4.2 Potential in a Rectangle 259
4.2 Potential in a Rectangle
One of the simplest and most important problems in mathematical physics is
Dirichlet’s problem in a rectangle. To take an easy case, we consider a problem
with just two nonzero boundary conditions:
∂^2 u
∂x^2+∂
(^2) u
∂y^2
= 0 , 0 <x<a, 0 <y<b, (1)
u(x, 0 )=f 1 (x), 0 <x<a, (2)
u(x,b)=f 2 (x), 0 <x<a, (3)
u( 0 ,y)= 0 , 0 <y<b, (4)
u(a,y)= 0 , 0 <y<b. (5)
It is not immediately clear that separation of variables will work. However,
we have a homogeneous partial differential equation and some homogeneous
boundary conditions, so we can try the method. If we assume thatu(x,y)has
a product formu=X(x)Y(y),thenEq.(1)becomes
X′′(x)Y(y)+X(x)Y′′(y)= 0.
This equation can be separated by dividing through byXYto yield
X′′(x)
X(x)
=−Y
′′(y)
Y(y). (6)
The nonhomogeneous conditions Eqs. (2) and (3) will not, in general, become
conditions onXorY, but the homogeneous conditions Eqs. (4) and (5), as
usual, require that
X( 0 )= 0 , X(a)= 0. (7)
Now, both sides of Eq. (6) must be constant, but the sign of the constant
is not obvious. If we try a positive constant (say,μ^2 ), Eq. (6) represents two
ordinary equations:
X′′−μ^2 X= 0 , Y′′+μ^2 Y= 0.The solutions of these equations are
X(x)=Acosh(μx)+Bsinh(μx), Y(y)=Ccos(μy)+Dsin(μy).In order to makeXsatisfy the boundary conditions Eq. (7), bothAandBmust
be zero, leading to a solutionu(x,y)≡0. Thus we try the other possibility for
sign, taking both members in Eq. (6) to equal−λ^2.