Mathematical Methods for Physics and Engineering : A Comprehensive Guide

17.5 SUPERPOSITION OF EIGENFUNCTIONS: GREEN’S FUNCTIONS

Now, the boundary conditions require thatB=0andsin

(√

1

4 −λ

)

π=0,andso

√

1

4 −λ=n, wheren=0,±^1 ,±^2 ,....

Therefore, the independent eigenfunctions that satisfy the boundary conditions are

yn(x)=Ansinnx,

wherenis any non-negative integer, and the corresponding eigenvalues areλn=^14 −n^2.
The normalisation condition further requires
∫π

0

A^2 nsin^2 nx dx=1 ⇒ An=

(

2

π

) 1 / 2

.

Comparison with (17.51) shows that the appropriate Green’s function is therefore given
by

G(x, z)=

2

π

∑∞

n=0

sinnxsinnz

1

4 −n

2.

Case (i). Using (17.50), the solution withf(x)=sin2xis given by

y(x)=

2

π

∫π

0

(∞

∑

n=0

sinnxsinnz

1

4 −n

2

)

sin 2zdz=

2

π

∑∞

n=0

sinnx

1

4 −n

2

∫π

0

sinnzsin 2zdz.

Now the integral is zero unlessn= 2, in which case it is
∫π

0

sin^22 zdz=

π

2

.

Thus

y(x)=−

2

π

sin 2x

15 / 4

π

2

=−

4

15

sin 2x

is the full solution forf(x)=sin2x. This is, of course, exactly the solution found by using
the methods of chapter 15.
Case (ii). The solution withf(x)=x/2isgivenby

y(x)=

∫π

0

(

2

π

∑∞

n=0

sinnxsinnz

1

4 −n

2

)

z

2

dz=

1

π

∑∞

n=0

sinnx

1

4 −n

2

∫π

0

zsinnz dz.

The integral may be evaluated by integrating by parts. Forn=0,
∫π

0

zsinnz dz=

[

−

zcosnz

n

]π

0

+

∫π

0

cosnz

n

dz

=

−πcosnπ

n

+

[

sinnz

n^2

]π

0

=−

π(−1)n

n

.

Forn= 0 the integral is zero, and thus

y(x)=

∑∞

n=1

(−1)n+1

sinnx

n

( 1

4 −n

2 ),

is the full solution forf(x)=x/2. Using the methods of subsection 15.1.2, the solution
is found to bey(x)=2x− 2 πsin(x/2), which may be shown to be equal to the above
solution by expanding 2x− 2 πsin(x/2) as a Fourier sine series.