Page 194 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 11: Sturm–Liouville’s Boundary Value Problems

Proof: Firstly, notice that the separated boundary condition (6) at x = a takes the form Notes
y (a) + y (a) = 0, y (a) + y (a) = 0. ...(11)
0 1 1 1 0 2 1 2
Taking the complex conjugate of the second of these gives
y*(a) + (y (a))* = 0. ...(12)
0 2 1 2
since and are real. For the pair of equations (11) and (12) to have a nontrivial solution, we
0 1
need
*
y (a)(y (a)) y (a)y (a) = 0.
1 2 1 2
A similar result holds at the other endpoint, x = b. This clearly shows that
x
y
x
y
p ( ){ ( , ) ( ( , )) * y ( , )( ( , )) } 0
y
x
x
x


as x a and x b, so that, from Green’s formula (7),
y ( , ) Sy ( , ) Sy ( , ), ( , )
x

x
x
x
y
If we evaluate this formula, we find that
b
x
r ( ) ( , ) ( , )dx 0
x
y
x
y
a
so that the eigenfunctions associated with the distinct eigenvalues and  are orthogonal with
respect to the weighting function r(x).
Example: Consider Hermite’s equation
2
d y 2x dy y 0
dx 2 dx ...(i)
for < x < . This is not in self-adjoint form. To do that let us define
x
x
x
p ( ) exp ( 2 )dx
...(ii)
exp ( x 2 )
Thus the equation (i) becomes
d dy x 2
p ( ) e y 0 ...(iii)
x
dx dx
By using the method of Frobenius, we showed in unit (3) that the solutions of equation (i) are
polynomials defined by H (x) when = 2n for n = 0, 1, 2, .... . The solutions of equation (iii), the
n
self-adjoint form of the equation, that are bounded at infinity for = 2n, then take the form
2
x
x
u n e 2 H n ( ) ...(iv)
and from theorem (2) satisfy the orthogonality condition
2
x
x
e H n ( )H m ( )dx 0 for n m
x
Self Assessment
3. Put the Laguerre’s equation
xy + (1 x) + y = 0, for 0 < x <
into self-adjoint form and deduce orthogonality condition for Laguerre’s polynomials.

LOVELY PROFESSIONAL UNIVERSITY 187