Myint Tyn U., Debnath L. Linear Partial Differential Equations for Scientists and Engineers

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9.1 Boundary-Value Problems 331

2. The Second Boundary-Value Problem

The Neumann Problem: Find a function u (x, y), harmonic in D,which

satisﬁes

∂u

∂n

= f (s)onB, (9. 1.7)

with



f (s) ds =0. (9.1.8)

The symbol ∂u/∂n denotes the directional derivative of u along the out-

ward normal to the boundary B. The last condition (9.1.8) is known as the

compatibility condition, since it is a consequence of (9.1.7) and the equa-

tion ▽

u = 0. Here the solution u may be interpreted as the steady-state

temperature distribution in a body containing no heat sources or heat sinks

when the heat ﬂux across the boundary is prescribed.

The compatibility condition, in this case, may be interpreted physically

as the heat requirement that the net heat ﬂ ux across the boundary be zero.

3. The Third Boundary-Value Problem

Find a function u (x, y) harmonic in D which satisﬁes

∂u

∂n

+ h (s) u = f (s)onB, (9.1.9)

where h and f are given continuous functions. In this problem, the solution

u may be interpreted as the steady-state temperature distribution in a body,

from the boundary of which the heat radiates freely into the surrounding

medium of prescribed temperature.

4. The Fourth Boundary-Value Problem

The Robin Problem: Find a function u (x, y), harmonic in D, which satisﬁes

boundary conditions of diﬀerent types on diﬀerent portions of the boundary

B. An example involving such boundary conditions is

u = f

(s)onB

, (9.1.10)

∂u

∂n

= f

(s)onB

where B = B

∪ B

Problems 1 through 4 are called interior boundary-value problems. These

diﬀer from exterior boundary-value problems in two respects:

i. For problems of the latter variety, part of the boundary is at inﬁnity.

ii. Solutions of exterior problems must satisfy an additional requirement,

namely, that of boundedness at inﬁnity.

332 9 Boundary-Value Problems and Applications

9.2 Maximum and Minimum Principles

Before we prove the uniqueness and continuity theorems for the interior

Dirichlet problem for the two-dimensional Laplace equation, we ﬁrst prove

the maximum and minimum principles.

Theorem 9.2.1. (The Maximum Principle) Suppose that u (x, y) is

harmonic in a bounded domain D and continuous in D = D ∪ B. Then u

attains its maximum on the boundary B of D.

Physically, we may interpret this as meaning that the temperature of a

body which was neither a source nor a sink of heat acquires its largest (and

smallest) values on the surface of the body, and the electrostatic potential in

a region which does not contain any free charge attains its maximum (and

minimum) values on the boundary of the region.

Proof. Let the maximum of u on B be M. Let us now suppose that

the maximum of u in D is not attained at any poi nt of B. Then it must

be attained at some point P

)inD.IfM

= u (x

) denotes the

maximum of u i n D,thenM

must also be the maximum of u in D.

Consider the function

v (x, y)=u (x, y)+

− M

(x − x

)

+(y − y

)

, (9.2.1)

where the point P (x, y)isinD and where R is the radius of a circle

containing D. Note that

v (x

)=u (x

)=M

We have v (x, y) ≤ M +(M

− M ) /2=

(M + M

) <M

on B.Thus,

v (x, y) like u (x, y) must attain its maximum at a point in D. It follows

from the deﬁnition of v that

+ v

= u

+ u

− M )

> 0. (9.2.2)

But for v to be a maximum in D,

≤ 0,v

≤ 0.

Thus,

+ v

≤ 0

which contradicts equation (9.2.2). Hence, the maximum of u must be at-

tained on B.

Theorem 9.2.2. (The Minimum Principle) If u (x, y) is harmonic in

a bounded domain D and continuous in D = D ∪ B, then u attains its

minimum on the boundary B of D.

9.3 Uniqueness and Continuity Theorems 333

Proof. The proof follows directly by applying the preceding theorem to

the harmonic function −u (x, y).

As a result of the above theorems, we see that u =constant which i s evi-

dently harmonic attains the same value in the domain D as on the boundary

9.3 Uniqueness and Continuity Theorems

Theorem 9.3.1. (Uniqueness Theorem) The solu tion of the Dirichlet

problem, if it exists, is unique.

Proof. Let u

(x, y)andu

(x, y) be two solutions of the Dirichlet prob-

lem. Then u

and u

satisfy

▽

=0, ▽

=0 inD,

= f, u

= f on B.

