FIRST ORDER PDES
CRISTIAN E. GUTIÉRREZ
AUGUST 9, 2015
Contents
1. Systems of 1st order ordinary differential equations
1.1. Existence of solutions
1.2. Uniqueness
1.3. Differentiability of solutions with respect to a parameter
2. Quasi-linear pdes
2.1. Step 1
2.2. Step 2
2.3. Cauchy problem
3. Degenerate case
4. Examples
4.1. Example 1
4.2. Example 2
4.3. Example 3
4.4. A maximum principle for a first order pde
5. Fully nonlinear case
5.1. Cauchy problem
5.2. Uniqueness
5.3. Solution of the eikonal equation
5.4. Higher dimensional case
6. A transversality problem
7. The Legendre transform
References
These are equations of the form
f (x, u, p) = 0,
1
2
2
4
6
10
10
11
11
12
14
14
14
14
15
15
17
21
22
23
24
25
27
2
C. E. GUTIÉRREZ, AUGUST 9, 2015
here f is a scalar function, x ∈ Rn , u ∈ C1 (Rn ), and p = Du the gradient of u.
These equations can be solved using the theory of systems of 1st order differential
equations. This is a self contained presentation showing how to do it.
1. Systems of 1st order ordinary differential equations
The problem we consider in this section is the following. Let F : (α, β) × Ω → Rn
be a continuous function, with Ω ⊂ Rn an open domain and (α, β) an interval of
real numbers, possibly unbounded. Fix t0 ∈ (α, β) and ξ0 ∈ Ω. We seek for a vector
valued function x(t) = (x1 (t), · · · , xn (t)), defined and continuously differentiable in
an interval [t0 − b, t0 + a] ⊂ (α, β), for some a, b > 0, such that x(t) ∈ Ω for all
t ∈ [t0 − b, t0 + a], and satisfying the initial value problem
(1.1)
x0 (t) = F(t, x(t))
(1.2)
x(t0 ) = ξ0 .
for all t0 − b ≤ t ≤ t0 + a, and
This gives a system of n odes with n unknowns.
1.1. Existence of solutions. There are several methods to show existence of solutions to the initial value problem (1.1) and (1.2). We present an elegant method
due to Tonelli.
Theorem 1. If F is continuous in (α, β) × Ω and |F(t, x)| ≤ M there, then the initial value
problem (1.1) and (1.2) has at least one solution.
Proof. For each k positive integer, and each t ∈ [t0 , t0 + a], with a > 0 that will be
determined in a moment, we define the following sequence of functions:

1



ξ0 ,
t0 ≤ t ≤ t0 + a


k

Z t− a
xk (t) = 
k
j
j+1



F(s, xk (s)) ds,
t0 + a ≤ t ≤ t0 +
a; j = 1, 2, · · · , k − 1.

ξ 0 +
k
k
t0
Notice that
in the sense that the values of xk on the
" xk is defined recursively
#
j
j+1
interval t0 + a, t0 +
a are defined in terms of the values of xk on the interval
k
"
# k
j−1
j
t0 +
a, t0 + a .
k
k
Since F(s, x) is defined for s ∈ (α, β), we then need a < β − t0 . We have, when
j
j+1
t0 + a ≤ t ≤ t0 +
a, 1 ≤ j ≤ k − 1, and k > 1, that
k
k
!
!
Z t−a/k
j
a
k−1
(1.3) |xk (t) − ξ0 | ≤
|F(s, xk (s))| ds ≤ M t − − t0 ≤ M
≤ Ma
.
k
k
k
t0
FIRST ORDER PDES August 9, 2015
3
If R ≤ dist(ξ0 , ∂Ω), then open ball BR (ξ0 ) ⊂ Ω, and if we let
(
)
dist(ξ0 , ∂Ω)
0 < a < min β − t0 ,
= A,
M
then we get from (1.3) that xk (t) ∈ Ω for t0 ≤ t ≤ t0 +a. The sequence xk is uniformly
Lipschitz and therefore equicontinuous. Indeed,
Z t−a/k
0
|xk (t) − xk (t )| ≤ F(s, xk (s)) ds ≤ M |t − t0 |,
t0 −a/k
for all t0 ≤ t, t0 ≤ t0 +a. In addition, xk (t) is uniformly bounded: |xk (t)| ≤ M a+|ξ0 | for
all k. Therefore by Arzelá-Ascoli’s theorem, xk contains a subsequence uniformly
convergent in [t0 , t0 + a] to a function x. Let us denote this subsequence also by xk .
We write
Z t
Z t
Z t
xk (t) = ξ0 +
F(s, xk (s)) ds −
F(s, xk (s)) ds := ξ0 +
F(s, xk (s)) ds − Ak (t).
t0
t−a/k
t0
We have |Ak (t)| ≤ Ma/k, and so it converges uniformly to zero in [t0 , t0 + a] as
k → ∞. Since a < A, then from (1.3) xk (t) belongs to the closed ball BMa (ξ0 ) and
Ma < dist(ξ0 , ∂Ω). Since F(s, x) is uniformly continuous in that ball and xk → x
uniformly, we get that
Z t
Z t
F(s, xk (s)) ds →
F(s, x(s)) ds
t0
t0
for each t ∈ [t0 , t0 + a]. Therefore we obtain that x(t) satisfies the integral equation
Z t
x(t) = ξ0 +
F(s, x(s)) ds
t0
for all t ∈ [t0 , t0 + a]. From the fundamental theorem of calculus x(t) satisfies (1.1)
in the interval [t0 , t0 + a] and also (1.2).
We next construct a solution for t < t0 . We use the result before changing the
right hand side of the equation and the initial time. Let G(t, x) = −F(t0 − t, x). The
function G is defined for t0 −β < t < t0 −α, x ∈ Ω, and satisfies the same hypotheses
than F in this new region. Since t0 −β < 0 < t0 −α, applying the previous argument
(with base point t0 = 0, i.e., on the interval [0, t0 − α]) there exists a solution y(t),
defined in some interval 0 ≤ t ≤ b with b any number satisfying
)
(
dist(ξ0 , ∂Ω)
= B,
0 < b < min t0 − α,
M
4
C. E. GUTIÉRREZ, AUGUST 9, 2015
to the integral equation
t
Z
y(t) = ξ0 +
G(s, y(s)) ds,
0
for 0 ≤ t ≤ b. Let x̄(t) = y(t0 − t). Then x̄(t) is defined for t0 − b ≤ t ≤ t0 and satisfies
there
Z t0 −t
x̄(t) = y(t0 − t) = ξ0 −
F(t0 − s, y(s)) ds
0
Z
t
= ξ0 +
Z
t
F(u, y(t0 − u)) du = ξ0 +
t0
F(u, x̄(u)) du.
t0
Since x̄(t0 ) = ξ0 , if we define



x(t)
x∗ (t) = 

x̄(t)
for t0 ≤ t ≤ t0 + a
for t0 − b ≤ t ≤ t0
we then obtain that x∗ is continuously differentiable in [t0 − b, t0 + a] and satisfies
there the integral equation
Z t
∗
x (t) = ξ0 +
F(s, x∗ (s)) ds
t0
and therefore is the desired solution.
1.2. Uniqueness. Under the assumptions of Theorem 1 the solution might not
be unique. A simple example is the scalar equation x0 = 3x2/3 with the initial
condition x(0) = 0. Obviously x = 0 is a solution, and given any a ≤ 0 ≤ b the
function


