Partial Differential Equations: Fourier Transform in Space and Time?

Question

Consider the one-dimensional wave equation: $$\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\psi(x, t) = \frac{\partial^2}{\partial x^2}\psi(x, t).$$ We can take the Fourier transform of $\psi(x,t)$ with respect to $x$: $$\mathcal{F}\{\psi(x, t)\}(k) = \hat{\psi}(k, t).$$ Therefore, the wave equation becomes: $$\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\hat{\psi}(k, t) = - k^2 \hat{\psi}(k, t).$$ Although the resulting equation doesn't make much sense, can we repeat this process and take the Fourier transform of $\hat{\psi}(k, t)$ with respect to $t$ to have $$\frac{l^2}{c^2}\hat{\hat{\psi}}(k, l) = k^2 \hat{\hat{\psi}}(k, l)?$$ In general, can we use the Fourier transform to eliminate all derivatives from any PDE?

Answer 1

I think your impression here is generally correct, but let's be specific: if we assume that the function $\psi(x,t)$ is a tempered distribution as a "function" of two variables, then, yes, we can take Fourier transforms, and a constant-coefficient PDE becomes an equation involving no derivatives.

For (linear) wave equations, the hypothesis that $\psi$ is tempered is not unreasonable.

For heat equations, there is a problem, both in physical terms and in mathematical terms. Physically, lots of scenarios cannot be run backward beyond some time $t_o$. Mathematically, in parallel, there are lots of (physically interesting/meaningful) set-ups where, as a function of $t$, $\psi(x,t)$ cannot be reasonably extended backward. In those cases, as in comments, one-sided things that accommodate boundare/initial conditions (Laplace...) are more relevant.