!!HEADER: 8-1
8.
Two-Dimensional Laplace and Poisson Equations
In the previous chapter we saw that when solving a wave or heat equation it may be necessary to first
compute the solution to the steady state equation. In the case of one-dimensional equations this steady state
equation is a second order ordinary differential equation. In this chapter will look at two-dimensional steady state
equations, that is equations of the form
??ÐBß CÑ ´ ?BB ÐBß CÑ  ?CC ÐBß CÑ œ 0 ÐBß CÑ, ÐBß CÑ − H,
where H is some region in the plane. If 0 is not identically zero this equation is called the Poisson Equation. If
0 ´ ! the equation is homogeneous and is called the Laplace Equation. Although the methods for solving these
equations is different from those used to solve the heat and wave equations, there is a great deal of similarity. The
method for solving these problems again depends on eigenfunction expansions. The eigenfunctions may be
functions of one variable, B or C, as they were in the previous chapter, or they might be functions of both
variables. Indeed, the eigenvalue problem for elliptic partial differential operators is of fundamental importance
and the Sturm-Liouville problem is merely the one dimensional version of this more general concept. Therefore,
besides the Laplace and Poisson equations we will also have to look at eigenvalue problems of the type
 ??ÐBß CÑ œ -?ÐBß CÑ, ÐBß CÑ − H,
together with certain prescribed homogeneous boundary conditions.
This chapter consists of four sections. In the first section we will look at the Laplace equation with
boundary conditions of various types. These boundary conditions will be linear and of the general form
U ?ÐBß CÑ ´ !ÐBß CÑ `?
`8 ÐBß CÑ  " ÐBß CÑ ?ÐBß CÑ œ 1ÐBß CÑß ÐBß CÑ − ` H,
where `Î`8 denotes the outward normal derivative. We always assume that !#  " #  !. If ! ´ ! the boundary
condition is a Dirichlet boundary condition, if " ´ ! it is a Neumann boundary condition, and if !ÐBß CÑ and
" ÐBß CÑ are both nonvanishing on the boundary then it is a Robin boundary condition. If either ! or " has the
property that it is zero on only part of the boundary then the boundary condition is sometimes referred to as
mixed. In other words, the boundary condition is mixed if it is of Dirichlet type on part of the boundary, of
Neumann type on another part of the boundary, and of Robin type on a third part of the boundary (where at most
one of these parts is the empty set). If 0 is not identically zero this equation is called the Poisson Equation. In
the second section we study the two-dimensional eigenvalue problem. In the third section we will use the results
on eigenfunctions that were obtained in section 2 to solve the Poisson problem with homogeneous boundary
conditions (the caveat about eigenvalue problems only making sense for problems with homogeneous boundary
conditions is still in effect). Finally, in the last section, we will look at several examples, including problems of
the form
??ÐBß CÑ œ 0 ÐBß CÑ, ÐBß CÑ − Hß
U ?ÐBß CÑ œ 1ÐBß CÑ, ÐBß CÑ − ` H.)
Problems such as these are solved by using superposition (i.e. the linearity of the operators U and ?); we let
? œ @  A where
?@ÐBß CÑ œ !ß ÐBß CÑ − Hß
U @ÐBß CÑ œ 1ÐBß CÑ, ÐBß CÑ − ` H,
and
?AÐBß CÑ œ 0 ÐBß CÑ, ÐBß CÑ − Hß
U AÐBß CÑ œ !, ÐBß CÑ − ` H.
In other words we do not really need to treat the most general problems; when confronted with a “monster"
problem we just “divide and conquer". In the first example that we shall look at we see that if the strategy of
“divide and conquer" is not good enough then we resort to “further divide and then conquer"
8.1
THE LAPLACE EQUATION.
We begin with the Laplace equation on a rectangle with homogeneous Dirichlet boundary conditions on three
sides and a nonhomogeneous Dirichlet boundary condition on the fourth side.
Example 1. Solve the problem
?? ´ ?BB  ?CC œ !
on the rectangle d : !  B  +ß !  C  ,ß with boundary conditions
?Ð!ß CÑ œ ?ÐBß !Ñ œ ?Ð+ß CÑ œ !ß
?ÐBß ,Ñ œ 0 ÐBÑÞ
8.1 Dirichlet boundary conditions along the rectangular plate.
Solution. We separate variables again. Let us look for solutions of the form \ÐBÑ ] ÐCÑÞ Substituting this into
the equation we get
\ ww ÐBÑ ] ÐCÑ œ  \ÐBÑ ] ww ÐCÑÞ
Upon dividing this equation through by \ ] we obtain
\ ww ÐBÑ
\ÐBÑ
œ 
] ww ÐCÑ
] ÐCÑ
œ  -Þ
The homogeneous boundary conditions at B œ ! and B œ + indicate that we should demand that
\Ð!Ñ œ \Ð+Ñ œ !.
