Given two arbitrary continuous functions $f,g:\mathbb{R} \to \mathbb{R}$, show that $u(x,t) = f(x + t) + g(x-t)$ is a weak solution of the wave equation $u_{tt} - u_{xx}$. That is, $u$ satisfies:
$$ \iint_{\mathbb{R^2}} u(x,t) (\varphi_{tt} - \varphi_{xx}) \mathrm{d}x \mathrm{d}t = \iint_{\mathbb{R^2}} [f(x +t ) + f(x-t)] (\varphi_{tt} - \varphi_{xx}) \mathrm{d}x \mathrm{d}t = 0 $$ For any smooth, compactly supported functions $\varphi$.
Anyone have any ideas for this? I tried proving it by changing variables inside the integral, i.e sending $x \to x+t$ and integrating, while combining terms into one derivative and using the fundamental theorem of calculus like is done in the derivation of weak solutions to the transport equation. However, I'm being thrown off by the fact that there are two derivatives in the integral, and moreover the fact that the change of variables gets awkward due to the there being $x+t$ and $x-t$ in the integral.
The key point is that in new coordinates $u = x+t$, $v=x-t$ the expression $\varphi_{tt} - \varphi_{xx}$ becomes $-2\varphi_{uv}$. Indeed, we have $x = (u+v)/2$ and $t=(u-v)/2$, hence $$\varphi_u = \varphi_x x_u + \varphi_t t_u = \frac12 (\varphi_x + \varphi_t)$$ $$\varphi_{uv} = \frac12 (\varphi_x + \varphi_t)_x x_v + \frac12 (\varphi_x + \varphi_t)_t t_v = \frac14(\varphi_{xx} + \varphi_{tx} - \varphi_{xt} - \varphi_{tt}) = \frac12(\varphi_{xx} - \varphi_{tt}) $$
In the $uv$ coordinates, $f(x+t)$ becomes $g(u)$, independent of $v$. The integral gets the factor of $2$ from the Jacobian and becomes $$ \iint_{\mathbb{R^2}} f(x +t ) (\varphi_{tt} - \varphi_{xx}) \mathrm{d}x \mathrm{d}t = -4 \int_\mathbb{R}\left( \int_\mathbb{R} \varphi_{uv}\,dv\right) g(u)\,du $$ where the inner integral is zero because $\varphi$ has compact support.