6.4 More Difficult Examples 385
The transformed equation and its general solution are
d^2 U
dx^2 =sU,^0 <x, U(^0 ,s)=F(s),
U(x,s)=Aexp(
−
√
sx)
+Bexp(√
sx)
.
We make two further assumptions about the solution: first, thatu(x,t)is
bounded asxtends to infinity and, second, that
√
smeans the square root
ofsthat has a nonnegative real part. Under these two assumptions, we must
chooseB=0andA=F(s),making
U(x,s)=F(s)exp(
−
√
sx)
.
In order to find the persistent part ofu(x,t), we apply the Heaviside inver-
sion formula to those values ofshavingnonnegativereal parts, because a value
ofswithnegativereal part corresponds to a function containing a decaying
exponential — a transient. To understand this fact, consider the pair
f(t)= 1 −e−βt+αsin(ωt),F(s)=^1
s−^1
s+β+ αω
s^2 +ω^2.
Now we return to the original problem with this choice forf(t).Thevalues
ofsfor whichU(x,s)=F(s)exp(−
√
sx)becomes infinite are 0,±iω,−β.The
last value is discarded, because it is negative. Thus, the persistent part of the
solution is given by
A 0 e^0 t+A 1 eiωt+A 2 e−iωt,and the coefficients are found from
A 0 =lims→ 0[
sF(s)exp(
−
√
sx)]
= 1 ,
A 1 =slim→iω[
(s−iω)F(s)exp(
−
√
sx)]
= 2 αiexp(
−
√
iωx)
,
A 2 =s→−limiω[
(s+iω)F(s)exp(
−
√
sx)]
=− 2 αiexp(
−
√
−iωx)
.
We also need to know that the roots of±iwith positive real part are
√
i=√^1
2( 1 +i),√
−i=√^1
2( 1 −i).Thus the function we seek is