I need to show that $H(x-ct)$ is a solution for the transport equation in the sense of distributions. I'm following the text Partial Differential Equations by Michael Shearer. Specifically, in Section 9.2.2, Example 7, Shearer has written:
Consider a jump discontinuity $u(x,t) = u_-$ for $x < ct$ and $u(x,t) = u_+$ for $x > ct$ where $u_\pm$ and $c$ are constants $\dots$ We rewrite this function using the Heaviside function: $u(x,t) = (u_+ - u_-)H(x - ct) + u_-$ $\dots$ Thus $\frac{du}{dt} = (u_+ - u_-) H'(x - ct) (-c)$.
My question is: How on earth does he obtain the mysterious $-c$ factor in his final expression for $\frac{du}{dt}$?
A little more background: It looks like Shearer is doing something like a chain rule, where differentiation of $H$ against the variable $t$ causes a factor of $-c$ to come out. In office hours, my professor essentially said not to use a chain rule. So I've tried all manner of integration and I just can't get it to work. Here's what one of my attempts looks like. In this attempt, I've tried to treat $u$ and $\phi$ as functions of the single variable $t$, since I am taking a partial derivative. I have no idea if this is the right approach, which is why I'm asking this question.
$$\begin{align*} \langle u', \phi \rangle &= - \langle u, \phi' \rangle \\ &= -\int_{-\infty}^\infty u(x - ct) \phi'(t) \; dt \\ &= \frac{1}{c} \int_{-\infty}^\infty u(z) \phi' \left(\frac{x}{c} - \frac{z}{c} \right) \; dz \\ &= \frac{1}{c} \int_{-\infty}^\infty H(z) \phi'\left(\frac{x}{c} - \frac{z}{c} \right) \; dz \\ &= \frac{1}{c} \int_0^\infty \phi' \left(\frac{x}{c} - \frac{z}{c} \right) \; dz \\ &= \left. \frac{1}{c} \cdot (-c) \cdot \phi\left(\frac{x}{c} - \frac{z}{c} \right) \right|_0^\infty \\ &= -\phi\left(\frac{x}{c}\right), \end{align*}$$ so that $\phi' = -\delta(t - \frac{x}{c}) = \delta(x - ct)$. So...no factor of $-c$ appears. I've tried every thing I can think of. This is one of a dozen attempts. I'm just missing something. Please help!
