The Homogeneous Problem

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

$$A\vec u = 0\,.$$

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

$$\left[ \frac{d^2}{dx^2} +Q (x,\lambda )\frac{d}{dx} + R(x,\lambda ) \right] u(x)=0$$

where $a<x<b$ and $\lambda$ is a parameter, with one of the following end point conditions:

1. $u(a)=0$             Dirichlet conditions
   $u(b)=0$
2. $u'(a)=0$             Neumann conditions
   $u'(b)=0$
3. $\left.\begin{array}{l} \alpha u(a)+\alpha 'u'(a)=0\\ \beta u(b)+\beta 'u'(b)=0\end{array}\right\}$             Mixed D. and N. conditions
4. $\left.\begin{array}{l} u(a)-u(b)=0\\ u'(a)-u'(b)=0\end{array}\right\}$             Periodic boundary conditions

More generally one has

$$B_1(u) \equiv \alpha_{1}u(a)+\alpha_{1}'u'(a)+\beta_{1}u(b)+\beta_{1}'u'(b) = 0$$

$$B_2(u) \equiv \alpha_{2} u(a)+\alpha_{2}'u'(a)+\beta_{2} u(b)+\beta_{2}' u'(b)=0\,,$$

which are the most general end point conditions as determined by the given $\alpha$ 's, $\alpha'$ 's, $\beta$ 's, and $\beta'$ 's, which are constants. These two boundary conditions $B_1$ and $B_2$ are supposed to be independent, i.e., there do not exist any non-zero numbers $c_1$ and $c_2$ such that

$$c_1B_1(u)+c_2B_2(u)=0~~\qquad~~\forall\, u(x)\,.$$

By contrast, if there does exist a non-zero solution $c_1$ and $c_2$ to this equation, then $B_1$ and $B_2$ are dependent.

Question: Can one give a clear vector space formulation of

$$B_1(u) =$$ 0

$$B_2(u) =$$ 0

in terms of subspaces?

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

Answer: The vector $4$ -tuples $\{ \alpha_{1},\alpha_{1}', \beta_{1},\beta_{1}' \}$ and $\{ \alpha_{2},\alpha_{2}',\beta_{2},\beta_{2}' \}$ 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

$$u(x)=eu_1(x,\lambda )+fu_2(x,\lambda )$$

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

$$0=e[\alpha u_1(a,\lambda )+\alpha 'u'_1(a,\lambda )]+f[\alpha u_2(a,\lambda )+ \alpha 'u'_2 (a,\lambda )]$$

and

$$0=e[\beta u_1(b,\lambda )+\beta 'u_1'(b,\lambda )]+f[\beta u_2(b,\lambda)+ \beta ' u_2'(b,\lambda) ]\,.$$

The content of the square brackets is known because $u_i(x,\lambda )$ , $\alpha ,\alpha '$ , and $\beta ,\beta '$ are known or given. The unknowns are $e$ and $f$ , or rather their ratio. Note that the trivial solution

$$e=f=0\Leftrightarrow u(x)=0~~\qquad~~\forall\, x$$

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

$$0= D(\lambda )=\left\vert \begin{array}{ll} [\alpha u_1(a,\lambda... ...beta u_2(b,\lambda)+ \beta ' u_2'(b,\lambda) \rbrack \end{array}\right\vert~~.$$

Values of $\lambda$ , if any, satisfying $D(\lambda )=0$ 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 $\lambda$ , and hence the nature of the solution is determined by these boundary conditions.

Example (Simple vibrating string): Solve

$$u'' +\lambda u =0$$

subject to the boundary conditions

$$u(a,\lambda ) =$$ 0

$$u(b,\lambda ) = 0 ~~.$$

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

$$u_1(x) = \sin\sqrt{\lambda} x$$

$$u_2(x) = \cos\sqrt{\lambda} x\,.$$

Consequently, the solution in its most general form is

$$u(x)=e\sin\sqrt{\lambda} x +f\cos\sqrt{\lambda} x\,.$$

The boundary conditions yield two equations in two unknowns:

$$e\sin a\sqrt{\lambda} +f\cos a\sqrt{\lambda} = 0$$

$$e\sin b\sqrt{\lambda} +f\cos b\sqrt{\lambda} = 0\,.$$

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

$$0=\left\vert\begin{array}{ll} \sin a\sqrt{\lambda} &\cos a\sqrt{\lambda}\\ \sin b\sqrt{\lambda} &\cos b\sqrt{\lambda} \end{array}\right\vert$$

or

$$\sin (a-b)\sqrt{\lambda} = 0$$

which implies

$$\lambda_n =\left(\frac{\pi n}{a-b} \right)^2~~\qquad~~ n=1,2,\dots\,.$$

Note that $n=0$ yields a trivial solution only. Why? Negative integers yield nothing different, as seen below.

What are the solutions corresponding to each $\lambda_n$ ? The boundary conditions demand that $e$ be related to $f$ , namely,

$$e\sin a\sqrt{\lambda _n} +f\cos a\sqrt{\lambda _n}=0~~,$$

or

$$f=-e\frac{\sin a\sqrt{\lambda _n}}{\cos a\sqrt{\lambda _n}}\,.$$

Thus

$$u(x)=\frac{e}{\cos a\sqrt{\lambda _n}} (\cos a\sqrt{\lambda _n}\sin \sqrt{\lambda _n} x-\sin a\sqrt{\lambda _n}\cos\sqrt{\lambda _n}x)$$

or

$$u_n(x)=c_n\sin\sqrt{\lambda_n}(x-a)~~.$$

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

$$\lambda_n = \left(\frac{n\pi}{a-b}\right)^2~~\qquad~~ n=1,2,\dots\,.$$

The negative integers give nothing new. (Why?)

Figure 3.1: First two eigenfunctions of an eigenvalue problem based on Dirichlet boundary conditions.

Comment: For $n=0$ , i.e. $\lambda =0$ , there does not exist a non-trivial solution. Why? Because the application of the boundary conditions to the $n=0$ solution,

$$u(x)=ex+f$$

yields only $e=f=0$ .

Lecture 19