Page 168 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 9: Adjoint and Self-Adjoint Equations

(ii) p(a) = p(b) 0, y (a) = y (b) and y ( ) y ( ). This is the case of periodic boundary conditions. Notes
b
a
i i i i
(iii) y (a) + y ( ) 0 and y (b) + y ' ( ) 0, with at least one of the and one of the
a
b
1 i 2 1 1 i 2 1 i i
non-zero. These conditions then have non-trivial solutions if and only if
b
y
y
b
b
b
y
a
y
y 1 ( ) ( ) y 1 ( ) ( ) 0, y 1 ( ) ( ) y 1 ( ) ( ) 0,
a
a
a
2
2
2
2
and hence (10) is satisfied.
Conditions (iii), each of which involves y and y at a single endpoint, are called unmixed or
separated. We have therefore shown that our linear differential operator is Hermitian with
respect to a pair of unmixed boundary conditions. The significance of this result becomes apparent
when we examine the eigenvalues and eigenfunctions of Hermitian linear operators.
As an example of how such boundary conditions arise when we model physical systems, consider
a string that is rotating or vibrating with its ends fixed. This leads to boundary conditions
y(0) = y(a) = 0 - separated boundary conditions. In the study of the motion of electrons in a crystal
lattice, the periodic conditions p(0) = p(l), y(0) = y(l) are frequently used to represent the repeating
structure of the lattice.
9.3 Eigenvalues and Eigenfunctions of Hermitian Linear Operators
The eigenvalues and eigenfunctions of a Hermitian linear operator L are the non-trivial solutions
of Ly = y subject to appropriate boundary conditions.
Theorem 1. Eigenfunctions belonging to distinct eigenvalues of a Hermitian linear operator are
orthogonal.
Proof: Let y and y be eigenfunctions that correspond to the distinct eigenvalues and . Then
1 2 1 2
Ly ,y y ,y y ,y
1 2 1 1 2 1 1 2
and
y
y 1 ,Ly 2 y 1 , 2 2 2 y 1 , y 2
so that the Hermitian property Ly y 2 y 1 ,Ly 2 gives
,
1
( 1 2 )(y 1 ,y 2 ) 0
Since , (y , y ) = 0, and y and y are orthogonal.
1 2 1 2 1 2
As we shall see in the next section, all of the eigenvalues of a Hermitian linear operator are real,
a result that we will prove once we have defined the notion of a complex inner product.
2
If the space of functions C [a, b] were of finite dimension, we would now argue that the orthogonal
eigenfunctions generated by a Hermitian operator are linearly independent and can be used as
2
a basis (or in the case of repeated eigenvalues, extended into a basis). Unfortunately, C [a, b] is
not finite dimensional, and we cannot use this argument. We will have to content ourselves with
presenting a credible method for solving inhomogeneous boundary value problems based
upon the ideas we have developed, and simply state a theorem that guarantees that the method
will work in certain circumstances.

9.4 Eigenfunction Expansions

In order to solve the inhomogeneous boundary value problem given by (4) with f C[a, b] and
unmixed boundary conditions, we begin by finding the eigenvalues and eigenfunctions of L.

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