How to show $|t-x|$ is a weak solution to advection equation On the wikipedia page on Weak solutions they give the example that $|t-x|$ is a weak solution to the 1st order wave equation $u_t +u_x=0$.
I tried to follow through the working but get stuck at $$\int^{\infty}_{-\infty}\int^{x}_{-\infty}\phi dtdx-\int^{\infty}_{-\infty}\int^{\infty}_{x}\phi dtdx=
\int^{\infty}_{-\infty}\int^{t}_{-\infty}\phi dxdt-
\int^{\infty}_{-\infty}\int^{\infty}_{t}\phi dxdt$$
I cant see anyway to proceed from here. I used so far integration by parts and the compact support of the test functions to simplify after splitting the integral apart above and below $x=t$. 
Can someone help me proceed from here or point out the error I have made?
Thanks
 A: We multiply $u_t + u_x = 0$ by $\phi$ and integrate over space and time. Using Fubini's theorem and integration by parts, the Wikipedia page on weak solutions states that we must verify
$$
\int_{-\infty}^\infty \int_{-\infty}^\infty u\, \phi_t \,\text d t\,\text dx +\int_{-\infty}^\infty \int_{-\infty}^\infty u\, \phi_x \,\text d t\,\text dx = 0 \, .
$$
Let us start with the first term for $u(x,t) = |t-x|$. Using integration by parts and the change of variables $\tau = t-x$,
\begin{aligned}
\int_{-\infty}^\infty \int_{-\infty}^\infty u\, \phi_t \,\text d t\,\text dx
&= \int_{-\infty}^\infty \int_{x}^\infty (t-x)\,\phi_t \,\text d t\,\text dx - \int_{-\infty}^\infty \int_{-\infty}^x (t-x)\,\phi_t \,\text d t\,\text dx \\
&= \int_{-\infty}^\infty \left[ -\int_{x}^\infty \phi \,\text d t + \int_{-\infty}^x \phi \,\text d t \right] \text dx \\
&= \int_{-\infty}^\infty \left[ -\int_{0}^\infty \phi|_{t = x+\tau} \,\text d \tau + \int_{-\infty}^0 \phi|_{t = x+\tau} \,\text d \tau \right] \text dx \, .
\end{aligned}
Similarly, using Fubini's theorem twice, the second term writes
\begin{aligned}
\int_{-\infty}^\infty \int_{-\infty}^\infty u\, \phi_x \,\text d x\,\text dt
&= -\int_{-\infty}^\infty \int_{t}^\infty (t-x)\,\phi_x \,\text d x\,\text dt + \int_{-\infty}^\infty \int_{-\infty}^t (t-x)\,\phi_x \,\text d x\,\text dt \\
&= -\int_{-\infty}^\infty \left[ \int_{t}^\infty \phi \,\text d x - \int_{-\infty}^t \phi \,\text d x \right] \text dt \\
&= -\int_{-\infty}^\infty \left[ \int_{-\infty}^0 \phi |_{x=t-\tau} \,\text d \tau - \int_{0}^{\infty} \phi|_{x=t-\tau} \,\text d \tau \right] \text dt \\
&= -\int_{-\infty}^0\int_{-\infty}^\infty \phi |_{x=t-\tau}\,\text dt \,\text d \tau + \int_{0}^\infty\int_{-\infty}^\infty \phi|_{x=t-\tau}\,\text dt \,\text d \tau \\
&= -\int_{-\infty}^0\int_{-\infty}^\infty \phi |_{t=x+\tau}\,\text dx \,\text d \tau + \int_{0}^\infty\int_{-\infty}^\infty \phi|_{t=x+\tau}\,\text dx \,\text d \tau \\
&= \int_{-\infty}^\infty \left[\int_{0}^\infty \phi|_{t = x+\tau} \,\text d \tau - \int_{-\infty}^0 \phi|_{t = x+\tau} \,\text d \tau \right] \text dx \, ,
\end{aligned}
which ends the proof. 
