One-dimensional Cavity Problem: Interior Boundary Value Problem

The process of solving the inhomogeneous boundary value problem

$$\frac{d}{dx}p\frac{du}{dx} +[q(x)+\lambda\rho(x)]u = -f(x)~\quad~a<x<b$$

$$B_1(u) = d$$

$$B_2(u) = e$$

is somewhat awkward from a numerical and even a conceptual point of view. Solving the differential equation is a local process: one determines the function and its properties at $x+dx$ from those at $x$ . One repeats this step-like process until one has found $u(x)$ for $a\le x\le b$ . Upon completion one checks whether the boundary conditions $B_1$ and $B_2$ have been satisfied. If not, one alters the function $u$ at the point where one started solving the differential equation and then starts all over again. Thus one might have to solve the differential equation many times before one finally obtains the solution to the desired degree of accuracy.

It is evident that this undersirable drudgery is due to the fact that the key property, boundary conditions, which determine the qualitatively important features of the solution $u$ , are stated separately and are not an intrinsic part of the differential equation.

This deficiency can be removed by recasting the boundary value problem in the form of an integral equation. The one-dimensional Sturm-Liouville system with, say, inhomogeneous Dirichlet boundary conditions,

$$\frac{d}{dx}p\frac{du}{dx} +[q(x)+\lambda\rho(x)]u =$$ 0

$$u(a) = e$$

$$u(b) = f\,,$$

illustrates the general principle. To convert this sytem into a single integral equation, one considers the corresponding Green's function problem

$$\left[\frac{d}{dx}p\frac{d}{dx} +q(x)\right] G(x;\xi ) = -\delta (x-\xi)$$

$$G(a;\xi ) =$$ 0

$$G(b;\xi ) = 0\,.$$

One transposes the term $\lambda\rho (x)u(x)$ to the right hand side of the S-L equation and considers it as an inhomogeneous equation. Multiply this equation by $G(x;\xi )$ , multiply the Green's function equation by $u(x)$ . One finds

$$G(x;\xi )\left[ \frac{d}{dx} p\frac{du}{dx}+q(x)u \right]= -\lambda G(x;\xi )\rho (x)u$$

and

$$u(x)\left[\frac{d}{dx}p\frac{d}{dx} +q(x)\right] G(x;\xi )= -\delta (x-\xi )u(x)~.$$

Upon subtracting one finds that the $q(x)$ -terms cancel and that the left hand side becomes a total derivative (Lagrange's identity!):

$$\textrm{l.h.s.} = G(x;\xi )\frac{d}{dx} p\frac{du}{dx} -u(x)\frac{d}{dx}p\frac{dG(x;\xi )}{dx}$$

$$= \frac{d}{dx} \left\{ G(x;\xi )p(x)\frac{du(x)}{dx}-u(x)p(x)\frac{dG(x;\xi )}{dx} \right\}~.$$

The r.h.s. becomes

$$\textrm{r.h.s.}= -\lambda G(x;\xi )\rho (x)u(x) +\delta (x-\xi )u(x)~.$$

Integration of l.h.s.=r.h.s. yields

$$p(x)\left[ G(x;\xi )\frac{du(x)}{dx}-\frac{dG(x;\xi )}{dx} u(x) \right]^{x=b}_{x=a} =-\lambda\int^b_a G(x;\xi )\rho (x)u(x)dx+u(\xi )\,.$$

The boundary value problem is self-adjoint. Consequently, $G(x;\xi)=G(\xi ;x)$ . Using this reciprocity relation, and then switching variables, one finds

$$u(x) = \lambda\int^b_a G(x;\xi )\rho (\xi )u(\xi )d\xi$$ (453)

$$+ p(a)u(a)\left.\frac{dG(\xi ;x)}{d\xi}\right\vert _{\xi =a}-p(b)u(b)\left. \frac{dG(\xi ;x)}{d\xi}\right\vert _{\xi =b}\,.$$

This is an integral equation for $u(x)$ . Note that the boundary conditions for $u(x)$ are an intrinsic part of the equation: the boundary conditions do not have to be stated separately. Also note that if $u(x)$ satisfies the homogeneous Dirichlet conditions $u(a)=0$ , $u(b)=0$ , then the integral equation becomes

$$u(x) = \lambda\int^b_a G(x;\xi )\rho (\xi )u(\xi )d\xi\,,$$ (454)

which is an eigenvalue equation for the function $u$ .