Chapter 4 PDE

A. G. Walton M2AA2 Multivariable Calculus: Partial Differential Equations 1

4 Partial differential equations

In this section we will consider second order partial differential equations (pdes). The three mainequations we will study are

2u

t2= c22u, Wave equation,

u

t= 2u, Heat equation,

2 = f(r), Laplace/Poisson equation.

4.1 The wave equation

For this discussion we will restrict ourselves to the one-dimensional version of the equation, namely

2u

t2= c2

2u

x2. (1)

4.1.1 Solution on a finite domain in x : the method of separation of variables

Consider the following problem. Solve (1) for u(x, t) subject to the boundary conditions

u(0, t) = u(L, t) = 0, (t 0) (2)and the initial conditions

u(x, 0) = f(x), (0 x L), (3)u

t(x, 0) = 0, (0 x L). (4)

We seek a solution to (1) of the separated-variables form

u(x, t) = X(x)T (t).

Substituting this expression into (1) we get

X(x)T (t) = c2X (x)T (t).

Dividing both sides by c2XT we can write this as

T (t)c2T (t)

=X (x)X(x)

.

We observe that the left-hand-side above only depends on t, while the right-hand-side is dependent onlyon x. The expression must hold for all t > 0 and x [L,L]. The only way this can be satisfied is if theleft and right hand sides are equal to a constant. In other words we must have

X

X= K =

T

c2T.

We have therefore reduced our pde to two second order ordinary differential equations which should beeasier to solve, particularly as they have constant coefficients. Now lets consider applying the boundaryconditions. These imply that

X(0) = X(L) = 0,

with X(x) satisfyingX KX = 0. (5)

The form of the general solution depends on whether K is positive, zero or negative. Lets consider thosecases in turn.

(i) K > 0. In this case the general solution of (5) is

X = A cosh(Kx) +B sinh(

Kx).


-3 -2 -1 1 2 3

-10

-5

5

10

Figure 1: sinh x versus x

However if we then apply the boundary conditions, we have

X(0) = 0 A = 0,X(L) = 0 B = 0 or sinh(

KL) = 0.

Neither of these options are acceptable: the first leads to X identically zero, while a consideration of thegraph of sinh x (figure 1) shows that no such positive value for K exists. We therefore conclude that forthese boundary conditions the constant K cannot be positive.

(ii) K = 0. In this case the solution of (5) is

X = ax+ b.

Again, if we apply X(0) = X(L) = 0 we see that no non-zero solution is possible.

(iii) K < 0. It is convenient to set K = 2. Now the general solution of (5) isX = A cosx+B sinx.

Applying the boundary conditionsX(0) = 0 A = 0,

as before, but now the condition X(L) = 0 leads to

sinL = 0,

for which there are an infinite number of solutions:

=n

L, n = 1,2, . . . . (6)

The solution for X is thereforeX = Bn sin(nx/L).

Now we turn to the corresponding equation for T which takes the form

T + c22T = 0,

and therefore has the general solution

T = C cos(ct) +D sin(ct).

The initial condition (4) implies that T (0) = 0 and this leads to D = 0. Substituting for from (6) weare left with

T = Cn cos(nct/L).


A solution that satisfies the wave equation and conditions (2), (4) is therefore

yn = XT = n sin(nx/L) cos(nct/L)

where we have introduced n = BnCn. Since the wave equation is linear it follows that any linearcombination of the solutions for different n is also a solution. The most general solution is therefore ofthe form

u(x, t) =

n=

n sin(nx/L) cos(nct/L),

which can be expressed more succinctly as

u(x, t) =

n=1

bn sin(nx/L) cos(nct/L), (7)

where we have written bn = n n. We have one more initial condition to apply - the condition thatu = f(x) when t = 0. Imposing this, we see that f(x) is related to the unknown coefficients bn in thefollowing way:

f(x) =

n=1

bn sin(nx

L

), (0 < x < L).