But this leads us to a familiar eigenvalue problem in the variable B. We know that the eigenpairs are
-8 œ Ð81Î+Ñ# and \8 ÐBÑ œ sin Ð81BÎ+ÑÞ
Since the equation is homogeneous the corresponding functions ] ÐCÑ are useful and so we compute them.
]8ww ÐCÑ œ Ð81Î+Ñ# ] ÐCÑÞ
Taking into account the homogeneous boundary condition at C œ ! we see that ]8 ÐCÑ œ sinh Ð81CÎ+Ñ. Therefore
we consider a solution of the form
_
?ÐBß CÑ œ ! -8 sinÐ81BÎ+Ñ sinhÐ81CÎ+Ñ.
8œ!
If we substitute this into the boundary condition at C œ , we see that we must have
(1)
_
0 ÐBÑ œ ! -8 sinÐ81BÎ+Ñ sinhÐ81bÎ+Ñ.
8œ!
This means that the quantities -8 sinhÐ81bÎ+Ñ are the Fourier coefficients of 0 , and so
-8 œ
#
'
+ sinhÐ81,Î+Ñ
!
+
0 ÐBÑ sinÐ81BÎ+Ñ .BÞ
(2)
Equations (1)-(2) give us the solution to the problem.
Example 2. Solve the problem ?@ œ ! in the rectangle !  B  +ß !  C  ,, with boundary conditions
@Ð!ß CÑ œ @Ð+ß CÑ œ @ÐBß ,Ñ œ !ß
@ÐBß !Ñ œ 1ÐBÑ.
Solution. We proceed as before by separating variables and finding that our solutions must be of the form
\8 ÐBÑ]8 ÐCÑ where \8 ÐBÑ œ sinÐ81BÎ+Ñ and where
]8ww ÐCÑ œ Ð81Î+Ñ# ]8 ÐCÑ,
but where ] must now satisfy the homogeneous boundary condition
]8 Ð,Ñ œ !Þ
The most convenient form for the solution is
]8 ÐCÑ œ sinh Ð81Ð,  CÑÎ+ÑÞ
This means that our solution has the form
_
@ÐBß CÑ œ ! ,8 sinh Ð81Ð,  CÑÎ+Ñ sinÐ81BÎ+ÑÞ
8œ"
Setting C œ ! we have
_
1ÐBÑ œ ! ,8 sinh Ð81,Î+Ñ sinÐ81BÎ+Ñ.
8œ"
This implies that we must have
,8 œ
#
'
+ sinhÐ81,Î+Ñ
!
+
1ÐBÑ sinÐ81BÎ+Ñ.BÞ
Remark. We observe that the superposition µ
? œ ?  @ satisfies the Laplace equation ?µ
? œ ! with the
boundary conditions
µ
? ÐBß !Ñ œ 1ÐBÑß µ
? ÐBß ,Ñ œ 0 ÐBÑß
µ
? Ð!ß CÑ œ µ
? Ð+ß CÑ œ !;
_
µ
? ÐBß CÑ œ ! Ò-8 sinh Ð81CÎ+Ñ  ,8 sinhÐ81Ð,  CÑÎ+ÑÓsinÐ81BÎ+Ñ.
8œ!
This form could also have been obtained directly without superposition of solutions to two separate problems.
This will be illustrated in the next example. However if we have nonhomogeneous boundary conditions on two
adjacent sides then superposition usually cannot be avoided. This can be seen by doing the next exercise.
Exercise 1. Solve the problem ?A œ ! on the rectangle !  B  +ß !  C  ,ß with boundary conditions
AÐBß !Ñ œ 1ÐBÑß AÐBß ,Ñ œ 0 ÐBÑß
AÐ!ß CÑ œ 9ÐCÑß AÐ+ß CÑ œ <ÐCÑÞ
Example 3. Solve the problem ?A œ ! on the rectangle !  B  +ß !  C  ,ß with boundary conditions
AB Ð!ß CÑ œ AB Ð+ß CÑ œ !ß
AÐBß !Ñ œ 0 ÐBÑß AC ÐBß ,Ñ œ 2ÐBÑÞ
Solution. We may, as indicated in the previous remark, split this problem into two problems, each with
homogeneous boundary conditions on three of the four sides. However, here we will show that this not really
necessary. Again we separate variables. This leads to the equations
] ww ÐCÑ
] ÐCÑ
œ 
\ ww ÐBÑ
\ÐBÑ
œ -,
with the boundary conditions \ w Ð!Ñ œ \ w Ð+Ñ œ !Þ This leads to the eigenfunctions
\! ÐBÑ ´ "ß \8 ÐBÑ œ cosÐ81BÎ+Ñß 8 œ "ß #ß ß Þ Þ Þ
For 8 € ! we see that the corresponding functions ]8 are given , for example, by
]8 ÐCÑ œ +8 expÐ81CÎ+Ñ  ,8 exp(  81CÎ+ÑÞ
However we can take any other linear combination of two linearly independent solutions. Any two of the
following set of functions are linearly independent solutions of the differential equation: expÐ81CÎ+Ñ,
exp(  81CÎ+Ñ, sinhÐ81CÎ+Ñ, coshÐ81CÎ+Ñ, sinhÐ81ÐC  ,ÑÎ+Ñ, cosh(81ÐC  ,ÑÎ+Ñ. We will choose the form
]8 ÐCÑ œ +8 sinhÐ81CÎ+Ñ  ,8 cosh(81ÐC  ,ÑÎ+ÑÞ
The reason for this choice will soon become obvious. For the case 8 œ ! we will choose
]! ÐCÑ œ +! C  ,! .