Since u

and u

are harmonic in D,(u

− u

)isalsoharmonicinD.

But

− u

=0 onB.

The maximum-minimum principle gives

− u

at all interior points of D.Thus,wehave

= u

Therefore, the solution is unique.

Theorem 9.3.2. (Continuity Theorem) The solution of the Dirichlet

problem depends continuously on the boundary data.

Proof. Let u

and u

be the solutions of

▽

=0 inD,

= f

on B,

and

▽

=0 inD,

= f

on B.

If v = u

− u

,thenv satisﬁes

334 9 Boundary-Value Problems and Applications

▽

v =0 inD,

v = f

− f

on B.

By maximum and minimum principles, f

− f

attains the maximum

and minimum of v on B .Thus,if|f

− f

| <ε,then

−ε<v

min

≤ v

max

<ε on B.

Thus, at any interior point in D ,wehave

−ε<v

min

≤ v ≤ v

max

<ε.

Therefore, |v | <εin D. Hence,

− u

| <ε.

Theorem 9.3.3. Let {u

} be a sequence of functions harmonic in D and

continuous in D.Letf

be the values of u

on B. If a sequence {u

} con-

verges uniformly on B, then it converges uniformly in D.

Proof. By hypothesis, {f

} converges uniformly on B.Thus,forε>0,

there exists an integer N such that everywhere on B

− f

| <ε for n, m > N.

It follows from the continuity theorem that for all n, m > N

− u

| <ε

in D, and hence, the th eorem is proved.

9.4 Dirichlet Problem for a Circle

1. Interior Problem

We shall now establish the existence of the solution of the Dirichlet problem

for a circle.

The Dirichlet problem is

▽

u = u

θθ

=0, 0 ≤ r<a, 0 <θ≤ 2π,(9.4.1)

u (a , θ)=f (θ) for all θ in [0, 2π] . (9.4.2)

By the method of separation of variables, we seek a solution in the form

u (r, θ)=R (r) Θ (θ) =0.

Substitution of this in equation (9.4.1) yields

9.4 Dirichlet Problem for a Circle 335

′′

+ r

′

= −

′′

= λ.

Hence,

′′

+ rR

′

− λR =0, (9.4.3)

′′

+ λΘ =0. (9.4.4)

Because of the periodicity conditions Θ (0) = Θ (2π)andΘ

′

(0) =

′

(2π) which ensure that the function Θ is single-valued, the case λ<0

does not yield an acceptable solution. When λ =0,wehave

u (r, θ)=(A + B log r)(Cθ + D) .

Since log r →−∞as r → 0+ (note that r = 0 is a singular point of

equation (9. 4.1)), B must vanish in order for u to be ﬁnite at r =0.C

must also vanish in order for u to be periodic with period 2π. Hence, the

solution for λ =0isu = constant. When λ>0, the solution of equation

(9.4.4) is

Θ (θ)=A cos

√

λθ+ B sin

√

λθ.

The periodicity conditions imply

√

λ = n for n =1, 2, 3,....

Equation (9.4.3) is the Euler equation and therefore, the general solution

R (r)=Cr

+ Dr

−n

Since r

−n

→∞as r → 0, D must vanish for u to be continuous at

r =0.

Thus, the solution is

u (r, 0) = Cr

(A cos nθ+ B sin nθ)forn =1, 2,....

Hence, the general solution of equation (9.4.1) may be written in the

form

u (r, θ)=

∞



n=1





cos nθ + b

sin nθ) , (9.4.5)

where the constant term (a

/2) represents the solution for λ =0,anda

and b

are constants. Letting ρ = r/a,wehave

u (ρ, θ)=

∞



n=1

cos nθ + b

sin nθ) . (9.4.6)

336 9 Boundary-Value Problems and Applications

Our next task is to show that u (r, θ)isharmonicin0≤ r<aand

continuous in 0 ≤ r ≤ a. We must also show that u satisﬁes the boundary

condition (9.4.2).

We assume that a

and b

are the Fourier coeﬃcients of f (θ), that is,



2π

f (θ)cosnθ dθ, n =0, 1, 2, 3,...,

(9.4.7)



2π

f (θ)sinnθ dθ, n =1, 2, 3,....

Thus, from their very deﬁnitions, a

and b

are bounded, that is, there

exists some number M>0 such that

| <M, |a

| <M, |b

| <M, n=1, 2, 3,....

