Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 2

Put $\Psi =U(r)V(\varphi )$ in the Laplace equation in polar coordinates. After multiplication by $r^2/(UV)$, separation of variables givesThe angular function is $2\pi$-periodic and single-valued, so its nonconstant modes are $\cos (n\varphi ),\sin (n\varphi )$ with integer $n\ge 1$. Their radial Euler differential equation has solutions $r^n,r^{-n}$; finiteness at the origin excludes $r^{-n}$.

The zero angular mode has constant angular part and radial solutions $1,\log r$. The logarithmic term is again excluded at the origin. Superposition therefore yieldsThis is the regular harmonic expansion inside the disk.

On the boundary, integrate the expansion from part (a) over one angular period. Every nonconstant sine and cosine has zero average, soAll positive-power radial terms vanish at $r=0$, leaving $\Psi (0)=A_0$. HenceThis is the mean value property for harmonic functions. It is an average over arclength as well as over angle, since $ds=Rd\varphi$. The usual continuity or integrable-boundary-data hypotheses justify taking this angular average.

Write the boundary Fourier series as $f=A_0+{\sum }_{n\ge 1}(a_n\cos n\varphi +b_n\sin n\varphi )$. Since $R=2$, its coefficients obey $a_n=2^n{A}_n$ and $b_n=2^n{B}_n$. The mean isFor $n\ge 1$, the integral of $\cos n\varphi$ on the first semicircle is zero. On the second semicircle, use ${\cos }^2\varphi =(1+\cos 2\varphi )/2$ and cosine orthogonality. This gives $a_2=1/4$ and every other $a_n=0$.

For the sine coefficients, every even $n$ gives zero on both semicircles. For odd $n$, product-to-sum formulas giveThus the requested radial-expansion coefficients are

For any interior point $\mathbf{p}$, choose a sufficiently small circle centred on $\mathbf{p}$ and wholly inside the disk. Translating the origin does not change the Laplace equation, so part (b) givesSuppose a real harmonic function, continuous on the closed disk, attains its global maximum $M$ at an interior point. Every value on each such circle is at most $M$, and their average is $M$. By continuity, every value on the circle must therefore equal $M$. Varying its radius shows that a neighbourhood of the point also has value $M$.

Thus the set of points attaining $M$ is open in the disk, and it is closed by continuity. The disk is connected, so this set must be the whole disk. The function is constant and has the same maximum on the boundary. Otherwise the maximum cannot be interior. Compactness of the closed disk guarantees existence of a maximum, proving the maximum principle for harmonic functions:Continuity up to the boundary is the standard hypothesis for this attained-maximum formulation.