Math 381 Fall 2006 Homework # 10 Solutions
Due: Wednesday, November 29, 2006
1. Let zmp and zmq be two different zeros of the Bessel function Jm (z). Show
that Jm (zmp ar ) and Jm (zmq ar ) are orthogonal with respect to the inner
Ra
product hf, gi = 0 f grdr. [Hint: show they are eigenfunctions of the SL
equation
df
m2
d
r
+ λr −
f =0
dr
dr
r
with boundary conditions f (0) finite and f (a) = 0.]
Recall that we obtained
√ the Bessel PDE from the SL equation above using
the substitution
z
=
λr where λ is the eigenvalue of the SL equation. In
√
this case λ equals zmp /a and zmq /a. Thus Jm (zmp ar ) and Jm (zmq ar ) are
eigenfunctions of the SL equation above. Since σ = r for this SL equation
this means the they are orthogonal with respect to our standard inner
product with weight σ = r.
2. Denote by Pn (x) = Pn0 (x) the Legendre polynomials. Recall that Pn (x) is
an eigenfunction to the differential equation
d
dg
(1 − x2 )
+ µn g = 0
dx
dx
where µn = n(n + 1). This is a Sturm-Liouville equation and hence the
R1
eigenfunctions are orthogonal with respect to hf, gi = −1 f gdx. Compute
P2 (x) by using the fact that it’s a polynomial of degree two and orthogonal
to P0 (x) = 1 and P1 (x) = x and normalized so that P2 (1) = 1. Note: in
general you can find Pn (x) in this way if you know P0 , . . . , Pn−1 .
P2 (x) = ax2 + bx+ c for some unknown a, b, c where a 6= 0 since we assume
P2 (x) has degree two. Since it’s orthogonal to P0 (x) = 1 and P1 (x) = x
we find that
Z
Z
1
1
P2 (x)dx = 0 =
−1
P2 (x)xdx.
−1
The first equality means that [a/3x3 + b/2x2 + cx]1−1 = 0 which gives
2a/3 + 2c = 0 or c = −a/3. The second equality means that [a/4x4 +
b/3x3 + c/2x2 ]1−1 = 0 which gives 2b/3 = 0 or b = 0. Thus P2 (x) =
ax2 − a/3. Since P2 (1) = 1 (convention) we get also a − a/3 = 1 or
a = 3/2 so that
1
P2 (x) = (3x2 − 1).
2
3. Consider again the differential equation
dg
d
(1 − x2 )
+ µg = 0.
dx
dx
Show that if p(x) is a solution which is a polynomial of degree two then
µ = 2 · 3. [Hint: Suppose p(x) = a0 + a1 x + a2 x2 where a2 6= 0 and solve
for the coefficients]. The same method can be generalized to explain why
a degree n polynomial is a solution only if µ = n(n + 1).
Assuming p(x) = a0 + a1 x + a2 x2 is a solution we have p′ (x) = a1 + 2a2 x
so
(1 − x2 )p′ (x) = a1 + 2a2 x − a1 x2 − 2a2 x3
and so
d
[(1 − x2 )p′ (x)] = 2a2 − 2a1 x − 6a2 x2 .
dx
Since this must equal −µp(x) = −µa0 − µa1 x − µa2 x2 we find µa2 = 6a2 .
Since a2 6= 0 this means µ = 2 · 3.