Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 3

For the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$, the product rule givesThe mixed dot products cancel. The divergence theorem therefore yields Green's second identity,The fields and boundary are assumed sufficiently regular. A singular Green function can be treated by excising a small ball around its source and taking the radius to zero, or using distributions.

Fix the evaluation point $r^{\prime }$ inside $V$ and put $\psi (r)=G(r,r^{\prime })$ in Green's second identity. The two $k^2$ terms cancel because ${\nabla }^2\varphi =-k^2\varphi$ and ${\nabla }^{2}G={\delta }^{(3)}-k^{2}G$. The remaining volume integral is $\int \varphi (r){\delta }^{(3)}(r-r^{\prime })dV=\varphi (r^{\prime })$. ThusThe Dirac delta has a positive sign in this problem's Green-function equation, fixing the order of the two terms. All boundary derivatives act on the integration variable $r$.

Impose a homogeneous Dirichlet boundary condition on the Green function in its first variable, $G(r,r^{\prime })=0$ for $r\in S$. The second boundary term then vanishes, givingIn a bounded domain, a Dirichlet Green function requires $k^2$ not to be a Dirichlet eigenvalue of $-\Delta$; otherwise a resonance can obstruct its existence. The representation assumes such a Green function exists.

In all space, put $\rho =|r-r^{\prime }|$ and try $G=A{e}^{ik\rho }/\rho$. For $\rho >0$, direct differentiation gives $({\rho }^2{G}^{\prime }{)}^{\prime }+k^2{\rho }^{2}G=0$, so the Helmholtz equation holds away from the source. Integrate its inhomogeneous version over a small source-centred ball. The $k^{2}G$ volume integral tends to zero, while the flux isIt must equal the integrated delta, one. ThereforeThis is the negative of the convention used in the outgoing Green function for the three-dimensional Helmholtz equation, where the source is $-\delta$. The sign here follows the present $+\delta$ convention. The factor $e^{ik\rho }$ chooses the outgoing solution, with time convention $e^{-i\omega t}$ and the Sommerfeld radiation condition; one could add homogeneous solutions without a radiation prescription.

Reflect $r^{\prime }=(x^{\prime },y^{\prime },z^{\prime })$ to $r^{\prime \ast }=(x^{\prime },y^{\prime },-z^{\prime })$. The method of images gives the Dirichlet image kernel for the three-dimensional Helmholtz equation:The two distances coincide on the plane, so its boundary value is zero; the image source is outside the domain. It therefore has the required positive delta source inside and the selected outgoing behaviour.

Let $h(R)=e^{ikR}/R$. At a boundary point $r=(\xi ,\eta ,0)$, the distances to the real and reflected sources have $z$ derivatives $-z^{\prime }/R$ and $+z^{\prime }/R$. ThusThe outward unit normal of the upper half-space is $-e_z$, and $z^{\prime }/R=\cos \theta$. HenceUsing the decay assumption to discard the distant hemisphere in the boundary representation givesThe qualification at infinity concerns the whole Green boundary term, not just the pointwise limit of $\varphi$. Sufficient decay of the field and its derivatives, as assumed, makes it vanish; compatible radiation conditions can also do so. As a sign check, taking $k\to 0$ in the kernel gives $z^{\prime }/(2\pi R^3)$, the positive Poisson kernel for the upper half-space.