1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

4.2 Potential in a Rectangle 263
It is evident thatu 1 +u 2 is the solution of the original problem Eqs. (13)–
(17). Also, each of the functionsu 1 andu 2 has homogeneous conditions on
parallel boundaries. We already have determined the form ofu 1. The other
function would be of the form

u 2 (x,y)=

∑∞

n= 1

sin(μny)

Ansinh(μnx)+Bnsinh(μn(a−x))

sinh(μna) , (18)

whereμn=nπ/band

An=^2

b

∫b

0

g 2 (y)sin(μny)dy,

Bn=

2

b

∫b

0

g 1 (y)sin(μny)dy.

In the individual problems foru 1 andu 2 , the technique of separation of
variable works because the homogeneous conditions on parallel sides of the
rectangle can be translated into conditions on one of the factor functions.
When the boundary conditions are not complicated functions, it may be
possible to satisfy some of them with a polynomial function. (See Exercises 1
and 2 of Section 4.1.) Then the difference betweenuand the polynomial is a
solution of the potential equation that satisfies some homogeneous boundary
conditions.

EXERCISES

1. Show that sinh(λy)and sinh(λ(b−y)) are independent solutions of
   Y′′−λ^2 Y=0 withλ=0. Thus a combination of these two functions may
   replace a combination of sinh and cosh as the general solution of this dif-
   ferential equation.


2. Show that the solution of the example problem may be written

u(x,y)=^8

π^2

∑∞

n= 1

sin

(nπ

2

)

n^2

cosh

(nπ

a(y−

1

2 b)

)

cosh

(nπb

2 a

) sin

(nπx

a

)

.

3. Use the form in Exercise 2 to computeuin the center of the rectangle in
   the three casesb=a,b= 2 a,b=a/2. (Hint: Check the magnitude of the
   terms.)


4. Verify that each term of Eq. (9) satisfies Eqs. (1), (4), and (5).