Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 3 / 29A / Solution

The heat kernel is $H_t(x)=(4\pi t{)}^{-1/2}{e}^{-x^2/(4t)}$, so the homogeneous solution for smooth bounded data is $u(t,x)=(H_t\ast g)(x)$. Differentiating under the integral gives the heat equation, and the approximate-identity property gives the initial value.

For a tempered distribution, its derivative is defined by $⟨T^{\prime },\psi ⟩=-⟨T,{\psi }^{\prime }⟩$ for every Schwartz test function. A fundamental solution of a constant-coefficient operator $P$ is a distribution $E$ satisfying $PE={\delta }_0$. Put $E(x)=e^{-|x|}/2$. Away from zero, $E^{\prime \prime }=E$; its first derivative has jump $-1$ at zero. Distributional differentiation therefore gives $E^{\prime \prime }=E-{\delta }_0$, andFor an ansatz $u=e^{t}v$, the forced equation becomes $(1-D^2)v=\varphi$. With Fourier convention $\hat{v}(k)=\int e^{-ikx}v(x)dx$, its unique solution in the Schwartz space isThe smooth multiplier and its derivatives have polynomial bounds, so this quotient is Schwartz. For uniqueness, multiplication of the difference's Fourier transform by $1+k^2$ gives zero, forcing the difference to vanish.

Subtract this particular solution and evolve the remaining initial data by the heat kernel:For smooth bounded $f$, the second term remains uniformly bounded, so $e^{-t}u(t,\cdot )\to v$ uniformly. If additionally $f-v$ is integrable, its uniform norm is $O(t^{-1/2})$, so the remainder itself decays. General bounded data need not have a decaying heat remainder, and no such stronger claim follows without a condition on the data. The homogeneous uniqueness assertion is understood in the bounded or usual tempered-growth solution class; the uniqueness of the particular Schwartz ansatz is unconditional within that stated class.