IV. Commentary

Partial Differential Equations and the Method of Characteristics

James V. Herod*

This document contains some commentary on James V. Herod's lecture notes.

*(c) Copyright 1994,1995 by Evans M. Harrell, II. All rights reserved.

version of 28 January 1996.

Logically, the t of this chapter is different from the one of chapters 1-2, where it was the time variable of motion along the characteristic curves. In a differential equation of the form a(t,x,y) u_t + b(t,x,y) u_x + c(t,x,y) u_y = d u, we can think of t as the time on the characteristic only if a = 1. Otherwise, you might prefer to call the new characteristic time something like tau.

The characteristic equations are then

    dt /d tau = a, dx /d tau = b, dy/d tau = c,

and these will give us a family of curves in space - a "spaghetti" of curves filling up some region of 3D space.

Here is an example, where a is not 1. Solve for u(t,x,y), where

    y u_t + x u_x - 2 u_y = u

    u(0,x,y) = x + y.

The characteristic equations are

    dt/d tau = y, t(0) = 0
    dx/d tau = x, x(0) = xi
    dy/d tau = -2, y(0) = eta

As usual, solving these in the right order will make life much easier. Don't try to solve for t before you know y(tau).

The x equation leads us to

    x(tau; xi) = xi exp(tau),

while the y-equation is solved by

    y(tau; eta) = eta - 2 tau.

Now we are ready for the t equation, which becomes

    dt/d tau = eta - 2 tau, t(0) = 0,

with solution

t(tau, eta) = eta tau - tau^2

As for z(tau,xi,eta) = u(t,x,y), it solves

    dz/d tau = z, z(0,xi,eta) = xi + eta,
    z(tau,xi,eta) = (xi + eta) exp(tau).

The next step is to invert the relationship between the sets of coordinates (tau,xi,eta) and (t,x,y). After some algebra, we get:     tau = -y/2 + (1/2) (y^2 + 4 t)^(1/2)
    eta = (y^2 + 4 t)^(1/2)
    xi = x exp(y/2 - (1/2) (y^2 + 4 t)^(1/2)).

(At one point there was a quadratic with two roots, but only one root had y > 0 for t > 0.) When we substitute into the formula for z, we find

    u(x,y)= ( (y^2 + 4 t)^(1/2) + x exp(y/2 - (1/2) (y^2 + 4 t)^(1/2)) exp(-y/2 + (1/2) (y^2 + 4 t)^(1/2)).

How simple!

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