We recognize this as a half-range Fourier sine series for f(x). Our Fourier series studies of section 2 haveshown us that

bn =2

L

L0

f(x) sin(nx

L

)dx, (8)

and so can be computed for a given f(x). The required solution to the wave equation is therefore givenby the infinite sum (7), with the coefficients bn calculated from (8).

Example

Solve the wave equation subject to

u(0, t) = u(L, t) = 0 for t 0,u(x, 0) = 0 for 0 x L,

u

t(x, 0) = g(x) for 0 x L.

4.1.2 Solution on an infinite domain: use of Fourier transforms

Suppose we have to solve the following problem:

2u

t2= c2

2u

x2for < x 0,

u(x, 0) = 4e5|x| for < x


using a result we saw on Problem Sheet 6. Hence we have

u(, t) =40

25 + 2cos(ct).

We can invert this by using the convolution theorem, since u is the product of two terms we know theindividual inverses of. We proceed as follows.

u(x, t) = F1{

40

25 + 2cos(ct)

}= F1

{40

25 + 2

} F1 {cos(ct)}

= 4e5|x| F1 {cos(ct)} . (11)Now to find the inverse transform of the second term we can use the result from section 3.5.3 that

F{cos(0x)} = ( + 0) + ( 0),where is the Dirac delta function. Then using the symmetry formula (property (vi) in section 3.2) itfollows that

F{(x+ 0) + (x 0)} = 2 cos(0),and hence

F1{cos(0)} = 12(x+ 0) +

1

2(x 0).

Using this result in (11):

u(x, t) = 4e5|x| 12((x+ ct) + (x ct))

= 2

e5|xs|(s+ ct) ds+ 2

e5|xs|(s ct) ds

= 2e5|x+ct| + 2e5|xct|,

with the last line following from the sifting property of the delta function (section 3.5.3).

4.1.3 DAlemberts solution for the wave equation

The particular solution we obtained by separation of variables can be rewritten as

u(x, t) =1

2

n=1

bn sin(nL(x+ ct)

)+1

2

n=1

bn sin(nL(x ct)

),

Similarly, the solution we obtained by Fourier transforms also depends on the combination of variablesx ct and x+ ct. The functional dependence on these quantities indicates that in both cases the solutionis the sum of a left-travelling (x+ ct) and right-travelling (x ct) wave, with both waves propagating atspeed c. This observation provides us with some motivation for the following study which results in thederivation of the general solution of the wave equation.

We introduce new variables = x+ ct, = x ct.

The partial derivatives transform as follows:

x=

x

+

x

=

+

,

t=

t

+

t

= c

c

.

We can then calculate

2u

x2=

x(u

x) =

(

+

)(u

+u

)=

2u

2+ 2

2u

+2u

2,


and

2u

t2=

t(u

t) =

(c

c

)(cu

cu

)= c2

(2u

2 2

2u

+2u

2

).

Under this transformation the wave equation

2u

t2= c2

2u

x2

becomes

4c2 2u

= 0,

The equation is said to be in its canonical form. This equation can be integrated once with respect to to give

u

= f ()

where f is an arbitrary function of . Integrating again, this time with respect to , we obtain

u = f() + g(),

with g an arbitrary function of . The general solution of the wave equation therefore has the form

u = f(x ct) + g(x+ ct),and so can always be written as the sum of right and left travelling waves.

4.1.4 Reduction to canonical form for other pdes

The same technique used for the wave equation can be used to simplify other partial differential equations.Consider the most general constant coefficient linear pde of 2nd order:

auxx + buxy + cuyy + rux + suy + qu = f(x, y). (12)

We introduce new variablesX = x+ y, Y = x+ y.

Here , are constants to be chosen later. The partial derivatives transform according to

u

x=

X

x

u

X+Y

x

u

Y=

u

X+

u

Y,

u

y=

X

y

u

X+Y

y

u

Y=

u

X+

u

Y,

and so

2u

x2=

(

X+

Y

)(u

X+

u

Y

)=

2u

X2+

2u

Y 2+ 2

2u

XY,

2u

y2=

(

X+

Y

)(u

X+

u

Y

)= 2

2u

X2+ 2

2u

Y 2+ 2

2u

XY,

2u

xy=

(

X+

Y

)(u

X+

u

Y

)=

2u

X2+

2u

Y 2+ (+ )

2u

XY.