The solution to the problem will be given by
_
AÐBß CÑ œ "# Ð+! C  ,! Ñ  ! Ð+8 sinhÐ81CÎ+Ñ  ,8 cosh(81ÐC  ,ÑÎ+ÑÑ cosÐ81 BÎ+Ñ.
8œ"
The boundary conditions require that
_
2ÐBÑ œ "# +!  ! +8 Ð81Î+Ñcosh(81,Î+Ñ cosÐ81BÎ+Ñ,
8œ"
_
0 ÐBÑ œ "# ,!  ! ,8 coshÐ81,Î+Ñ cosÐ81BÎ+Ñ.
8œ"
This means that the coefficients are
+8 œ
,8 œ
#
'
81 coshÐ81,Î+Ñ
!
+
#
'
+ coshÐ81,Î+Ñ
!
+
2ÐBÑcosÐ81BÎ+Ñ.B ß 8  !ß
1ÐBÑcosÐ81BÎ+Ñ.B ß 8 € !.
If we integrate the series expansion we obtained for 2ÐBÑ from ! to + we obtain
+! œ
#'
+
!
+
2ÐBÑ .B ß 8  !ß
Exercise 2. Solve the problem ?? œ ! on the rectangle !  B  +ß !  C  ,, with boundary conditions
?B Ð!ß CÑ œ 9ÐCÑß ?B Ð+ß CÑ œ <ÐCÑß
?ÐBß !Ñ œ 0 ÐBÑß ?C ÐBß ,Ñ œ gÐBÑ.
Remark. The problem ?? œ ! with nonhomogeneous Neumann boundary conditions on the entire boundary
( `?
œ 9ÐBß CÑ on the boundary) usually does not have a solution (for example, look at the steady state equation
`8
for exercise 11 of the previous chapter). If it does happen to have a solution ?, then that solution is not unique
since ?  - is another solution for any constant -.
Next we look at the Laplace equation on a disk. The appropriate independent variables to use in this
case are of course the polar coordinates Ð<ß )Ñ. The Laplacian operator in polar coordinates is
?? œ "< Ð<?< Ñ< 
"
<# ?)) .
Exercise 3. Prove the above formula by using the chain rule.
Example 4. Solve the problem
?? œ !ß !  <  Vß
?ÐVß )Ñ œ 0 Ð)ÑÞ
We again try separation of variables and substitute a solution of the form eÐ<Ñ@Ð)Ñ. This yields

<Ð<ew Ð<ÑÑw
eÐ<Ñ
œ
@ww Ð)Ñ
@Ð ) Ñ
œ  -Þ
The functions @Ð)Ñ must be periodic of period #1:
@Ð1Ñ œ @Ð  1Ñß @w Ð1Ñ œ @w Ð  1ÑÞ
These are implicit boundary conditions. They are homogeneous boundary conditions and therefore we can solve
the eigenvalue problem
 @ww Ð)Ñ œ -@Ð)Ñß
with the above boundary conditions. But we already solved this problem at the beginning of chapter 5. The
eigenfunctions are /38) with corresponding eigenvalues 8# . Let us now solve the equation for e which we expect
to give useful information since the partial differential equation is homogeneous.
<Ð<ew Ð<ÑÑw
eÐ<Ñ
œ 8# Þ
This is a Cauchy-Euler equation,
<# eww  <ew  8# e œ !.
The solutions of the Cauchy-Euler equation are known to be of the form <: . Substituting this into the equation we
obtain the characteristic equation :#  8# œ !. If 8 Á ! then the general solution is of the form -<8  .<8 . If
8 œ ! then the general solution is of the form -  . j8 < (if the characteristic equation for the Cauchy-Euler
equation has a repeated root / then, besides the solution </ there is the solution B/ j8 <). This means that we have
found the following solutions to the Laplace equation:
<±8± /38) ß <±8± /38) ß 8 œ „ "ß „ #ß Þ Þ Þ ß
"ß j8 <Þ
Consequently we want to try to obtain a solution of the form
_
_
?Ð<ß )Ñ œ ! -8 <±8± /38)  .! j8 <  ! .8 <±8± /38) .
8œ_
(3)
8œ_
8Á!
