Page 20 - 6099

It follows that
−x
u(x, y) = f(e y) (2.16)
is the general solution of this PDE, where again f is an arbitrary function of only a single
variable. This is easily checked by differentiation using the chain rule. Geometrically,
the ”picture” of the solution u(x, y) is that it is constant on each characteristic curve in
Figure 3.

3
Example 2.1 Find the solution of (2.13) that satisfies the condition u(0, y) = y .
3
3
−0
Indeed, putting x = 0 in (2.16), we get y = f(e y), so that f(y) = y .Therefore,
3
−x
y .
u(x, y) = (e y) = e −3x 3
Example 2.2 Solve the PDE
2
u x + 2xy u y = 0. (2.17)
2
2
The characteristic curves satisfy the ODE dy/dx = 2xy /1 = 2xy . To solve the ODE,
2
2
we separate variables: dy/y = 2xdx; hence −1/y = x − C, so that
2 −1
y = (C − x ) . (2.18)
These curves are the characteristics. Again, u(x, y) is a constant on each such curve.
(Check it by writing it out.) So u(x, y) = f(C), where f is an arbitrary function.
Therefore, the general solution of (2.17) is obtained by solving (2.18) for C. That is

1
2
u(x, y) = f(x + ) (2.19)
y
0
2
Again this is easily checked by differentiation, using the chain rule: u x = 2x · f (x +
0
2
2
2
1/y) and u y = −(1/y ) · f (x + 1/y), whence u x + 2xy u y = 0.
In summary, the geometric method works nicely for any PDE of the form a(x, y)u x +
b(x, y)u y = 0. It reduces the solution of the PDE to the solution of the ODE dy/dx =
b(x, y)/a(x, y). If the ODE can be solved, so can the PDE. Every solution of the PDE is
constant on the solution curves of the ODE.

2.5 First Order Quasi-Linear Equations

Consider the following quasi-linear equation.

u t + a(x, t, u)u x = 0 (2.20)

We will solve this equation with the method of characteristics. We differentiate the
solution along a path x(t).

d 0
u(x(t), t) = u t + x (t)u x (2.21)
dt

13