ABSTRACT
Boundary problems with one closed boundary
Laplace and Poisson equations
It is well known that a vector eld
K having a vanishing curl
K can be
represented by a potential function U
K
x
x y z
U
x
K
y
x y z
U
y
K
z
x y z
U
z
Setting up the divergence of such a eld results in
K
x
x
K
y
y
K
z
z
div
K
x
U
x
y
U
y
z
U
z
U
There are many practical examples in physics and engineering for the Laplace
equation U In hydrodynamics in problems of heat conduction in elec
tromagnetism and in many other elds the Laplace equation or its inhomo
geneous counterpart the Poisson equation appears as
U x y
Let us now solve a simple twodimensional boundary value problem for the
Laplace equation The closed domain should be a rectangle of dimensions a
and b see Figure
x
a
a
g
y g
y
f
x
f
x
b b
y
Figure
Rectangular boundary
Ux y f
x x a
Ux y b f
x x a
Ux y g
y y b
Ux a y g
y y b
So that this problem is well dened the functions must be continuous at
the corners but it is not necessary that the boundary value functions satisfy
U Let be the values of the boundary functions at the corner
then they have to satisfy
f
g
f
a g
b
f
a g
f
g
b
Using the usual setup Ux y Xx Y y for linear partial dierential
equations with constant coecients one gets
X
x p
Xx
Y
y p
Y y
where p is the separation constant The general solutions of these equations
read
Xx
X
m
A
m
sin p
m
x C
m
Y y
X
n
B
n
sinh p
n
y C
n
but other solutions like harmonic polynomials section or functions of a
complex variable like in exist too The partial amplitudes A
m
B
n
the separation constants p
m
and the constants C
m
C
n
are still unknown
They have to be determined by the boundary conditions Since u
is linear the sum over particular solutions is again a solution superposition
principle Since the partial dierential equation u is homogeneous each
particular solution may be multiplied by a factor partial amplitude The
phase constants C
m
may be used to nd a solution with two functions
A
m
sin p
m
x C
m
A
m
sin p
m
x cosC
m
A
m
cos p
m
x sinC
m
and analogously for sinh
Ux y
X
k
D
k
sin p
k
x C
k
sinh
p
k
y
C
k
To simplify calculations we choose
U y g
y f
Ua y g
y f
a
Ux b f
x g
b
but
Ux f
x
and for continuity f
a g
b must be valid To satisfy the
boundary condition by the solution the expression
U y
X
k
D
k
sin C
k
sinh
p
k
y
C
k
must be valid This can be satised by C
k
for all k The boundary
condition demands
Ua y
X
k
D
k
sin p
k
a sinh
p
k
y
C
k
From this condition the separation constants p
k
will immediately be deter
mined
p
k
a k p
k
k a k
Like in many boundary value problems the separation constants are deter
mined by the boundary conditions Next we consider the boundary condition
Ux b
X
k
D
k
sin
k x
a
sinh
k b
a
C
k
which can be satised by
k b
a
C
k
C
k
k b
a
Finally condition has to be taken into account
Ux f
x
X
D
k
sin
k x
a
sinh
k b
a
f
is satised automatically
As one sees represents a Fourier series expansion of the given
function f
x To nd the Fourier expansion coecients we multiply
by sinm xam k and integrate over x
a
Z
f
x sin
m x
a
dx
a
Z
X
k
D
k
sin
k x
a
sin
m x
a
sinh
k b
a
d
x
In section we discussed the orthogonality of some functions compare
The attribute of orthogonality is very important It allows to ex
pand given functions like f
x and to express them by a series according to
orthogonal functions In one has
a
Z
b
sin
k x
a
sin
m x
a
dx for k m
This is easy to prove The simple transformation x a modies
into
a
Z
sin k sinmd
a
for k m
for k m
m k
Thus we get
a
Z
f
x sin
k x
a
dx
a
D
k
sinh kba
which gives the Fourier expansion coecients D
k
The solution of our
boundary value problem dened by to is then given by
Ux y
X
k
sinhk ba k ya
a sinh kba
a
Z
f
sink a d sink xa
But what about the more general problem to We will come
back to it in problem of this section
The boundary problem we just solved is inhomogeneous because f
x
Why could we not homogenize the problem as we learnt in section There
we have shown that an inhomogeneous problem consisting of an homogeneous
equation together with an inhomogeneous condition may be converted into an
inhomogeneous equation of the type with an homogeneous boundary
condition
There are two methods to solve linear inhomogeneous partial dierential
and determine free parameters like expansion coecients within U in
such a way that U satises the inhomogeneous partial dierential equa
tions This delivers a particular solution of the inhomogeneous equation
that however does not satisfy the homogeneous equation
First solve the matching homogeneous partial dierential equation and
expand the inhomogeneous term according to the solutions of the
homogeneous equation
We start with method The dierential equation is inhomogeneous
and the boundary conditions should be homogeneous With regard to our
experience we set up
Ux y
X
c
y sin
x
a
where c
y are the expansion coecients Now should rst satisfy
the boundary conditions We multiply by sin xa and integrate
over x Due to we then obtain
a
Z
sin
x
a
sin
x
a
dx
a
and
c
y
a
a
Z
Ux y sin
x
a
dx
Due to the homogeneous boundary conditions
U y Ua y Ux Ux b
the c
must satisfy c
c
b for all We now set up
c
y
X
sin
y
b
so that
b
b
Z
c
y sin
y
b
dy
and the expression satisfying the homogeneous boundary conditions reads
Ux y
X
X
sin
x
a
sin
y
b
where now
ab
a
Z
b
Z
Ux y sin
x
a
sin
y
b
dydx
must satisfy the inhomogeneous equation too To nd a particular solution
of the inhomogeneous partial dierential equation expand the inhomogeneous
term with respect to the solutions of the matching homogeneous equation
But wait we do not yet have these solutions So we just have to write down
a setup compatible with the boundary conditions
x y
X
X
d
sin
x
a
sin
y
b
where
d
ab
a
Z
b
Z
x y sin
y
b
sin
x
a
dydx
But from where do we get x y We have to insert which con
tains the homogeneous boundary conditions into the inhomogeneousPoisson
equation
x y U
xx
U
yy
X
X
sin
x
a
sin
y
b
a
b
X
X
d
sin
x
a
sin
y
b
Now we can read o the
from
a
b
a
b
d
and the boundary problem is solved and has the solution
Ux y
X
X
a
b
d
a
b
nu
sin
x
a
sin
y
b
But remember this only is a particular solution of and does not solve
the homogeneous equation
Now let us look how method works We rst have to solve the matching
homogeneous equation U and then expand with respect to the func
tions solving U We demonstrate the procedure on a threedimensional
spherical problem We use and the ordinary dierential equations
The solutions to had been given in section
!
