# Characteristics for nonlinear waves governed by $u_t + c(u)u_x = 0$

I have some confuse about the characteristic of nonlinear wave propagation. I read the PDF from A. Salih at IIST about the Inviscid Burgers’ Equation, where the original PDF can be find at the (https://www.iist.ac.in/sites/default/files/people/IN08026/Burgers_equation_inviscid.pdf)

So, I am not fully understand some statement about the characteristic line in this PDF.

Consider the 1D nonlinear advection equation $$u_t + c(u)u_x = 0$$ where the wave speed is not constant but a nonlinear term $$c(u)$$.

As above PDF state, we defined the characteristic curve as $$\frac{dx}{dt} = c(u).$$ Then Let $$x = x(t)$$, we have $$\frac{d}{dt}u(x(t),t) = \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x}\frac{dx}{dt} = u_t +c(u)u_x = 0$$ Therefore the $$u$$ is constant long the characteristic curve, and the characteristic curve is straight line since $$\frac{d^2x}{dt^2} = \frac{d}{dt}(\frac{dx}{dt}) = \frac{dc(u)}{dt} = c'(u)\frac{du}{dt} = 0$$

I didn't understand 3 places,

(1) why we can just assume the $$x$$ is dependence of $$t$$.

(2) why we say the $$u$$ is constant long the characteristic curve. For sure, the $$\frac{d}{dt}u(x(t),t) = 0$$ shows that the solution $$u$$ does not change along time, but I don't understand what logic shows that the $$u$$ is constant along $$\frac{dx}{dt} = c(u)$$.

(3) why the characteristic curve is straight line because $$\frac{d^2x}{dt^2} = c'(u)\frac{du}{dt} = 0$$. How do I know the derivative of $$c(u)$$ is equal to zero?

Could someone can help me?

First things first, we write the initial-value problem for the quasi-linear conservation law $$u_t + c(u) u_x =0 , \qquad u(x,0) = \phi(x).$$ The dependent variables are position $$x$$ and time $$t$$, and the unknown is $$u(x,t)$$. The method of characteristics consists in seeking a parametrisation $$s \mapsto \big(x(s),t(s),u(s)\big)$$ of these quantities, in such a way that the PDE transforms into ordinary differential equations which we might be able to solve. Note in passing that the dependence of $$u$$ w.r.t. $$s$$ can be expressed as $$u = u(x(s), t(s))$$, and similar notation can be used for the partial derivatives. Using the chain rule, the evolution of $$u$$ is governed by \begin{aligned} \frac{d}{ds}u(s) &= x'(s) u_x(s) + t'(s) u_t(s) \\ &= \left[x'(s) - c(u(s))t'(s)\right] u_x(s) \end{aligned} where we have used the PDE. As shown in Wikipedia, we may write the system $$\frac{dt}{ds} = 1, \quad \frac{dx}{ds} = c(u), \quad \frac{du}{ds} = 0$$ with initial condition $$t(0) = 0$$, $$x(0) = x_0$$ and $$u(0) = \phi(x_0)$$, which resolution can be tackled by hand. Here, we find $$t=s, \quad x = x_0 + c(\phi(x_0))\, s,\quad u = \phi(x_0).$$
Now let us go back to OP's questions. Given that $$t=s$$, this can be rewritten as $$x = x_0 + c(\phi(x_0))\, t,\quad u = \phi(x_0),$$ as a consequence of the choice $$x' = c(u)$$ with $$u'=0$$. Now, we can express the unknown as $$u = u(x(t), t)$$, and similar notation can be used for the partial derivatives. We note that $$u = \phi(x_0)$$ is constant along the characteristic curve $$t \mapsto x(t)$$ starting at $$(x_0, 0)$$, and that those curves are straight lines in the $$x$$-$$t$$ plane. In fact, computation of the derivative of $$u$$ along those lines gives $$\frac{d}{dt}u(t) = x'(t) u_x(t) + u_t(t) = 0$$ according to the definition of the characteristic curves and the PDE itself. Since the slope $$x' = c(u)$$ of these lines is a function of $$u$$ with $$u$$ constant, we can conclude that $$x'$$ is constant too.