ

0= 12 πi∫C

❑ F (ξ )ξ−R2/w

dξ

Subtracting

F (w )= 12πi∫C

❑ F (ξ ) (R2−ww )ξ (R2+ww )−wξ2−w R2 dξ

As ξ lies on C and w inside C, set

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ξ=Re iθ ,w=r eiϕ , ,r<R

Then,

F (ℜiϕ )= 12π ∫0

2π

F (ℜiθ ) R2−r 2

R2+r2−2rR cos (θ−ϕ)dθ

Taking the real part on both sides, we get

u ( x , y )= R2−r2

2π ∫0

2π f (θ)dθR2+r 2−2r R cos (θ−ϕ )

……………….(9)

Where r2=x2+ y2, tanϕ =y / x

Equation (9) is called Poisson’s integral formula in two dimensions. This

completes the proof of the existence of the solution.

Definition 2:3

In Rm the solutions v (r ) of the potential equation Δmu=0, which

depend only on the distance (r ≠0) of a point x from a fixed point a, say, are

given by the equation.

d2vdr2 +m−1

r=0

r=|x−a|=√(xα−aα)(xα−aα)} ……….(10)

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This equation has solutions

v (r )=c1+c2 r2−m ,m>2

¿c1+c2 logr ,m=2 } ……………….(11)

Where c1 and c2 are arbitrary constants. These solutions exhibit the so

called characteristic singularity at r=0. We call

s (a , X )¿

L(m−2 )wm

|a−X|2−m,m>2

¿−12 π

log|a−X|,m=2 }……………….(12)

The singularity function for Δmu=0, where wm is the surface area of the unit

sphere in m dimensions given by

wm=2¿ ……………….(13)

Note 2.1

s(a , x) has the property that s∈C∞ and ∆m s=0 for x≠a, with a

singularity at x ¿a. For m=3, s(a , x) corresponds physically to the

gravitational potential at the point x of a unit mass concentrated at the

point a.

Definition 2:4

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Every solution of the potential equation ∆mu=0 in D of the form

γ (a , x )=s (a , x )+ϕ (X ) , a ∊ D ……………….(14)

Where ϕ (X ) ∊C2 in D and ϕ (X ) ∊C1 in the closed region D+∂ D and ∆mϕ=0 in

D, is called a fundamental solution relative to D with a Singularity at a.

Definition 2:5

Green’s formulae

Let u andv be two functions defined in a domain D bounded by a piecewise

smooth surface ∂D. Then

∫D

❑

∇mu∇mυdx+∫D

❑

υ Δmudx=∫∂ D

❑

υ ∂u∂ v

dS ……………….(15)

And

∫D

❑

(u ∆mυ−υ Δmu )dx=∫∂d

❑

⟨u ∂υ∂V

−υ ∂υ∂V ⟩ dS ..…………….(16)

Where ∂∂V represents differentiation in the direction of the outward drawn

normal to S.

Note 2.2

If △m u=0 and v=1 we obtained from (15)

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𝜕D⨜uvds=0 …………(17)

where the suffix v denotes normal derivative to the surface 𝜕D. That

means, if a function satisfies ∆ mu=0 in a bounded region D and is

continuously differentiable in D+∂ D thenthe surface integral of its normal

derivative is zero. Then we get a Neumann problem

That is,

Green’s formulas undergo an important modification if for v we substitute

a function having the characteristic singularity of potential function at an

interior point. Let us choose v to be a fundamental solution

V=ɣ(a,x)= s(a,x) +ɸ(x)

Theorem 2.1

If u(x)єC1 in D + 𝜕D and єC2 in D and u(x ) is a solution of the potential

equation then for an arbitrary point aє D

u(a)=∫∂ D

❑

γ (a , X )u ( x )−u(x)γ v (a , x)dS

Proof

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Since D is open ,the ball V; |a-x|⩽𝛒 about any point a in D is contained in

D for sufficiently small 𝛒. Since ɣ(a,x) becomes singular at x= a, We

remove V from D and apply Green’s formula (16) for ɣ and u on D-V.

