 Solution

Put $s=\lambda t>0$. In the heat-kernel convolution, complete the real square asThe Gaussian integral therefore givesAt $t=0$ the same formula gives $\beta =\alpha$. Equivalently, ${\beta }^{-2}={\alpha }^{-2}+4\lambda t$: the variance of this normalized Gaussian function grows from $1/(2{\alpha }^2)$ to $1/(2{\alpha }^2)+2\lambda t$.

The initial profiles have unit integral and concentrate at zero as $\alpha \to \infty$. Precisely, for every smooth compactly supported test function $\phi$,Thus the limit is the Dirac delta distribution, rather than an ordinary function: pointwise the profiles tend to zero away from the origin and diverge at the origin. At every fixed $t>0$,The limiting evolved profile is the heat kernel for a unit point source.