James V. Herod*
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,
so
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!