# Solution to the one-dimensional wave equation

We are told according to http://mathworld.wolfram.com/WaveEquation1-Dimensional.html (and other sources as well) that the general solution the the 1-D wave equation; $$\frac{\partial^2 \psi}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 \psi}{\partial t^2}$$ is $$\psi(\xi, \eta) = f(\eta) + g(\xi) = f(x+vt) + g(x-vt).$$

So the general solution is a linear combination of equations of functions which are functions of $x+vt$ and $x-vt.$

But can't I write any function I want as a function of $x \pm vt$? For example, consider the function $$f(x-vt) = x^2$$ For every element in the domain (i.e. $\mathbb{R}$) this assigns some unique element in the range ($\mathbb{R}$), so it is a valid function. (What I am not doing here is saying that $v=0$.) But taking this as a general solution $$\psi (x,t) = x^2$$ clearly doesn't work (it doesn't satisfy the original differential equation).

So my question is, what does it really mean for a function to be a function of the variable $x \pm vt?$ Where is the error in my reasoning?

f is parameterized by $\eta$ so you want $f(\eta)=\eta^{2}=(x-v\,t)^{2}$ which should give you the correct answer - a function that is shifted in time by distance $v\,t$. Same thing goes for g.
• My point thought is that I could write $f(\eta) = x^2 \neq \eta ^2$. What is stopping me from doing this? It is not a solution but it is still a function of $\eta.$ – user501217 Nov 12 '17 at 20:53
• Your solution does work for $v=0$ as it implies you have $\frac{\partial^{2} \psi}{\partial t^{2}}=0$ which is true. – user3071643 Nov 14 '17 at 3:33