Substituting these expressions into the pde (12) and gathering together like terms we get

(a+ b+ c2)2u

X2+ (a+ b + c2)

2u

Y 2+ (2a+ (+ )b+ 2c)

2u

XY(13)

+(r + s)u

X+ (r + s)

u

Y+ qu = F (X,Y ),


where F (X,Y ) = f(x, y). We can simplify this equation by eliminating the coefficients of uXX and uY Y .This can be achieved by choosing , to be the roots of the quadratic equation

c2 + b+ a = 0.

The previous analysis is only consistent if the roots are real and distinct, i.e. we require our pde to havethe property

b2 4ac > 0.We will come back to this point shortly. Assuming that this condition is satisfied our pde (13) can bewritten in the simpler form

(2a+ (+ )b+ 2c)2u

XY+ (r + s)

u

X+ (r + s)

u

Y+ qu = F (X,Y ).

This is known as the reduction to canonical form. In some cases we may be able to solve the resultingequation exactly, as we managed for the wave equation.

Example

Reduce the pde3uxx + 4uxy + uyy + 2ux + 2uy = 0

to canonical form and find its general solution. Find also the particular solution that satisfies the followingconditions on y = 0 :

u = 1 + xex, uy = (x 3)ex.We see that in this case a = 3, b = 4, c = 1. Following the method described above we set

X = x+ y, Y = x+ y,

with , the roots ofc2 + b+ a = 0.

For this choice of a, b, c we have

2 + 4+ 3 = 0 (+ 3)(+ 1) = 0,so that we can take = 1, = 3 (or vice versa). The pde becomes

(6 + 4(+ ) + 2)uXY + (2 + 2)uX + (2 + 2)uY = 0.

Substituting for , we obtain 4uXY 4uY = 0, and hence the canonical form isuXY + uY = 0.

To solve this we set = uY so that

X+ = 0.

The variables can be separated, and can be found to have the form

= G(Y )eX

where G is an arbitrary function of Y. Integrating again, this time with respect to Y, we obtain

u = G(Y )eX + F (X)

with F an arbitrary function of X. Writing back in terms of x and y we obtain the general solution

u = G(x 3y)e(xy) + F (x y). (14)We now apply the boundary conditions to determine the arbitrary functions F,G. Applying u = 1+xex

on y = 0 we have1 + xex = G(x)ex + F (x).


which must hold for all x. Replacing x by x y in this equation and rearranging we haveF (x y) = 1 + (x y)e(xy) G(x y)e(xy).

We can then substitute for F (x y) in (14) to obtainu = G(x 3y)e(xy) + 1 + (x y)e(xy) G(x y)e(xy). (15)

The function G can then be determined by applying the second boundary condition that uy = (x3)exon y = 0. Firstly we calculate from (15) that

uy = 3G(x3y)e(xy)+G(x3y)e(xy)e(xy)+(xy)e(xy)+G(xy)e(xy)G(xy)e(xy),so that

(uy)y=0 = (2G(x) 1 + x)ex.Then, applying the boundary condition

(x 3)ex = (2G(x) 1 + x)ex,from which we see that G(x) = 1 and hence

G(x) = x+ C,

with C an arbitrary constant. From this it follows that G(xy) = xy+C and G(x3y) = x3y+C.Substituting these expressions into (15) we obtain the particular solution

u = (x 3y + C)e(xy) + 1 + (x y)e(xy) (x y + C)e(xy).The constant C cancels out and we are left with

u = (x 3y)e(xy) + 1.Obviously we could go back and check that this satisfies the pde and the boundary conditions.

