Chapter 6 Partial Differential Equations

22 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS
Proof. Our regularity assumptions guarantee that the initial value problem
dx
dt
dy
dt
du
dt
= a(x, y, u), x(s,
= b(x, y, u), y(s,
= c(x, y, u), u(s,
0) = x 0(s)
0) = y 0(s)
0) = u 0(s)
has a unique local solution that is C 1 in t and s. By our hypotheses
⎛ ⎞
∂x ∂x
|J| = det ⎜ ∂t ∂s
⎟
⎝∂y
∂y⎠
=
∣ a(x 0(s),y 0 (s),u 0 (s)) x ′ 0(s)
b(x 0 (s),y 0 (s),u 0 (s)) y 0(s)
′
∣
∂t ∂s t=0
∣ ≠0
Thus |J| ≠ 0 in a neighborhood of the initial curve C 0 and by the inverse function theorem
we can solve for
s = s(x, y), t = t(x, y)
and s = s(x 0 (s),y 0 (s)),
0=t(x 0 (s),y 0 (s)), so we can define
u = u(x, y) =u(s(x, y),t(x, y)).
We first note that u satisfies the initial conditions:
u(x 0 (s),y 0 (s)) = u(s, 0) = u 0 (s).
Furthermore, by the chain rule
(
)
∂s
au x + bu y = a u s
∂x + u ∂t
t
∂x
(
∂s
+ b u s
=(as x + bs y )u s +(at x + bt y ))u t .
)
∂y + u ∂t
t
∂y
Now by our construction of x, y, and since x t = a, y t = b, wehave
and
as x + bs y = s x x t + s y y t = ∂s
∂t =0,
at x + bt y = s x x t + t y y t = ∂t
∂t =1