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# The Homogeneous Problem

The most basic linear problem consists of finding the null space of

The simplest nontrivial extension to differential equations consists of the homogeneous boundary value problem based on the second order differential equation

where and is a parameter, with one of the following end point conditions:
1.             Dirichlet conditions
2.             Neumann conditions
3.             Mixed D. and N. conditions
4.             Periodic boundary conditions

More generally one has

which are the most general end point conditions as determined by the given 's, 's, 's, and 's, which are constants. These two boundary conditions and are supposed to be independent, i.e., there do not exist any non-zero numbers and such that

By contrast, if there does exist a non-zero solution and to this equation, then and are dependent.

Question: Can one give a clear vector space formulation of

 0 0

in terms of subspaces?

Question: What geometrical circumstance is expressed by independence''?

Answer: The vector -tuples and point into different directions.

Question: What, if any, is the (or a) solution to the homogeneous boundary value problem?

Answer: The general solution to the d.e. is

where and are integration constants. Let us consider the circumstance where satisfies the mixed D.-N. boundary conditions (3.) at each end point. These conditions imply

and

The content of the square brackets is known because , , and are known or given. The unknowns are and , or rather their ratio. Note that the trivial solution

is always a solution to the homogeneous system. Our interest lies in a non-trivial solution. For certain values of this is possible. This happens when

Values of , if any, satisfying are called eigenvalues.

KEY PRINCIPLE: A differential equation is never solved until boundary conditions have been imposed.

We note that the allowed value(s) of , and hence the nature of the solution is determined by these boundary conditions.

Example (Simple vibrating string): Solve

subject to the boundary conditions
 0

Solution: Two independent solutions to the d.e. are

Consequently, the solution in its most general form is

The boundary conditions yield two equations in two unknowns:

In order to obtain a nontrivial solution, it is necessary that

or

which implies

Note that yields a trivial solution only. Why? Negative integers yield nothing different, as seen below.

What are the solutions corresponding to each ? The boundary conditions demand that be related to , namely,

or

Thus

or

Here we have introduced subscript to distinguish the solutions associated with the different allowed values

The negative integers give nothing new. (Why?)

Comment: For , i.e. , there does not exist a non-trivial solution. Why? Because the application of the boundary conditions to the solution,

yields only .

Lecture 19

Next: Sturm-Liouville Systems Up: Sturm-Liouville Theory Previous: Three Archetypical Linear Problems   Contents   Index
Ulrich Gerlach 2010-12-09