32
BRUCE K. DRIVER
†
3.
Fully nonlinear first order PDE
In this section let
U
⊂
o
R
n
be an open subset of
R
n
and
(
x, z, p
)
∈
¯
U×
R
n
×
R
→
F
(
x, z, p
)
∈
R
be a
C
2
— function. Actually to simplify notation let us suppose
U
=
R
n
.
We are now looking for a solution
u
:
R
n
→
R
to the fully nonlinear PDE,
(3.1)
F
(
x, u
(
x
)
,
∇
u
(
x
)) = 0
.
As above, we “reduce” the problem to solving ODE’s. To see how this might be
done, suppose
u
solves (3.1) and
x
(
s
)
is a curve in
R
n
and let
z
(
s
)=
u
(
x
(
s
))
and
p
(
s
)=
∇
u
(
x
(
s
))
.
Then
z
0
(
s
)=
∇
u
(
x
(
s
))
·
x
0
(
s
)=
p
(
s
)
·
x
0
(
s
)
and
(3.2)
p
0
(
s
)=
∂
x
0
(
s
)
∇
u
(
x
(
s
))
.
(3.3)
We would now like to
f
nd an equation for
x
(
s
)
which along with the above system
of equations would form and ODE for
(
x
(
s
)
,z
(
s
)
,p
(
s
))
.
The term,
∂
x
0
(
s
)
∇
u
(
x
(
s
))
,
which involves two derivative of
u
is problematic and we would like to replace it by
something involving only
∇
u
and
u.
In order to get the desired relation, di
f
erentiate
Eq. (3.1) in
x
in the direction
v
to
f
nd
0=
F
x
·
v
+
F
z
∂
v
u
+
F
p
·
∂
v
∇
u
=
F
x
·
v
+
F
z
∂
v
u
+
F
p
·
∇
∂
v
u
=
F
x
·
v
+
F
z
∇
u
·
v
+(
∂
F
p
∇
u
)
·
v,
wherein we have used the fact that mixed partial derivative commute. This equation
is equivalent to
(3.4)
∂
F
p
∇
u

(
x,u
(
x
)
,
∇
u
(
x
))
=
−
(
F
x
+
F
z
∇
u
)

(
x,u
(
x
)
,
∇
u
(
x
))
.
By requiring
x
(
s
)
to solve
x
0
(
s
)=
F
p
(
x
(
s
)
,z
(
s
)
,p
(
s
))
,
we
f
nd, using Eq. (3.4) and
Eqs. (3.2) and (3.3) that
(
x
(
s
)
,z
(
s
)
,p
(
s
))
solves the
characteristic equations
,
x
0
(
s
)=
F
p
(
x
(
s
)
,z
(
s
)
,p
(
s
))
z
0
(
s
)=
p
(
s
)
·
F
p
(
x
(
s
)
,z
(
s
)
,p
(
s
))
p
0
(
s
)=
−
F
x
(
x
(
s
)
,z
(
s
)
,p
(
s
))
−
F
z
(
x
(
s
)
,z
(
s
)
,p
(
s
))
p
(
s
)
.
Wew
i
l
linthefutures
imp
lyabbrev
iatetheseequat
ionsby
x
0
=
F
p
z
0
=
p
·
F
p
(3.5)
p
0
=
−
F
x
−
F
z
p.
The above considerations have proved the following Lemma.
Lemma 3.1.
Let
A
(
x, z, p
):=(
F
p
(
x, z, p
)
,p
·
F
p
(
x, z, p
)
,
−
F
x
(
x, z, p
)
−
F
z
(
x, z, p
)
p
)
,
π
1
(
x, z, p
)=
x
and
π
2
(
x, z, p
)=
z.
If
u
solves Eq. (3.1) and
x
0
∈
U,
then
e
sA
(
x
0
,u
(
x
0
)
,
∇
u
(
x
0
)) = (
x
(
s
)
,u
(
x
(
s
))
,
∇
u
(
x
(
s
)))
and
u
(
x
(
s
)) =
π
2
◦
e
sA
(
x
0
,u
(
x
0
)
,
∇
u
(
x
0
))
(3.6)
where
x
(
s
)=
π
1
◦
e
sA
(
x
0
,u
(
x
0
)
,