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238 Chapter 3 The Wave Equation

Example.
Estimate the first eigenvalue of

φ′′+λ^2 φ= 0 , 0 <x< 1 ,

φ( 0 )=φ( 1 )= 0.

Let us tryy(x)=x( 1 −x), which satisfies the boundary conditions. Then
y′(x)= 1 − 2 xand

N(y)=

∫ 1

0

[

y′(x)

] 2

dx=

∫ 1

0

( 1 − 2 x)^2 dx=^1

3

,

D(y)=

∫ 1

0

y^2 (x)dx=

∫ 1

0

x^2 ( 1 −x)^2 dx=

1

30.

Therefore,N(y)/D(y)=10. We know, of course, thatφ 1 (x)=sin(πx),and

N(φ 1 )=

∫ 1

0

π^2 cos^2 (πx)dx=π

2

2

, D(φ 1 )=

∫ 1

0

sin^2 (πx)dx=^1

2

,

soN(φ 1 )/D(φ 1 )=λ^21 =π^2 <10, confirming Eq. (4).

Example.
Estimate the first eigenvalue of

(xφ′)′+λ^2

1

xφ=^0 ,^1 <x<^2 ,

φ( 1 )=φ( 2 )= 0.

The integrals to be calculated are

N(y)=

∫ 2

1

x(y′)^2 dx, D(y)=

∫ 2

1

1

x

y^2 dx.

The tabulation gives results for several trial functions. It is known that the first
eigenvalue and eigenfunction are

λ^21 =

(π

ln 2

) 2

∼= 20. 5423 ,

φ 1 (x)=sin

(πlnx

ln 2

)

.

The error for the best of the trial functions is about 1.44%

y(x)

√

x( 2 −x)(x− 1 ) ( 2 −x)(x− 1 ) (^2 −x)(xx−^1 )

N(y)

D(y)^23.^750022.^134920.^8379