1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

Miscellaneous Exercises 395

so that

α^2 −β^2 + 2 iαβ=

1

4 +iω,

or {

α^2 −β^2 =^14 ,

2 αβ=ω.

(To solve these equations: (i) solve the second forβ; (ii) substitute the

expression found into the first; (iii) solve the resulting biquadratic forα.)

30.Noting that the persistent part isu 1 =Im(f(iω))(see Exercise 28), de-
termine
u 1

(

x ̄, ̄t

)

=e(^12 −α)x ̄sin

(

ω ̄t−βx ̄

)

,

whereαandβareasinExercise29.