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The Method of Characteristics

The Method of Characteristics

\[\newcommand{\p}{\partial}\]

Introduction

The essential point of the method of characteristics is as follows. Let us say we have a equation

\[ \p_t u + A(x,t,u) \p_x u = 0 \]

This is a nonlinear first order differential equation. We looks for paths \(t(s), x(s)\) such that \(u\) is constant all along the path. Using the chain rule, we find:

\[ \p_s u = \p_s t \p_t u + \p_s x \p_x u = 0 \]

If we choose \(\p_s t = 1\), \(\p_s x = A\), then \(\p_s u = 0\) by the equations of motion. he game, then, is to solve these equations to find a set of lines in \((t,x)\) on which \(u\) is constant. If we can do so, then if for example the line \(t = 0\) crossed these lines transversally, then we can propagate the initial data along the characteristics.

This can be generalized to the case of multiple variables (we will study a case with two). In general, the quantities conserved along the characteristics will be called Riemann invariants and there is a procedure for at leas trying to identify them. A more general discussion can be found in Chapter 3 of [Chorin et al., 1990]; we will actually discuss the case of 2 variables (\(u\) and \(h\) for the shallow water equations.
In this more general case, there are separate families of characteristics for each invariant and one will not be constant along the other’s characteristics.

Note also:

  • There is no guarantee that we can find the characteristics.

  • If the characteristics cross, we hit a discontinuity as the invariants will jump across the locus where hey meet; in other words we will hit a shock. This is a case we will discuss.

Example: linear equations

Let us consider the simple case

\[ \p_t u + a(x,t) \p_x u = b(x,t) \]

where \(a,b\) are known. Here we demand

(44)\[\begin{align} & & \p_s t = 1\\ & & \p_s x = a(x,t) \end{align}\]

These are first order (nonlinear) ODEs which can in principle be solved to find the characteristics. In this case,

\[ \p_s u(t(s),x(s)) = b \]

so that \(v + \int_{s_0}^s d\tau b(x(\tau),t(\tau))\) is conserved along characteristics.

In the simple case that \(b= 0, a = v\), the characteristics are: \(t = s\) (here we choose \(s\) to remove the integration constant), \(x = v s + x_0\), that is, they are straight lines \(x - v t = x_0\). Given any initial condition \(v(x,0) = v(x_0)\), the solution is then \(v(x - vt)\).

Example: shallow water equations

Here we return to the SWEs for one hoizontal direction:

(45)\[\begin{align} & & \p_t u + u \p_x u = - g \p_x h\\ & & \p_t h + \p_x (u h) = 0 \end{align}\]

Here we have two equations. In fact we can unpack them by defining \(c = \sqrt{g h}\) (for linearized surface gravity waves in the shallow water limit, this is the phase velocity), for which the equations can be written

(46)\[\begin{align} & & \p_t u + u \p_x u + 2c \p_x c = 0\\ & & \p_t (2c) + u \p_x (2c) + c \p_x u = 0 \end{align}\]

The sum and difference give:

\[ \left[ \p_t + (u \pm c) \p_x \right] (u \pm 2c) = 0 \]

For these cases, we have two families of characteristics,

(47)\[\begin{align} & & \p_s t = 1\\ & & \p_s x = u \pm c \end{align}\]

along which the Riemann invariants \(\Gamma_{\pm} = u \pm 2c\) are conserved. The game now is to solve the characteristic equations, which we will do in two specific situations.

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