4.1.5 Classification of second order pdes

In the last section we saw that our method of reduction to canonical form was only valid for second orderpdes of the form

auxx + buxy + cuyy + rux + suy + qu = f(x, y)

with the propertyb2 4ac > 0.

We define pdes for which this inequality holds to be hyperbolic. Note that the classification only involvesthe coefficients of the second order derivatives. Clearly the wave equation is hyperbolic, since b = 0 inthat case and a and c are of different signs. However for Laplaces equation b = 0 and a = c = 1, andso b2 4ac < 0. Such pdes are known as elliptic. For the heat equation which we study next we haveb = c = 0 and so b2 4ac = 0. Such a pde is said to be parabolic.

4.2 The Heat Equation

In one dimension this is the partial differential equation

u

t=

2u

x2. (16)

Again we will look at some straightforward methods of solution for simple geometries.


4.2.1 Solution on a finite domain: separation of variables

Suppose we have boundary conditions of the form

u(0, t) = u(L, t) = 0 for t > 0, (17)

and an initial conditionu(x, 0) = f(x) for 0 x L. (18)

In a similar way to the wave equation we can seek a separated-variables solution of the form

u(x, t) = X(x)T (t).

Substitution into (16) leads toXT = X T,

so thatX

X=

T

T.

As in the case of the wave equation we now see that the left-hand-side is a function of x only, and theright-hand-side depends only on t. Therefore we must have, for some constant 2 :

X

X=

T

T= 2.

The sign of the separation constant 2 depends on the type of boundary conditions we impose. In ourparticular case since we are imposing (17) we require X(0) = X(L) = 0 which can only be accomplishedif 2 > 0 (otherwise the solutions for X will be of exponential form). We therefore have

X(x) = A cosx+B sinx

with A = 0 and = n/L. The corresponding solution for T arises from

T

T= 2 = n

22

L2,

and henceT (t) = Cen

22t/L2 .

Putting the two components of our solution together and summing over all modes we obtain

u(x, t) =n=1

bn sin(nx

L

)exp

(n

22t

L2

). (19)

Finally, applying the initial condition (18) we have

f(x) =n=1

bn sin(nx

L

), (0 < x < L),

which we recognize as a half-range Fourier sine series for f(x) with coefficients

bn =2

L

L0

f(x) sin(nx

L

)dx. (20)

Therefore the required solution of the heat equation subject to (17) and (18) is given by (19) and (20).The method can be adapted to accommodate boundary conditions on u/x, rather than u [see problemsheet 7].


4.2.2 Solutions on an infinite or semi-infinite domain

Again, as with the wave equation we can use Fourier transforms to help us obtain a solution. Considerfor example the problem

u

t=

2u

x2for 0 < x 0,

u(0, t) = 0 for t 0, (21)u(x, 0) = f(x) for 0 < x


Figure 2: A volume V bounded on its exterior by the surface S0.

4.3.2 Solution using separation of variables and transform techniques

The same techniques used for the wave and heat equations will also work on Laplaces equation. Themain limitations of these methods are the simple shapes of domain to which they can be applied, andthe necessity for the variables in the boundary conditions to separate. There are examples of the use ofthese techniques on problem sheets 7 and 8.

In what follows we shall consider the three-dimensional form of the equations and apply some of thetechniques and theorems we have learned to formulate solutions. We will use a number of results fromvector calculus and will also generalize the Dirac delta function we encountered in section 3.

4.3.3 Uniqueness of solution for interior problems

Let satisfy Poissons equation2 = f(r)

in a volume V. The volume is bounded on its exterior by a surface S0 (figure 2). On the surface we havea Dirichlet boundary condition, i.e. u = p(r) on S0. Then the solution for is unique.

Proof. We will suppose there are two solutions and seek a contradiction. Let the two solutions be1 and 2. They must both satisfy Poissons equation and the same boundary conditions. Forming thedifference

1 2,we must therefore have that

2 = 0 in V with = 0 on the boundary S0.Now recall from section 1.8.9, Greens first identity with = = :

S0

ndS =

V

2+ ||2 dV.

