= Solution
For
$$
\theta(x,t)=t^{-1/2}e^{-x^2/(4Dt)},
$$
direct <partial derivative>[differentiation] gives
$$
\theta_t
=\left(-\frac1{2t}+\frac{x^2}{4Dt^2}\right)\theta
$$
and
$$
D\theta_{xx}
=D\left(-\frac1{2Dt}+\frac{x^2}{4D^2t^2}\right)\theta
=\left(-\frac1{2t}+\frac{x^2}{4Dt^2}\right)\theta.
$$
Thus $\theta_t=D\theta_{xx}$.
Its total mass is found from the <Gaussian integral>:
$$
\int_{-\infty}^{\infty}\theta(x,t)\,dx
=2\sqrt{\pi D}.
$$
The unit-mass solution converging to the <Dirac delta function> as $t\downarrow0$ is therefore the <heat kernel>
$$
\boxed{
K(x,t)=\frac1{\sqrt{4\pi Dt}}
\exp\left(-\frac{x^2}{4Dt}\right)
}.
$$
For the second initial condition, the <heat-kernel solution> is the <convolution>
$$
\theta(x,t)
=\int_{-1}^1
\frac1{\sqrt{4\pi Dt}}
\exp\left(-\frac{(x-y)^2}{4Dt}\right)\,dy.
$$
With $u=(x-y)/(2\sqrt{Dt})$ and the definition of the <error function>, this becomes
$$
\boxed{
\theta(x,t)=\frac12\left[
\operatorname{Erf}\left(\frac{x+1}{2\sqrt{Dt}}\right)
-\operatorname{Erf}\left(\frac{x-1}{2\sqrt{Dt}}\right)
\right]
}.
$$
This 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 $t\to\infty$.
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