We obtain

∫D−v

❑

( γ ∆mu−u∆m γγ )dX=∫∂D

❑

( γ uv−u γv )dS+ ∫|X−a|= ρ

❑

(γ uv−uγ v)dS

Since △m ɣ=0, △mu=0,in D-V in the left hand side is zero

Consider the integral ∫|X−a|=ρ

❑

(γ uv−uγ v )dS

and using the fact ɣ(a,x)=s(a,x)+ɸ(x)

Then equation (20) expressed as a sum of two integrals I1 and I2

I1 =⨜(ɸuv-uɸv)ds

At any point on the surface of V, x=a +𝛒v and dS=𝛒m-1dw

Therefore

|X-a|=𝛒⨜suv ds =𝛒m-1|v|=1⨜s(a.a+𝛒v) uv (a+𝛒v)dw

= ρm−1

m−2wm∫

|v|=1

❑

|a−X|2-muv(a+𝛒v)dw

= ρm−1 ρ2−m

m−2wm∫

|v|=1

❑

uv(a+ρv)dw

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=ρ

(m−2)wm∫

¿v∨¿¿

❑uv(a+ρv)dw

This tends to zero as ρ 0

Further,

- ∫|X−a|=ρ

❑

svudS=−ρm-1 ∫|v|=1

❑

u(a+v ρ) ρv(a,a+ρv)dw

=- −ρm−1

(m−2)wm∫

|v|=1

❑

u(a+v ρ) ∂∂v

ρ2-mdw

Here ∂∂v= −∂

∂ρ . Therefore

- ∫|X−a|=ρ

❑

svudS= ρm−1

(m−2)wm∫|v|

❑

u(a+v ρ) ∂∂ s

¿¿) dw

= ρm−1

(m−2)wm∫

|v|=1

❑

u(a+v ρ)(2−m) ρ1−mdw

=−1wm

∫|v|=1

❑

u (a+v ρ )dw

As ρ 0, This is tends to –u(a). This proves the results (19)

That is, u(a)= ∫∂ D

❑

{γ (a , X )uv (X )−u (X ) γ v (a ,X )dS

This formula holds for any a𝞊 D

Note 2.3

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If instead a line on ∂D, then we get

u(a) = 2∫∂ D

❑

{γ (a , X)uv(X) – u(X)γv (a , X)} dS

Where as if a lies outside D, then

∫∂ D

❑

{γ (a , X )uv (X )−u(X )γ v (a , X )}dS =0

In particular in 3-dimensions m=3, if∅ (x )=0, hen we have for a∈D

u(a)=14 ∫

∂D

❑ {∂u∂v−u ∂∂v ( 1

r )}dS

u can be considered as the potential of adistribution consisting of a

single layer surface distribution of density ( 14 π ) ∂u∂v anddouble layer dipole

distribution of density −u4 π on the boundary surface∂ D. We had taken u be

an arbitrary function, then for a∈D the formula .

u(a)=∫∂ D

❑

{γ (a , x )uv (X )−u(X)γ v (a , X )}dS

is modified as

∫D

❑

γ ∆mudX+∫∂ D

❑

(γ ∂u∂ v

−u ∂γ∂v )dS

Formally, using the Greens formula of the equation.