In view of the boundary conditions, the left-hand side is zero, and since satisfies Laplaces equationthroughout V , the first term on the right-hand-side is also zero. This leaves

V

||2 dV = 0.

The volume integral of a positive quantity can only be zero if the integrand is in fact identically zero,and so this implies

= 0 throughout V.This means that is at most a constant throughout V, but because is zero on the boundaries, it followsthat is identically zero throughout V. Hence 1 = 2 and the solution is unique.

Much the same reasoning applies if is a complex-valued function of position, if we have Neumannrather than Dirichlet boundary conditions, and also if the volume V has holes in it [problem sheet 8].


Figure 3: An unbounded volume V with an inner boundary S1. The dashed line representing the outerboundary is to be considered at infinity.

Example

Solve 2 = 2 inside the unit sphere r 1 with = 1 on r = 1.We note that the right-hand-side, the geometry and the boundary conditions have radial symmetry.

We therefore seek a solution = (r) independent of , in spherical polar coordinates. Referring backto section 1.9.10, the form for the Laplacian in spherical polars is

2 = 1r2 sin

{

r

(r2 sin

r

)+

(sin

)+

(1

sin

)}=

2

r2+2

r

r+cot

r2

+

1

r22

2+

1

r2 sin2

2

2,

so that our equation reduces under the conditions of radial symmetry to

1

r2d

dr

(r2d

dr

)= 2.

Integrating we obtain

r2d

dr=

2

3r3 + C

and hence

=1

3r2 C

r+D.

For to be finite at r = 0 we require C = 0. Applying = 1 on r = 1 gives D = 2/3 and so the requiredsolution is

=1

3r2 +

2

3.

Due to the uniqueness theorem, we know this is the only possible solution.

4.3.4 Uniqueness of solution for exterior problems

Now suppose we have a situation where there is an inner boundary S1, but the outer boundary S0 istaken to infinity so that V is now an unbounded volume (figure 3). Suppose

2 = f(r)throughout V. Suppose in addition that = O(1/r), /r = O(1/r2) as r . Then the solution for in V is unique.1

1If f = O(r) as r , this means that limr rf(r) = K 6= 0.


Proof

Lets start by considering the surface S0 to be a large sphere of radius R. Suppose there are two solutions1, 2 and form the difference . Proceeding as in the previous proof, using Greens first identity, wehave

V

||2 dV =1i=0

Si

ndS

=

S0

ndS,

since the integral over S1 is zero due to the boundary condition that vanishes on S1. Since S0 is asphere of radius R we can write dS = R2 sin dd in spherical polar coordinates, and /n = /r.We therefore have

V

||2 dV = R2 20

0

rsin d d.

Because of the assumed behaviour of as r , the right-hand-side above is of order 1/R and hencetends to zero as R. We therefore see that for the exterior problem

V

||2 dV = 0,

and hence, arguing as in the previous proof, = 0 and hence the solution is unique. The proof extendsto Neumann boundary conditions as before, and also to the situation where the volume V has any finitenumber of inner boundaries [problem sheet 8].

Example

Solve 2 = f(r) for 0 r a.

We will assume that is bounded throughout the region and that and d/dr 0 as r toguarantee a unique solution. As before we can seek a solution with radial symmetry and so

1

r2d

dr

(r2d

dr

)= f(r). (24)

Solving for r a first, we obtain (after some elementary calculus):

=1

6f0r

2 Ar+B.

We need A = 0 so that the solution is finite at r = 0. Next solving (24) for r > a we obtain

=C

r+D.

We require 0 as r and so we deduce that D must be zero. To find the remaining constantsB,C we impose continuity of and d/dr at r = a. This gives

1

6f0a

2 +B =C

a,1

3f0a = C

a2.

Substituting for B and C we find that the resulting solution for is

=

{16f0r

2 12f0a2, r a, 13f0a3/r, r > a.


