2The Cauchy–Kovalevskaya theorem
III Analysis of Partial Differential Equations
2.2 Reduction to first-order systems
In nature, very few equations come in the form required by the Cauchy–
Kovalevskaya theorem, but it turns out a lot of PDEs can be cast into this form
after some work. We shall demonstrate this via an example.
Example. Consider the problem
u
tt
= uu
xy
− u
xx
+ u
t
u|
t=0
= u
0
u
t
|
t=0
= u
1
,
where u
0
, u
1
are some real analytic functions near the origin. We define
f = u
0
+ tu
1
.
This is then real analytic near 0, and f|
t=0
= u
0
and f
t
|
t=0
= u
1
. Set
w = u − f.
Then w satisfies
w
tt
= ww
xy
− w
xx
+ w
t
+ fw
xy
+ f
xy
w + F,
where
F = ff
xy
− f
xx
+ f
t
,
and
w|
t=0
= w
t
|
t=0
= 0.
We let (
x, y, t
) = (
x
1
, x
2
, x
3
) and set
u
= (
w, w
x
, w
y
, w
t
). Then our PDE becomes
u
1
t
= w
t
= u
4
u
2
t
= w
xt
= u
4
x
u
3
t
= w
yt
= u
4
y
u
4
t
= w
tt
= u
1
u
2
x
2
− u
2
x
1
+ u
4
+ fu
2
x
2
+ f
xy
u
1
+ F,
and the initial condition is
u
(
x
1
, x
2
,
0) = 0. This is not quite autonomous, but
we can solve that problem simply by introducing a further new variable.
Let’s try to understand this in more generality. In certain cases, it is not
possible to write the equation in Cauchy–Kovalevskaya form. For example, if
the equation has no local solutions, then it certainly cannot be written in that
form, or else Cauchy–Kovalevskaya would give us a solution! It is thus helpful
to understand when this is possible.
Note that in the formulation of Cauchy–Kovalevskaya, the derivative
u
x
n
is
assumed to depend only on
x
0
, and not
x
n
. If we want
u
x
n
to depend on
x
n
as
well, we can introduce a new variable
u
n+1
and set (
u
n+1
)
x
n
= 1. So from now
on, we shall ignore the fact that our PDE only has x
0
on the right-hand side.
Let’s now consider the scalar quasi-linear problem
X
|α|=k
a
α
(D
k−1
u, . . . , Du, u, x)D
α
u + a
0
(D
k−1
u, . . . , u, x) = 0,
where u : B
r
(0) ⊆ R
n
→ R, with initial data
u =
∂u
∂x
n
= ··· =
∂
k−1
u
∂x
k−1
n
= 0.
whenever |x
0
| < r, x
n
= 0.
We introduce a new vector
u =
u,
∂u
∂x
1
, . . . ,
∂u
∂x
n
,
∂
2
u
∂x
2
1
, . . . ,
∂
n−1
u
∂x
k−1
n
= (u
1
, . . . , u
m
)
Here
u
contains all partial derivatives of
u
up to order
k−
1, for
j ∈ {
1
, . . . , m−
1
}
,
we can compute
∂u
j
∂x
n
in terms of u
`
or
∂u
`
∂x
p
for some ` ∈ {1, . . . , m} and p < n.
To express
∂u
m
∂x
n
in terms of the other variables, we need to actually use
the differential equation. To do so, we need to make an assumption about our
equation. We suppose that
a
(0,...,0,k)
(0
,
0) is non-zero. We can then rewrite the
equation as
∂
k
u
∂x
k
n
=
−1
a
(0,...,0,k)
(D
k−1
u, . . . , u, x)
X
|α|=k,α
n
<k
a
α
D
α
u + a
0
,
where at least near
x
= 0, the denominator can’t vanish. The RHS can then be
written in terms of
∂u
k
∂x
p
and ub for p < n.
So we have cast our original equation into the form we previously discussed,
provided that the
a
α
’s and
a
0
’s are real analytic about the origin, and that
a
(0,...,0,k)
(0
, . . . ,
0) = 0. Under these assumptions, we can solve the equation by
Cauchy–Kovalevskaya.
It is convenient to make the following definition: if
a
(0,...,k)
(0
, . . . ,
0)
6
= 0, we
say {x
n
= 0} is non-characteristic. Otherwise, we say it is characteristic.
Often times, we want to specify our initial data on some more exotic surface.
Unfortunately, they cannot be too exotic. They have to be real analytic in some
sense for our theory to have any chance of working.
Definition
(Real analytic hypersurface)
.
We say that Σ
⊆ R
n
is a real analytic
hypersurface near
x ∈
Σ if there exists
ε >
0 and a real analytic map Φ :
B
ε
(
x
)
→
U ⊆ R
n
, where U = Φ(B
ε
(x)), such that
– Φ is bijective and Φ
−1
: U → B
ε
(x) is real analytic.
– Φ(Σ ∩B
ε
(x)) = {x
n
= 0} ∩ U and Φ(x) = 0.
We think of this Φ as “straightening out the boundary”.
