I am given the following question:
Let $\phi, \psi\in C^0(\mathbb{R})$ and $$u(x,t)=\frac{1}{2}(\phi(x+t)+\phi(x-t)+\frac{1}{2}\int_{x-t}^{x+t}\psi(y)dy$$
Show that $u$ is a solution to the wave-equation in the distributional sense.
Based on my understanding of the exercise, I am asked to show that for any $f\in C^\infty_c(\mathbb{R}^2)$, we have
$$\int_{\mathbb{R}^2}u(x,t)(\partial_{tt}f(x,t)-\partial_{xx}f(x,t))d\mathcal{L}^2(x,t)=0$$
If $\phi,\psi$ were sufficiently differentiable, then one could apply Gauss and use the fact that $f$ is compactly supported to remove any integrals over boundaries. However, since $\phi$ is not even once differentiable, neither is $u$, so that doesn't work. Any tips on how to proceed?
Best Answer
Using Hint 1 provided by Michał Miśkiewicz, we first choose R>0 such that $\text{supp}(f)\subseteq B(0,R)$ and we set $K:=B(0,2R)$. We now define $$\phi^*:=\chi_K\phi,~~ \psi^*:=\chi_K\psi,~~u^*(x,t):=\frac{1}{2}(\phi^*(x+t)+\phi^*(x-t))+ \frac{1}{2}\int_{x-t}^{x+t}\psi^*(y)dy$$ so that $$\int_{\mathbb{R}^2}u^*(x,t)(\partial_{tt}f(x,t)-\partial_{xx}f(x,t))d\mathcal{L}^2(x,t)=\int_{\mathbb{R}^2}u(x,t)(\partial_{tt}f(x,t)-\partial_{xx}f(x,t))d\mathcal{L}^2(x,t)$$ due to the support of $f$. Since $\phi^*, \psi^*$ and $u^*$ are compactly supported, they are integrable and thus from density of $C^\infty_c(\mathbb{R}^2)$ in $L^1(\mathbb{R}^2)$, it follows that we can find $\phi^*_n\rightarrow \phi^*, \psi^*_n\rightarrow \psi^*$ and $u^*_n\rightarrow u^*$, with $u^*_n$ defined from $\phi^*_n$ and $\psi^*_n$ as above. Using dominated convergence and the fact that the statement is true for $\phi, \psi\in C^\infty(\mathbb{R}^2)$ we have, for $T_u, T_{u^*}$ and $T_{u^*_n}$ as the induced distributions: $$(\partial_{tt}T_u-\partial_{xx}T_u)(f)=(\partial_{tt} T_{u^*}-\partial_{xx}T_{u^*})(f) = (\partial_{tt} T_{\lim_{n\rightarrow\infty}u^*_n}-\partial_{xx} T_{\lim_{n\rightarrow\infty}u^*_n})(f)=\lim_{n\rightarrow\infty}(\partial_{tt}T_{u^*_n}-\partial_{xx}T_{u^*_n})(f)=0$$