Heat-kernel solution

For initial data $u_0$ on the real line, Fourier transformation or convolution with the fundamental solution gives

For a bounded function $u_0$, the convolution is well defined at every positive time because the heat kernel has integral one. The error-function diffusion front uses a Heaviside step function initial condition and gives one smoothed interface. The heat equation with interval-indicator initial data instead has two interfaces; subtracting their two error functions produces the finite-interval profile.

Let $\phi$ be continuous and suppose that the displayed improper integral converges separately at both infinite endpoints for some $T>0$. Absolute convergence is not required. Then its heat-kernel solution exists, is smooth and solves the heat equation for $0<t<T$, and tends to $\phi (x)$ as $t\downarrow 0$ at each fixed $x$.

To prove the smoothing claim, take a primitive $H$ of $\phi (\xi )e^{-{\xi }^2/(4T)}$. The assumed convergence makes $H$ bounded. For $q(x,\xi ,t)=K_t(x-\xi )e^{{\xi }^2/(4T)}$, integration by parts givesThe endpoint term vanishes. The coefficient of ${\xi }^2$ in the exponent of $q$ is $-(T-t)/(4tT)<0$. Its derivatives therefore have Gaussian decay on every compact subset of $\mathbb{R}\times (0,T)$, justifying differentiation under the integral sign in the absolutely convergent right-hand expression. Since $q_t=q_{xx}$, it satisfies the heat equation.

For the initial limit, apply the same construction to $\phi (\xi )-\phi (x)$ and split at $|\xi -x|=\delta$. Continuity controls the local part by its small oscillation times the unit kernel mass. On each remaining tail, $q$ is monotone for sufficiently small $t$: its maximum lies at $\xi =xT/(T-t)$, approaching $x$. If the relevant primitive has absolute bound $M$, integration by parts bounds that tail by $2M$ times the value of $q$ at its finite endpoint. For either endpoint $\xi =x\pm \delta$, that value tends to zero like $t^{-1/2}{e}^{-{\delta }^2/(4t)}$ times a fixed constant. Both tails vanish, proving the limit. Separate endpoint convergence matters here; merely assuming a symmetric Cauchy principal value does not provide the bounded primitive used in this proof.

Suppose $\phi$ is continuous at $x$ and, for some $T>0$,For $0<t<T$, its convolution with the heat kernel $K_t(s)=(4\pi t{)}^{-1/2}{e}^{-s^2/(4t)}$ is smooth and solves the heat equation. On every compact set in $\mathbb{R}\times (0,T)$, spatial or time derivatives of the heat kernel are bounded by the displayed integrable weight times a constant: their polynomial factors are absorbed by the strictly narrower Gaussian. This justifies differentiation under the integral sign.

Moreover, $\int K_t=1$, soInside $|\xi -x|<\delta$, continuity makes the difference uniformly small. Outside that interval, for $t\le T/4$, the inequality ${\xi }^2\le 2(\xi -x{)}^2+2x^2$ givesThe right side tends to zero, proving that the weighted tail contribution vanishes; the constant term $\phi (x)$ is controlled by the vanishing Gaussian tail mass. Thus $(K_t\ast \phi )(x)\to \phi (x)$ as $t\downarrow 0$. In particular, continuous data of polynomial growth satisfy this hypothesis, even when they are unbounded. This extends the Gaussian approximate identity argument beyond bounded data.

For $u_t=D{u}_{xx}$ and $u(x,0)=1_{[-a,a]}(x)$, convolution with the heat kernel gives