13eigen.html

Partial Differential Equations PowerTool

by Dr. Jim Herod

Section 3.3 Eigenvalues and Eigenfunctions

Maple Packages in Section 3.3

> restart;

> with(LinearAlgebra):

>

We have found that having an orthogonal family in a vector space is useful for the purposes of representing other elements of the space. If one were looking for a basis for the space, an orthogonal family would be the ideal choice. We have one method for generating an orthogonal family: perform the Gramm-Schmidt process. In this worksheet we introduce another method for getting an orthogonal family.

We compute eigenvectors for self-adjoint, linear transformations.

Definition: A function A on a vector space is linear if

for all numbers and all vectors x and y .

Examples.

1 . Let A be a square matrix defined on a finite dimensional vector space.

2. Let the space be C ''([0, )) and A( f ) = f '' for any f in the space.

Definition: A linear function A is self-adjoint on the space { E, < *, * > } if , for all x and y in the domain of the operator

Example 3.

One class of self-adjoint linear functions consists of matrices which have real number entries and which are symmetric about the main diagonal. To be specific, we illustrate with a two dimensional example.

First, we choose all the entries to be real numbers, not complex numbers.

> assume(alpha,real):
assume(beta,real):
assume(delta,real):

Now, we define a symmetric matrix.

> A:=Matrix([[alpha,beta],[beta,delta]]);

Take two vectors, X and U.

> X:=<x,y>;
U:=<u,v>;

Compute the dot product of AX with U and the dot product of X with AU.

> AXU:=DotProduct(A.X,U);
XAU:=DotProduct(X,A.U);

Show that these two are equal.

> simplify( AXU-XAU );

>

Example 4

As a second example, we take A to be a differential operator on a subset of C ''([0, 1]) with the usual dot product. The subset of C ''([0, 1]) consists of functions f such that f (0) = f (1) = 0. The definition of A is A( f ) = f ''. We establish that this transformation is self adjoint. That is, we show that

<A(f(x)), y(x)>=<f(x), A(y(x))>, or .

To see how this is done the reader needs to remember integration-by-parts:

We must consider

Using the integration by parts with u = f '' and v = g , the first integral is

f '(1) g (1) - f '(0) g (0) - .

On the other hand, using integration by parts with u = g '' and v = f we have that the second integral is

g '(1) f (1) - g '(0) f (0) - .

When we subtract the second integral from the first, we have

f '(1) g (1) - f '(0) g (0) - g '(1) f (1) + g '(0) f(0).

This last line will be zero if f and g satisfy the boundary conditions to be in the domain of A, that is, if

f (1) = 0 = f (0) and g (1) = 0 = g (0).

We have established that this A -- take two derivatives, with boundary conditions -- is self adjoint. (The reader might observe in passing that there are other boundary conditions that would achieve this same condition. We will see some of these in future problems.)

Definition: The number is an eigenvalue and the vector v is an eigenvector of the linear transformation A if

A( v ) = v .

Example 5

The numbers -1 and -2 are eigenvalues corresponding to the eigenvectors

and

for the matrix A given as

A =

We use Maple to give those eigenvalues and eigenvectors

> A:=Matrix([[-3/2,1/2],[1/2,-3/2]]);

Maple produces a list. The first entry is interpreted as follows: -1 is an eigenvalue of multiplicity 1 and corresponding to the eigenvector [1, 1]. The second entry is interpreted asfollows: -2 is an eigenvalue of multiplicity 1 and corresponding to the eigenvector [-1, 1].

> Eigenvectors(A, output='list');

>

Example 6

Each of the numbers , , , ..., ... is an eigenvalue corresponding to the eigenfunction

for the linear operator of Example 4 above. We verify this here.

> assume(n,integer):
u:=x->sin(n*Pi*x);
lambda:=-n^2*Pi^2;

> is(diff(u(x),x,x)=lambda*u(x));
is(u(0)=0);
is(u(1)=0);

>

Remark : We have presented the eigenvalues and eigenvectors for these two linear transformations as faits accompli . The process for getting eigenvalues and eigenvectors for a matrix is the subject of another course. The process of getting the eigenvalues and eigenfunctions for this differential operator is a subject for this course, and we illustrate how to derive the results here.

Question: What are all the eigenvalues of the selfadjoint, Sturm-Liouville Problem

y '' = y , with y (0) = y (1) = 0?

We break the problem into two cases. First, suppose that is positive. To remind us of this, take . Thus, we seek numbers such that

y '' = y with y (0) = y (1) = 0.

The differential equation is a second order constant coefficient equation. We know how to solve it. Solutions are of the form . In this case, the general solution is of the form

y ( x ) = A exp( x ) + B exp( x ).

To ask that y (0) = 0 asks that 0 = A + B. To ask that y (1) = 0 asks that 0 = A exp( ) + B/exp( ) . The only solution is A = 0 = B. This is the trivial solutions and will not be of interest to us.

The second case is that is negative. To remind us of this, take . Thus, we seek numbers such that

y '' = y with y (0) = y (1) = 0.

We also know how to solve this differential equation. Solutions are of the form

y ( x ) = A sin( x ) + B cos( x ).

To ask that y (0) = 0 asks that 0 = B. To also ask that y (1) = 0 asks that 0 = A sin( ). We don't want to have the trivial solutions again, so it must be that 0 = sin( ) . But we know all the places where the sine function is zero. It must be that , and .

The answer to the question is that the eigenvalues are and the eigenfuntions are sin( ).

Somehow, this is no surprise.

We do not forget the purpose of this development.

Fact 1: Eigenvalues corresponding to self-adjoint transformations are real numbers - not complex numbers.

Fact 2: Eigenvectors for self adjoint transformations corresponding to different eigenvalues are orthogonal.

Revisit Examples 3 and 4. Note that the eigenvectors are orthogonal. In the matrix example, this is seen quickly. For the differential operator, recall that the dot product is an integral.

> assume(m,integer):
int(sin(n*Pi*x)*sin(m*Pi*x),x=0..1);

>

Remark . Concerning Fact 2 above, suppose we have two eigenvectors corresponding to one eigenvalue. The eigenvectors found may not be orthogonal. Not to worry! Perform the Gramm-Schmidt process to these eigenvectors.

Most of the partial differential equations that arise in this set of lecture notes will generate a special type of differential equation that is called a Sturm-Liouville problem.

Definition of a regular Sturm-Liouville problem: Suppose that s , s ', and q are continuous and for x in [a,b]. In the presence of appropriate boundary conditions, the differential operator

L(f) = ( s f ') ' - q f

is self adjoint in C ''([a,b]). Example of appropriate boundary conditions are that functions f in the domain of L should satisfy are

(1) f (0) = f (1) = 0, or

(2) f '(0) = f '(1) = 0, or

(3) f (0) = f '(0) and f (1) = f '(1), or

(4) if s (0) = s (1), then f (0) = f (1) and f '(0) = f '(1). (These last are called periodic boundary conditions.)

Example 7

We provide two eigenvectors (eigenfunctions) for each eigenvalue of the Sturm-Liouville problem

f '' = f , f (-1) = f (1), f '(-1) = f '(1).

In this example, there are an infinite number of eigenvalues, and for each one, there are two eigenfunctions. We will see that the eigenvalues are , for each integer n.