(t − a)3 ,
−∞ < t ≤ a




x(t) = 
0,
a<t<b




(t − b)3 ,
b≤t<∞
is a solution.
The following construction illustrates the problem of uniqueness. Suppose
there are two solutions ϕ1 and ϕ2 to the problem (1.1) and (1.2). Then by integration
Z t
(1.4)
ϕi (t) = ξ0 +
F(s, ϕi (s)) ds, i = 1, 2.
t0
Let Φ(t) = |ϕ1 (t) − ϕ2 (t)| and suppose the set E = {t : t0 < t ≤ t0 + a, Φ(t) , 0} is
non empty. Let ω = inf E. We have Φ(t) = 0 for t0 ≤ t ≤ ω by continuity. Let
FIRST ORDER PDES August 9, 2015
5
Ψ(t) = |F(s, ϕ1 (s)) − F(s, ϕ2 (s))|. Then we have Ψ(t) = 0 for t0 ≤ t ≤ ω, and from
(1.4)
Z t
(1.5)
Φ(t) ≤
Ψ(s) ds,
for all t0 ≤ t ≤ t0 + a.
t0
Let h > 0 be small such that ω + h ≤ t0 + a. Consider
µ = max{Ψ(t) : ω ≤ t ≤ ω + h}.
We claim that µ , 0. Because if µ = 0, then from (1.5) we obtain for ω ≤ t ≤ ω + h
that
Z
Z
Z
Φ(t) ≤
ω
t0
t
Ψ(s) ds +
ω
t
Ψ(s) ds =
ω
Ψ(s) ds ≤ µ h = 0,
and so ω would not be the infimum.
Then there exists ω < t0 ≤ ω + h such that
R t0
µ = Ψ(t0 ). From (1.5), Φ(t0 ) ≤ ω Ψ(s) ds ≤ µ h = Ψ(t0 ) h. That is, for each h > 0
sufficiently small, we find t0 ∈ [t0 , t0 + a] such that
1
|F(t0 , ϕ1 (t0 )) − F(t0 , ϕ2 (t0 ))| ≥ |ϕ1 (t0 ) − ϕ2 (t0 )| > 0.
h
If the function F satisfies the Lipschitz condition
|F(s, x) − F(s, y)| ≤ K|x − y|
for x, y ∈ Ω and all s ∈ (α, β) we obtain a contradiction. Therefore we have
uniqueness of the initial value problem when F is Lipschitz.
Remark 2 (Exercise). Suppose F : (α, β) × Ω → Rn is such that there exists a
function g(t) ≥ 0 continuous in (α, β) with |F(t, x) − F(t, y)| ≤ g(t) |x − y| for all
t ∈ (α, β), and x, y ∈ Ω. Let t0 ∈ (α, β) and ξ0 ∈ Ω. Prove that the initial value
problem x0 = F(t, x), x(t0 ) = ξ0 has a unique solution in a neighborhood of t0 .
We shall prove a uniqueness result due to Osgood.
Theorem 3. Let g be a continuous function defined in [0, +∞) non decreasing with
g(0) = 0, g(u) > 0 for u > 0, and satisfying
Z δ
1
du = +∞
0 g(u)
for some δ > 0. Suppose F(t, x) satisfies
|F(t, x) − F(t, y)| ≤ g(|x − y|)
for all t ∈ (α, β) and for all x, y ∈ Ω. Then the solution to (1.1) and (1.2) is unique.
6
C. E. GUTIÉRREZ, AUGUST 9, 2015
Proof. With the notation in this subsection, suppose Φ(t) is not zero in [t0 , t0 + a].
Let M(x) = maxt0 ≤t≤x Φ(t). So as before, with ω = inf E, we have M(x) > 0 for
ω < x ≤ t0 +a and so for each such x, there exists ω < x1 ≤ x such that M(x) = Φ(x1 ).
We have from (1.5) and since g is non decreasing that
Z x1
Z x1
M(x) = Φ(x1 ) ≤
Ψ(s) ds ≤
g(Φ(s)) ds
ω
ω
Z x
Z x
≤
g(Φ(s)) ds ≤
g(M(s)) ds := h(x).
ω
ω
We have h(ω) = 0, M(x) ≤ h(x), and h (x) = g(M(x)) ≤ g(h(x)) for ω ≤ x ≤ t0 + a.
h0 (x)
≤ 1 in ω < x < t0 + a. Hence
Also g(M(x)) > 0 for x > ω. Thus
g(h(x))
Z t0 +a 0
h (x)
dx ≤ (t0 + a − ω),
g(h(x))
ω
0
that is,
Z
h(t0 +a)
1
du < ∞
g(u)
0
a contradiction, and the uniqueness is then proved.
1.3. Differentiability of solutions with respect to a parameter. We assume F :
(α, β) × Ω × (a, b) → Rn , F = F(t, x, λ) = (F1 (t, x, λ), · · · , Fn (t, x, λ)) is continuous and
Dx Fi and (Fi )λ are also continuous, 1 ≤ i ≤ n, where Ω is an open subset of Rn . We
consider the problem
x0 = F(t, x, λ)
x(t0 ) = x0 ,
here t0 ∈ (α, β) and x0 ∈ Ω. So the solution x = x(t, λ) = (x1 (t, λ), · · · , xn (t, λ)) and it
is our goal to show that ∂λ xi (t, λ) is continuous for 1 ≤ i ≤ n.
We have
Z
t
F(s, x(s, λ), λ) ds,
x(t, λ) = x0 +
t0
so
(1.6)
Z
t
xi (t, λ) − xi (t, λ ) =
(Fi (s, x(s, λ), λ) − Fi (s, x(s, λ0 ), λ0 )) ds
0
t0
t
Z
=
Dx Fi (s, ξi , µi ) · (x(s, λ) − x(s, λ0 )) + (Fi )λ (s, ξi , µi )(λ − λ0 ) ds,
t0
FIRST ORDER PDES August 9, 2015
7
1 ≤ i ≤ n, by the mean value theorem where ξi is an intermediate point between
x(s, λ) and x(s, λ0 ), and µi an intermediate point between λ and λ0 ; λ , λ0 . From
the continuity assumption on Dx F and Fλ we have that
!
Z t
0
0
x(t, λ) − x(t, λ ) x(s, λ) − x(s, λ ) (1.7)
M
≤
+ A ds,
λ − λ0
λ − λ0
t0
for some positive constants M and A and for all t0 ≤ t ≤ t0 + h with h > 0 and for
all λ, λ0 in an interval sufficiently small.
Lemma 4 (Gronwall). Let φ be continuous on [t0 , t0 + h], h > 0 such that
Z t
φ(t) ≤
M φ(s) + A ds,
t0
for t ∈ [t0 , t0 + h] with M, A ≥ 0. Then
φ(t) ≤ A h eM(t−t0 )
for t ∈ [t0 , t0 + h].
Proof. Let ψ(t) = φ(t) eM(t0 −t) . We have max[t0 ,t0 +h] ψ(t) = ψ(t1 ) for some t1 ∈ [t0 , t0 +h].
So for t ∈ [t0 , t0 + h]
Z t
M(t−t0 )
ψ(t) e
≤
M ψ(s) eM(s−t0 ) + A ds,
t0
and so at t = t1
Z
ψ(t1 ) e
M(t1 −t0 )
t1
≤
Z
M(s−t0 )
M ψ(s) e
+ A ds ≤ ψ(t1 )
t0
t1
M eM(s−t0 ) ds + A(t1 − t0 )
t0
= ψ(t1 ) eM(t1 −t0 ) − 1 + A(t1 − t0 )
and we get
ψ(t1 ) ≤ A(t1 − t0 ) ≤ A h
and the lemma follows.
4 to (1.7) with λ, λ0 fixed and λ , λ0 , i.e., with φ(t) =
Applying Lemma
0 x(t, λ) − x(t, λ ) we obtain
λ − λ0
0
x(t, λ) − x(t, λ ) 0
(1.8)
z(t, λ, λ ) := ≤ A h eM(t−t0 ) ,
0
λ−λ
uniformly in λ, λ0 and t0 ≤ t ≤ t0 + h. In particular, this shows that x(t, λ) is
continuous in λ for each t.
8
C. E. GUTIÉRREZ, AUGUST 9, 2015
Let us now consider the linear system in Y
Yi0 (t) = Dx Fi (t, x(t, λ), λ) · Y(t) + (Fi )λ (t, x(t, λ), λ),
(1.9)
1 ≤ i ≤ n,
Y(t0 ) = 0.
From the assumptions on F, the right hand side of the ode is Lipschitz in the variable Y, and therefore the solution Y = Y(t, λ) exists and is unique for t sufficiently
close to t0 and λ in some interval.
We claim that
∂λ x(t, λ) = (∂λ x1 (t, λ), · · · , ∂λ xn (t, λ)) = Y(t, λ).
(1.10)
We have that
Z
t
Yi (t, λ) =
(Dx Fi (s, x(s, λ), λ) · Y(t) + (Fi )λ (s, x(s, λ), λ)) ds
t0
and so from (1.6)
xi (t, λ) − xi (t, λ0 )
− Yi (t, λ)
λ − λ0
!
#
Z t"
x(s, λ) − x(s, λ0 )
=
+ (Fi )λ (s, ξi , µi ) ds
Dx Fi (s, ξi , µi ) ·
λ − λ0
t0
Z t
(Dx Fi (s, x(s, λ), λ) · Y(t, λ) + (Fi )λ (s, x(s, λ), λ)) ds
−
t0
Z t"
=
Z
t0
t
+
!#
x(s, λ) − x(s, λ0 )
− Y(s, λ) ds
Dx Fi (s, ξi , µi ) ·
λ − λ0
Dx Fi (s, ξi , µi ) − Dx Fi (s, x(s, λ), λ) · Y(t, λ) + (Fi )λ (s, ξ, µ) − (Fi )λ (s, x(s, λ), λ) ds.
t0
Set
Bi := Dx Fi (s, ξi , µi ) − Dx Fi (s, x(s, λ), λ) · Y(t, λ) + (Fi )λ (s, ξi , µi ) − (Fi )λ (s, x(s, λ), λ) .