Let γ be the unit normal to Σ, and suppose u solves
X
|α|=k
a
α
(D
k−1
u, . . . , u, x)D
α
u + a
0
(D
k−1
u, . . . , u, x) = 0
subject to
u = γ
i
∂
i
u = ··· , (γ
i
∂
i
)
k−1
u = 0
on Σ.
To do so, we define w(y) = u(Φ
−1
(y)), so that
u(x) = w(Φ(x)).
Then by the chain rule, we have
∂u
∂x
i
=
n
X
j=1
∂w
∂y
i
∂ψ
j
∂x
i
.
So plugging this into the equation, we see w satisfies an equation of the form
X
b
α
D
α
w + b
0
= 0,
as well as boundary conditions of
w =
∂w
∂y
n
= ··· =
∂
k−1
w
∂y
k−1
n
= 0.
So we have transformed this to a quasi-linear equation with boundary conditions
on
y
n
= 0, which we can tackle with Cauchy–Kovalevskaya, provided the surface
y
n
= 0 is non-characteristic. Can we relate this back to the a’s?
We can compute b
(0,...,0,k)
directly. Note that if |α| = k, then
D
α
u =
∂
k
w
∂y
k
n
(DΦ
n
)
α
+ terms not involving
∂
n
w
∂y
k
n
.
So the coefficient of
∂
k
w
∂y
k
n
is
b
(0,...,k)
X
|α|=k
a
α
(DΦ
n
)
α
.
Definition
((Non-)characteristic surface)
.
A surface Σ is non-characteristic at
xΣ provided
X
|α|=k
a
α
(DΦ
n
)
α
6= 0.
Equivalently, if
X
|α|=k
a
α
ν
α
6= 0,
where
ν
is the normal to the surface. We say a surface is characteristic if it is
not non-characteristic.
We focus on the case where our PDE is second-order. Consider an operator
of the form.
Lu =
n
X
i,j=1
a
ij
∂
i
u
∂x
i
∂x
j
where
a
ij
∈ R
. We may wlog assume
a
ij
=
a
ji
. For example the wave equation
and Laplace’s equation are given by operators of this form. Consider the equation
Lu = f
u = v
i
∂u
∂x
i
= 0 on Π
ν
= {x · ν = 0}.
Then Π
ν
is non-characteristic if
n
X
i,j
a
ij
ν
i
ν
j
6= 0.
Since
a
ij
is diagonalizable, we see that if all eigenvalues are positive, then
P
a
ij
ν
i
ν
j
is non-zero, and so the problem has no characteristic surfaces. In this
case, we say the operator is elliptic. If (
a
ij
) has one negative eigenvalue and the
rest positive, then we say L is hyperbolic.
Example. If L is the Laplacian
L = ∆ =
n
X
i=1
∂
2
∂x
2
i
,
then L is elliptic.
If L is the wave operator
L = −∂
2
t
+ ∆,
then L is hyperbolic.
If we consider the problem
Lu = 0,
and forget the Cauchy data, we can look for solutions of the form
e
ik·x
, as a
good physicist would do. We can plug this into our operator to compute
L(e
ik·x
) = −
n
X
i,j=1
a
ij
k
i
k
j
e
ik·x
.
So if
L
is elliptic, the only solution of this form is
k
= 0. If
L
is hyperbolic, we
can have non-trivial plane wave solutions provided k ∝ ν for some ν with
n
X
i,j=1
a
ij
ν
i
ν
j
= 0.
So if we set
u
λ
(
x
) =
e
iλν·x
for such a
ν
(with
|ν|
= 1, wlog). By taking
λ
very large, we can arrange this solution to have very large derivative in the
ν
direction. Vaguely, this says the characteristic directions are the directions where
singularities can propagate. By contrast, we will see that this is not the case for
elliptic operators, and this is known as elliptic regularity. In fact, we will show
that if L is elliptic and u satisfies Lu = 0, then u ∈ C
∞
.
While Cauchy–Kovalevskaya is sometimes useful, it has a few issues:
– Not all functions are real analytic.
– We have no control over “how long” a solution exists.
– It doesn’t answer the question of well-posedness.
Indeed, consider the PDE
u
xx
+ u
yy
= 0.
This admits a solution
u(x, y) = cos kx cosh ky
for some
k ∈ R
. We can think of this coming as coming from the Cauchy problem
u(x, 0) = cos kx, u
y
(x, 0) = 0.
By Cauchy–Kovalevskaya, there is a unique real analytic solution, and we’ve
found one. So this is the unique solution.
Let’s think about what happens when
k
gets large. In this case, it seems
like nothing is very wrong with the initial data. While the initial data oscillates
more and more, it is still bounded by 1. However, we see that the solution at
any
y
=
ε >
0 grows exponentially. We might say that the derivatives of the
initial condition grows to infinity as well, but if we do a bit more work (as you
will on the example sheet), we can construct a sequence of initial data all of
whose derivatives tend to 0, but the solution still blows up.
This is actually a serious problem. If we want to solve the PDE for a more
general initial condition, we may want to decompose the initial data into Fourier
modes, and then integrate up these solutions we found. But we cannot do this
in general, if these solutions blow up as k → ∞.