Ngô Quốc Anh

March 31, 2010

Wave equations: Shock Formation

Filed under: PDEs — Ngô Quốc Anh @ 0:01

So far once we have discontinuous initial data, we might have shocks. This also occurs even the given initial data is smooth. Let us still consider the following 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 $c(u)>0$, $c'(u)>0$ and $u_0 \in C^1$.

From a discussion of this entry we consider the case when $u_0$ is non-increasing as we need singularities (because $c'>0$). Moreover, the solution can be given implicitly as $\displaystyle\left\{ \begin{gathered} u(x,t) = {u_0}(\xi ), \hfill \\ x - \xi = c({u_0}(\xi ))t. \hfill \\ \end{gathered} \right.$

Since $u_0$ is non-increasing, $u'_0(x)<0$ on $\mathbb R$. We now consider the characteristics which are straight lines, issuing from two points $\xi_1$ and $\xi_2$ on the $x$ axis with $\xi_1 < \xi_2$ have speeds $c(u_0(\xi_1))$ and $c(u_0(\xi_2))$ respectively. Because $u_0$ is decreasing and $c$ is increasing, it follows that $\displaystyle c(u_0(\xi_1))>c(u_0(\xi_2))$.

In other words, the characteristic emanating from $\xi_1$ is faster than the one emanating from $\xi_2$. Therefore the characteristics cross so a smooth solution cannot exist for all $t>0$.

We are now interested in calculating the breaking time. To this purpose, we calculate $u_x$ along a characteristic which has equation $x-\xi=c(u_0(\xi))t$.

Let $g(t)=u_x(x(t),t)$ denote the gradient of $u$ along the characteristic $x=x(t)$ given as above. Then $\displaystyle \frac{dg}{dt}=u_{tx}+c(u)u_{xx}$.

By differentiating the PDE with respect to $x$ we also have $\displaystyle u_{tx}+c(u)u_{xx}+c'(u)u_x^2=0$.

Comparing gives $\displaystyle \frac{dg}{dt}=-c(u)g^2$

along the characteristic. Solving this ODE gives us $\displaystyle g=\frac{g(0)}{1+g(0)c'(u_0(\xi))t}$

where $g(0)$ is the initial gradient at $t=0$. Thus $\displaystyle {u_x} = \frac{{{{u'}_0}(\xi )}}{{1 + {{u'}_0}(\xi )c'({u_0}(\xi ))t}}$.

The fact that $u'_0$ and $c'$ have different sign implies that $u_0$ will blow up at a finite time along the characteristic. What we need to do is to examine $u_0$ along all characteristics to find such $\xi$ so that $u_x$ first blows up.

A simple calculation show by denoting $F(\xi)=c(u_0(\xi))$ that the time of the first breaking is $\displaystyle t_b=-\frac{1}{F'(\xi_b)}$

where $\xi_b$ is such that $|F'(\xi)|$ is maximum. An in-depth observation shows that if the initial data is not monotone, breaking will first occur on the characteristic $\xi=\xi_b$ for which $F'(\xi)<0$ and $|F'(\xi)|$ is maximum.

Source: J.D. Logan, An introduction to nonlinear partial differential equations, 2nd, 2008; Section 3.3.

1. em thua thay theay co the post len may cai de thi cao hoc cua truong minh cac nam 2008 va 2009 khong ah ,em cam on thay

Comment by nguyenvanphan — April 2, 2010 @ 14:18

• Hi bạn, đề CH thì chắc là vẫn thế thôi, cứ ôn luyện đề thi của các năm trước nữa là ổn thôi mà. Chúc may mắn nhé.

Comment by Ngô Quốc Anh — April 2, 2010 @ 14:25

This site uses Akismet to reduce spam. Learn how your comment data is processed.