I am trying to prove that $u(x,y)=H(x-y)$ ($H$ denotes the heaviside function) is a solution of the partial differential equation
$$\frac{ \partial^2 u}{\partial x^2} -\frac{ \partial^2 u}{\partial y^2}=0.$$
This is what I've done so far:
Let $\varphi \in C_{0}^{\infty}(\mathbb{R}^2)$. Then
$$\langle \frac{ \partial^2 u}{\partial x^2} -\frac{ \partial^2 u}{\partial y^2} , \varphi \rangle = \iint_{\mathbb{R}^2} H(x-y) \left(\frac{ \partial^2 \varphi}{\partial x^2} -\frac{ \partial^2 \varphi}{\partial y^2}\right)dxdy=\iint_{\Delta} \left(\frac{ \partial^2 \varphi}{\partial x^2} -\frac{ \partial^2 \varphi}{\partial y^2}\right)dxdy,$$
where
$$\Delta=\{ (x,y) \in \mathbb{R}^2 \quad | \quad y\le x\}.$$
But now I don't know what to do, is there an easier way to do this?
You could factor the differential operator $$ \frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2} = \left(\frac{\partial}{\partial x} - \frac{\partial}{\partial y}\right)\left(\frac{\partial u}{\partial x} +\frac{\partial u}{\partial y}\right) $$ and show $$ \left(\frac{\partial u}{\partial x} +\frac{\partial u}{\partial y}\right)u=0. $$ The resulting integral is $$ \left\langle\left(\frac{\partial u}{\partial x} +\frac{\partial u}{\partial y}\right)u,\phi\right\rangle = - \int_\Delta \left(\frac{\partial \phi}{\partial x} +\frac{\partial \phi}{\partial y}\right). $$ The support of $\phi$ is contained in a square $(-M,M)^2$ for some large $M>0$. Then $$ - \int_{\Delta \cap [-M,M]^2}\left(\frac{\partial \phi}{\partial x} +\frac{\partial \phi}{\partial y}\right) = \int_{-M}^M \int_{-M}^x \left(\frac{\partial \phi}{\partial x}(x,y) +\frac{\partial \phi}{\partial y}(x,y)\right) dy \ dx\\ = \int_{-M}^M \int_{-M}^x \frac{\partial \phi}{\partial y}(x,y)dy \ dx + \int_{-M}^M \int_y^M \frac{\partial \phi}{\partial x}(x,y)dx \ dy\\ = \int_{-M}^M \phi(x,x) dx + \int_{-M}^M (-\phi(y,y)) \ dy= 0.\\ $$