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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.

Fordirect differentiation givesandThus ${\theta }_t=D{\theta }_{xx}$.

Its total mass is found from the Gaussian integral:The unit-mass solution converging to the Dirac delta function as $t\downarrow 0$ is therefore the heat kernel

For the second initial condition, the heat-kernel solution is the convolutionWith $u=(x-y)/(2\sqrt{Dt})$ and the definition of the error function, this becomesThis is the heat equation with interval-indicator initial data for $a=1$.

At $t=0$ the graph is the rectangle of height one on $[-1,1]$. For every $t>0$ it is smooth, positive and even, with its maximum at $x=0$. As time increases the graph broadens and its maximum falls, while its total area remains $2$; pointwise it tends to zero as $t\to \infty$.

Take the Fourier transform in $x$. Since $\widetilde{{\theta }_{xx}}=-k^2\tilde{\theta }$, the transformed heat equation isThereforeThe given transform pair and the convolution theorem yield the heat-kernel solution

Insert the linear initial profile into the transformed integral from (b):The constant part integrates to $ax$. The other part is an absolutely integrable odd function, so its integral vanishes. HenceAs a direct check, $u_t=u_{xx}=0$ and $u(x,0)=ax$. Thus a linear temperature profile is stationary on the whole line. More generally, every affine function is preserved by this heat-kernel solution.

The convolution theorem turns the product $e^{-k^{2}t}\tilde{\theta }(k,0)$ into a spatial convolution. Its kernel isComplete the square:After rescaling the variable, the supplied translated Gaussian integral givesThis is the heat-kernel solution for diffusivity one. Its kernel integrates to one, preserving the total temperature integral, and approaches a Dirac delta function as $t\downarrow 0$. Its variance is $2t$; the factor is appropriate to ${\theta }_t={\theta }_{xx}$, rather than the alternative convention with diffusivity $1/2$.

For the Gaussian initial data for the heat equation, substitute the initial profile into the heat-kernel solution. Completing the square in $y$ givesThe remaining Gaussian integral yieldsThe centre stays at $x_0$ and the squared width grows linearly with time.

For the interval pulse, the convolution reduces to $B{\int }_a^{b}G(x-y,t)dy$. Using $u=(y-x)/(2\sqrt{t})$ and the error function givesThis is the translated and scaled heat equation with interval-indicator initial data. At its jump points the $t\downarrow 0$ limits are $B/2$; away from them the initial data are recovered pointwise.

For fixed $a<x<b$ at late time, both error function arguments are small. Since $\mathrm{erf}u=2u/\sqrt{\pi }+O(u^3)$,The estimate is uniform for $x\in [a,b]$. Equivalently, on this fixed interval the Gaussian in the convolution is $1+O((b-a{)}^2/t)$.

The spreading distance is more apparent on the growing spatial scale. For the Gaussian pulse the exponent is order one when $|x-x_0|=O(\sqrt{t})$. For the interval pulse, set $x=c+\xi \sqrt{t}$ with $c=(a+b)/2$ and fixed $\xi$. Uniformly for bounded $\xi$, its convolution givesThus both nonzero pulses spread over a distance $O(\sqrt{t})$, while their heights are $O(t^{-1/2})$ and their total integrals remain constant. More quantitatively, the variances of their normalized profiles are $1/(2a_0)+2t$ and $(b-a{)}^2/12+2t$, respectively, confirming the same width law.