However the terms which are not defined at < œ ! are unacceptable since we expect a perfectly well- defined
solution at the origin. Therefore we eliminate these terms and arrive at the form
_
?Ð<ß )Ñ œ ! -8 <±8± /38) Þ
(4)
8œ_
Next we impose the boundary condition:
_
0 Ð)Ñ œ ! -8 V±8± /38) Þ
8œ_
This is recognized as the complex Fourier series for the function 0 . This means that
-8 œ
" '
#1V±8±
1
1
0 Ð9Ñ/389 .9Þ
(5)
Although equations (4)-(5) provide the solution to the problem, we will proceed to simplify our answer. First
substitute the integral in equation (5) for the coefficient -8 in equation (4) and interchange the integral and the
sum. It is easy to see that by the Weierstrass M-test this interchange is justified for ! Ÿ <  V. We obtain
?Ð<ß )Ñ œ
" '
#1
1
1
_
±8±
0 Ð9Ñ ! ˆ V< ‰ /38Ð)9Ñ .9 Þ
8œ_
The sum inside the integral can be broken up into two geometric series which can be summed
_
_
_
8œ_
8œ!
8œ"
! ˆ < ‰±8± /38Ð)9Ñ œ ! ˆ < ‰±8± /38Ð)9Ñ  ! ˆ < ‰±8± /38Ð)9Ñ
V
V
V
œ
"
"Ð<ÎVÑ/3Ð)9Ñ
œ

Ð<ÎVÑ/3Ð)9Ñ
"Ð<ÎVÑ/3Ð)9Ñ
V# <#
V# #<V cos Ð)9Ñ<#
,
where we used the fact that /3Ð)9Ñ  /3Ð)9Ñ œ #cos Ð)  9Ñ. The solution can therefore be simply written as
?Ð<ß )Ñ œ
" '
#1
1
1
ÐV# <# Ñ 0 Ð9Ñ
V# #<V cos Ð)9Ñ<#
.9Þ
This is the Poisson Integral Formula for the Dirichlet problem on the disk. The factor
V# <#
#1ÐV# #<V cos Ð)9Ñ<# Ñ
which appears in this integral is called the Poisson kernel and it is closely related to the concepts of Green's
functions and fundamental solutions, topics that will be investigated in chapter 12.
Remark on the exterior problem. An interesting situation arises if we attempt to solve the Dirichlet problem in
the region <  V with the boundary condition ?ÐVß )Ñ œ 0 Ð)Ñ. In this case we can use all the terms in the series
(3). But that will mean that we cannot determine all the coefficients. This indicates that if we want a well-posed
problem then we must impose some additional auxiliary conditions. These are sometimes loosely called
boundary conditions at infinity. Such conditions should, ideally, come from the application that is being
modeled. For example, if we are looking at a heat equation then the total heat energy of the steady state should be
finite. Hence we might require that the ?Ð<ß )Ñ be absolutely integrable over the region <  V. This might lead us
to require that each of the terms in the series be absolutely integrable. However this leads to the series
_
?Ð<ß )Ñ œ ! .8 <±8± /38)
8œ_
±8±"
which would in general not allow us to fit the boundary condition since the Fourier series for 0 will be missing
three terms:
_
0 Ð)Ñ œ !
8œ_
-8 V±8± /38) Þ
±8± "
This equality cannot be satisfied if we pick boundary values 0 Ð)Ñ whose complex Fourier series contains nonzero
coefficients for 8 œ !ß „ ". Hence we should make our requirement on the solution less stringent. In fact, it is
clear that if we only eliminate terms that are unbounded then we arrive at a well-posed problem whose solution is
of the form
_
?Ð<ß )Ñ œ -!  ! .8 <±8± /38) .
8œ_
8Á!
This kind of solution seems unacceptable on the basis of physical considerations since it requires an infinite
amount of heat energy. The conclusion that we must draw is that the heat equation in the exterior region will
never attain a steady state. A little reflection upon the problem may lead us to conclude that this is reasonable.
Suppose for example that the initial temperature is ! and that 0 Ð)Ñ ´ ". Since the region is unbounded it will
never receive enough heat energy to achieve the steady state ? ´ ". At any time, no matter how far in the future,
the temperature far away from the origin will still be nearly zero.
Exercise 4. Solve the Laplace equation in the anulus 3  <  V with boundary conditions ?Ð3ß )Ñ œ 1Ð)Ñ and
?ÐVß )Ñ œ 0 Ð)Ñ.
Exercise 5. Solve the Laplace equation in the region !  <  V, !  )  1Î3, with boundary conditions
?ÐVß )Ñ œ 0 Ð)Ñ, ?Ð<ß !Ñ œ !, ?) Ð<ß 1Î$Ñ œ !Þ
We can also use the methods of separation of variables and eigenfunction expansion to solve certain
boundary value problems on semi-bounded domains. Let us illustrate this with an example.
Example 5. Solve the Laplace equation on the region !  C  ,ß !  B, with boundary conditions ?ÐBß !Ñ œ !,
?ÐBß ,Ñ  #?C ÐBß ,Ñ œ !, ?Ð!ß CÑ œ 0 ÐCÑ, where #  !.