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∫D

❑

(u ∆mv−v ∆mu )dX+∫∂ D

❑

(γ ∂υ∂ v

−v ∂u∂υ )dS

For the right hand side, we have

u(a)0= −∫D

❑

u∆m γ dX ≡−∆m γ (u (X ) )

Definition 2.6

We define a grees function G(a,X) of the differential expression ∆mu for

the region D as a specific fundamental solution of ∆mu=0, depending on the

parameter a, of the form

G(a, X) =G(a1, a 2, ….,a n,x1,x2,xn)

=s(a,X)+ϕ

Which vanishes at all points x on 𝟃D and for which the component 𝟇 is

continuous in D + 𝟃D and harmonic in D. Assuming the existence of a

Green’s functions G, we replace γ in equation

u(a)= ∫∂ D

❑

{γ (a , X ) uv (X )−u ( X ) γ v (a , X ) }dS

by G and we get the solution of a Dirichlet boundary value problem at

appoint a𝞊D as

u(a)=-∫∂ D

❑

f (x) ∂G∂ v

(a , X )dS --------------->(21)

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Example 2.2

Greens function for asphers of radius R,withcentre at origin in m-

dimensions.

Here G(a,X) = Ψ (r )=¿𝛹[¿a∨ ¿R

r ¿1]

Where

Ψ (r )= 1(m−2)wm

∨X−a∨¿2-m , m>2

= 12π log 1

¿ X−a∨¿¿ m=2

|a2| =a𝛼a𝛼, r12=(x𝛼- R2

¿a∨¿2¿a𝛼)( x𝛼 - R2

¿a∨¿2¿a𝛼)

=¿¿

r1denotes the distance of the point x from the reflected image of the

point in the sphere. This function satisfies all the requirement of the

Green’s function since(i) it is of the form s(a , x) +ϕ (x ), where ϕ (x ), where

ϕ (x )is regular in D and continuous inD+∂ D and (ii) it vanishes on ∂ D, since

r= ¿a∨ ¿R

r ¿1 on ∂ D

Example 2.3

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Green’s function for a positive half plane bounded by x1 =0, m > 2

G(a , x )= 1

(m−2)wm¿ X−a∨¿2−m−ϕ ( x ) ,¿m >2

Where

ϕ (x )=1

(m−2)wm¿(2-m)/2

ϕ (x )is obtained by taking the image of ‘a’ in the boundary by x1=0 .We can

find the solution at x = aby

u(a) =∫x1

❑

f (X ) { ∂∂ x1

G (a ,X )}dS

Example 2.4

Green’s function for a circle and poisson’s formula,m=2. This is a

special case of example 2.2

G(a,X) =1

2πlog 1

¿ X−a∨¿− 12π

log ∨X− R2

¿a∨¿2a∨¿¿¿

Of special interest is the case of the Green’s function in two dimension ,for

a domine D which can be mapped conformlly onto the unit circle.

Theorem 2.2

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Let F(x+iy)= u+ iv represent a mapping of D+∂ D on the unit circle in the

u-v plane, where F(x+iy) is a simple analytic function of the complex

variable x+ iy .Then the Greens function for D is given by

G(a1,a2 ;x,y)= - 12π

Relog¿

Where Re denotes the real part of a complex quantity.

Proof

To show that G=0 on the boundary ∂ D , we denote that ∂ D is

mapped by F onto boundary of the unit circle in the z-plane.

Therefore

F(x+iy)=e iθ

When (x,y) lies on ∂ D. Set z= x+iy, α=¿a1+ia2

On ∂ D, G(a1 ,a2 ;x,y) =−12π Relog[ F (α )−e iθ

F (α )e iθ−1 ]

=−12π Relog[ F (α )−e iθ

(F (α )−e iθ)e iθ ] =−1

2π log 1

=0

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To show that G is a fundamental solution of the Laplace equation,

We have to show that it satisfies ∆2u =0 except at (x,y)=(a1,a2)and in the

neighbourhood of this point .

2πG=−log|( x , y )−(a1 , a2 )|+aharmonic function

Since F is a one-one mapping

For any two points (x,y), (a1,a2), the function

log[ F (α )−F (z)F (α )−F ( z )−1 ]

is analytic in z and its real part satisfies Laplace’s euation. In the

neighbourhood of α

F(z)-F(α)= (z-α)F’(α)-¿¿F”(α)+…….

= (z-α)¿…….

