2. FIRST ORDER QUASILINEAR PARTIAL DIFFERENTIAL EQUA-
TIONS
We restrict our exposition to first order quasilinear partial differential equations (FO-
QPDE) with two variables, since this case affords a real geometric interpretation.
However, the treatment can be extended without difficulty to higher order spaces.
The general form of FOQPDE with two independent variables is
a(x, y, u)
∂u
∂x
+ b(x, y, u)
∂u
∂y
= c(x, y, u) (2.1)
where a, b, and c are continuous functions with respect to the three variables x, y, u.
Let u = u(x, y) be a solution to equation (2.1). If we identify u with the third
coordinate z in R
3
, then u = u(x, y) represents a surface S. The direction of the
normal to S is the vector (u
x
, u
y
, −1), where u
x
, u
y
are short hand notation of
∂u
∂x
and
∂u
∂y
, respectively. On the other hand, equation (2.1) can be written as the inner
product
(u
x
, u
y
, −1) · (a, b, c) = 0. (2.2)
Thus, (a, b, c) is perpendicular to the normal to S and consequently, must lie in the
plane tangent to S.
Let us consider a path γ(s) on S. The rate of variation of u as we move along γ is
du
ds
=
∂u
∂x
dx
ds
+
∂u
∂y
dy
ds
. (2.3)
The vector (dx/ds, dy/ds, du/ds) is naturally the tangent to the curve γ(s). Equation
(2.3) can be rewritten as the inner product
(u
x
, u
y
, −1) · (dx/ds, dy/ds, du/ds) = 0. (2.4)
Comparing (2.2) and (2.4), we see that there is a particular family of curves on the
surface S, defined by
dx/ds
a
=
dy/ds
b
=
du/ds
c
. (2.5)
These curves are called characteristics and will be denoted by C(s), or simply C.
We note that there are actually only two independent equations in the system (2.5),
therefore, its solutions comprise in all a two-parameter family of curves in space.
Theorem 1: Any one parameter subset of the characteristics generates a solution of
the first order quasilinear partial differential equation (2.1).
Proof: Let u = u(x, y) be the surface generated by a one-parameter family C(s) of the
characteristics whose differential equations are (2.5). By taking the rate of variation
of u along a characteristic curve C(s), we get
du
ds
= u
x
dx
ds
+ u
y
dy
ds
,
or
(u
x
, u
y
, −1) · (dx/ds, dy/ds, du/ds) = 0. (2.6)
3