Chapter 6 Partial Differential Equations

6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 23
and so
au x + bu y = c.
Recall that in the case of a differential equation in R 2 a solution surface is a surface in
R 3 which (at least locally) can be parameterized by a two parameter family. Thus what
we have shown is that the collection of all characteristic curves thorugh C 0 gives a solution
surface. Uniqueness will follow by arguing that any solution surface is essentially a collection
of characteristic curves. In particular let u(x, y) be a solution of the equation and fix a point
P 0 (x 0 ,y 0 ,z 0 ) on the surface. Let γ :(x(t),y(t),z(t)) be the curve through P 0 determined by
dx
dt = a(x, y, u(x, y)), x(0) = x 0
dy
dt = b(x, y, u(x, y)), y(0) = y 0
z(t) =u(x, y), z(0) = z 0 .
Then along this curve
dz
dt = u dx
x
dt + u dy
y
dt = au x + bu y = c
since u is a solution. Thus we see that γ is the characteristic curve through P 0 . In other
words, a solution is always a union of characteristic curves. Through any point on a solution
surface there is a unique characteristic curve.
Therefore if C 0 is not a characteristic curve, there is a unique solution surface that
contains it. If, on the other hand, C 0 is a characteristic curve then
x ′ 0(s) =a(x 0 (s),y 0 (s),ϕ ( s))
y 0(s) ′ =b(x 0 (s),y 0 (s),ϕ ( s))
which contradicts |J| ≠0.
Remark 6.2.15. The above discussion shows that if C 0 were a characteristic curve we could
construct infinitely many solutions containing C 0 . Namely, take any curve C 1 that meets C 0
in a point P 0 and such that |J| ≠0on C 1 . Then construct the solution surface through C 1 .
As discussed above, this solution surface must contain the characteristic curve C 0 .