Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 2

Suppose ${\Psi }_1$ and ${\Psi }_2$ have the same boundary data, and put $u={\Psi }_1-{\Psi }_2$. Then $\Delta u=0$ in $V$ and $u=0$ on its boundary. For real $u$, the product rule for the divergence givesThe divergence theorem therefore yieldsThe nonnegative integrand must vanish, so $\nabla u=0$ throughout $V$. On each connected component $u$ is constant, and its zero boundary value makes that constant zero. Hence $\boxed{{\Psi }_1={\Psi }_2}$, proving uniqueness. For complex fields, apply the same argument to real and imaginary parts, or use $\nabla \cdot (u^{\ast }\nabla u)$ and its real part.

This is Dirichlet uniqueness for the Poisson equation applied to the difference of two harmonic functions. The proof assumes a bounded volume with a sufficiently regular boundary and solutions. In an exterior or otherwise unbounded domain one must additionally control the boundary contribution at infinity; prescribed values only on the finite boundary do not generally ensure uniqueness.

For $\Phi =X(x)Y(y)$, separation of variables gives $X^{\prime \prime }/X+Y^{\prime \prime }/Y=-(m^2+p^2)$. With the standard separation constants $m^2$ and $p^2$, the zero-endpoint solutions are sines. Thus, provided $mp\ne 0$,is nonzero, has zero value on all four sides, and satisfies $\Delta \Phi =-(m^2+p^2)\Phi$.

There is a missing qualification in the printed request for arbitrary integer $m,p$. The Dirichlet and Neumann spectra of a square show that a nonzero Dirichlet solution exists precisely whenWhen one of $m,p$ is zero, the displayed sine product vanishes identically, and a nonzero solution need not exist: for example $(m,p)=(0,0)$ or $(0,1)$ admits none. On the other hand $(0,5)$ does admit $\sin (3x)\sin (4y)$. Thus the natural intended case is two nonzero integer indices, with the stated spectral criterion covering the exceptional cases.

Choose the separated cosine solutions instead:Its derivatives are ${\Phi }_x=-m\sin (mx)\cos (py)$ and ${\Phi }_y=-p\cos (mx)\sin (py)$. They vanish on the appropriate opposite sides of the square for integer $m,p$, so the outward normal derivative is zero everywhere on the boundary. Also $\Delta \Phi =-(m^2+p^2)\Phi$.

This is nonzero for every pair of integer indices, including a zero index. For $m=p=0$ it is the constant solution, illustrating the zero mode allowed by Neumann boundary conditions. Negative indices give the same cosine modes. The construction is the Neumann part of the Dirichlet and Neumann spectra of a square.