# 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.

## 2 Comments »

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

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