1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

272 Chapter 4 The Potential Equation

The solution of the second problem is somewhat different. Again we seek
solutions in the product formu 2 (x,y)=X(x)Y(y). The homogeneous bound-
ary condition aty=0 and the boundedness condition become conditions on
Y(y):

Y( 0 )= 0 , Y(y)bounded asy→∞.

Then the potential equation becomes

X′′(x)

X(x)+

Y′′(y)

Y(y) =^0 , (8)

and both ratios must be constant. IfY′′/Yis positive, the auxiliary conditions
forceYto be identically 0. Thus, we takeY′′/Y=−μ^2 ,orY′′+μ^2 Y=0, and
find that the solution that satisfies the auxiliary conditions is

Y(y)=sin(μy),

for anyμ>0. Then the general solution of the equationX′′/X=μ^2 is

X(x)=Asinh(μx)

sinh(μa)

+Bsinh(μ(a−x))

sinh(μa)

.

We have chosen this special form on the basis of our experience in solving the
potential equation in the rectangle.
Sinceμis a continuous parameter, we combine our product solutions by
means of an integral, finding

u 2 (x,y)=

∫∞

0

[

A(μ)

sinh(μx)

sinh(μa)+B(μ)

sinh(μ(a−x))

sinh(μa)

]

sin(μy)dμ. (9)

The nonhomogeneous boundary conditions atx=0andx=aare satisfied if

u 2 ( 0 ,y)=

∫∞

0

B(μ)sin(μy)dμ=g 1 (y), 0 <y,

u 2 (a,y)=

∫∞

0

A(μ)sin(μy)dμ=g 2 (y), 0 <y.

Obviously these two equations are Fourier integral problems, so we know how
to determine the coefficientsA(μ)andB(μ). An example of this kind of prob-
lem is shown on the CD.
The potential equation can also be solved in a strip (0<x<a,−∞<y
<∞), a quarter-plane (0<x,0<y), or a half-plane (0<x,−∞<y<∞).
Along each boundary line, a boundary condition is imposed, and the solution
is required to remain bounded in remote portions of the region considered. In
general, a Fourier integral is employed in the solution, because the separation
constant is a continuous parameter, as in the second problem here.