# Fourier transform to the wave equation

Consider a solution to the wave equation $$\psi\left(x,t\right)$$, then using Fourier transform, we can represent:

$$\psi\left(x,t\right)=\left(\frac{1}{2\pi}\right)^{2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\widetilde{\psi}\left(k,\omega\right)e^{i\left(kx+wt\right)}dkdw$$

Now if we'll apply this form into the wave equation $$\frac{\partial^{2}\psi}{\partial x^{2}}-\frac{1}{c^{2}}\frac{\partial^{2}\psi}{\partial t^{2}}=0$$

We'll get:

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\widetilde{\psi}\left(k,\omega\right)e^{i\left(kx+wt\right)}\left(-k^{2}+\frac{\omega^{2}}{c^{2}}\right)dkdw=0$$

Now according to my book, this obligates the term $$\left(-k^{2}+\frac{\omega^{2}}{c^{2}}\right)$$ to be $$0$$.

I really cant understand why. My knowledge in Fourier transform is very low, we've just learned maybe $$2$$ hours just getting familier with the equations and applying it to some basic physics exercise. But I want to understand in a more profound way.

As I understand (with my basic knowledge of just year of math learnings), taking a fourier transform is equvivalent to representing a vector in a vector space using orthogonal basis. Which means $$e^{i\left(kx+wt\right)}$$ those are forming the orthogonal vector basis and the inner product is probably the integrals $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} d\omega dk$$.

So why does $$\left(-k^{2}+\frac{\omega^{2}}{c^{2}}\right)$$ has to be $$0$$ in order for the equation to make sense? as I see it maybe the term $$\widetilde{\psi}\left(k,\omega\right)$$ can also cause everything to be $$0$$.

If someone can explain in a simple way which dosent require profound understanding of the math behind the scenes, I'll be greatful. Thanks

So if the integral you give is to be zero, then $$\tilde \Psi(k,\omega)(\omega^2-c^2k^2)$$ has to be zero for every $$\omega$$ and $$k$$. Thus either $$\tilde \Psi$$ is identically zero or $$(\omega^2-c^2k^2)$$ is zero.