Properties and Utility of a Green's Function

More generally, a unit force applied to a general linear system yields a response which is governed by the equation

$$\frac{d}{dx}p(x)\frac{d}{dx} G_\varepsilon +\gamma (x)G_\varepsilon =-\delta _\varepsilon (x-\xi)~~.$$

Integrate both sides and obtain

$$\int^{\xi+\varepsilon /2}_{\xi-\varepsilon /2} \left( \frac{d}{dx}p\frac{d}{dx} G_\varepsilon +\gamma G_\varepsilon \right) dx=-1$$ (414)

What happens to $G_\varepsilon$ as $\varepsilon \to 0$ ? The physical properties of the linear system imply that the response $G_\varepsilon$ remain a continuous function of $x$ , and its mathematical formulation should reflect this fact. Indeed, this continuity is guaranteed by the fact that the equation

$$\frac{d}{dx}p(x)\frac{d}{dx} G +\gamma (x)G=-\delta (x-\xi)$$

be satisfied. If $G\equiv \lim_{\varepsilon \to 0} G_\varepsilon$ were not continuous, then the first term of the differential equation,

$$p\frac{d^2G}{dx^2}~~,$$

would yield the derivative of a Dirac delta function, and there is no such expression on the right hand side.

The continuity of $G(x;\xi )$ and the evaluation of the integral Eq.(4.14) lead to the two key conditions which the unit impulse response $G$ must satisfy,

$$\boxed{ G(\xi^+)-G(\xi^-)=0}~~~~~~~\textrm{\lq\lq continuity~for~all~$a<x<b$''}~.$$

and

$$\boxed{ \left. \frac{dG}{dx}\right\vert ^{\xi^+}-\left. \frac{dG}... ...rt ^{\xi^-} =-\frac{1}{p(\xi)} } ~~~~\textrm{\lq\lq jump~condition~at~$x=\xi$''~}~.$$

A more careful statement of these properties is provided by the following

Theorem 44.1 (Fundamental Theorem for Green's Functions)

Let $G(x;\xi )$ be a function which

(a)

considered as a function of $x$ , satisfies the differential equation

$$\left[ \frac{d}{dx}p(x)\frac{d}{dx} +\gamma (x)\right]G(x;\xi) \equiv LG(x;\xi)=0$$

in $(a,b)$ except at the point $x=\xi$ ,

(b)

satisfies the given homogeneous boundary conditions,

(c)

for fixed $\xi$ is continuous, even at $x=\xi$ ,

(d)

has continuous 1$^{st}$ and 2$^{nd}$ derivatives everywhere in $(a,b)$ , except at $x=\xi$ , where it has a jump discontinuity given by

$$\left. \frac{d}{dx}G(x;\xi)\right\vert^{\xi^+}_{\xi^-} =\frac{-1}{p(\xi)} ~~.$$

Conclusion:

$$u(x) =\int^b_a G(x;\xi)~f(\xi)~d\xi \Longleftrightarrow \begin{ar... ...~f~\textrm{is~piecewise}\\ \textrm{continuous~in}~(a,b) \end{array} \end{array}$$ (415)

Comment. A function which satisfies properties (a)-(d) is, of course, the Green's function for the boundary value problem stated in the conclusion, equivalently given by Eq.(4.9). Even though there is more than one way of constructing such a function (if it exists), the result is always the same. In other words, one has the following

Theorem 44.2 (Uniqueness of a Green's function)
The Green's function of a given linear system is unique.

It is easy to verify the validity of this theorem. If there were two such functions:

$$LG_1(x;\xi) = -\delta(x-\xi)$$

$$LG_2(x;\xi) = -\delta(x-\xi)~,$$

then their difference satisfies the homogeneous equation

$$L(G_1 -G_2)=0~.$$

Consider the Green's function $H$ adjoint to either $G_1$ or $G_2$ . It satisfies Eq.(4.10),

$$L^*H(x;\xi')=-\delta(x-\xi')~.$$

Consequently,

$$=\langle H,L(G_1 -G_2)\rangle$$ 0

$$=\langle L^*H,(G_1 -G_2)\rangle$$

$$=\int_a^b (-)\delta(x-\xi')\left( G_1(x;\xi)-G_2(x;\xi)\right)dx$$

$$=G_2(\xi';\xi)-G_1(\xi';\xi)~.$$

Thus the Greens function is unique:

$$G_1(\xi';\xi)=G_2(\xi';\xi)~.$$

It is informative to restate this calculation algebraically: Starting with the fact that the difference satisfies $G_1-G_2$ satisfies the homogeneous problem, one recalls that such a problem furnishes us with only two alternatives:

