Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 4 / 17B (Methods)

Take $K>0$. For $t>0$, the diffusion fundamental solution of a linear differential operator $F$ satisfies $({\partial }_t-K{\partial }_x^2)F=0$ and has initial value $\delta (x)$ in the sense of distributions:for every smooth compactly supported test function $\psi$. With the usual spatial decay it has unit mass. Equivalently the causal extension $\Theta (t)F(x,t)$ is a retarded fundamental solution satisfying $({\partial }_t-K{\partial }_x^2)(\Theta F)=\delta (x)\delta (t)$.

The causal Green function is zero for $t<\tau$ and satisfiesFor $t>\tau$ it solves the homogeneous diffusion equation, and its initial jump is $G(x,{\tau }^{+};y,\tau )=\delta (x-y)$, again distributionally. The natural spatial decay and a standard bounded or integrable solution class specify the usual kernel, avoiding unrestricted growing solutions. The PDE acts on $x,t$; $y,\tau$ label the location and time of the unit impulse. Values of the Heaviside step function exactly at the jump do not affect these distribution identities.

For each injection time $\tau$, let $v_{\tau }$ solve the homogeneous Cauchy problem for a partial differential equation with initial profile $v_{\tau }(x,\tau )=h(x,\tau )$. Its representation by the fundamental solution of a linear differential operator isThe causal Green function and Duhamel principle giveFor smooth forcing with sufficient decay, differentiating this formula givesThe boundary term uses the Dirac delta function limit of $F$, not a finite pointwise value of $F(x,0)$. The same formula extends to weaker forcing in an appropriate integrable or distribution sense. Physically, each short time interval adds $h(\cdot ,\tau )d\tau$ to the temperature profile; that addition subsequently undergoes homogeneous diffusion equation. The total solution is the superposition of all such diffusing injections.

The insulated boundary imposes the Neumann boundary condition $T_{x_1}(0,x_2,t)=0$. Use the method of images by reflecting the initial data evenly across $x_1=0$. At the origin, the original sector and its reflection contribute equally. Using the given two-dimensional heat kernel,Differentiation givesThis vanishes exactly when $e^{3/(4Kt)}=4$. Its sign is positive before that time and negative afterwards. Moreover $T\to 0$ as $t\downarrow 0$ and as $t\to \infty$. Therefore this is the unique global maximum:The factor two from the insulated-boundary reflection is essential; applying the full-plane heat kernel only to the unreflected sector would give half the temperature.