= (z-α)H(z)

Where H(z) is analytic . Due to the conformal nature of the

mapping, H(z) is non zero in a suitable neighbourhood of 𝛼. Therefore

2𝜋G(𝛼,z) = -Re log( −H (z )F (α )F ( z )−1 )

= -log|z-(α )| +harmonic function

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Therefore, G is the desired Green’s function

Theorem 2.3: Poisson’s Theorem

This theorem gives the solution of Dirichlet’s problem in sphere of

radious R about the origin.

If f(x) C0 on |x|=R, then for m≥2

u(x)= {R2−|x|2

Rwm∫

|y|=R

❑ f ( y )¿ x− y ¿m dS for|x|<R

f ( x ) for|x|=R

Belongs to C0 in |x|< R and u(x) if a solution of the problem

∆mu=0 for|x|<R ,u= f (x ) for|x|=R

We can further show that in |x| <R, u∈C∞

Proof

Step 1: We shall first show that u given by(22) satisfies ∆mu=0.

If |x|<R, then |x-y|≠0 in the integrand and (23) can be differentiated under

the integral sign arbitrarily often

∆mu ( X )= 1R wm

∫|y|=R

❑

f ( y )∆m¿

(22)

(23)

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Since∆m¿

In particular u(x)≡1 is a solution of ∆mu=0 in D and u ∈ C2 in D+∂D.

Applying (21) in this case,we get

I=R2−¿X∨¿2

R wm∫

|y|=R

❑ dS¿ X− y∨¿m ¿¿………………………………….(25)

Step 2: To show that on approaching the boundary ∂D,u (x) us given by

(22) tends to the prescribed boundary value f(x).

Let 0<p<a, x0 is an arbitrary point on ∂D and x is such that |x-x0|<0/2. Then

form (22) and (25) we get

u(x)-f(x0)=R2−¿ X∨¿2

R wm∫

|y|=R

❑ f ( y )−f (x0)dS¿ X− y∨¿m ¿¿

about x0 construct a sphere with radiousρand the part of ∂D which lies in

this sphere we denote by S1

That is S1 ; |y|=R,|y-x0| ≤ ρ,S2 : |y|=R|, |y-x0| >ρ ,∂D=S1 +S2

∫|y|=R

❑ f ( y )−f (x0)dS¿ X− y∨¿m =¿¿

R

S2

29 | P a g e

Fig 2.2. Dividing the boundary into two parts S1 and S2 for an arbitrary

point x0on the boundary.

Consider each of these integrals in turn

¿

≤ R2 –|X|Rwm

∫S1

❑ f ( y )−f (X0 )|X− y|m

dS

≤maxy∈S1

∣f ( y )−f (X0 ) ∣ R2−∣ X ∣2Rwm

∫∣v ∣=R

❑ dS∣ X− y ∣m

= maxy∈S1

|f(y)-f(X 0)| from(25)

Next maxy∈S1

R2−¿X ¿2

∣X− y ∣m = maxy∈S1

¿¿

≤2R¿¿

Since |x-y| = |y-x0 + x0−x∨¿

= |y-x0|-|x0−x|

S1

X❑0

ρ

30 | P a g e

=ρ− ρ2 = ρ

2

If we set maxy∈∂ D

|f ( y )|=M ,we obtain the estimate for the second term.

APPLICATIONS

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

31 | P a g e

– This bit is NOT examined

Laplace’s Equation

∂2∅∂ x2 + ∂2∅

∂ y2 =0

In the vector calculus course, this appears as∇2∅=0 where

∇=[ ∂∂ x∂∂ x ]

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

variablesx,y. This means that Laplace’s Equation describes steady state

situations 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, …

32 | P a g e

Stress analysis example: Dirichlet conditions

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

w=w0 sin πxa

w = 0

w = 0 a

b

w = 0

Y

X

33 | P a g e

To Solve:∂2w

∂ x2 + ∂2w∂ y2 =¿0

Solution by separation of variables

w(x,y)=X(x)Y(y)

from which X”Y +XY”=0

and so X } over {X} + {YY

=0

as usual X } over {X} = {-YY =k

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

Case k=0

X(x)= (Ax+ B),Y( y) = (Cy+ D)