Solution. We again separate variables. This leads to the equations
 ] ww ÐCÑ œ -] ÐCÑ,
\ ww ÐBÑ œ -\ÐBÑ,
together with the homogeneous boundary conditions
] Ð!Ñ œ !ß ] Ð,Ñ  #] w Ð,Ñ œ !Þ
We can solve the eigenvalue problem for ] . There are no negative eigenvalues. One also easily verifies that !
cannot be an eigenvalue. Hence we are left with eigenvalues - œ .# ß .  !. In view of the boundary condition
at C œ ! we see that ] œ sin .C. The second boundary condition yields the equation
tan ., œ  #.
which has positive solutions ."  .#  Þ Þ Þ Þ The corresponding functions \8 must be of the form
\8 ÐBÑ œ +8 /.8 B  ,8 /.8 B .
We now argue that an appropriate solution should not tend to infinity as B Ä _ and that we must therefore
demand that the coefficients ,8 be zero. The solution should therefore be of the form
_
?ÐBß CÑ œ ! +8 /.8 B sin .8 C.
8œ"
The boundary condition at B œ ! requires that
_
0 ÐCÑ œ ! +8 sin .8 C.
8œ"
This means that we must have
,
+8 œ
' 0 ÐCÑ sin .8 C .C
!
,
' sin# .8 C .C
.
!
Remark. If the boundary conditions in the above problem are changed to ?Ð!ß CÑ œ !ß ?ÐBß !Ñ œ !ß
?ÐBß ,Ñ œ 1ÐBÑ then we have a problem which cannot be solved with the tools we have available at this point. In
this case there is a nonhomogeneous boundary condition at C œ , and hence we must search for an eigenvalue
problem for \ . If we look at the equation for \ , that is  \ ww œ -\ , we see that the solutions are of the form
E/Ð!3"ÑB  F/Ð!3"ÑB , where Ð!  3" Ñ# œ -. If we impose the requirements that \Ð!Ñ œ ! and that \ÐBÑ be
bounded we are left with \ÐBÑ œ sin " B. There is no further boundary condition that will lead to a discrete set
(i.e. a countable set) of allowable values "8 . If \ÐBÑ œ sin " B then we easily see that we must have ] œ sinh " C.
This means that we must consider the solution of the boundary value problem to have the form
_
?ÐBß CÑ œ ' Y Ð" Ñ sinh " C sin " B ." .
!
This is an example of a Fourier transform and it will be studied in the next chapter.
8.2a. THE EIGENVALUE PROBLEM IN THE RECTANGLE.
In this section we will look at an eigenvalue problem
 Ð?BB  ?CC Ñ œ -?
on the rectangle d œ ÖÐBß CÑÀ !  B  +ß !  C  ,×Þ This problem makes sense with homogeneous boundary
conditions. We will choose the boundary conditions
?ÐBß !Ñ œ ?ÐBß ,Ñ œ ?B Ð!ß CÑ œ ?B Ð+ß CÑ œ !.
Since all boundary conditions are homogeneous, it makes sense to separate variables and look at one-dimensional
eigenvalue problems in both B and C. If we try separation of variables we obtain
 Ð\ ww ]  \] ww Ñ œ -\] ,
or
\ ww
\
ww
œ  -  ]] .
This implies that both of the ratios in the above equation must be constant, leading us to two one-dimensional
eigenvalue problems:
 \ ww œ -" \
\ Ð!Ñ œ \ w Ð+Ñ œ !
w
 ] ww œ -# ]
] Ð!Ñ œ ] Ð,Ñ œ !.
These are familiar eigenvalue problems and we know that the corresponding eigenfunctions are
\! ÐBÑ ´ "ß \8 ÐBÑ œ cos 81+B ß 8 œ "ß #ß ß Þ Þ Þ ß
]7 ÐCÑ œ sin 7,1C ß 7 œ "ß #ß Þ Þ Þ Þ
The eigenfunctions to the two-dimensional eigenvalue problem are therefore
F87 ÐBß CÑ œ \8 ÐBÑ]7 ÐCÑß
and the corresponding eigenvalues are
-87 œ ˆ 8+1 ‰ +ˆ 7,1 ‰ .
#
#
The eigenfunctions F87 ÐBß CÑ can be shown to form a complete orthogonal family, or basis, on the linear space
P# Ðd Ñ. This space is endowed with the inner product
,
+
¢0 ß 1£ œ ' ' 0 ÐBß CÑ 1ÐBß CÑ .B.CÞ
!
!
A function 0 ÐBß CÑ in this space may therefore be expanded as an eigenfunction expansion
_ _ µ
0 ÐBß CÑ µ ! ! J 87 F87 ÐBß CÑ œ
8œ! 7œ"
_
_
_
! " J!7 sin ˆ 71C ‰  ! ! J87 sin 71C cos 81B ,
#
,
,
+
8œ!