Given > 0, by the continuity of Dx Fi and (Fi )λ we have that
|Bi | < ,
for 1 ≤ i ≤ n,
when |λ − λ0 | is sufficiently small. Therefore we obtain
!
Z t
0
0
x(s, λ) − x(s, λ )
x(t, λ) − x(t, λ )
− Y(t, λ) ≤
M
− Y(s, λ) + ds
λ − λ0
λ − λ0
t0
and by Gronwall’s lemma
0
x(t, λ) − x(t, λ )
− Y(t, λ) ≤ heMh
0
λ−λ
FIRST ORDER PDES August 9, 2015
9
when |λ − λ0 | is sufficiently small, i.e., λ0 → λ, and so ∂λ x(t, λ) = Y(t, λ).
From this we deduce the differentiability of the solution with respect to the
initial conditions. Let
x0 = F(t, x)
x(t0 ) = λ = (λ1 , · · · , λn ).
So the solution x = x(t, λ). Let Y(t, λ) = x(t, λ) − λ, so x = Y + λ. Then
Y0 = x0 = F(t, x) = F(t, Y + λ) := G(t, Y, λ)
Y(t0 , λ) = 0,
and therefore Y is differentiable with respect to each λi , 1 ≤ i ≤ n, and so is x.
Remark 5. We show now that the solution Y(t, λ) to the linear system (1.9) is
continuous as a function of λ. From (1.10) this implies that x(t, λ) is continuously
differentiable in λ. Let Dx F(t, x(t, λ), λ) denote the n × n matrix having rows
Dx Fi (t, x(t, λ), λ), and Fλ (t, x(t, λ), λ) the vector with components (Fi )λ (t, x(t, λ), λ).
Since Y(t, λ) solves (1.9), we have
Z t
Y(t, λ) =
{Dx F(s, x(s, λ), λ) Y(s, λ) + Fλ (s, x(s, λ), λ)} ds.
t0
We write
Z
t
Y(t, λ) − Y(t, λ0 ) =
[Dx F(s, x(s, λ), λ) Y(s, λ) − Dx F(s, x(s, λ0 ), λ0 ) Y(s, λ0 )] ds
t0
t
Z
+
[Fλ (s, x(s, λ), λ) − Fλ (s, x(s, λ0 ), λ0 )] ds
t0
= A + B.
We have
Z
t
Dx F(s, x(s, λ), λ) [Y(s, λ) − Y(s, λ0 )] ds
A=
t0
Z
t
+
[Dx F(s, x(s, λ), λ) − Dx F(s, x(s, λ0 ), λ0 )] Y(s, λ0 ) ds.
t0
From (1.8) x(s, λ) is continuous in s and λ, and since Dx F is continuous, we get that
kDx F(s, x(s, λ), λ)k ≤ C1 , and kDx F(s, x(s, λ), λ)−Dx F(s, x(s, λ0 ), λ0 )k < for λ close to
λ0 . Also since Fλ is continuous, we also have |Fλ (s, x(s, λ), λ) − Fλ (s, x(s, λ0 ), λ0 )| ≤ 10
C. E. GUTIÉRREZ, AUGUST 9, 2015
for λ close to λ0 . Hence
Z
t
|Y(t, λ) − Y(t, λ0 )| ≤
(C1 |Y(s, λ) − Y(s, λ0 )| + C2 ) ds
t0
for λ close to λ0 and t0 ≤ t ≤ t0 + h with h sufficiently small, and so from Lemma
4 we get
|Y(t, λ) − Y(t, λ0 )| ≤ C2 h eC1 (t−t0 )
for t0 ≤ t ≤ t0 + h and λ close to λ0 . This proves the desired continuity.
2. Quasi-linear pdes
For simplicity, we shall first consider n = 2 and the case in which the equation
is quasi-linear. In this case the pde has the form
(2.11)
a(x, y, u)ux + b(x, y, u)u y = c(x, y, u),
where the coefficients a, b, c ∈ C1 (Ω), Ω ⊂ R3 open. We think that each point
(x, y, z) ∈ Ω has assigned a vector (a, b, c), i.e.,
(2.12)
(x, y, z) 7→ a(x, y, z), b(x, y, z), c(x, y, z) ,
and then the solution z = u(x, y) describes a surface in R3 whose normal vector
(ux , u y , −1) is perpendicular to the given vector field. Or in other words, the
tangent plane to the surface at each point contains the vector field. The solution
to (2.11) will be find using the so called method of characteristics, that is, finding
the solutions of the following system of odes
(2.13)
dx
= a(x, y, z),
dt
dy
= b(x, y, z),
dt
dz
= c(x, y, z).
dt
From the theory of odes and since a, b, c ∈ C1 (Ω), for each point (x0 , y0 , z0 ) ∈ Ω
there exists a unique solution (x(t), y(t), z(t)) to the system (2.13) defined for |t| < and satisfying the initial condition (x(0), y(0), z(0)) = (x0 , y0 , z0 ). The idea is to
prove that these solutions to the pde (2.11) are “union” of characteristic curves,
that is, solutions to odes (2.13).
2.1. Step 1. Suppose z = u(x, y) describes a C1 surface S such that is “union” of
characteristic curves, then u solves (2.11). Indeed, given (x0 , y0 , z0 ) ∈ S there is
a characteristic curve C given by (x(t), y(t), z(t)) such that z(t) = u(x(t), y(t)) and
(x(0), y(0), z(0)) = (x0 , y0 , z0 ). Then z0 (t) = x0 (t)ux (x(t), y(t)) + y0 (t)u y (x(t), y(t)), and
letting t = 0 we get c(x0 , y0 , z0 ) = a(x0 , y0 , z0 )ux (x0 , y0 ) + b(x0 , y0 , z0 )u y (x0 , y0 ). That
is, for each (x0 , y0 , z0 ) ∈ S, u solves (2.11).
FIRST ORDER PDES August 9, 2015
11
2.2. Step 2. Suppose z = u(x, y) is a C1 solution to (2.11) and let S its graph. If
P0 = (x0 , y0 , z0 ) ∈ S and C is a characteristic curve passing through P0 , then C ⊂ S.
Indeed, if C has equation (x(t), y(t), z(t)) and (x(0), y(0), z(0)) = P0 , then we show
that z(t) = u(x(t), y(t)). Let U(t) = z(t) − u(x(t), y(t)), then we have
dU
= z0 (t) − x0 (t) ux (x(t), y(t)) − y0 (t) u y (x(t), y(t))
dt
= c(x(t), y(t), z(t)) − a(x(t), y(t), z(t)) ux (x(t), y(t)) − b(x(t), y(t), z(t)) ux (x(t), y(t))
= c x(t), y(t), U(t) + u(x(t), y(t))
− a(x(t), y(t), U(t) + u(x(t), y(t))) ux (x(t), y(t))
− b(x(t), y(t), U(t) + u(x(t), y(t))) ux (x(t), y(t)) := F(U, t).
dU
= F(U, t) since u solves (2.11) (notice that
The function U = 0 solves the ode
dt
F(U, t) depends on the curve and also of u). Also U(0) = 0. Since the coefficients
a, b, c are C1 , u is also C1 , it follows from the mean value theorem and the form of
F(U, t) that |F(U, t) − F(V, t)| ≤ C|U − V|, then by the uniqueness theorem for odes1
it follows that U ≡ 0 and we are done.
2.3. Cauchy problem. Given a curve Γ parameterized by ( f (s), g(s), h(s)) we are
looking for a solution u to (2.11) passing through Γ, that is, h(s) = u( f (s), g(s)).
Let us assume the curve Γ is C1 , we want to solve (2.11) in a neighborhood of the
point (x0 , y0 , z0 ) = ( f (s0 ), g(s0 ), h(s0 )). For each point in P ∈ Γ close to (x0 , y0 , z0 ),
the idea is to take the characteristic curve passing through P and then glue all
these curves together. This will give the desired solution. More precisely, let
|s − s0 | < δ, take ( f (s), g(s), h(s)) ∈ Γ, and solve (2.13) with the initial condition
( f (s), g(s), h(s)). That is, we have a solution to (2.13), called (x(s, t), y(s, t), z(s, t))
with (x(s, 0), y(s, 0), z(s, 0)) = ( f (s), g(s), h(s)), and let us say this solution is defined
for all |t| < . We then have a transformation
Φ(s, t) = (x(s, t), y(s, t)),
defined for |s − s0 | < δ and |t| < . The Jacobian matrix of Φ is


