Fourier decomposition of solutions of the wave equation with respect to the spatial variable

Say I have a wave equation of the form $$\nabla^{2}f(t,\mathbf{x})=\frac{1}{v^{2}}\frac{\partial^{2}f(t,\mathbf{x})}{\partial t^{2}}$$ which is clearly a partial differential equation (PDE) in $\mathbf{x}$ and $t$.

Is it valid to consider a solution in which one Fourier decomposes the spatial part of the function $f(t,\mathbf{x})$ and not its temporal part. That is, is it ok to express the solution in the form $$f(t,\mathbf{x})=\int_{-\infty}^{+\infty}\frac{d^{3}k}{(2\pi)^{3}}\tilde{f}(t,\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}}$$ And if so, why is it ok to do this?

[Is it simply that one Fourier decomposes the solution at a particular fixed instant in time, $t$ and then requires that the solution has this form for all $t$, hence the mode functions $\tilde{f}(t,\mathbf{k})$ must satisfy the PDE $$\frac{\partial^{2}\tilde{f}(t,\mathbf{k})}{\partial t^{2}}+v^{2}\mathbf{k}^{2}\tilde{f}(t,\mathbf{k})=0$$ or is there some other reasoning behind it?]

When considering a function $f(t,\mathbf x)$ on $[0,\infty)\times \mathbb{R}^n$, we can focus on one time slice at a time, fixing $t$ and dealing with a function of $\mathbf x$ only. The Fourier transform can be applied to this slice, since it's just a function on $\mathbb{R}^n$. The result can be denoted $\tilde f(t,\mathbf k)$, and is usually called the Fourier transform with respect to spatial variable (or sometimes, "partial Fourier transform"). As long as $t$ is fixed, nothing new happens: for example, $$f(t,\mathbf{x})=\int_{\mathbb{R}^n}\frac{d\mathbf k}{(2\pi)^{3}}\tilde{f}(t,\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}}\tag{1}$$ is just the inversion formula for Fourier transform.
• Thanks for the link. So is what I put correct then, that we want the solution to be of this form for all values of $t$ and hence the Fourier coefficient functions $\tilde{f}(t,\mathbf{x})$ must evolve according to the corresponding differential equation in time? (Otherwise I find it confusing since we have Fourier transformed at a fixed instant in time and hence the solution isn't time dependent) – user35305 May 3 '16 at 17:44
• Sorry to reiterate things, but just to check that I've understood the reasoning correctly. Is the point that we Fourier expand at a given instant in time (providing a solution on a given time slice), but then we require that the solution be a Fourier expansion of this form for every instant in time. We do so by allowing the Fourier coefficients to evolve in time and in particular insisting that they are solutions to the differential equation $\frac{\partial^{2}f(t,\mathbf{k})}{\partial t^{2}}+v^{2}\mathbf{k}^{2}f(t,\mathbf{k})=0$. Would this be the correct reasoning? – user35305 May 3 '16 at 18:12
• We apply the Fourier transform to each, fixed, moment of time. Thus obtaining a new function of $t$ and $k$. The transform can be inverted, so for each fixed $t$, the solution is the inverse Fourier transform of $\tilde f(t,k)$ with respect to $k$. (So, this is not a special requirement, the Fourier transform can be inverted.) – user147263 May 3 '16 at 18:15
• Ah OK, and so the requirement that the Fourier coefficients satisfy the differential equation in time ensures that the Fourier transforms carried out at each fixed instant in time describe a single solution to the wave equation (heuristically, the fact that the Fourier coefficients satisfy the differential equation in time ensures that the each of the Fourier transforms, at each instant in time, are "connected" continuously to one another, describing a continuous solution in position $\mathbf{x}$ and time $t$. – user35305 May 3 '16 at 18:46