Given the heat equation, $$ u_{t} = ku_{xx}, $$ how do we modify the solution below (when $k$ is a constant) $$ u(x,t) = \frac{1}{\sqrt{4\pi kt}}\int\limits_{-\infty}^{\infty} g(y)e^{\frac{-(x-y)^{2}}{4kt}} dy $$ to solve the equation also for $k = k(t)$? Thank you.

  • $\begingroup$ Keep in mind that the heat equation is only well-posed for $k>0$. So, you need to keep in mind that you time-dependent coefficient $k(t)>0$ for all $t$. $\endgroup$ – D.B. Jan 8 at 2:50

The solution does not change much, denote $K(t) = \int_0^t k(z)\,dz$, then the solution is

$$u(x,t) = g(x) \ast \frac{e^{\frac{-x^2}{4K(t)}}}{\sqrt{4 \pi K(t)}}$$

I omit the details because this is obviously a homework question, but my hint is to notice how the Fourier Transform does not care about $t$.


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