 ∂x (s, t) ∂x (s, t)



 ∂t
∂s


JΦ (s, t) = 
 .


∂y

 ∂y
(s, t)
(s, t)
∂t
∂s
1
If F(U, t) is Lipschitz in U (uniformly in t), then from Theorem 3 the initial value problem
dU
= F(U, t), U(0) = U0 has at most one solution.
dt
12
C. E. GUTIÉRREZ, AUGUST 9, 2015
At s = s0 and t = 0 we have


 ∂x (s , 0) ∂x (s , 0) 

 a(x0 , y0 , z0 ) f 0 (s0 )
0
 ∂t 0



∂s

 

 = 
JΦ (s0 , 0) = 
 .
 




0
 ∂y

∂y
b(x
0 , y0 , z0 ) g (s0 )
 (s0 , 0)

(s0 , 0)
∂t
∂s
If det JΦ (s0 , 0) , 0, then by the inverse function theorem there exists an inverse
Φ−1 (x, y) = (s, t) defined locally and we set
u(x, y) = z Φ−1 (x, y) .
Every characteristic curve of (2.13) is contained in the graph of u, because
u(x(s, t), y(s, t)) = z Φ−1 (x(s, t), y(s, t)) = z(s, t),
so by Subsection 2.1 u solves the pde (2.11). Also from Subsection 2.2 the solution
u is unique. Because if u1 and u2 are two solutions, then they both contain any
characteristic curve passing by each point in Γ. So we have proved the following
theorem.
Theorem 6. If the coefficients of (2.11) are C1 in a neighborhood of (x0 , y0 , z0 ) and Γ is a
C1 curve given by ( f (s), g(s), h(s)) such that ( f (s0 ), g(s0 ), h(s0 )) = (x0 , y0 , z0 ) and


a(x0 , y0 , z0 ) f 0 (s0 )


 , 0,
(2.14)
det 



0
b(x0 , y0 , z0 ) g (s0 )
then there exists a unique solution u to the pde (2.11) defined in a neighborhood of (x0 , y0 )
such that the graph of u contains Γ, i.e., h(s) = u( f (s), g(s)) for |s − s0 | < δ with δ
sufficiently small.
3. Degenerate case
This is when
(3.15)


a(x0 , y0 , z0 ) f 0 (s0 )


 = 0.
det 



0
b(x0 , y0 , z0 ) g (s0 )
We will show that in this case the Cauchy problem might not have solutions.
We shall prove that if there is a solution u to the Cauchy problem then this
prescribes the value of the tangent to the initial curve Γ when s = s0 , actually, the
initial curve Γ will be characteristic at s0 . Indeed, if there is a solution u, then
we have h(s) = u( f (s), g(s)) for |s − s0 | < δ. Differentiating this with respect to
FIRST ORDER PDES August 9, 2015
13
s yields h0 (s) = f 0 (s)ux + g0 (s)u y and since u is a solution we also have c(P0 ) =
a(P0 )ux (x0 , y0 ) + b(P0 )u y (x0 , y0 ). So together with (3.15) we have the system
b f 0 − a g0 = 0
f 0 ux + g0 u y − h0 = 0
a ux + b u y − c = 0.
If we make a(second eq)- f 0 (third eq) and use the first equation we get −a f 0 + f 0 c =
0. Also if we make b(second eq)-g’(third eq) and use the first equation we obtain
b h0 − g0 c = 0. Therefore we obtain the equivalent system
b f 0 − a g0 = 0
−c f 0 + a h0 = 0
−c g0 + b h0 = 0.
The tangent vector to Γ at s0 is τ = ( f 0 (s0 ), g0 (s0 ), h0 (s0 )), and if we set


 b −a 0


A = −c 0 a ,


0 −c b
then Aτ = 0. If (a, b, c) (x0 ,y0 ,z0 ) is not zero, then rank(A) = 2 and therefore
the dimen
sion of the space of solutions of Aw = 0 is one and so τ = λ (a, b, c) (x0 ,y0 ,z0 ) . Therefore, if there is a solution
u, then the tangent to Γ at s0 is determined and it must be
0
0
0
a multiple of
(a, b, c) (x0 ,y0 ,z0 ) . Consequently, if (3.15) holds, and ( f (s0 ), g (s0 ), h (s0 ))
and (a, b, c) (x0 ,y0 ,z0 ) are not linearly dependent, then there is no solution to the
Cauchy problem for the equation (2.11) with the condition h(s) = u( f (s), g(s)).
We also remark that if the initial curve Γ is characteristic, then the Cauchy
problem has infinitely many solutions. Indeed, if ( f (s), g(s), h(s)) is characteristic
and passes through (x0 , y0 , z0 ) when s = s0 , then f 0 = a, g0 = b and h0 = c. Let
Γ̄ = ( f¯(s), ḡ(s), h̄(s)) be defined by f¯(s) = α(s − s0 ) + x0 , ḡ(s) = β(s − s0 ) + y0 and h̄(s)
is any C1 function such that h̄(s0 ) = z0 . So




a(x0 , y0 , z0 ) f¯0 (s0 )
a(x0 , y0 , z0 ) α




det 
 = det 
 = aβ − bα.




