Cauchy problem for the heat equation in Fourier space

Question

I'm trying to understand how to work with heat equations and I'm stuck on a step that is probably very basic.

We have the Cauchy problem$$u_t- \Delta u=0, (x,t)\in R^n \times R_+ $$ $$u(x,0)=g(x), x \in R^n$$

I understand the initial part, aka Fourier transform. We assume that the time derivative doesn't affect it. Furthermore, we consider that $F[{d^nu \over dx^n}]=(i\xi)^n \hat u$, so what I end up with is $${\partial \over \partial t} \hat u(\xi,t)+|\xi|^2 \hat u(\xi,t)=0$$ $$\hat u(\xi,0)=\hat g(\xi)$$ What leaves me hanging is that at this point my book concludes with "thus $\hat u(\xi,t)=\hat g(\xi)e^{-|\xi|^2t}$ ".

I can see that it is right, what I can't see is how to get to that expression. Thanks!

Answer 1

The new equation is now an ODE in time for each $\xi$. Hint: What is the solution to $$ y'(t) = -ky(t)$$ $$ y(0) = a$$ where $k,a$ are constants?

Answer 2

$$ \hat u'(\xi,t)+|\xi|^2 \hat u(\xi,t)=0$$ This equaation is separable: $$ \frac {\hat u'(\xi,t)}{\hat u(\xi,t)}=-|\xi|^2 $$ $$ \left (\ln {\hat u(\xi,t)} \right )'=-|\xi|^2 $$ Integrate: $$ \left (\ln {\hat u(\xi,t)}\right )=-\int |\xi|^2dt $$ $$ \left (\ln {\hat u(\xi,t)}\right )=- |\xi|^2t+K(\xi) $$ $$ \hat u(\xi,t) =e^{- |\xi|^2t+K(\xi) }$$ $$ \hat u(\xi,t) =C(\xi) e^{- |\xi|^2t}$$ Since we have that: $$\hat u(\xi,0)=\hat g(\xi) \implies C(\xi) =\hat u(\xi,0) =\hat g(\xi) $$ Finally: $$ \boxed{\hat u(\xi,t) =\hat g(\xi) e^{- |\xi|^2t}}$$