W(x,0)=0 =>ADx= 0

Either A=0 or D=0

Continue with w(x,y) = ACxy

W(a,y) = 0 => ACay = 0 =>A =0 or C= 0

=>w(x,y)≡0

34 | P a g e

That is, the case k=0is not possible

Case k>0

Suppose that k2=α2 , so that

W(x,y)= (A cosh αx +Bsinh αx )(C cos αy+ sin αy )

Recall that cosh 0=0,sinh0=0

w(o,y)=0 => A(Ccos αy +Dsin αy)=0

C = D= 0 => w(x,y)≡0

Continue with A=0 =>w(x,y)= Bsinfαx(Ccos αy+Dsin αy)

W(x,0)=0 => BCsinh αx =0

B= 0 => w(x,y)≡0

Continue with C=0 =>w(x,y)=BD sinhαxsinαy

w(a,y)= => BDsinhαa sinαy=0

So either B =0 or D =0 => w(x,y)≡0

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

Final case k<0

Suppose that k = -α2

35 | P a g e

W(x,y) = (A cos αx + B sin αx) (C cosh αy + D sinh αy)

W(0,y)= 0 A(C cosh αy + D sinh αy) = 0

As usual, C = D = 0 w ≡ 0

Continue with A = 0 w(x,y) = B sin αx(C cosh αy + D sinh αy)

W(x,0) = 0 BC sin αx = 0

B = 0 w ≡ 0

Continue with C = 0 w(x,y)= BD sin αx sinh αy

W(a,y)=0α BD sinαa sinh αy = 0

B = 0 or D = 0 w ≡ 0

sin αa = 0 α = n πa w (x,y) = BD sin n π

a x sinh n π2a

y y

Solution

Applying the first three boundary conditions,

we have w ( x , y )=∑n=1

∞

(kn sin (2n−1)∏2a

x sinh (2n−1)πy2a )

The final boundary condition is : w(x,b) = w0 sin∏x2a

36 | P a g e

Which gives: w0 sin∏x2a

=∑n=1

∞

(kn sin (2n−1)∏2a

x sinh (2n−1 )πb2a )

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

K1= W 0

sinh ∏b2a

And the final solution to the stress distribution is

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*

A problem in electrostatics

w(x,y) = w0

sinh ∏b2a

sin ∏xa sinh ∏y

a

37 | P a g e

This is a cross section of a charged cylindrical rod.

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

∇2V = ∂2v

∂ r2 + 1r∂V∂r + 1

r2∂2 v∂rѳ2 + ∂

2v∂ z2 = 0 ….. brief time out while I derive this

2D Laplace’s Equation in Polar Coordinates

x Ѳ r y

∇2u = ∂2u

∂x2 + ∂2u

∂ y2 =0 where x= x(r,ѳ) , y = y(r,ѳ)

X = r cos Ѳ

Y = r sin Ѳ

r = √ x2+ y2

Ѳ = tan−1¿

Thin strip of insulating material

V(rѲ,z) = U on the upper half

Radius a

38 | P a g e

U(x,y) = u(r,ѳ)

So, Laplace’s Equation is ∇2u(r,ѳ) = 0

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

Recall the chain rule :∂u∂x = ∂u∂r∂r∂ x + ∂u∂ѳ

∂ѳ∂ x

Use the product rule to differentiate again

∂2u∂x2 = ∂u∂r

∂2u∂x2 + ∂

∂x (∂u∂r ) ∂r∂ x + ∂u∂ѳ∂2ѳ∂ x2 + ∂

∂x ( ∂u∂ r ) ∂ѳ∂ x (*)