0
b(x0 , y0 , z0 ) ḡ (s0 )
b(x0 , y0 , z0 ) β
If a2 + b2 , 0 at (x0 , y0 , z0 ), then we can choose α and β such that aβ − bα , 0 and so
(2.14) holds and so by Theorem 6, there is a unique solution ū containing Γ̄. Since
Γ is a characteristic curve, from Subsection 2.2, Γ must be contained in the graph
14
C. E. GUTIÉRREZ, AUGUST 9, 2015
of ū. Therefore, varying α, β or h̄ we obtain infinitely many curves Γ̄ satisfying
(2.14) and so infinitely many solutions ū to the pde (2.11) all of them containing Γ.
4. Examples
4.1. Example 1. Solve cux + u y = 0 with c=constant and satisfying u(x, 0) = h(x)
where h ∈ C1 . The curve Γ is given by (s, 0, h(s)), condition (2.14) is then


 c 1


 = −1.
det 



1 0
Solve x0 = c, y0 = 1, z0 = 0 with (x(s, 0), y(s, 0), z(s, 0)) = (s, 0, h(s)). We get x(s, t) =
s + ct, y(s, t) = t, and z(s, t) = h(s). Inverting t = y and s = x − cy. So the solution is
u(x, y) = h(x − cy).
4.2. Example 2. Solve βxux + u y = βu with c ∈ R and satisfying u(x, 0) = h(x)
where h ∈ C1 . The curve Γ is given by (s, 0, h(s)), condition (2.14) is then


βx 1


 = −1.
det 



1 0
Solve x0 = βx, y0 = 1, z0 = βz with (x(s, 0), y(s, 0), z(s, 0)) = (s, 0, h(s)). We get
x(s, t) = s eβt , y(s, t) = t and z(s, t) = h(s) eβt . Inverting yields t = y, s = x e−βy , and so
the solution is u(x, y) = h x e−βy eβy .
4.3. Example 3. Solve ux + u y = u2 satisfying u(x, 0) = h(x) where h ∈ C1 . The
curve Γ is given by (s, 0, h(s)), condition (2.14) is then


1 1


 = −1.
det 



1 0
Solve x0 = 1, y0 = 1, z0 = z2 with (x(s, 0), y(s, 0), z(s, 0)) = (s, 0, h(s)). We get
1
x(s, t) = s + t, y(s, t) = t and z(s, t) = −
. Inverting yields t = y, s = x − y,
t − 1/h(s)
1
.
and so the solution is u(x, y) = −
y − 1/h(x − y)
FIRST ORDER PDES August 9, 2015
15
4.4. A maximum principle for a first order pde. Let u(x, y) ∈ C1 B1 (0) be a
solution to the equation
a(x, y)ux + b(x, y)u y = −u,
in B1 (0).
Suppose that
a(x, y)x + b(x, y)y > 0, 2
for x2 + y2 = 1.
Then u ≡ 0.
Proof. Let M = maxB1 (0) u and m = minB1 (0) u. We shall prove that M ≤ 0 and
m ≥ 0. If the maximum is attained in the interior, then from the equation M = 0.
Similarly, if the minimum is attained in the interior m = 0.
Suppose the maximum M is attained at some point (x0 , y0 ) on the boundary. If
ν = (ν1 , ν2 ) is any vector such that
(4.16)
π/2 < angle ν, (x0 , y0 ) ≤ π.
The point (x0 , y0 )+tν ∈ B1 (0) for all t > 0 sufficiently small. Let g(t) = u (x0 , y0 ) + tν .
Since at (x0 , y0 ) the maximum of u is attained, then
n
o
0 ≥ g(t) − g(0) = g0 (ξ) t = ν1 ux (x0 , y0 ) + ξν + ν2 u y (x0 , y0 ) + ξν t,
with 0 ≤ ξ ≤ t. Since u is C1 up to the boundary, letting t → 0+ yields
(4.17)
ν1 ux x0 , y0 + ν2 u y x0 , y0 ≤ 0.
By the assumption a(x0 , y0 )x0 + b(x0 , y0 )y0 = (a(x0 , y0 ), b(x0 , y0 )) · (x0 , y0 ) > 0, that
is angle (a(x0 , y0 ), b(x0 , y0 )), (x0 , y0 ) < π/2. So ν = −(a(x0 , y0 ), b(x0 , y0 )) satisfies
(4.16), and from (4.17) and the pde we get u(x0 , y0 ) ≤ 0 and we are done. The
argument to show m ≥ 0 is completely similar, in this case the function g satisfies
g(t) − g(0) ≥ 0.
5. Fully nonlinear case
Very beautiful references for this part are the books by Constantin Carathéodory
[Car82, Chapter 3], also containing historical references; the wonderful book by
Fritz John [Joh82, pp. 19-31] containing also examples and exercises, and the
fundamental treatise by Courant and Hilbert [CH62, Vol.2, pp. 75-103]. Another
2
This means that the field (a, b) points always towards the outer side of the tangent plane to
the boundary of the disc. The result also holds when the disc is replaced by domains having this
property.
16
C. E. GUTIÉRREZ, AUGUST 9, 2015
worthy reference is the book by Sneddon [Sne06, Chapters 1 and 2] for readers
interested in finding solutions of many particular 1st order pdes.
We consider an equation of the form
F(x, y, u, ux , u y ) = 0,
(5.18)
or more generally in any dimensions
F(x, u, Du) = 0.
(5.19)
We will write as before p = ux and q = u y . Suppose we have a twice differentiable solution z = u(x) to the pde (5.19) and set pi (x) = uxi (x), p(x) = Dx u(x).
Differentiating (5.19) with respect to x j yields
(5.20) Fx j (x, u(x), Du(x))+Fu (x, u(x), Du(x))ux j (x)+
n
X
Fpi (x, u(x), Du(x))uxi x j (x) = 0.
i=1
Now consider any n-dimensional curve x = x(t), insert it in u(x) and ux j (x) and let
z(t) = u(x(t)) and p j (t) = ux j (x(t)). Differentiating with respect to t yields
ż(t) =
n
X
uxi (x(t))ẋi (t) =
i=1
n
X
pi (t)ẋi (t)
i=1
and
ṗi (t) =
n
X
uxi x j (x(t))ẋ j (t).
j=1
Suppose now the curve x(t) is chosen so that
ẋi (t) = Fpi (x(t), u(x(t)), Du(x(t))) = Fpi (x(t), z(t), p(t))
FIRST ORDER PDES August 9, 2015
17
and substituting this into (5.20) yields
0 = Fx j (x(t), u(x(t)), Du(x(t))) + Fu (x(t), u(x(t)), Du(x(t)))ux j (x(t))
+
n
X
Fpi (x(t), u(x(t)), Du(x(t)))uxi x j (x(t))
i=1
= Fx j (x(t), u(x(t)), Du(x(t))) + Fu (x(t), u(x(t)), Du(x(t)))ux j (x(t))
+
n
X
uxi x j (x(t))ẋi (t)
i=1
= Fx j (x(t), u(x(t)), Du(x(t))) + Fu (x(t), u(x(t)), Du(x(t)))ux j (x(t))
+ ṗ j (t)
= Fx j (x(t), z(t), p(t)) + Fu (x(t), z(t), p(t))p j (t)
+ ṗ j (t).
This means
ṗ j (t) = −Fx j (x(t), z(t), p(t)) − Fu (x(t), z(t), p(t))p j (t).
So we obtain the system of 2n + 1 odes:
(5.21)
ẋ(t) = Dp F(x(t), z(t), p(t))
(5.22)
ṗ(t) = −Dx F(x(t), z(t), p(t)) − Fu x(t), z(t), p(t) p(t)
(5.23)
ż(t) = p(t) · Dp F(x(t), z(t), p(t)).
In other words, we have proved that if u is a C2 solution of the pde (5.19) and
for a C1 curve x(t) we set z(t) = u(x(t)), p(t) = Du(x(t)), and x(t) is chosen so that
ẋ(t) = Dp F(x(t), z(t), p(t)), then (5.22) and (5.23) are satisfied.
5.1. Cauchy problem. Suppose we want to find a solution z = u(x, y) of the pde
(5.18) such that passes through a curve Γ parameterized by Γ(s) = ( f (s), g(s), h(s)).
Assume that Γ is C1 for |s − s0 | < δ and
Γ(s0 ) = (x0 , y0 , z0 ).
18
C. E. GUTIÉRREZ, AUGUST 9, 2015
We want to construct u using the system of odes (5.21), (5.22), and (5.23), which
in this particular case is the system of five odes in (x(t), y(t), z(t), p(t), q(t))



