# Ngô Quốc Anh

## March 29, 2010

### Wave equations: Nonlinear equations in 1D

Filed under: PDEs — Ngô Quốc Anh @ 23:36

It’s time to consider nonlinear equations in 1D that we need to run out next parts. To begin, let us consider the simple nonlinear initial value problem

$\displaystyle u_t+c(u)u_x=0,\quad x\in \mathbb R, t>0$,

subject to the following initial condition

$\displaystyle u(x,0)=u_0(x), \quad x\in \mathbb R$

where $c$ is a given smooth function of $u$.

To analyze this kind of problems motivated by the approach for linear equations, we define characteristic curves by the differential equation

$\displaystyle \frac{dx}{dt}=c(u(x,t))$.

Since $u$ is still unknown, the characteristics cannot be determined in advance. However, along the characteristics, the PDE becomes

$\displaystyle u_t+c(u)u_x=u_t+u_x\frac{dx}{dt}=\frac{du}{dt}=0$

which implies $u$ is constant along these curves. Since

$\displaystyle \frac{d^2x}{dt^2}=\frac{dc(u)}{dt}=c'(u)\frac{du}{dt}=0$

then the characteristic curves are straight lines. That means once we have a point $(x_0,t_0)$ in spacetime we can draw a characteristic back in time to a point $(\xi, 0)$ on the $x$ axis. The equation of this characteristic is then given by

$\displaystyle \frac{x-\xi}{x_0-\xi}=\frac{t-0}{t_0-0}$

which is equivalent to

$\displaystyle x -\xi =\underbrace {\frac{{{x_0} - \xi }}{{{t_0}}}}_{c({u_0}(\xi ))}t$.

This is because $u$ is constant on the entire characteristic line. It now follows that

$u(x,t)=u(\xi, 0)=u_0(\xi)$.

Therefore, if a solution exists for $t>0$, it must be given by

$\displaystyle\left\{ \begin{gathered} u(x,t) = {u_0}(\xi ), \hfill \\ x - \xi = c({u_0}(\xi ))t. \hfill \\ \end{gathered} \right.$

To verify this fact, clearly

$\displaystyle u_t=u'_0(\xi)\xi_t, \quad u_x=u'_0(\xi)\xi_x$.

The partial derivatives $\xi_t$ and $\xi_x$ can be calculated by implicit differentiation from $x-\xi=c(u_0(\xi))t$. Eventually, we get

$\displaystyle {u_t} = - \frac{{c({u_0}(\xi )){{u'}_0}(\xi )}}{D},{u_x} = \frac{{{{u'}_0}(\xi )}}{D}$

where

$\displaystyle D = 1 + c'({u_0}(\xi )){{u'}_0}(\xi )t$.

Clearly

$u_t+c(u)u_x=0$.

Obviously, if $c'$ and $u'$ have the same sign then $u_t$ and $u_x$ are finite.

Example. Consider the problem

$\displaystyle u_t+uu_x=0,\quad x\in \mathbb R, t>0$,

subject to the following initial condition

$\displaystyle {u_0}(x) = \left\{ \begin{gathered} 0, \quad x \leq 0, \hfill \\ e^{-\frac{1}{x}}, \quad x > 0. \hfill \\ \end{gathered} \right.$

Here $c(u)=u$. The solution is nothing but

$\displaystyle u(x,t) = \left\{ \begin{gathered} 0, \quad x \leqslant 0, \hfill \\ {e^{ - \frac{1}{\xi }}},\quad x - \xi = t{e^{ - \frac{1}{\xi }}},x > 0. \hfill \\ \end{gathered} \right.$