Page 534 - Advanced engineering mathematics

514 CHAPTER 15 Special Functions and Eigenfunction Expansions

Then
y(0) = c 1 + c 2 = 0
so c 2 =−c 1 .And
−α
−2
α
−2
y(1) = c 1 e (e − e ) =−2c 1 e sinh(α).
Since α> 0, sinh(α) > 0, so c 2 = c 1 = 0 and again y is the trivial solution. There is no eigenvalue
λ< 1.
Case 3: 1 − λ< 0
2
Write 1 − λ =−α , with α> 0. Now
y(x) = c 1 e −2x cos(αx) + c 2 e −2x sin(αx).
Then
y(0) = c 1 = 0.
This leaves us with y(x) = c 2 e −2x sin(αx).Now
−2
y(1) = c 2 e sin(α) = 0.
To have a nontrivial solution, we need c 2 =0, so we must have sin(α)=0. We can choose α =nπ
for n any positive integer. The eigenvalues are
2
2
2
λ n = 1 + α = 1 + n π .
The eigenfunctions are
ϕ n (x) = e −2x sin(nπx)
for n = 1,2,···.
To compute the coefficients in the expansion of f (x) = x in a series of these eigenfunctions,
we need to know the weight function p. For this, we must write the differential equation in
Sturm-Liouville form. Multiply it by
4dx 4x
e = e
to obtain
4x
4x
4x
e y + 4e y + (3 + λ)e y = 0.

This is
4x 4x 4x
e y + (3e + λe )y = 0.
4x
This is in Sturm-Liouville form with r(x) = p(x) = e 4x and q(x) = 3e . To write f (x) = x =

∞ c n ϕ n (x), choose
n=1
1 −2x

p(x)e sin(nπx)dx
0
1 2
c n =
p(x)e −4x sin (nπx)dx
0
1 2x
e sin(nπx)dx
0
1 2
=
0 sin (nπx)dx
2 3
3
−8nπ − 2e n π (−1) n
= .
2 2
(4 + (nπ) )
Figure 15.2 compares graphs of f (x) = x with the 70th partial sum of this expansion. This
particular expansion converges fairly slowly to f (x), compared to Example 15.3. The Gibbs
phenomenon can be seen at x = 1. This behavior applies to general eigenfunction expansions,
not just to Fourier series.
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October 14, 2010 15:20 THM/NEIL Page-514 27410_15_ch15_p505-562