4.3.5 Point sources and the Dirac delta function

Lets consider the result of our previous example in the limit as a 0. This means that the right-handside is becoming concentrated at the origin r = 0. In addition we will let f0 in such a way that(4/3)a3f0 remains equal to a constant - call this K. In this limit the solution we obtained above for for r > a becomes

= K4r

.

We have therefore obtained a solution to2 = f(r),

where the right-hand-side has the properties

f(r) =

{0, r 6= 0,, r = 0,

Another property of the function f(r) can be seen from the following calculation, in which we use thedivergence theorem over a sphere radius a centred at the origin:

V

f(r) dV =

V

2 dV =V

dV

=

r=a

rdS

=

20

0

a0

K

4r2r2 sin d d

= K.

The function f/K is the extension of the Dirac delta function studied in section 3 to three dimensionsand is denoted by (r). Moreover we can extend this definition so that the argument is a vector, i.e. thefunction (r). We define

(r) = 0 for r 6= 0,and

V

(r) dV =

{1, if V contains the origin0, otherwise.

We claim that the solution of2 = K(r) (25)

is

= K4 |r| . (26)

This can be verified in the following way. Firstly we have already seen in section 1.4.6 that 2(1/r) = 0for r 6= 0. To check the solution for r = 0 we integrate both sides of (25) over a volume V containingr = 0 to get

V

2 dV=KV

(r) dV = K.

Using the divergence theorem, the left-hand-side can be rewritten asS

n dS.

where the origin is interior to the closed surface S which bounds V. But if = K/4 |r| , then =(K/4)r/(1/r) = (K/4)r/r2 = (K/4)r/r3, and so this can be rewritten as

K

4

S

r nr3

dS.

The integral here should be familiar as it is equal to 4 using Gauss flux theorem (1.8.10). The left-hand-side is therefore also equal to K, and we have therefore verified that (26) is indeed the solution to(25), and is the only solution since we can invoke the uniqueness theorem.


If we move the source to r = r0 then we can easily modify our analysis to show that

The solution of 2 = K(r r0) is = K/(4 |r r0|).It can also be shown that the sifting property of the delta function carries over to three dimensions, i.e.

V

g(r)(r r1) dV = g(r1),

for any continuous function g.

4.3.6 Greens function

The Greens function G(r; r0) is defined as the solution of

2G = (r r0), (27)subject to some appropriate boundary conditions. For the three-dimensional problems we have studiedwe have seen that the so-called free-space Greens function, i.e. the function that satisfies (27) and tendsto zero as r is given by

G(r; r0) = 14 |r r0|

= 14[(x x0)2 + (y y0)2 + (z z0)2]1/2 ,

in Cartesian coordinates. As we shall see, knowledge of the Greens function will enable us to write downsolutions to Poissons and Laplaces equation in closed form.

4.3.7 Solutions to Poissons equation using Greens functions

Suppose that satisfies2 = f(r)

throughout some volume V. We will denote the boundary of V by the surface V. (If the volume isunbounded we will assume that = O(1/r) as r , so that a unique solution is guaranteed). Wesuppose that the boundary condition is of Dirichlet-type, i.e.

= p(r) on V.

Consider the following associated problem for G(r; r0) :

2G = (r r0),in V with

G = 0 on V.

Now take Greens second identity (section 1.8.9) and apply it to the functions and G :V

(2GG2) dV =V

(G

nG

n) dS.

The right hand side can be rewritten as V

p(r)G

n(r; r0) dS,

while after substituting for 2G and 2, the left-hand-side isV

(r)(r r0)G(r; r0)f(r) dV,


Figure 4: The points r0 and r0 and their position with respect to the plane z = 0.

which, upon use of the sifting property, becomes

(r0)V

G(r; r0)f(r) dV.

We can therefore write the solution for in the form

(r0) =

V

G(r; r0)f(r) dV +

V

p(r)G

n(r; r0) dS. (28)

Thus, in principle, if we can find the Greens function we can solve Poissons equation for .