Thus, if we consider the sequence of functions {u

} deﬁned by

(ρ, θ)=ρ

cos nθ + b

sin nθ) , (9.4.8)

we see that

| < 2ρ

M, 0 ≤ ρ ≤ ρ

< 1.

Hence, in any closed circular region, series (9.4.6) converges uniformly.

Next, diﬀerentiate u

with respect to r. Then, for 0 ≤ ρ ≤ ρ

< 1,



∂u

∂r



n−1

cos nθ + b

sin nθ)



< 2





n−1

Thus, the series obtained by diﬀerentiating series (9.4.6) term by term

with respect to r converges uniformly. In a similar manner, we can prove

that the series obtained by twice diﬀerentiating series (9.4.6) term by term

with respect to r and θ converge uniformly. Consequently,

▽

u = u

θθ

∞



n=1

n−2

cos nθ + b

sin nθ)

n (n − 1) + n − n

=0, 0 ≤ ρ ≤ ρ

< 1.

Since each term of series (9.4.6) is a harmonic function, and since the

series converges uniformly, u (r, θ) is harmonic at any interior point of the

region 0 ≤ ρ<1. It now remains to show that u satisﬁes the boundary

data f (θ).

Substitution of the Fourier coeﬃcients a

and b

into equation (9.4.6)

yields

9.4 Dirichlet Problem for a Circle 337

u (ρ, θ)=

2π



2π

f (θ) dθ +

∞



n=1



2π

f (τ)

×(cos nτ cos nθ +sinnτ sin nθ) dτ

2π



2π



1+2

∞



n=1

cos n (θ − τ )



f (τ) dτ. (9.4.9)

The interchange of summation and integration is permitted due to the

uniform convergence of the series. For 0 ≤ ρ ≤ 1

1+2

∞



n=1

[ρ

cos n (θ − τ )]=1+

∞



n=1

in(θ−τ )

+ ρ

−in(θ−τ )

=1+

ρe

i(θ−τ )

1 − ρe

i(θ−τ )

ρe

−i(θ−τ )

1 − ρe

−i(θ−τ )

1 − ρ

1 − ρe

i(θ−τ )

− ρe

−i(θ−τ )

+ ρ

1 − ρ

1 − 2ρ cos (θ − τ )+ρ

Hence,

u (ρ, θ)=

2π



2π

1 − ρ

1 − 2ρ cos (θ − τ )+ρ

f (τ) dτ. (9.4.10)

The integral on the right side of (9.4.10) is called the Poisson integral

formula for a circle.

Now if f (θ) = 1, then, according to series (9.4.9), u (r, θ) = 1 for 0 ≤

ρ ≤ 1. Thus, equation (9.4.10) gives

2π



2π

1 − ρ

1 − 2ρ cos (θ − τ )+ρ

dτ.

Hence,

f (θ)=

2π



2π

1 − ρ

1 − 2ρ cos (θ − τ )+ρ

f (θ) dτ, 0 ≤ ρ<1.

Therefore,

u (ρ, θ) − f (θ)=

2π



2π



1 − ρ



[f (τ) − f (θ)]

1 − 2ρ cos (θ − τ )+ρ

dτ. (9.4.11)

Since f (θ) is uniformly continuous on [0, 2π], for given ε>0, there exists

a positive number δ (ε) such that |θ − τ| <δimplies |f (θ) − f (τ )| <ε.If

|θ − τ|≥δ so that θ − τ =2nπ for n =0, 1, 2,...,then

338 9 Boundary-Value Problems and Applications

lim

ρ→1−

1 − ρ

1 − 2ρ cos (θ − τ )+ρ

=0.

In other word s, there exists ρ

such that if |θ − τ|≥δ,then

1 − ρ

1 − 2ρ cos (θ − τ )+ρ

<ε,

for 0 ≤ ρ ≤ ρ

< 1. Hence, equation (9.4.10) yields

|u (r, θ)|−f (θ)|≤

2π



2π

|0−τ|≥δ



1 − ρ



|f (τ) − f (θ)|

1 − 2ρ cos (θ − τ)+ρ

dτ

2π



2π

|θ−τ |<δ



1 − ρ



|f (θ) − f (τ )|

1 − 2ρ cos (θ − τ)+ρ

dτ

≤

2π

(2πε)



2max

0≤θ≤2π

|f (θ)|



2π

· 2π

= ε



1+2



max

0≤θ≤2π

|f (θ)|



which implies that

lim

ρ→1−

u (r, θ)=f (θ)

uniformly in θ. Therefore, we state the following theorem:

Theorem 9.4.1. There exists one and only one harmonic function u (r, θ)

which satisﬁes the continuous boundary data f (θ). This function is either

given by

u (r, θ)=

2π



2π

− r

− 2ar cos (θ − τ)+r

f (τ) dτ, (9.4.12)

u (r, θ)=

∞



n=1

cos nθ + b

sin nθ) , (9.4.13)

where a

and b

are the Fourier coeﬃcients of f (θ).

