1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

6.3 Partial Differential Equations 379

only wheresor sinh(

√

s)is zero. Because sinh(

√

s)=0 has no real root besides
zero, we seek complex roots by setting√s=ξ+iη(ξandηreal).
The addition rules for hyperbolic and trigonometric functions remain valid
for complex arguments. Furthermore, we know that

cosh(iA)=cos(A), sinh(iA)=isin(A).

By combining the addition rule and these identities we find

sinh(ξ+iη)=sinh(ξ )cos(η)+icosh(ξ )sin(η).

This function is zero only if both the real and imaginary parts are zero. Thusξ
andηmust be chosen to satisfy simultaneously

sinh(ξ )cos(η)= 0 , cosh(ξ )sin(η)= 0.

Of the four possible combinations, only sinh(ξ )=0 and sin(η)=0produce
solutions. Thereforeξ=0andη=±nπ(n= 0 , 1 , 2 ,...); whence
√
s=±inπ, s=−n^2 π^2.

Recall that only the value ofs,notthevalueof√s, is significant.
Finally then, we have locatedr 0 =0, andrn=−n^2 π^2 (n= 1 , 2 ,...). We
proceed to find theAn(x)by the same method used in Section 6.2. The com-
putations are done piecemeal and then the solution is assembled.

Part a. (r 0 =0.) In order to findA 0 we multiply both sides of our proposed
partial fractions development

U(x,s)=

∑∞

n= 0

An(x)^1

s−rn

bys−r 0 =sand take the limit assapproachesr 0 =0. The right-hand side
goes toA 0. On the left-hand side we have

lims→ 0 ssinh(

√sx)+sinh(√s( 1 −x))

ssinh(

√

s)

=x+ 1 −x= 1 =A 0 (x).

Thus the part ofu(x,t)corresponding tos=0is1·e^0 t=1, which is easily
recognized as the steady-state solution.

Part b. (rn=−n^2 π^2 ,n= 1 , 2 ,....) For these cases, we find

An=

q(rn)

p′(rn),