Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 3 / 6D (Methods)

Assume real $k\ne 0$, and put $a=|k|$. Away from the source point the Green function solves $G^{\prime \prime }-a^{2}G=0$. The two decay conditions select $G=A{e}^{a(x-\xi )}$ on $x<\xi$ and $G=B{e}^{-a(x-\xi )}$ on $x>\xi$. Continuity at $\xi$ requires $A=B$: a jump in $G$ would create a derivative of the Dirac delta that is absent from the equation. Integrating the equation across the source gives $G^{\prime }(\xi +)-G^{\prime }(\xi -)=1$, so $-2aA=1$. Thus the negative of the one-dimensional modified Helmholtz Green function isIts distributional derivative verifies the delta equation, including its sign.

For a source for which the integral converges and the resulting function has the required decay, defineApplying the differential operator under the integral, or in the sense of distributions, gives $\mathcal{L}y=\int \delta (x-\xi )S(\xi )d\xi =S(x)$. A continuous integrable source, for example, makes this convolution vanish at both infinities and gives a classical solution. If there were two solutions, their difference would be $A{e}^{ax}+B{e}^{-ax}$; decay at positive infinity forces $A=0$, and at negative infinity forces $B=0$. Thus the decaying solution is unique whenever it exists.

The assumptions needed for existence should not be dropped: an arbitrary source need not admit a decaying solution, as $S=1$ demonstrates. Also the displayed formula presupposes $k\ne 0$; for $k=0$ no Green function for a point source can decay at both infinities, since its derivative must jump by one while both exterior linear pieces would have to be zero.