I have a question about the heat equation $\frac{\partial \varphi}{\partial t} = \frac{\partial^2 \varphi}{\partial x^2}$ with the conditions that $\varphi(x,t=0) = f_0(x)$ and $\lim_{x \rightarrow\pm \infty}\varphi(x,t) = 0$ for every $t \in \mathbb{R}$. The usual way of solving this equation is by separation of variables, i.e., $\varphi(x,t) = A(x)B(t)$. Then one gets the two ordinary differential equations
\begin{equation}
\frac{1}{B}\frac{dB}{dt} = \frac{1}{A}\frac{d^2A}{dx^2} = -\gamma \mbox{,}
\end{equation}
where $\gamma > 0$ in order that the condition $\lim_{x \rightarrow\pm \infty}\varphi(x,t) = 0$ is satisfied. The thing is that I can also Fourier
transform the solution: $\widehat{\varphi}(k,t) = \widehat{A}(k)B(t)$, i.e., the Fourier transformed solution is also a product of a function only of $k$ and one only of $t$. However, when I Fourier transform the equation ($\partial/\partial x \leftrightarrow ik$) I get $\frac{\partial\widehat{\varphi}}{\partial t} = -k^2 \widehat{\varphi}(k,t)$, which can be readily solved as $\widehat{\varphi}(k,t) = \widehat{f_0}(k)e^{-k^2 t}$. But this solution in Fourier space cannot be represented as a product of two functions – one only of $k$ and the other one only of $t$ although it should. I oversee something. Can someone help me?
Best Answer
You should not mix up the two methods (a) separation of variables and (b) Fourier transform.
The final solution of your problem will not be a product of two functions, but a superposition of such products. You use separation of variables in order to determine a sufficient supply of basis functions that can be used for the superposition.
Fourier transform with respect to the variable $x$ gives you immediately the Fourier transform $\hat\phi(k,t)$ of the final solution. Now you have to transform back. By the rules of Fourier transform the product in Fourier space will be transformed into a convolution.