= Solution
For the Gaussian ansatz, <partial differentiation> gives
$$
T_t=(F'-x^2FH')e^{-x^2H},\qquad
T_{xx}=F(-2H+4x^2H^2)e^{-x^2H}.
$$
Cancel the positive exponential and compare the constant and quadratic powers of $x$ in the <diffusion equation>. The two coupled <ordinary differential equations> are
$$
\boxed{F'=-2\kappa FH,\qquad H'=-4\kappa H^2}.
$$
The initial profile requires $F(0)=T_0$ and $H(0)=L^{-2}$. The second equation is a <separable differential equation>:
$$
\frac{d}{dt}\left(\frac1H\right)=4\kappa,
\qquad \boxed{H(t)=\frac1{L^2+4\kappa t}}.
$$
Substitute this into the first equation and integrate its <logarithmic derivative>:
$$
\ln\frac{F(t)}{T_0}=-2\kappa\int_0^t\frac{ds}{L^2+4\kappa s}
=-\frac12\ln\frac{L^2+4\kappa t}{L^2}.
$$
Thus
$$
\boxed{F(t)=\frac{T_0L}{\sqrt{L^2+4\kappa t}}},
$$
and the solution is
$$
\boxed{T(x,t)=\frac{T_0L}{\sqrt{L^2+4\kappa t}}\exp\left[-\frac{x^2}{L^2+4\kappa t}\right]},\qquad t\geq0.
$$
This is <Gaussian diffusion of a smooth initial profile>: its width grows while its peak decreases. The <Gaussian integral> gives $\int_{-\infty}^{\infty}T(x,t)\,dx=\sqrt\pi T_0L$, independent of time, which checks conservation of the total amount under the <diffusion equation>.
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