Page 167 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

INTRODUCTION TO PERTURBATION METHODS 157
that as ε → 0

x(ε)
→ constant (10.24)
ε p
We than define a new variable

p
x = ε v(ε) (10.25)

such that v(0) = 0. Substitution into (10.20) yields

2p 2
ε v − 2ε p+1 v − ε = Q (10.26)

where Q must be identically zero. Note that since Q must also be zero no matter how small ε
ε
becomes, as long as it is not identically zero.
p
2p
Now, if p > 1/2, 2p − 1 > 0 and in the limit as ε → 0 ε v(ε) − 2ε v(ε) − 1 →−1,
2p is
which cannot be true given that Q = 0 identically. Next suppose p < 1/2. Again, Q
ε
2
identically zero for all ε including the limit as ε → 0. In the limit as ε → 0, v(ε) − ε 1−p v(ε) −
ε 1−2p → v(0) = 0. p = 1/2 is the only possibility left, so we attempt a solution with this value.
Hence
x = ε 1/2 v(ε) (10.27)

Substitution into (10.20) gives

√
2
v − 2 εv − 1 = 0 (10.28)
√
and this can now be solved by a regular perturbation assuming β = ε
1. Hence,
2 3
v = v 0 + a 1 β + a 2 β + a 3 β + ··· (10.29)
√
Inserting this into (10.28) with β = ε

2 2
v 0 − 1 + (2v 0 a 1 − 2v 0 )β + a + 2v 0 a 2 − 2a 1 β + ··· = 0 (10.30)
1
Thus

v 0 =±1

a 1 = 1 (10.31)
1 1
a 2 =+ or −
2 2