1. [(i)] the trivial solution,which is the zero solution. In this case the difference between the two Green's functions vanishes identically. This means the Green's function is unique.
2. [(ii)] a nontrivial solution, which implies that the nullspace of the homogeneous adjoint problem is nonzero. In this case the inner product of this solution with the inhomogeneity, the Dirac delta function, does not vanish. Hence the existence of a solution to the inhomogeneous problem is impossible. In other words, the Green's function does not exist.

The two possibilities (i) and (ii) are mutually exclusive and jointly exhaustive. They illustrate the so-called Fredholm alternatives of a linear operator.

Proof of the Fundamental Theorem: The implication `` $\Longleftarrow$ '' has already been demonstrated with Eq.(4.11). To show `` $\Longrightarrow$ '' compute the various derivatives and then form the linear combination $Lu$ . The fact that the slope of $G$ makes a jump at $x=\xi$ demands that the integral for $u$ be split at that point,

$$u(x) = \int^{x-0}_a G(x;\xi)~f(\xi)~d\xi~+~\int^b_{x+0} G(x;\xi)~f(\xi)~d\xi$$

$$u'(x) = \int^{x-0}_a \frac{dG(x;\xi)}{dx} ~f(\xi)~d\xi~ +~\int^b_{x+0} \frac{dG(x;\xi)}{dx}~f(\xi)~d\xi$$

$$~~~~~~ + ~G(x;x-0)~f(x-0)~~-~~G(x;x+0)~f(x+0)$$

By hypothesis (c) the last two terms cancel for all $x$ where $f(x)$ has no jump discontinuity. (If $f$ does have a jump discontinuity at, say, $x_0$ then consider $u(x)$ for the case $x<x_0$ separately from the case $x>x_0$ .) Finally, take the second derivative,

$$u''(x) = \int^{x-0}_a \frac{d^2G(x;\xi)}{dx^2} ~f(\xi)~d\xi~ +~\int^b_{x+0} \frac{d^2G(x;\xi)}{dx^2}~f(\xi)~d\xi$$

$$~~~~~~ + ~\left. \frac{dG(x;\xi)}{dx}\right\vert^{\xi =x-0}~f(x-0)~~-~~ \left. \frac{dG(x;\xi)}{dx}\right\vert ^{\xi =x+0}~f(x+0)$$

Combine these derivatives to form

$$Lu(x) = \int^{x-0}_a \left[ p \frac{d^2G}{dx^2} + p'\frac{dg}{dx} + \gamma G\right] f(\xi)~d\xi$$

$$~ + \int^b_{x+0} \left[ p \frac{d^2G}{dx^2} + p'\frac{dg}{dx} + \gamma G\right] f(\xi)~d\xi$$

$$~ + ~p(x)f(x)~\left[ \frac{dG(x;x-0)}{dx}~-~\frac{dG(x;x+0)}{dx} \right]$$

The first two integrals are zero because of hypothesis (a). Compare the last term with the jump discontinuity stipulated by (d),

$$\frac{dG(\xi^+;\xi)}{dx}~-~\frac{dG(\xi^-;\xi)}{dx}=\frac{-1}{p(\xi)}~~.$$

Next compare the first term in this difference with the first term in the square bracket on the right hand side of $Lu(x)$ . Note that the first argument (``point of observation'') is to the right of the second argument (``source point'') in both of these first terms.

Comparing the second terms, one finds the same thing, except that the ``point of observation'' is to the left of the ``source point''. This agreement implies that

$$\frac{dG(x;x-0)}{dx}~-~\frac{dG(x;x+0)}{dx}=\frac{-1}{p(x)}~~.$$

Insert this expression into the right hand side of $Lu(x)$ and obtain

$$Lu(x)=-f(x)~~.$$

This verifies that $u(x)$ as given in the conclusion satisfies the inhomogeneous differential equation indeed.

Lecture 30