Mathematical Methods for Physicists: A concise introduction - Site Map

PARTIAL DIFFERENTIAL EQUATIONS
bring this regular singular point to the origin, so we make the substitution
u ˆ 1 x; U…u† ˆP…x†. Then Eq. (10.32) becomes
" #
d dU
u…2 u†
‡ m2
U ˆ 0:
du du u…2 u†
When we solve this equation by a power series: U ˆ P1
nˆ0 a nu n‡ , we ®nd that the
indicial equation leads to the values m=2 for . For the point x ˆ1, we make
the substitution v ˆ 1 ‡ x, and then solve the resulting di€erential equation by the
power series method; we ®nd that the indicial equation leads to the same values
m=2 for .
Let us ®rst consider the value ‡m=2; m 0. The above considerations lead us
to assume
P…x† ˆ…1 x† m=2 …1 ‡ x† m=2 y…x† ˆ…1 x 2 † m=2 y…x†; m 0
as the solution of Eq. (10.32). Substituting this into Eq. (10.32) we ®nd
…1 x 2 † d2 y
dx
Solving this equation by a power series
dy
2…m ‡ 1†x
2
dx
y…x† ˆX1
nˆ0
‡ ‰ m…m ‡ 1† Šy ˆ 0:
c n x n‡ ;
we ®nd that the indicial equation is … 1† ˆ0. Thus the solution can be written
y…x† ˆ X
c n x n ‡ X c n x n :
n even n odd
The recursion formula is
c n‡2 ˆ
…n ‡ m†…n ‡ m ‡ 1†
c
…n ‡ 1†…n ‡ 2† n :
Now consider the convergence of the series. By the ratio test,
R n ˆ cnx n
…n ‡ m†…n ‡ m ‡ 1†
c n2 x n2 ˆ …n ‡ 1†…n ‡ 2† jxj2
:
The series converges for jxj < 1, whatever the ®nite value of may be. For
jxj ˆ1, the ratio test is inconclusive. However, the integral test yields
Z
Z
Z
…t ‡ m†…t ‡ m ‡ 1† …t ‡ m†…t ‡ m ‡ 1†
dt ˆ
dt
…t ‡ 1†…t ‡ 2†
…t ‡ 1†…t ‡ 2† …t ‡ 1†…t ‡ 2† dt
M
and since
Z
M
M
…t ‡ m†…t ‡ m ‡ 1†
dt !1 as M !1;
…t ‡ 1†…t ‡ 2†
400
M