Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 3 / 29A / Solution

Let $H_t(x)=(4\pi t{)}^{-n/2}{e}^{-|x{|}^2/(4t)}$ be the heat kernel, and write $P_{t}g=H_t\ast g$. The homogeneous solution is $u(t,x)=P_{t}g(x)$: differentiation verifies the heat equation, and the approximate-identity property recovers the smooth initial data.

The Duhamel principle givesEach source impulse at time $s$ evolves by the homogeneous heat flow for duration $t-s$. More formally, differentiating the time integral gives $f(t,x)+{\int }_0^t\Delta P_{t-s}f(s)(x)ds$. Subtracting its Laplacian leaves $f$. The integral tends to zero as $t\downarrow 0$, since its absolute value is at most $t{\sup }_{s\le t}\Vert f(s){\Vert }_{\infty }$. Thus the initial condition is also correct. Smoothness permits differentiation locally; standard cutoff or semigroup arguments justify it for bounded smooth data without global bounds on all derivatives.

For $n=4$ and time-independent Schwartz $f$, integrate the kernel in time. For $r>0$, ${\int }_0^t(4\pi s{)}^{-2}{e}^{-r^2/(4s)}ds=e^{-r^2/(4t)}/(4{\pi }^2{r}^2)$. ConsequentlyThe forcing term converges by dominated convergence to $W(x)=(4{\pi }^2|\cdot {|}^2{)}^{-1}\ast f$. Its singularity is locally integrable in four dimensions, and the Schwartz decay controls infinity. The supplied fundamental solution identity gives $-\Delta W=f$, first in distributions and then classically by regularity.

The asserted large-time limit is not guaranteed for arbitrary bounded smooth $g$. Take $f=0$ and $g(x)=(1+|x{|}^2{)}^i$, which is smooth and has modulus one. For a standard four-dimensional Gaussian $Z$,Dominated convergence gives the asymptotic, and the gamma factor is nonzero. The phase $\log t$ rotates indefinitely, so there is no limit. This is an example of heat evolution of bounded data need not converge at large times.

Under the natural additional assumption $g\in C_0({\mathbb{R}}^4)$, $P_{t}g\to 0$: split the convolution into a large ball, whose heat-kernel mass tends to zero, and its complement, where $g$ is uniformly small. Then $w=W$ and $-\Delta w=f$. More generally, if $g(x)\to L$ at spatial infinity, the limit is $w=L+W$, satisfying the same equation. These are the intended qualified stationary-limit conclusions.