# Meaning to the wave equation solution

Mathematically, every $$\mathcal C^2$$ function given by: $$$$c^2\nabla^2F=\dfrac{\partial^2 F}{\partial t^2}$$$$ Can be decomposed into two other $$\mathcal C^2$$ functions $$g,h$$ in the following manner: $$$$F=g(x+ct)+h(x-ct)$$$$ Constituting the solution to the wave equation.

Usually, one takes to interpret that as being possible to decompose every wave into right and left travelling components. Well, mathematically, it's not hard to derive an analytical solution in the latter form.

But, physically, what does it mean exactly for every wave to have this inner property of being able to be described by left and right travelling components?

Going even out of my way here, Schroendiger's Equation is obviously a wave equation that arose essentially by De Broglie's work on matter waves and Maxwell's derivation of the Wave Equations for the EM field, even there, in subatomic scale, this pure mathematical reasoning holds?

• I think that the property fundamentally is associated with the symmetries of the differential equation. The equation remains invariant if $t\rightarrow -t$, $x\rightarrow -x$, etc., and this fact leads to the equal footing of left- and right-going waves. Commented Jul 18, 2022 at 10:51
• Yes, I agree, as I said, it's not hard to derive an analytical explanation for why wave equations have their solutions in a symmetrical form. What I wish to understand is, what does it means, physically, for the waves to behave, always, in all scales, in such a way. Commented Jul 28, 2022 at 15:07
• This decomposition only holds in 1+1 dimension... interesting link. In my opinion, this decomposition is mathematical. I would relate it to the factorization of differential operators, decomposition of the kernel as direct sum and Bezout's identity... Commented Jun 19, 2023 at 10:14

It isn't true that both waves are always present. The amplitude of one can be $$0$$.