(6)
8œ! 7œ"
µ
µ
where the coefficients J 87 œ J87 if 8  ! and #J !7 œ J!7 . The coefficients are computed in the usual
manner by taking advantage of the orthogonality. They are
J87 œ
, +
% ' '
0 ÐBß CÑ sin 7,1C
+,
! !
cos 81+B .B.CÞ
(7)
Series of the form (6) are called double Fourier series. In this case it is a Fourier sine series in C and a Fourier
cosine series in B. In general such Fourier series can occur in a multitude of other forms: let Ö98 ÐCÑ× be an
orthogonal basis, derived perhaps from some self-adjoint regular Sturm-Liouville problem on ! Ÿ C Ÿ ,, and
similarly let Ö<7 ÐBÑ× be such an orthogonal basis defined on the interval ! Ÿ B Ÿ +. Then the products
Ö<7 ÐBÑ98 ÐCÑ× form an orthogonal basis for P# Ðd Ñ and can be used to form generalized double Fourier series for
square integrable functions defined on the rectangle d . Indeed, the concept of generalized Fourier series on two-
dimensional regions is not constrained to regions and boundary conditions for which we may obtain
eigenfunctions for the Laplacian operator by means of separation of variables. It can be shown that for bounded
domains W whose boundaries are not too bizarre the eigenfunctions G5 ÐBß CÑ which satisfy ?G5 œ -5 G5 in W
and G5 ÐBß CÑ œ ! on ` W form an orthogonal basis for the space of square integrable functions on W. This is also
true in higher dimensions as well as for other boundary conditions and more general elliptic equations.
Exercise 6. Find the eigenfunctions and eigenvalues for the problem
 ?? œ -? in the rectangle !  B  +ß !  C  ,ß
?ÐBß !Ñ œ ?Ð!ß CÑ œ ?C ÐBß ,Ñ œ ?Ð+ß CÑ œ !Þ
Also find the double Fourier series of the function
0 ÐBß CÑ œ š ! "
if !B+Î#ß !C,Î#
elsewhere
8.2b.
THE EIGENVALUE PROBLEM ON THE DISK.
Before we study the eigenvalue problem on the disk we need to look at some special functions that will
be needed. The first of these is the gamma function >Ð7 Ñ. This function allows us to extend the definition of the
factorial of a number to numbers other than nonnegative integers. The gamma function is defined by
_
>Ð7 Ñ œ ' /B B7 " .B.
Ð7  !Ñ
!
One can, without too much difficulty, prove the following properties of the Gamma function:
>Ð"Ñ œ "
>Ð7  "Ñ œ 7>Ð7 Ñ 1
>Ð8  "Ñ œ 8x if n is a nonnegative integer
>Ð "# Ñ œ È1.
Fig. 8.2 Graph of the gamma function >Ð7 Ñ 2Þ
1
2
When 5 is a positive integer we will sometimes use the notation
Ð7 Ñ5 œ 7 † Ð7  "Ñ † Ð7  #Ñ † † † † † Ð7  5  "Ñ œ
>Ð7 5Ñ
>Ð7 Ñ Þ
Ð7 Ñ! œ "Þ
Next we we take a brief look at Bessel's equation:
Cww ÐBÑ 
"
B
Cw ÐBÑ  Š" 
/#
CÐBÑ
B# ‹
œ !,
where we take / € !Þ It can be shown that this equation has a solution of the form
_
! -5 B/ 5 Þ
5œ!
The first coefficient -! is arbitrary. Once it has been assigned, the value of the others may be determined by
substituting the above expression into the differential equation and balancing coefficients of like powers of B.
The Bessel function of the first kind of order / is obtained by choosing the first coefficient to be
-! œ Ò#/ >Ð/  "ÑÓ" . This function is denoted by N/ and is given by the series
N/ ÐBÑ œ
B/
#/ >Ð/ "Ñ
_
!
5œ!
ˆ " B# ‰5
%
.
5x Ð/ "Ñ5
Fig 8.3 The graphs for the Bessel functions of the first kind of orders 0, 1, and 2.
We note that the function N/ ÐBÑ ¸ -! B/ for small values of B. For large values of B it can be shown that this
function approaches the damped sinusoid É 1#B cosÐB  #" 81  %" Ñ. The graphs for the functions N! ß N" ß and N#
are shown below.
Since Bessel's equation is of second order we know that there must be a second, linearly independent
solution. If / is not an integer then a second solution may be found which is of the form
_
! -5 B/ 5 Þ
5œ!
If / is not an integer and the first coefficient is chosen as -! œ Ò#/ >Ð  /  "ÑÓ" then we obtain the function N/ .
This function behaves like -! B/ for small values of B. If / is an integer then, since the gamma function is not
defined for nonpositive integers, N/ makes no sense and we must find a second solution by some other means.
