SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
singular points, whereas any singular point not satisfying both these criteria is
termed anirregularoressentialsingularity.
Legendre’s equation has the form
(1−z^2 )y′′− 2 zy′+(+1)y=0, (16.8)
whereis a constant. Show thatz=0is an ordinary point andz=± 1 are regular singular
points of this equation.
Firstly, divide through by 1−z^2 to put the equation into our standard form (16.7):
y′′−
2 z
1 −z^2
y′+
(+1)
1 −z^2
y=0.
Comparing this with (16.7), we identifyp(z)andq(z)as
p(z)=
− 2 z
1 −z^2
=
− 2 z
(1 +z)(1−z)
,q(z)=
(+1)
1 −z^2
=
(+1)
(1 +z)(1−z)
.
By inspection,p(z)andq(z) are analytic atz= 0, which is therefore an ordinary point,
but both diverge forz=±1, which are thus singular points. However, atz=1wesee
that both (z−1)p(z)and(z−1)^2 q(z) are analytic and hencez= 1 is a regular singular
point. Similarly, atz=−1 both (z+1)p(z)and(z+1)^2 q(z) are analytic, and it too is a
regular singular point.
So far we have assumed thatz 0 is finite. However, we may sometimes wish to
determine the nature of the point|z|→∞. This may be achieved straightforwardly
by substitutingw=1/zinto the equation and investigating the behaviour at
w=0.
Show that Legendre’s equation has a regular singularity at|z|→∞.
Lettingw=1/z, the derivatives with respect tozbecome
dy
dz
=
dy
dw
dw
dz
=−
1
z^2
dy
dw
=−w^2
dy
dw
,
d^2 y
dz^2
=
dw
dz
d
dw
(
dy
dz
)
=−w^2
(
− 2 w
dy
dw
−w^2
d^2 y
dw^2
)
=w^3
(
2
dy
dw
+w
d^2 y
dw^2
)
.
If we substitute these derivatives intoLegendre’s equation (16.8) we obtain
(
1 −
1
w^2
)
w^3
(
2
dy
dw
+w
d^2 y
dw^2
)
+2
1
w
w^2
dy
dw
+(+1)y=0,
which simplifies to give
w^2 (w^2 −1)
d^2 y
dw^2
+2w^3
dy
dw
+(+1)y=0.
Dividing through byw^2 (w^2 −1) to put the equation into standard form, and comparing
with (16.7), we identifyp(w)andq(w)as
p(w)=
2 w
w^2 − 1
,q(w)=
(+1)
w^2 (w^2 −1)
.
Atw=0,p(w) is analytic butq(w) diverges, and so the point|z|→∞is a singular point
of Legendre’s equation. However, sincewpandw^2 qare both analytic atw=0,|z|→∞
is a regular singular point.