# Modified Heat Equation: k is not a constant

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.

• 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 $t$. – 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 $$t$$.