and the chain rule again to get these derivatives

∂∂x (∂u∂ r )= ∂

∂r (∂u∂r ) ∂r∂ x + ∂∂ѳ (∂u∂ r ) ∂ѳ∂ x = ∂

2u∂x2

∂r∂ x+ ∂2u

∂ѳ∂r∂ѳ∂ x

∂∂x (∂u∂ѳ ) = ∂

∂r ( ∂u∂ѳ ) ∂ r∂ x + ∂∂ѳ ( ∂u∂ѳ ) ∂ѳ∂x = ∂2u

∂r ∂ѳ∂r∂ x

+ ∂2u

∂ѳ2∂ѳ∂ x

THE REQUIRED PARTIAL DERIVATIVES

X= r cosѲ y = r sin Ѳ r =√ x2+ y2 Ѳ = tan−1( y / x)

39 | P a g e

r2 = x2 + y2 2r ∂r∂x= 2x ∂r∂x = xr similarly, ∂r∂ y = yr ,

∂2r∂x2 = y2/r3 , ∂

2 r∂ y2 = x2 /r3

in like manner…..

∂ѳ∂x = -

yr2 ,

∂ѳ∂ y = x

r2∂2ѳ∂x2 =

2xyr 4 , ∂

2ѳ∂ y2 =

2xyr 4

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:

∂2u∂x2 = ∂

2u∂r2 x2/r 2 + ∂u∂r y2 /r3+ ∂2u− ¿

∂r ∂ѳ2 xyr3 ¿ + ∂u∂ѳ

2 xyr4 + ∂

2u∂ѳ2 y2/r4

Similarly, ∂2u

∂ y2 = ∂2u

∂r2 y2/r2+∂u∂r

x2/r3+ ∂2u∂r ∂ѳ

2 xyr3 -

∂u∂ѳ

2 xyr4 + ∂

2u∂ѳ2 x

2/r4

So Laplace’s Equation in polars is

∂2u∂x2 +

∂2u∂ y2 =

∂2u∂r2 +1

r∂u∂r+ 1

r2∂2u∂ѳ2=0

∂2u∂x2 +

∂2u∂ y2 =0

40 | P a g e

Is equivalent to

∂2u∂r2 +1

r∂u∂ r + 1

r2∂2u∂ѳ2=0

Example of Laplace in cylindrical Polar Coordinates (r, Ѳ, z)

Consider a cylindrical capacitor z

θ y

0v

Laplace’s Equation in cylindrical polars is: Boundary conditions

∇2v = ∂2 v

∂ r2 +1r∂v∂r + 1

r2∂2 v∂ѳ2+

∂2 v∂ z2=0 v (a,ѳ¿ = U

∀ ѳ :0≤ѳ≤π

R θ

Thin strip of insulating mmaterial

V(rѲ,z) = U on the upper half

Radius a

41 | P a g e

v (a,ѳ¿ = 0 ∀ѳ : π≤ѳ≤2π

In the polar system, note that the solution must repeat itself every Ѳ = 2π V should remain finite at r = 0

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

This means we can treat it as a 2D problem ∂2 v

∂ r2 +1r∂v∂ r + 1

r2∂2 v∂ѳ2 = 0

Using separation of variables

V = R(r) Θ (Ѳ)

R″Θ + 1/r R′Θ + 1/r2 RΘ″ = 0

- Θ″ = R″+ R′/r

Θ R/r2

As before, this means

- Θ″ = R″+ R′/r = k, a constant

Θ R/r2

THE CASE K=0

Θ″ = 0 Θ(Ѳ) = a Ѳ + b

R″ + R′/r = 0 R( r )= (C lnr +D)

and so V(r, Ѳ) = (a Ѳ+ b) (C ln r+ D)

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

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

THE CASE K< 0

V(r,Ѳ) = bd = g, a constant

42 | P a g e

Suppose that k = -m2

Ѳ″ + m2 Ѳ = 0 Θ (Ѳ)= (Amcoshm Ѳ+ Bmsinhm Ѳ)

R″ + R′/r + m2/r2 R=0 R(r) = (Cmr-m + Dmrm)

