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.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at WordPress.com.

%d bloggers like this: