Ngô Quốc Anh

March 20, 2010

Wave equations: Discontinuous solutions

Filed under: PDEs — Ngô Quốc Anh @ 16:42

Characteristics play a fundamental role in understanding how solutions to first order PDEs propagate. So far our study (see this and this) presupposed that solutions are smooth, or, at the worst, piecewise smooth and continuous. Now we take up the question of discontinuous solutions. As the following example shows,  linear equations propagate discontinuous initial or boundary data into the region of interest along characteristics.

Example 1. Consider the advection equation $\displaystyle u_t+cu_x=0, \quad x \in \mathbb R, t>0, c>0$

subject to the initial condition $u(x,0)=u_0(x)$ where $\displaystyle {u_0}(x) = \left\{ \begin{gathered} 1, \quad x < 0, \hfill \\ 0, \quad x > 0. \hfill \\ \end{gathered} \right.$

Because the solution to the advection equation is $u(x)=u_0(x-ct)$, the initial condition is propagated along the characteristics $x - ct = {\rm const}$, and the discontinuity at $x = 0$ is propagated along the line $x= ct$, as shown in the following picture. Thus the solution to the initial value problem is given by $\displaystyle {u_0}(x) = \left\{ \begin{gathered} 0, \quad x >ct, \hfill \\ 1, \quad x

We now examine a simple nonlinear problem with the same initial data. In general, a hyperbolic system with piecewise constant initial data is called a Riemann problem.

Example 2. Consider the advection equation $\displaystyle u_t+ u u_x=0, \quad x \in \mathbb R, t>0$

subject to the initial condition $u(x,0)=u_0(x)$ where $\displaystyle {u_0}(x) = \left\{ \begin{gathered} 1, \quad x < 0, \hfill \\ 0, \quad x > 0. \hfill \\ \end{gathered} \right.$

Because $u$ is constant on the straight line characteristics having speed $u$. The characteristics emanating from the $x$ axis have speed $0$ (vertical) if $x > 0$, and they have speed unity ( $1$) if $x < 0$. The characteristic diagram is shown below Immediately, at $t > 0$, the characteristics collide and a contradiction is implied because $u$ must
be constant on characteristics. One way to avoid this impasse is to insert a straight line $x = mt$ of nonnegative speed along which the initial discontinuity at $x = 0$ is carried. The characteristic diagram is now changed to For $x > mt$ we can take $u = 0$, and for $x < mt$ we can take $u = 1$, thus giving a solution to the PDE on both sides of the discontinuity. The only question is the choice of $m$; for any $m \geq 0$ it appears that a solution can be obtained away from the discontinuity and that solution also satisfies the initial condition. Shall we give up uniqueness for this problem? Is there some other solution that we have not discovered? Or, is there only one valid choice of $m$?

The answers to these questions lie at the foundation of what a discontinuous solution is. As it turns out, in the same way that discontinuities in derivatives have to propagate along characteristics, discontinuities in the solutions themselves must propagate along special loci in spacetime. These curves are called shock paths, and they are not, in general, characteristic curves. So the answer to the questions posed in the last example is that there is a special value of $m$ (in fact, $m =\frac{1}{2}$, this can be derived from the so-called jump conditions, or, the Rankine-Hugoniot conditions) for which conservation holds along the discontinuity.

Source: J.D. Logan, An introduction to nonlinear partial differential equations, 2nd, 2008; Section 3.1.