4.4 Potential in Unbounded Regions 271
In order to make the separation of variables work, we must break this
up into two problems. Following the model of Sections 4.2 and 4.3 we set
u(x,y)=u 1 (x,y)+u 2 (x,y)and require that the parts satisfy these two solv-
able problems:
∇^2 u 1 = 0 , ∇^2 u 2 = 0 , 0 <x<a, 0 <y,
u 1 (x, 0 )=f(x), u 2 (x, 0 )= 0 , 0 <x<a,
u 1 ( 0 ,y)= 0 , u 2 ( 0 ,y)=g 1 (y), 0 <y,
u 1 (a,y)= 0 , u 2 (a,y)=g 2 (y), 0 <y.
We attack the problem foru 1 by assuming the product form and separating
variables:
u 1 (x,y)=X(x)Y(y), X′′(x)
X(x)=−Y
′′(y)
Y(y)=−λ^2.The sign of the constant−λ^2 is determined by the boundary conditions at
x=0andx=a, which become homogeneous conditions on the factorX(x):
X( 0 )= 0 , X(a)= 0. (5)(We also can see that the condition to be satisfied alongy=0demandsfunc-
tions ofxthat permit a representation of an arbitrary function.)
The boundary conditions, Eq. (5), together with the differential equation
X′′+λ^2 X=0(6)
that comes from the separation of variables, constitute a familiar eigenvalue
problem, whose solution is
Xn(x)=sin(nπx
a)
,λ^2 n=(nπ
a) 2
, n= 1 , 2 , 3 ,....The equation forYis
Y′′−λ^2 Y= 0 , 0 <y.In addition to satisfying this differential equation,Ymust remain bounded as
y→∞. The solutions of the equation areeλyande−λy. Of these, the first is
unbounded, so
Yn(y)=exp(−λny).Finally, we can write the solution of the first problem as
u 1 (x,y)=∑∞
n= 1ansin(nπx
a)
exp(−nπy
a)
. (7)
The constantsanaretobedeterminedfromtheconditionaty=0.