Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-2/2/solution

Integral identity and representation. Define and take to be the outward unit normal. The product rule gives
The terms cancel. Integrating and applying the divergence theorem proves the Helmholtz equation version of Green's second identity:
Choose the Dirichlet Green function with the sign convention
where is the three-dimensional Dirac delta. Substitute , use , and evaluate the Dirac delta integral. The boundary term containing vanishes and on , leaving
Hence
The choice of sign in the Green function equation matters: it fixes both the source normalization and these signs. For a bounded region this construction presumes that the chosen is not a Dirichlet eigenvalue of ; at such an eigenvalue a Green inverse for arbitrary data does not exist.
Infinite space. Put . Away from the source, spherical symmetry reduces the Helmholtz equation to
For its general radial solution is
Integrate over a ball of radius about the source. The divergence theorem yields
Since , the volume term is and the flux tends to . Therefore , while remains free:
The term extends regularly through the source and solves the homogeneous Helmholtz equation, so it cannot change the unit Dirac delta normalization. A condition at infinity is needed to select it. For example, with time dependence , the outgoing Green function for the three-dimensional Helmholtz equation is , corresponding to .
Zero wavenumber. For fixed finite , as , and
This is the decaying Green function of the Laplacian. Directly at , the general radial solution is ; source normalization fixes , and decay at infinity removes . If the free parameter depends on as , the limit instead includes that arbitrary homogeneous constant. Thus the usual Laplace limit holds for a fixed parameter, or after a suitable condition at infinity has been selected.

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