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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