Choosing which variable to take the Fourier Transform with respect to

Question

Why here do we take the fourier transform with respect to $x$ as opposed to $t$?

Answer 1

When you use Fourier's separation of variables technique, then everything is clear. That's where all of this machinery originated, and I can't understand why authors avoid it. $$ \frac{T'}{\kappa T} = -\lambda = \frac{X''}{X},\\ X(0) = 0. $$ You select the bounded solutions only. That means $\lambda \ge 0$ for $T$ equation, and you end up with $X(x) = A\sin(\sqrt{\lambda}x)$. The solution has the form $$ u(x,t)=\int_{0}^{\infty}a(\lambda)e^{-\lambda \kappa t}\sin(\sqrt{\lambda}x)d\lambda, $$ which may also be written as $$ u(x,t)=\int_{0}^{\infty}b(s)e^{-\kappa s^{2}t}\sin(sx)ds. $$ The final solution is obtained by solving for $b$ such that $$ f(x)=u(x,0)=\int_{0}^{\infty}b(s)\sin(sx)ds \\ \implies b(s)=\frac{2}{\pi}\int_{0}^{\infty}f(x)\sin(sx)dx. $$ You can unravel this to see why the technique stated in your problem will work to give the same result, and to see what techniques work in what variable. $$ u(x,t) = \frac{2}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}f(x)\sin(sx)dx\right)e^{-\kappa s^{2}t}ds. $$ If you change the variable back to $s=\sqrt{\lambda}$, you can see that the outer integral is a Laplace transform in $t$. Finally, verify the solution directly under suitable conditions on $f$, which essentially comes down to knowing that $\sin(sx)e^{-\kappa s^{2}t}$ is a separated solution of the original PDE for fixed $s$.