Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-2/3/solution

Write . A rotationally symmetric fundamental solution of the Laplace equation is harmonic for . The Laplace equation in polar coordinates reduces to , so . Integrate over a disk of radius around the source. The divergence theorem gives
Taking therefore gives
Changing the reference length inside the logarithm merely adds a constant. Notice the sign convention: the potential equation in this question is , whereas this Green function satisfies . Thus the point-source potential used below is .
For the two insulating walls, successive Euclidean reflections in and generate sources of strength at for every integer . Only lies inside the physical strip. This is the Neumann image lattice for a point source in a strip. Set
The raw sum of logarithmic potentials diverges by additive constants. A well-defined potential is obtained by pairing opposite images and subtracting their constant reference terms:
Each paired summand is on compact subsets away from the source. The series therefore has local uniform convergence, as do the required derivative series. It gives
In the corresponding -derivative, the images pair as and at , and as and at . Their opposite horizontal separations make the derivatives cancel. Thus both homogeneous Neumann boundary conditions hold. The constructed solution is even in ; without a condition at infinity the side conditions alone would allow further homogeneous harmonic functions.
For fixed , uniform convergence on permits integration term by term. The th term has integral
Here is the ordinary real inverse trigonometric function, taking values in . Summing from to makes a telescoping series:
The source flux splits equally between the upper and lower parts of the strip. The next two parts give the single-wall constructions that underlie the image lattice.
Figure 1.
Equal-sign image sources and the physical strip between two insulating boundaries
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