Chapter 6 Partial Differential Equations

6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 21
Example 6.2.13. Let us now consider an example in which data is prescribed on characteristics.
In this linear example we seek a function u = u(x, y) such that
We first seek solutions of
subject to
Thus we obtain
dx
dt = x,
x ∂u
∂x + y ∂u
∂y = u + 1 (6.2.14)
u(x, x) =x 2 . (6.2.15)
dy
dt = y,
du
dt = u +1,
x(s, 0) = s, y(s, 0) = s, u(s, 0) = s 2 .
x = se t , y = se t , u = s 2 e t + e t − 1.
In this example it is not possible to solve for s and t in terms of x and y. Not that A(x, y) =
(x, y) and the characteristic manifold
Char (x,y) (L) ={ξ =(ξ 1 ,ξ 2 ) ≠0:A(x, y) · ξ =0}.
We note that the initial curve S in this case is x = y with constant normal vector ν =(1, −1).
In the present case we see that
so the curve is a characteristic.
A(x, x) · ν = x − x =0
Within the context of these examples in R 2
solvability condition (6.2.7) as
we can restate Theorem 6.2.8 with the
Theorem 6.2.14. Suppose a(x, y, u), b(x, y, u), c(x, y, u) are C 1 in Ω ⊂ R 3 , C 0 is C 1 initial
curve given by (x 0 (s),y 0 (s),u 0 (s)) ⊂ Ω and
( )
a(x0 (s),y
det
0 (s),u 0 (s)) x ′ 0(s)
b(x 0 (s),y 0 (s),u 0 (s)) y 0(s)
′ ≠0.
Then there exists a unique C 1 solution of
a(x, y, u)u x + b(x, y, u)u y = c(x, y, u),
with
u(x 0 (s),y 0 (s)) = u 0 (s).