$u(x, t)$ of \begin{equation} \begin{cases} u_t = k u_{xx} + u \\ u(x, 0) = f(x) \end{cases} \end{equation}

Question

How do I use a Fourier transform to find a formula for the solution $u(x, t)$ of

\begin{equation} \begin{cases} u_t = k u_{xx} + u \\ u(x, 0) = f(x) \end{cases} \end{equation}

where $−∞ < x < ∞$, $t > 0$, in terms of $f(x)$ and $k$ ?

Answer 1

Let

$$U(k,t) = \int_{-\infty}^{\infty} dx \, u(x,t) e^{i k x}$$

Then, taking Fourier transforms

$$u_t = \kappa u_{xx} + u \implies U_t = -(\kappa k^2-1) U \implies U(k,t) = U(k,0)e^{-(\kappa k^2-1) t}$$ $$U(k,0) = F(k) \implies U(k,t) = F(k) e^{-(\kappa k^2-1) t}$$

To invert the FT, use the convolution theorem. We of course need the FT of the gaussian term:

$$\frac1{2 \pi} \int_{-\infty}^{\infty} dk \, e^{-\kappa t k^2} e^{-i k x} = \frac1{2 \pi}\sqrt{\frac{\pi}{\kappa t}} e^{-x^2/(4 \kappa t)}$$

$$u(x,t) = e^t \frac1{2 \pi} \sqrt{\frac{\pi}{\kappa t}}\int_{-\infty}^{\infty} dx' \, f(x') \, e^{-(x-x')^2/(4 \kappa t)} $$