For example a second solution may be obtained by the technique of reduction of order ( look for a solution of the
form ?ÐBÑN/ ÐBÑ ). Once we have found two linearly independent solutions the other solutions may be formed by
linear combinations. A standard second solution in case the order is an integer is either Weber's function or
Neumann's function. We will not supply a formula, however it should be pointed out that these solutions tend to
 _ as B Æ !. Hence we see that in all cases the second solution becomes undefined (singular) at B œ !Þ These
singular solutions also approach damped sinusoids as B gets large. The graphs of Weber functions of orders 0, 1,
and 2 are shown below.
Fig. 8.4 The graphs of Weber functions of orders 0, 1, and 2.
We are now ready to look at the eigenvalue problem on the disk:
 <Ð<?< Ñ<  ?)) œ -<# ?ß Ð<ß )Ñ − W ³ ÖÐ<ß )ÑÀ ! Ÿ <  V×ß
?ÐVß )Ñ œ !.
Separation of variables, ?Ð<ß )Ñ œ eÐ<Ñ@Ð)Ñ, leads to
-<# 
<Ð<ew Ð<ÑÑw
eÐ<Ñ
œ 
@ww Ð)Ñ
@Ð ) Ñ Þ
As before, since the function @Ð)Ñ must be periodic with period #1 so that it must be a linear combination of the
functions
@8 Ð)Ñ œ /38) ß
8 œ !ß „ "ß „ #ß Þ Þ Þ Þ
This leads us to an equation for e:
eww  <" ew  Ð-  8# Î<# Ñe œ !Þ
(8)
This is nearly Bessel's equation. First we will show that - must be positive. To do this we note that the above
equation may be written as
 Ð<ew Ñw  8# e œ -<eÞ
(9)
We see that this is a Sturm-Liouville problem. Although it is not regular and only has a boundary condition at
< œ V, namely eÐVÑ œ ! we can still use the same proofs as were used in Chapter 6 to prove that the eigenvalues
are real and positive, and that eigenfunctions corresponding to different eigenvalues are orthogonal with respect to
the weight function <. However, in order to carry out the various manipulations we need to adjoin the auxiliary
condition requiring that the eigenfunctions are bounded.
Exercise 7. Show that the solutions to (9) which are bounded and satisfy eÐVÑ œ ! must correspond to positive
eigenvalues. Show that eigenfunctions corresponding to different eigenvalues are orthogonal with respect to the
weight function AÐ<Ñ ´ <.
Next we make a change of variables in equation (8). Let = œ È- < and let f be the function of = defined by
f Ð=Ñ œ eÐ=ÎÈ-Ñ. With this change of variables equation (8) turns into a Bessel's equation:
f ww  "= f w  Š" 
8#
=# ‹f
œ !Þ
The only bounded solutions of this equation are multiples of N8 Ð=Ñ. The required boundary condition at < œ V
translates into the requirement that N8 ÐÈ- VÑ œ !. This means that È- V must correspond to a root of N8 . Let
us use the notation that these roots be labeled as "87 . We assume that they are indexed consecutively in
increasing order:
N8 Ð"87 Ñ œ !ß "8"  "8#  "8$  † † † .
Therefore we see that È- V œ "87 for some index 7 so that the eigenvalues must be
-87 œ Ð"87 ÎVÑ#
and the corresponding eigenfunctions are
F87 Ð<ß )Ñ œ N8 Ð"87 <ÎVÑ /„38) .
In order to simplify notation we will define N8 to be the same as N8 whenever 8 is an integer:
./0
N8 Ð=Ñ œ N8 Ð=Ñ, "87 œ "87
so that the eigenfunctions may be labeled as
F87 Ð<ß )Ñ œ N8 Ð"87 <ÎVÑ /38) , 8 œ !ß „ "ß „ #ß Þ Þ Þ and 7 œ "ß #ß Þ Þ Þ.
It can be shown that for each fixed nonnegative integer n the functions ÖN8 Ð"87 <ÎVÑ×_
7œ" form an
orthogonal basis for the space P#A Ð!ß <Ñ with weight function AÐ<Ñ ´ <. What this means is that the
functions F87 Ð<ß )Ñ œ N8 Ð"87 <ÎVÑ /38) form an orthogonal basis for P# ÐWÑ. We will not prove the completeness
of this family, and the orthogonality is left as an exercise for the reader.
Exercise 8. Show that
' F87 F45 .E œ !
W
unless 8 œ 4 and 7 œ 5Þ Note that the area differential .E is <.<.) in polar coordinates, thus providing exactly
the correct weight function.
Although it is possible to consider eigenvalue problems on the disk with other kinds of boundary
conditions, this will take us too far afield. However it is possible, with the information given above to solve
eigenvalue problems of the form
 ??Ð<ß )Ñ œ -?Ð<ß )Ñß !  <  Vß !  )  !ß
?ÐVß )Ñ œ !ß
+?) Ð<ß !Ñ  ,?Ð<ß !Ñ œ !ß
-?) Ð<ß !Ñ  .?Ð<ß !Ñ œ !ß
with +, Ÿ !ß -. € !Þ
Exercise 9. Solve the above problem with ! œ 1Î$ß + œ !ß . œ !Þ
8.3
THE POISSON PROBLEM.
