Chapter 6 Partial Differential Equations

6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 19
Such a curve is called a characteristic curve. To construct a solution surface we first find
x = x(s, t), y = y(s, t), u = u(s, t)
and then solve the two equations x = x(s, t) and y = y(s, t) for s and t in terms of x and y.
In order to guarantee that this can be done requires a result from advanced calculus – the
inverse function theorem which states:
If x = x(s, t) and y = y(s, t) are C 1 maps in a neighborhood of a point (s 0 ,t 0 ),
the Jacobian
⎛ ⎞
∂x ∂x
⎜ ⎟∣
|J| = det ⎜ ∂t
⎝∂y
∂t
∂s⎟
∂y⎠
∣
∂s
∣
(s0 ,t 0 )
≠0.
and, in addition, x 0 = x(s 0 ,t 0 ) and y 0 = y(s 0 ,t 0 ), then there exists a neighborhood
R of (s 0 ,t 0 ) and there exists unique C 1 mappings
and
s = s(x, y), t = t(x, y)
s = s(x 0 (s),y 0 (s)),
0=t(x 0 (s),y 0 (s))
With this we can construct our solution surface as
and
u = u(s, t) =u(s(x, y),t(x, y)) = u(x, y),
u 0 (s) =u(s, 0) = u(s(x 0 (s),y 0 (s)),t(x 0 (s),y 0 (s))) = u(x 0 ,y 0 ).
Example 6.2.11. Solve
We first seek solutions of
xu x +(x + y)u y = u +1, u(x, 0) = x 2 .
dx
dt = x,
In this case we can parameterize C 0 by
dy
dt = x + y,
du
dt = u +1.
x 0 (s) =s, y 0 (s) =0, u 0 (s) =s 2