ẋ(t) = Fp (x, y, z, p, q)






 ẏ(t) = Fq (x, y, z, p, q)



(5.24)
ż(t) = pFp (x, y, z, p, q) + qFq (x, y, z, p, q)







ṗ(t) = −Fx (x, y, z, p, q) − p Fz x, y, z, p, q




q̇(t) = −F y (x, y, z, p, q) − q Fz x, y, z, p, q .
As in the quasilinear case, we want to find a curve (x(s, t), y(s, t), z(s, t)) satisfying
the system of five odes and such that (x(s, 0), y(s, 0), z(s, 0)) = ( f (s), g(s), h(s)). But
in this case we have two more unknowns p(s, t) and q(s, t) and so to solve the
system of odes we need to prescribe two additional conditions p(s, 0) = φ(s) and
q(s, 0) = ψ(s). These conditions φ and ψ must be compatible with the first order
pde and also with Γ. In fact, since the prospective solution z = u(x, y) passes
through Γ we must have h(s) = u( f (s), g(s)) and so differentiating with respect to s
h0 (s) = ux ( f (s), g(s)) f 0 (s) + u y ( f (s), g(s))g0 (s).
In addition, the curve must satisfy the pde F(x, y, z, p, q) = 0, that is,
F( f (s), g(s), h(s), ux ( f (s), g(s)), u y ( f (s), g(s))) = 0.
So we need the following compatibility conditions for φ and ψ3
h0 (s) = φ(s) f 0 (s) + ψ(s)g0 (s)
(5.25)
F( f (s),g(s), h(s), φ(s), ψ(s)) = 0.
In general, there might not be solutions to this system of equations, and if there
are solutions they might not be unique. So we assume that there exist p0 , q0 such
that
h0 (s0 ) = p0 f 0 (s0 ) + q0 g0 (s0 )
F(x0 ,y0 , z0 , p0 , q0 ) = 0,
3
Notice the first equation means that the tangent vector ( f 0 (s), g0 (s), h0 (s)) is perpendicular to the
vector (φ(s), ψ(s), −1). That is, at a point ( f (s), g(s), h(s)) on the curve, the plane through this point
with normal (φ(s), ψ(s), −1) is tangent to the curve. When s moves this tangent plane also moves
and describes a ribbon or strip. It is for this reason that a set of functions f (s), g(s), h(s), φ(s), ψ(s)
satisfying the first equation is called a strip.
FIRST ORDER PDES August 9, 2015
19
and


Fp (x0 , y0 , z0 , p0 , q0 ) f 0 (s0 )


 , 0.
(5.26)
det 



Fq (x0 , y0 , z0 , p0 , q0 ) g0 (s0 )
Let H(s, p, q) = H1 (s, p, q), H2 (s, p, q) = p f 0 (s) + qg0 (s) − h0 (s), F( f (s), g(s), h(s), p, q)
which is defined for |s−s0 | < δ and for (p, q) close to (p0 , q0 ). We have H(s0 , p0 , q0 ) = 0
∂(H1 , H2 )
and the determinant of the Jacobian matrix
is different from zero at
∂(p, q)
(s0 , p0 , q0 ). Then by the implicit function theorem there exist unique C1 functions
p = φ(s), q = ψ(s) such that H(s, φ(s), ψ(s)) = 0 for |s − s0 | < δ with δ sufficiently
small. This means that the system (5.25) can be uniquely solved. Therefore with
this choice of φ(s) and ψ(s) let
x(s, t), y(s, t), z(s, t), p(s, t), q(s, t)
be the solution to the system (5.24) satisfying
x(s, 0), y(s, 0), z(s, 0), p(s, 0), q(s, 0) = f (s), g(s), h(s), φ(s), ψ(s) .
We first observe that
(5.27)
G(s, t) := F x(s, t), y(s, t), z(s, t), p(s, t), q(s, t) = 0.
Indeed, from the second equation in (5.25) G(s, 0) = 0. Differentiating G(s, t) with
respect to t and using the system (5.24) yields
∂G
=Fx ẋ + F y ẏ + Fz ż + Fp ṗ + Fq q̇
∂t
= Fx Fp + F y Fq + Fz pFp + qFq + Fp −Fx − pFz + Fq −F y − qFz = 0,
and so (5.27) follows from the fundamental theorem of calculus.
Next and with the aid of x(s, t), y(s, t), z(s, t), p(s, t), q(s, t) we construct the solution z = u(x, y) to the 1st order pde. Let us first invert x(s, t), y(s, t). We
have x(s0 , 0) = f (s0 ) and y(s0 , 0) = g(s0 ). Also the Jacobian of the transforma"
#
∂(x, y)
xs (s, t) xt (s, t)
tion Φ(s, t) = (x(s, t), y(s, t)) is
=
. If t = 0 and s = s0 , then
∂(s, t)
ys (s, t) yt (s, t)
"
# " 0
#
xs (s0 , 0) xt (s0 , 0)
f (s0 ) Fp (x0 , y0 , z0 , p0 , q0 )
= 0
which from assumption (5.26) has
ys (s0 , 0) yt (s0 , 0)
g (s0 ) Fq (x0 , y0 , z0 , p0 , q0 )
determinant different from zero. Therefore from the inverse function theorem
there is a neighborhood of (s0 , 0) such that Φ can be inverted, that is, we can write
(s, t) = Φ−1 (x, y) with Φ−1 a C1 transformation.
20
C. E. GUTIÉRREZ, AUGUST 9, 2015
We claim that the desired solution of the 1st order pde is
u(x, y) = z(Φ−1 (x, y)).
(5.28)
From (5.27) we have
0 = F x(s, t), y(s, t), z(s, t), p(s, t), q(s, t)
= F x(Φ−1 (x, y)), y(Φ−1 (x, y)), z(Φ−1 (x, y)), p(Φ−1 (x, y)), q(Φ−1 (x, y))
= F x, y, u(x, y), p(Φ−1 (x, y)), q(Φ−1 (x, y)) ,
so if we prove that
p(Φ−1 (x, y)) = ux (x, y),
q(Φ−1 (x, y)) = u y (x, y),
then u solves the 1st order pde. Or equivalently we need to show that
(5.29)
p(s, t) = ux (x(s, t), y(s, t)),
q(s, t) = u y (x(s, t), y(s, t)),
for |s − s0 | < δ and |t| < . From (5.28) we have
z(s, t) = u(x(s, t), y(s, t)),
and differentiating we get

∂y

∂z
∂x


 (s, t) = ux (x(s, t), y(s, t)) (s, t) + u y (x(s, t), y(s, t)) (s, t)

∂t
∂t
∂t
(5.30)


∂y
∂z
∂x


 (s, t) = ux (x(s, t), y(s, t)) (s, t) + u y (x(s, t), y(s, t)) (s, t).
∂s
∂s
∂s
We claim that
(5.31)

∂y

∂z
∂x



 ∂t (s, t) = p(s, t) ∂t (s, t) + q(s, t) ∂t (s, t)