The Poisson problem may be solved by looking for a solution which is expressed as a generalized
Fourier series in terms of the eigenfunctions of the Laplacian, The method is easily described in general.
Suppose that we have obtained all the eigenfunctions for
 ?? œ -? in Hß
U ? œ ! on ` HÞ
Let us denote the eigenpairs by Ð-5 ß F5 Ñ -- we are using a single index although generally these arise with double
indices for planar regions and triple indices for three-dimensional regions. Suppose we need to solve the Poisson
problem
??ÐBß CÑ œ 0 ÐBß CÑ in Hß
?ÐBß CÑ œ ! on ` HÞ
We express 0 as an eigenfunction expansion
_
0 œ ! J8 F8
8œ"
where the coefficients are computed, as usual, by the formula
' 0 F8 .E
J8 œ
H
' ±F8 ±# .E .
(10)
H
Next we suppose that ? may also be expressed as a series in terms of the eigenfunctions
_
? œ ! Y 8 F8 .
(11)
8œ"
If we substitute the series expressions for ? and 0 into the Poisson equation and use the fact that ?F8 œ  -8 F8 ,
we see that
_
_
8œ"
8œ"
!  -8 Y8 F8 œ ! J8 F8 ,
and hence the coefficients for the series expansion of the solution ?are given by
Y8 œ 
J8
-8 Þ
(12)
Equations (10)-(12) provide a formal solution to the Poisson problem. In some cases the solution of the Poisson
problem may be given in a form much like the Poisson Integral formula:
?ÐBß CÑ œ ' ' KÐBß Cà 0 ß (Ñ 0 Ð0 ß (Ñ .0 .(.
H
The function K is called Green's function. A formal calculation seems to indicate that
_
KÐBß Cà 0 ß (Ñ œ ! 
8œ"
F8 ÐBßCÑF8 Ð0ß( Ñ
-8
Þ
The question that arises is in what sense does this series converge? To answer that would take us far outside the
scope of this chapter. For certain problems such Green's functions can be cleverly constructed and can be used to
give the solution of the Poisson problem in a much more convenient form than the above series representation.
One such construction is the so-called method of images that is often seen in the study of electrostatics.
We conclude this chapter with an easy example.
Example 6. Solve the Poisson problem
?? œ "ß !  B  1ß !  C  1ß
?ÐBß !Ñ œ ?Ð1ß CÑ œ ?ÐBß 1Ñ œ ?Ð!ß CÑ œ !Þ
This problem represents the vertical displacement of a membrane due to a uniform downward force, such as
gravity.
Solution. It is easy to see that the eigenfunctions are sin 7B sin 8C . Applying the Laplacian to them we see that
?sin 7B sin 8C =  Ð7#  8# Ñsin 7B sin 8C. The function 0 ÐBß CÑ ´ " can be expanded as a double Fourier
series with coefficients
J78 œ
œ
% '
1#
!
%
781# Ò"
1
1
' sin 7B sin 8C.B.C
!
 Ð  "Ñ7 ÓÒ"  Ð  "Ñ8 ÓÞ
Therefore
Y78 œ
%Ò"Ð"Ñ7 ÓÒ"Ð"Ñ8 Ó
Þ
Ð7# 8# Ñ781#
We have
_
_
?ÐBß CÑ œ ! ! Y78 sin 7B sin 8C
7œ" 8œ"
œ 
)
1#
˜sin B sin C 
"
"& sin B sin $C

"
"& sin $B sin C

"
)" sin $B sin $C
 Þ Þ Þ ™.
Just a few terms give a good approximation since the series converges rapidly.
Exercise 10. Solve the problem
?? œ "ß !  B  1ß !  C  1ß
?C ÐBß !Ñ œ ?Ð1ß CÑ œ ?ÐBß 1Ñ œ ?Ð!ß CÑ œ !Þ
Exercise 11. Solve the problem ?? œ  -, - a constant, on !  B  1, !  C  1, with boundary conditions
?ÐBß !Ñ œ !, ?ÐBß 1Ñ œ !, ?Ð!ß CÑ œ !, ?Ð1ß CÑ œ CÐ1  CÑ.
Exercise 12. Solve the problem ?? œ ! in the semi-infinite strip C  !, !  B  P, with boundary conditions
?Ð!ß CÑ œ ?ÐPß CÑ œ !, and ?C ÐBß !Ñ œ 1ÐBÑ.
Exercise 13. Obtain a solution to the problem in example 6 as follows: note that @ÐBß CÑ ³ B# Î# is a solution to
the PDE, so that A ³ ?  @ satisfies the Laplace equation. Solve for A and obtain @. In chapter 9 we will obtain
a method for finding particular solutions to the Poisson problem so that w3e may reduce Poisson equations to
Laplace equations.