Method of Characteristics

The method of characteristics is an important part of the course; we have used it to solve linear first order partial differential equations.

Specifics of the method

To supplement what we’ve met in the lectures, here are some videos of worked examples. As in the lectures, they’re split into three cases arguably in increasing order of complexity. Each case requires different treatment, yet the ideas of the method are similar (in fact cases 1 and 2 are just special cases of case 3).

Case 1: Constant coefficients

This is the case

\[ au_x+bu_y=0, \]

where \(a,b\in\mathbb{R}\). A summary of the method would be as follows:

  • Rewrite the PDE in the form \((a,b)\cdot\nabla u=0\).

  • Recognise that this expression is telling us that the directional derivative in the direction \((a,b)\) is zero; thus \(u\) is constant in that direction.

  • Solve the relevant ODE to get equations for characteristic curves - they will be straight lines in \((x,y)\)-space. Each different characteristic line is described by an arbitrary constant \(c\) (its \(y\)-intercept).

  • Deduce that since \(u\) is constant along characteristic lines, \(u=f(c)\), where \(f\) is a function of one variable. Then write \(c\) in terms of \(x\) and \(y\) by rearranging the equations for the characteristic curves.

  • Finally apply any boundary condition to determine the hitherto arbitrary function \(f\).

Characteristics - Case 1

  • Method of Characteristics 1: Constant Coefficients.

Case 2: Variable coefficients

This is the case

\[a(x,y)u_x+b(x,y)u_y=0,\]

where \(a\) and \(b\) are now functions of \(x\) and \(y\). A summary of the method would be as follows:

  • Rewrite the PDE in the form \((a(x,y),b(x,y))\cdot\nabla u=0\).

  • Recognise that this expression is telling us that the directional derivative in the direction \((a(x,y),b(x,y))\) is zero; thus \(u\) is constant in that direction.

  • Solve the relevant ODE to get equations for characteristic curves - they will be genuine curves rather that straight lines in \((x,y)\)-space in this variable coefficients case.

  • Each different characteristic curve is described by an arbitrary constant \(c\).

  • Deduce that since \(u\) is constant along characteristic curves, \(u=f(c)\), where \(f\) is a function of one variable. Then write \(c\) in terms of \(x\) and \(y\) by rearranging the equations for the characteristic curves.

  • Finally apply any boundary condition to determine the hitherto arbitrary function \(f\).

Characteristics - Case 2

  • Method of Characteristics 2: Variable Coefficients.

Case 3: The general case

This is the case

\[a(x,y)u_x+b(x,y)u_y+c(x,y)u=f(x,y),\]

where \(a,b,c,f\) are functions of \(x\) and \(y\) and is the most technically challenging of the types of problem we will meet that use the method of characteristics. A summary of the method would be as follows:

  • We parameterise the characteristic curves by a parameter \(t\) (I’ve noticed some of you aren’t too confident with paramertised curves - this seems to be a good link explaining what we mean by parameterising a curve). This involves solving the ODEs

\[\frac{\mathrm{d}x}{\mathrm{d}t}=a(x,y),\qquad \frac{\mathrm{d}y}{\mathrm{d}t}=b(x,y),\]

with suitable boundary conditions. After this step, we have expressions for the curves, with \(x\) and \(y\) given as functions of \(t\) with an extra parameter \(s\), say.

  • Recognise that along these curves, the PDE reduces to an ODE.

  • Solve the ODE along characteristic curves.

  • Substitute back to the original coordinates.

  • Finally apply any boundary condition to determine any hitherto arbitrary functions.

Characteristics - Case 3

  • Method of Characteristics 3: The General Case.