The solution has to be peiodic in Ѳ, with period 2π. This implies that Am=Bm= 0 V(r, Ѳ) ≡ 0

THE CASE K>0

Suppose that k = n2

Ѳ″ + m2 Ѳ = 0 Θ (Ѳ)= (Ancosn Ѳ + Bnsin n Ѳ)

R″ + R′/r – n2 /r2 R = 0 R(r)= (Cnrn + Dnr –n)

V(r, Ѳ) = (Ancos n Ѳ + Bn sin n Ѳ) (Cnrn + Dnr –n)

Evidently, this is periodic with period 2π

To remain finite as r 0 Dn= 0

THE SOLUTION

V(r, Ѳ) = g + ∑nrn (Ancosn Ѳ + Bnsinn Ѳ)

Notice that we have not yet applied the voltage boundary condition!! Now is the time to do so g + ∑n{an (Ancosn Ѳ + Bnsinn Ѳ) = V (a, Ѳ) = U 0≤ Ѳ≤π

0 π<Ѳ≤2π

integrating V from 0 to 2π: ∫0

2π

v (a, Ө) d Ө = π U

Left hand side :∫0

2π

g + ∑n an (Ancosn Ѳ + Bnsinn Ѳ)d Ѳ = 2πg and so g =

U/2

SOLVING FOR Am and Bm

so far, the solution is V (a, Ө) = U/2 + ∑n an (Ancosn Ѳ + Bnsinn Ѳ)

43 | P a g e

we apply the orthogonality relationships:

∫0

2π

V ¿¿Ѳ)cosmѲdѲ = u2∫02π

cosmѲdѲ +∫0

2π

❑∑nan(An cosnѲ+BnsinnѲ)cosmѲdѲ

U∫0

x

cos mѲdѲ=0+ Am amπ

0 = Amamπ, and so Am = 0, for all m

∫0

2π

V ¿¿Ѳ)sinmѲdѲ=u2∫0

2π

sinmѲdѲ+∫0

2π

❑∑n

❑

an(An cosnѲ+BnsinnѲ)sin nѲdѲ

U∫0

x

sinmѲdѲ❑= u2 ¿] + Bnamπ

2um = Bnamπ, for odd m = (2n-1)

V (r ,Ѳ) = u2 + ∑n

❑ 2u(2n−1 )π a(2n−1) r (2n-1)sin(2n-1)Ѳ

V (r ,Ѳ) = u2 +2uπ ∑

n−1

∝ r(2n−1)

(2n−1 )a(2n−1)sin(2n-1)Ѳ

Check for r = a, Ѳ= π2 :

V (r ,Ѳ) = u2 +2uπ ∑

n−1

∝ a(2n−1 )

(2n−1 )a(2n−1)sin(2n-1)π2

= u2 +2uπ [sin π

2+ 1

3sin 3 π

2+1

5sin 5 π

2+…]

= u2 +2uπ [1−1

3+ 1

5−1

7+…]

= u2 +2uπ [ π4 ]= 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

44 | P a g e

The next few slides hint at the application of what we have done sofar in image analysis

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

Laplace’s Equation in image analysis

How do we compute the edges?

Image fragment Edge map

Gaussian smoothing

Remove the noise by smoothing Find places where the second derivative

of the image is zero

45 | P a g e

Blurring with a Gaussian filter is one way to tame noise

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

∇2Ismooth = 0

An application of Laplace’s Equation!

Limits of isotropic Gaussian blurring

46 | P a g e

A noisy image Gaussian blurring

Guassian is isotropic – takes no account of orientation of image features – so it gives crap edge features ∇2(Gσ * I)= 0

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

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

47 | P a g e

Anisotropic diffusion

∂tI = ∇T (g (x,t) ∇I)

g(x;t) = e|∇ I σ|2

k2 , for some constant k, or

g(x;t)= 1

1+|∇ I σ|2/k2

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