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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?

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BlackSith
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1 Answers1

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There is some confusion with the derivation's steps and notation, which OP presents in the wrong logical order. Below, the proper way is shown (see also this related post).

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.

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