Chapter 6 Partial Differential Equations

6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 13
is tangent to the graph of y = u(x) at any point (since it is orthogonal to ⃗v).
This suggests that we look at the integral curves of the vector field A(x, u) inR n+1 given
by solving the equations
dx j
dt = a j(x, y), j=1, ··· ,n,
dy
dt
= b(x, y). (6.2.6)
As you will see, any graph y = u(x) inR n+1 which is the union of an (n − 1)-parameter
family of these integral curves will define a solution of (6.2.5). Conversely, suppose that u is
a solution of (6.2.5). If we solve the equations
dx j
dt = a j(x, u(x)),
x j (0)=(x 0 ) j
to obtain a curve x(t) passing through x 0 , and then set y = u(x(t)), we have
dy
dt =
n∑
j=1
∂ j u dx j
dt =
n∑
a j (x, u)∂ j u = b(x, u).
j=1
Thus if the graph y = u(x) intersects an integral curve of A in one point (x 0 ,u(x 0 )), it
contains the whole curve.
Suppose we are given intial data u = ϕ on a hypersurface S in R n . If we form the
submanifold
S ∗ = {(x, ϕ(x)) : x ∈ S}
of R n+1 , the graph of the solution should be the hypersurface (in R n+1 ) generated by the integral
curves of A passing through S ∗ . Again, we need to assume that S is non-characteristic
in some sense. This is more complicated than the linear case because a j depend on u as well
as x. We need the following geometric interpretation:
For x ∈ S, the vector field A(x, ϕ(x)) = ( a 1 (x, ϕ(x)), ··· ,a n (x, ϕ(x)) ) should not
be tangent to S at x. Note that this condition involves ϕ as well as S.
If S is represented parametrically by a mapping g : R n−1 → R n and we take coordinates
s =(s 1 , ··· ,s n−1 ) ∈ R n−1 , so that g(s) =(g 1 (s), ··· ,g n (s)) , then the above condition can
be expressed as
⎛
⎞
∂g 1 ∂g 1
··· a 1 (g(s),φ(g(s)))
∂s 1 ∂s n−1
det
⎜ . .
.
⎟ ≠0. (6.2.7)
⎝∂g n ∂g n
⎠
··· a n (g(s),φ(g(s)))
∂s 1
∂s n−1