∂y
∂x
 ∂z

 (s, t) = p(s, t) (s, t) + q(s, t) (s, t).
∂s
∂s
∂s
#
xs (s0 , 0) xt (s0 , 0)
Suppose the claim is proved. Since by assumption (5.26), det
,
ys (s0 , 0) yt (s0 , 0)
0, the linear system (5.30) has a unique solution for |s − s0 | < δ and |t| < and therefore (5.29) follows. So it remains to prove (5.31). From the first three equations
in (5.24) the first equation in (5.31) immediately follows. To prove the second
equation in (5.31) let
"
r(s, t) :=
∂y
∂z
∂x
(s, t) − p(s, t) (s, t) − q(s, t) (s, t).
∂s
∂s
∂s
FIRST ORDER PDES August 9, 2015
21
We shall prove that r ≡ 0. First notice that r(s, 0) = h0 (s) − φ(s) f 0 (s) + ψ(s)g0 (s) = 0
by (5.25). On the other hand,
∂r
= zst − pt xs − pxst − qt ys − qyst
∂t
= (zt − pxt − qyt )s + ps xt + qs yt − pt xs − qt ys
= ps Fp + qs Fq + xs Fx + Fz p + ys F y + Fz q
∂G
− Fz (zs − pxs − qys ) = −Fz r,
∂s
Rt
from (5.24) and (5.27). So r(s, t) = r(s, 0) exp − 0 Fz dt = 0 and we are done.
It remains to verify that h(s) = u( f (s), g(s)). But
x(s, 0) = f (s)
and y(s, 0) = g(s) so
−1
−1
Φ ( f (s), g(s)) = (s, 0). That is, u( f (s), g(s)) = z Φ ( f (s), g(s)) = z(s, 0) = h(s).
=
5.2. Uniqueness. Notice that the solution to the pde F(x, y, u, ux , u y ) = 0 and
passing through the curve Γ, that is, u( f (s), g(s)) = h(s) might not be unique. The
number of solutions depends of the number of ways we have to complete Γ to a
strip f (s), g(s), h(s), φ(s), ψ(s), that is, the number of solutions φ and ψ of the system
(5.25). However once we choose φ and ψ the solution u is unique. Notice this
amounts to prescribe ux and u y on Γ. As an example of non uniqueness of the
Cauchy problem we will show that the equation
1 2
1
u(x, 0) = (1 − x2 ),
ux + u2y + xux + yu y = u,
2
2
has solutions
1
u(x, y) = ±y + (1 − x2 ).
2
1 2 2
In this case we have F(x, y, z, p, q) = (p + q ) + xp + yq − z, and so the system (5.24)
2
becomes



ẋ(t) = x + p






ẏ(t) = y + q




ż(t) = p(x + p) + q(y + q) .







ṗ(t) = 0




q̇(t) = 0.
The curve Γ is s, 0, 12 (1 − s2 ) and then the compatibility conditions (5.25) in this
case are
1
1
−s = φ(s),
φ(s)2 + ψ(s)2 + sφ(s) − (1 − s2 ) = 0,
2
2
22
C. E. GUTIÉRREZ, AUGUST 9, 2015
that is,
φ(s) = −s,
ψ(s) = ±1.
So p(s, t) = −s, and q(s, t) = ±1. Also the Jacobian condition (5.26) holds because


p + x 1


 = −(y + q)
det 



q+y 0
which at q = ±1 and y = 0 is different from zero. Next we need to solve ẋ = x − s,
ẏ = y ± s with initial conditions x(s, 0) = s and y(s, 0) = 0, which yields x(s, t) = s
and y(s, t) = ±(et − 1). Then ż = et and so z(s, t) = et − 21 − 12 s2 . Therefore s = x and
±y + 1 = et . So
1 1 2
1
− x = ±y + (1 − x2 ).
2 2
2
5.3. Solution of the eikonal equation. The eikonal equation in dimension two
when the index of refraction is constant is given by
c2 (ux )2 + (u y )2 = 14,
u(x, y) = ±y + 1 −
1 2 2
where c is a constant. Therefore F(x, y, z, p, q) =
c (p + q2 ) − 1 , (written in this
2
way for convenience in the calculations). The system (5.24) then becomes



ẋ(t) = c2 p




2


 ẏ(t) = c q



ż(t) = c2 (p2 + q2 ) .







ṗ(t) = 0




q̇(t) = 0.
Let us fix an initial curve Γ = ( f (s), g(s), h(s)) and let us complete it to a strip by
adding φ(s) and ψ(s). Then the compatibility conditions (5.25) become
h0 (s) = φ(s) f 0 (s) + ψ(s)g0 (s),
φ(s)2 + ψ(s)2 = c−2 .
2
This imply that h0 (s)2 = (φ(s), ψ(s)) · ( f 0 (s), g0 (s)) ≤ |(φ(s), ψ(s))|2 |( f 0 (s), g0 (s))|2 ≤
c−2 f 0 (s)2 + g0 (s)2 . That is, there are no real solutions φ and ψ if
h0 (s)2 > c−2 f 0 (s)2 + g0 (s)2 .
4
u(x, y) = constant represent the ”wave fronts”, that is, if t represents time, then the collection
of all (x, y) such that u(x, y) = t is the wave front. The orthogonal trajectories to these level curves
are the trajectories of the light rays. In the general case, c2 is replaced by a function 1/n(x, y) where
n(x, y) is the refractive index of the media at the point (x, y).
FIRST ORDER PDES August 9, 2015
23
So if h0 (s)2 ≤ c−2 f 0 (s)2 + g0 (s)2 , then there are two pairs of solutions φ(s), ψ(s),
(the line intersects the circle a2 + b2 = c−2 in two points).
Suppose the curve Γ lies on the plane x, y, that is, Γ = ( f (s), g(s), 0). Clearly
in this case, the compatibility condition has two pairs of solutions (φ, ψ) and
(−φ, −ψ). Solving the system of odes yields
x(s, t) = f (s) + c2 tφ(s),
p(s, t) = φ(s),
y(s, t) = g(s) + c2 tψ(s),
z(s, t) = t,
q(s, t) = ψ(s).
Condition (5.26) reads in this case
 2

c φ(s) f 0 (s)


 = c2 φ(s)g0 (s) − ψ(s) f 0 (s) .
det 

 2

0
c ψ(s) g (s)
From the compatibility conditions 0 = φ(s) f 0 (s) + ψ(s)g0 (s), φ(s)2 + ψ(s)2 = c−2 . If
we assume that f 0 (s)2 + g0 (s)2 > 0, then φ(s)g0 (s) − ψ(s) f 0 (s) , 0. Therefore the map
(s, t) 7→ ( f (s) + c2 tφ(s), g(s) + c2 tψ(s)) is invertible and the solution u can be found
with the inverse Φ−1 of this map setting u(x, y) = z(Φ−1 (x, y)).
5.4. Higher dimensional case. Now with the same method we can solve the
equation
F(x, u, Du(x)) = 0.
We write p = Du(x) and z = u and the Cauchy problem now is to find an ndimensional surface z = u(x) passing through an (n − 1)-dimensional manifold Γ
parameterized by
Γ(s1 , · · · , sn−1 ) = f1 (s1 , · · · , sn−1 ), · · · , fn (s1 , · · · , sn−1 ), h(s1 , · · · , sn−1 )
with Γ(s01 , · · · , s0n−1 ) = (x01 , · · · , x0n , z0 ) = P0 (we set s0 = (s01 , · · · , s0n−1 ), s = (s1 , · · · , sn−1 ),
and x0 = (x01 , · · · , x0n )). We then need to construct n-functions
φ1 (s1 , · · · , sn−1 ), · · · , φn (s1 , · · · , sn−1 ),
compatible with the curve Γ and the equation. That is, in analogy with (5.25), we
need
n
∂h X ∂ f j
=
φj
,
i = 1, · · · , n;
∂si
∂si
j=1
and
F( f1 , · · · , fn , h, φ1 , · · · , φn ) = 0.
24
C. E. GUTIÉRREZ, AUGUST 9, 2015
To solve this system of equation we assume as before the existence of a point
p0 = (p01 , · · · , p0n ) such that
n
X
∂ fj
∂h
(s0 ) =
p0j
(s0 ),
∂si
∂si
j=1
i = 1, · · · , n;
and
F( f1 (s0 ), · · · , fn (s0 ), h(s0 ), p01 , · · · , p0n ) = 0,
and in analogy with condition (5.26), we also assume


F (x , , z0 , p ) ∂ f1 (s ) · · · ∂ f1 (s )
0
0
0 
 p1 0

∂s1
∂sn−1




..
..
..
det 
 , 0.
.
.
.