A similar approach can be taken if Poissons equation is subject to Neumann conditions on theboundary [Problem Sheet 8].

4.3.8 The method of images

Often it is possible to find the Greens function explicitly by the method of images as shown in thefollowing Laplace equation example.

Suppose we wish to solve2 = 0

in the region z > 0 with the boundary condition

(x, y, 0) = p(x, y).

(We will assume = O(1/r) as z to ensure uniqueness).We tackle this problem by introducing an associated Dirichlet Greens function that satisfies

2G = (r r0) for z > 0, (29)G = 0 on z = 0. (30)

If the boundary condition at z = 0 is absent we know that the Greens function is

G(r; r0) = 14 |r r0| .

This solution is referred to as a source singularity of strength 1 situated at r = r0 = (x0, y0, z0). Nowlets add another singularity of opposite strength at a location the same distance below the x y plane,i.e. at r0 = (x0, y0,z0) (a mirror image). The modified Greens function is therefore

G(r; r0) = 14 |r r0| +

1

4 |r r0|. (31)


Now, when z = 0 we have|r r0| = |r r0|

(figure 4), and so we see that G = 0 on z = 0 as required. Thus the Greens function (31) satisfiesequation (29) and the boundary condition (30). Now that we have the Greens function we can applythe result of the previous section (equation (28) with f = 0) to obtain the solution for as

(r0) =

V

p(x, y)G

n(r, r0) dS.

In this example V is the plane z = 0 and /n = /z (since n is the outward normal to the volumeV ). Using our expression for G :

G = (4)1{((x x0)2 + (y y0)2 + (z z0)2

)1/2 ((x x0)2 + (y y0)2 + (z + z0)2)1/2} ,we find that

G

z= (4)1

{(z z0)

((x x0)2 + (y y0)2 + (z z0)2

)3/2+ (z + z0)

((x x0)2 + (y y0)2 + (z + z0)2

)3/2}.

Evaluating this quantity on z = 0 we have

G

z

z=0

= (4)12z0((x x0)2 + (y y0)2 + z20

)3/2,

and so the solution for can be expressed in the form

(x0, y0, z0) =z0

2

p(x, y)((x x0)2 + (y y0)2 + z20

)3/2dx dy.

Example

Suppose we wish to solve Laplaces equation for z > 0 with = p(x, y) on z = 0 and p(x, y) givenexplicitly as

p(x, y) =

{1, x2 + y2 1,0, x2 + y2 > 1.

Using the result derived above with this specific form for p :

(x0, y0, z0) =z0

2

x2+y21

((x x0)2 + (y y0)2 + z20

)3/2dx dy.

=z0

2

20

10

(2 + x20 + y20 + z

20 2x0 cos 2y0 sin )3/2 d d,

where we have switched to plane polar coordinates (, ). In particular, the solution along the zaxis is

(0, 0, z0) =z0

2

20

10

(2 + z20)3/2 d d

= z0[(2 + z20)1/2]10= 1 z0

(1 + z20)1/2

.

We can apply the method of images in a similar fashion to solve the same problem but with a Neumanncondition on z = 0 [Problem Sheet 8].


Figure 5: A two-dimensional Poisson problem.

4.3.9 Poissons equation in two dimensions

This same approach can also be used for two-dimensional problems, although the Greens function has adifferent (logarithmic) form in this case [see problem sheet 8]. For a Dirichlet problem in which we wishto solve

2 = f(r)in a region R of the x y plane, with

= p(x)

on the boundary C of R (figure 5), we consider the Greens function problem

2G = (r r0)in R, with

G = 0 on C.

Applying Greens second identity as in the three-dimensional case we find that

(r0) =

R

G(r; r0)f(r) dx dy +

C

p(x)G

n(r; r0) ds

where now r0 = (x0, y0). A similar expression can be derived for Neumann boundary conditions. Byusing the method of images as we did in three dimensions, Greens functions can be found explicitly forcertain problems. There is an example on the final problem sheet.

Chapter 4 PDE

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