For ρ = 0, the Poisson integral formula (9.4.10) becomes

u (0,θ)=u (0) =

2π



2π

f (τ) dτ. (9.4.14)

This result may be stated as follows:

Theorem 9.4.2. (Mean Value Theorem) If u is harmonic i n a circle,

then the value of u at the center is equal to the mean value of u on the

boundary of the circle.

9.4 Dirichlet Problem for a Circle 339

Several comments are in order. First, the Continuity Theorem 9.3.2

for the Dirichlet problem for the Laplace equation is a special example of

the general result that the Dirichlet problems for all elliptic equations are

well-posed. Second, the formula (9.4.12) represents the u niqu e continuous

solution of the Laplace equation in 0 ≤ r<aeven when f (θ) is discontin-

uous. This means that, for Laplace’s equation, discontinuities in boundary

conditions are smoothed out in the interior of the domain. This is a re-

markable contrast to linear hyperb olic equations where any discontinuity

in the data propagates along the characteristics. Third, the integral solution

(9.4.12) can be written as

u (r, θ)=



−π

P (r, τ − θ) f (τ) dτ,

where P (r, τ − θ) is called the Poisson kernel given by

P (r, τ − θ)=

2π



− r



− 2ar cos (τ − θ)+r

]

Clearly, P (a, τ − θ) = 0 except at τ = θ.Also

f (θ) = lim

r→a

−

u (r, θ)=



−π

lim

r→a

−

P (r, τ − θ) f (τ) dτ.

This implies that

lim

r→a

−

P (r, τ − θ)=δ (τ −θ) ,

where δ (x) is the Dirac delta function.

As in the preceding section, the exterior Di richlet problem for a circle

can readily be solved. For the exterior problem u must be bounded as

r →∞. The general solution, therefore, is

u (r, θ)=

∞



n=1



−n



cos nθ + b

sin nθ) . (9.4.15)

Applying the boundary condition u (a, θ)=f (θ), we obtain

f (θ)=

∞



n=1

cos nθ + b

sin nθ) .

Hence, we ﬁnd



2π

f (τ)cosnτ dτ, n =0, 1, 2,..., (9.4.16)



2π

f (τ)sinnτ dτ, n =0, 1, 2,.... (9.4.17)

340 9 Boundary-Value Problems and Applications

Substituting a

and b

into equation (9.4.15) yields

u (r, θ)=

2π



2π



1+2

∞



n=1





cos n (θ − τ)



f (τ) dτ.

Comparing this equation with (9.4.9), we see that the only diﬀerence

between the exterior and interior problem is that ρ

is replaced by ρ

−n

Therefore, the ﬁnal result takes the form

u (ρ, θ)=

2π



2π

− 1

1 − 2ρ cos (θ − τ)+ρ

f (τ) dτ, for ρ>1.(9.4.18)

9.5 Dirichlet Problem for a Circular Annulus

The natural extension of the Dirichlet problem for a circle is the Dirichlet

problem for a circular annulus, that is

∇

u =0,r

<r<r

, (9.5.1)

u (r

,θ)=f (θ) ,u(r

,θ)=g (θ) . (9.5.2)

In addit ion, u (r, θ) must satisfy the periodicity condition. Accordingly, f (θ)

and g (θ) must also be periodic with period 2π.

Proceeding as in the case of th e Dirichlet problem for a circle, we obtain

for λ =0

u (r, θ)=(A + B log r)(Cθ + D) .

The periodicity condition on u requires that C = 0. Then, u (r, θ) becomes

u (r, θ)=

log r,

where a

=2AD and b

=2BD.

The solution for the case λ>0is

u (r, θ)=



√

+ Dr

−

√



A cos

√

λθ+ B sin

√

λθ



where

√

λ = n =1, 2, 3,.... Thus, the general solution is

u (r, θ)=

+ b

log r)+

∞



n=1



+ b

−n



cos nθ



+ d

−n



sin nθ

, (9.5.3)

where a

,andd

are constants.

Applying the boundary conditions (9.5.2), we ﬁnd that the coeﬃcients

are given by