F (x , , z0 , p ) ∂ fn (s ) · · · ∂ fn (s )
pn 0
0
0
0
∂s1
∂sn−1
Then as in the three dimensional case, by the implicit function theorem we obtain
the desired functions φ1 (s1 , · · · , sn−1 ), · · · , φn (s1 , · · · , sn−1 ). With these functions as
initial conditions, we solve the system of odes (5.21), (5.22) and (5.23) obtaining
the solutions
x(s, t), z(s, t), p(s, t)
such that (x(s, 0), z(s, 0), p(s, 0)) = ( f1 (s), · · · , fn (s), h(s), φ(s), · · · , φn (s)). Now the
map (s, t) 7→ x = x(s, t) has an inverse Φ−1 (x) = (s, t) and the solution u is given by
u(x) = z(Φ−1 (x)).
6. A transversality problem
Suppose that in a domain of R3 we have a C1 vector field
(x, y, z) 7→ F(x, y, z) = (a(x, y, z), b(x, y, z), c(x, y, z)).
We seek for a C2 surface in R3 given by u(x, y, z) = 0 such that at each point the
normal to this surface is parallel to (a, b, c), that is, the surface is transversal to the
vector field at each point. That is, there is function λ(x, y, z) such that
(ux (x, y, z), u y (x, y, z), uz (x, y, z)) = λ(x, y, z)(a(x, y, z), b(x, y, z), c(x, y, z)).
We shall prove that the existence of such a surface implies that
F · (D × F) = 0.
FIRST ORDER PDES August 9, 2015
25
Indeed, we have



ux = λa




u y = λb





uz = λc
.
Since u is C2 , uxy = u yx , uxz = uzx , uzy = u yz , and using the equations we get



λ(a y − bx ) = λx b − λ y a




λ(bz − c y ) = λ y c − λz b





λ(az − cx ) = λx c − λz a
.
Multiplying the first equation by c, the second by a, the third by −b and adding
the resulting equations we obtain
λc(a y − bx ) + λa(bz − c y ) − λb(az − cx ) = 0,
and we are done.
7. The Legendre transform
We work in dimension three, the extension to other dimensions is straighforward. A surface S in 3d can be described by a function z = u(x, y). Suppose u is
differentiable in a domain D ⊂ R2 , that is, the surface has a tangent plane at each
point (x, y) ∈ D having equation, in the variables x0 , y0 , z0 ,
z0 = u(x, y) + ux (x, y)(x0 − x) + u y (x, y)(y0 − y),
that is
(7.32)
− z0 + ux (x, y)x0 + u y (x, y)y0 = −u(x, y) + ux (x, y)x + u y (x, y)y.
This means that we can describe the surface S be prescribing the tangent plane
at each point, and think of the surface as the envelope of the family of tangent
planes, that is, prescribing ux (x, y), u y (x, y) and −u(x, y) + ux (x, y)x + u y (x, y)y, we
get the equation of the tangent plane at each point by means of (7.32). Suppose
that ∇u(x, y) = (ux (x, y), u y (x, y)) maps D into Ω and is bijective. Then there is a
inverse map Φ : Ω → D with
Φ(ξ, η) = (φ(ξ, η), ψ(ξ, η)),
26
C. E. GUTIÉRREZ, AUGUST 9, 2015
and so ux (φ(ξ, η), ψ(ξ, η)) = ξ and u y (φ(ξ, η), ψ(ξ, η)) = η. Suppose in addition,
that the map Φ is differentiable in the domain Ω.5
Let us define the function
(7.33)
w(ξ, η) = −u(x, y) + xξ + yη, with x = φ(ξ, η),
y = ψ(ξ, η).
The differentiating w with respect to ξ yields
wξ (ξ, η) = −ux (φ(ξ, η), ψ(ξ, η))φξ (ξ, η) − u y (φ(ξ, η), ψ(ξ, η))ψξ (ξ, η)
+ φξ (ξ, η)ξ + φ(ξ, η) + ψξ (ξ, η)η
= −ux (φ(ξ, η), ψ(ξ, η))φξ (ξ, η) − u y (φ(ξ, η), ψ(ξ, η))ψξ (ξ, η)
+ φξ (ξ, η)ξ + φ(ξ, η) + ψξ (ξ, η)η
= −ξφξ (ξ, η) − ηψξ (ξ, η)
+ φξ (ξ, η)ξ + φ(ξ, η) + ψξ (ξ, η)η = φ(ξ, η).
Analogously, differentiating w with respect to η yields
wη (ξ, η) = ψ(ξ, η).
Therefore the gradient of the function w defined by (7.33) is the inverse of the
gradient of ∇u, i.e.,
∇u(∇w(ξ, η)) = (ξ, η); and ∇w(∇u(x, y)) = (x, y),
(7.34)
for (ξ, η) ∈ Ω and (x, y) ∈ D.
On the other hand, it we have a function w(ξ, η), differentiable in Ω with gradient
that is bijective from Ω to D, and with inverse map from D to Ω also differentiable
in D, then the function u defined by (7.33) satisfies (7.34).
If we assume that u ∈ C2 (D), and the Jacobian of the map ∇u(x, y) = (ux (x, y), u y (x, y))
is not zero at a point (x̄, ȳ) ∈ D, then by the inverse function theorem there exist
open sets U and V such that ∇u is an homomorphism from U onto V, with U
neighborhood of (x̄, ȳ) and V neighborhood of ∇u(x̄, ȳ). In addition, the inverse
map Φ(ξ, η) = (φ(ξ, η), ψ(ξ, η)) is differentiable in V and satisfies
Φ0 (∇u(x, y)) (∇u)0 (x, y) = Id,
that is,
"
5
#"
# "
#
φξ (∇u(x, y)) φη (∇u(x, y)) uxx (x, y) uxy (x, y)
1 0
=
,
ψξ (∇u(x, y)) ψη (∇u(x, y)) uxy (x, y) u yy (x, y)
0 1
Of course, by the inverse function theorem if the Jacobian of ∇u is not zero in D, then ∇u can
be locally inverted. But this requieres existence of second derivatives of u, which we do not want
to assume.
FIRST ORDER PDES August 9, 2015
27
for all (x, y) ∈ U. Since φ = wξ and ψ = wη , we obtain the formula
"
#"
# "
#
wξξ (∇u(x, y)) wξη (∇u(x, y)) uxx (x, y) uxy (x, y)
1 0
=
.
wξη (∇u(x, y)) wηη (∇u(x, y)) uxy (x, y) u yy (x, y)
0 1
Therefore
#−1
# "
"
uxx (x, y) uxy (x, y)
wξξ (∇u(x, y)) wξη (∇u(x, y))
=
uxy (x, y) u yy (x, y)
wξη (∇u(x, y)) wηη (∇u(x, y))
#
"
u yy (x, y) −uxy (x, y)
1
=
,
det D2 u(x, y) −uxy (x, y) uxx (x, y)
for all (x, y) ∈ U.
References
[Car82] C. Carathéodory. Calculus of variations and partial differential equations of the first order.
Originally published as Variationsrechnung und partialle differentialgleichungen erster
ordnung, by B. G. Teubner, Berlin, 1935. AMS Chelsea Publishing, 3rd edition, 1982.
[CH62] R. Courant and D. Hilbert. Methods of Mathematical Physics. John Wiley & Sons, New York,
1962.
[Joh82] Fritz John. Partial differential equations and applications. Springer-Verlag, 4th edition, 1982.
[Sne06] Ian N. Sneddon. Elements of partial differential equations. Dover, 2006.
Department of Mathematics, Temple University, Philadelphia, PA 19122
E-mail address: gutierre@temple.edu