7More partial differential equations

IB Methods 7.3 Characteristics for 2nd order partial differential equa-
tions
Whether the method of characteristics works for a 2nd order partial differential
equation, or even higher order ones, depends on the “type” of the differential
equation.
To classify these differential equations, we need to define
Definition
(Symbol and principal part)
.
Let
L
be the general 2nd order differ-
ential operator on R
n
. We can write it as
L =
n
X
i,j=1
a
ij
(x)
2
x
i
x
j
+
n
X
i=1
b
i
(x)
x
i
+ c(X),
where a
ij
(x), b
i
(x), c(x) R and a
ij
= a
ji
(wlog).
We define the symbol σ(k, x) of L to be
σ(k, x) =
n
X
i,j=1
a
ij
(x)k
i
k
j
+
n
X
i=1
b
i
(x)k
i
+ c(x).
So we just replace the derivatives by the variable k.
The principal part of the symbol is the leading term
σ
p
(k, x) =
n
X
i,j=1
a
ij
(x)k
i
k
j
.
Example. If L =
2
, then
σ(k, x) = σ
p
(k, x) =
n
X
i=1
(k
i
)
2
.
If
L
is the heat operator (where we think of the last coordinate
x
n
as the time,
and others as space), then the operator is given by
L =
x
n
n1
X
i=1
2
x
i2
.
The symbol is then
σ(k, x) = k
n
n1
X
i=1
(k
i
)
2
,
and the principal part is
σ
p
(k, x) =
n1
X
i=1
(k
i
)
2
.
Note that the symbol is closely related to the Fourier transform of the
differential operator, since both turn differentiation into multiplication. Indeed,
they are equal if the coefficients are constant. However, we define this symbol
for arbitrary differential operators with non-constant coefficients.
In general, for each x, we can write
σ
p
(k, x) = k
T
A(x)k,
where
A
(
x
) has elements
a
ij
(
x
). Recall that a real symmetric matrix (such as
A) has all real eigenvalues. So we define the following:
Definition
(Elliptic, hyperbolic, ultra-hyperbolic and parabolic differential
operators). Let L be a differential operator. We say L is
elliptic at x if all eigenvalues of A(x) have the same sign. Equivalently, if
σ
p
( ·, x) is a definite quadratic form;
hyperbolic at x if all but one eigenvalues of A(x) have the same sign;
ultra-hyperbolic at x if A(x) has more than one eigenvalues of each sign;
parabolic at x if A(x) has a zero eigenvalue, i.e. σ
p
( ·, x) is degenerate.
We say L is elliptic if L is elliptic at all x, and similarly for the other terms.
These names are supposed to remind us of conic sections, and indeed if we
think of k
T
Ak as an equation in k, then we get a conic section.
Example. Let
L = a(x, y)
2
x
2
+ 2b(x, y)
2
x∂y
+ c(x, y)
2
y
2
+ d(x, y)
x
+ e(x, y)
y
+ f(x, y).
Then the principal part of the symbol is
σ
p
(k, x) =
k
x
k
y
a(x, y) b(x, y)
b(x, y) c(x, y)
k
x
k
y
Then
L
is elliptic at
x
if
b
2
ac <
0; hyperbolic if
b
2
ac >
0; and parabolic if
b
2
ac = 0.
Note that since we only have two dimensions and hence only two eigenvalues,
we cannot possibly have an ultra-hyperbolic equation.
Characteristic surfaces
Definition (Characteristic surface). Given a differential operator L, let
f(x
1
, x
2
, ··· , x
n
) = 0
define a surface C R
n
. We say C is characteristic if
n
X
i,j=1
a
ij
(x)
f
x
i
f
x
j
= (f)
T
A(f) = σ
p
(f, x) = 0.
In the case where we only have two dimensions, a characteristic surface is just a
curve.
We see that the characteristic equation restricts what values
f
can take.
Recall that
f
is the normal to the surface
C
. So in general, at any point, we
can find what the normal of the surface should be, and stitch these together to
form the full characteristic surfaces.
For an elliptic operator, all the eigenvalues of
A
have the same sign. So there
are no non-trivial real solutions to this equation (
f
)
T
A
(
f
) = 0. Consequently,
elliptic operators have no real characteristics. So the method of characteristics
would not be of any use when studying, say, Laplace’s equation, at least if we
want to stay in the realm of real numbers.
If
L
is parabolic, we for simplicity assume that
A
has exactly one zero
eigenvector, say
n
, and the other eigenvalues have the same sign. This is the
case when, say, there are just two dimensions. So we have
An
=
n
T
A
=
0
. We
normalize n such that n · n = 1.
For any f , we can always decompose it as
f = n(n · f) + [f n · (n · f)].
This is a relation that is trivially true, since we just add and subtract the same
thing. Note, however, that the first term points along
n
, while the latter term is
orthogonal to n. To save some writing, we adopt the notation
f = f n(n · f ).
So we have
f = n(n · f) + f
.
Then we can compute
(f)
T
A(f) = [n(n · f ) +
f]
T
A[n(n · f) +
f]
= (
f)
T
A(
f).
Then by assumption, (
f
)
T
A
(
f
) is definite. So just as in the elliptic case,
there are no non-trivial solutions. Hence, if
f
defines a characteristic surface,
then
f
= 0. In other words,
f
is parallel to
n
. So at any point, the normal to
a characteristic surface must be
n
, and there is only one possible characteristic.
If
L
is hyperbolic, we assume all but one eigenvalues are positive, and let
λ
be the unique negative eigenvalue. We let
n
be the corresponding unit
eigenvector, where we normalize it such that
n ·n
= 1. We say
f
is characteristic
if
0 = (f)
T
A(f)
= [n(n · f) +
f]
T
A[n(n · f) +
f]
= λ(n · f)
2
+ (
f)
T
A(
f).
Consequently, for this to be a characteristic, we need
n · f = ±
r
(
f)
T
A(
f)
λ
.
So there are two choices for
n · f
, given any
f
. So hyperbolic equations
have two characteristic surfaces through any point.
This is not too helpful in general. However, in the case where we have two
dimensions, we can find the characteristic curves explicitly. Suppose our curve
is given by
f
(
x, y
) = 0. We can write
y
=
y
(
x
). Then since
f
is constant along
each characteristic, by the chain rule, we know
0 =
f
x
+
f
y
dy
dx
.
Hence, we can compute
dy
dx
=
b ±
b
2
ac
a
.
We now see explicitly how the type of the differential equation influences the
number of characteristics — if
b
2
ac >
0, then we obtain two distinct differential
equations and obtain two solutions; if
b
2
ac
= 0, then we only have one equation;
if b
2
ac < 0, then there are no real characteristics.
Example. Consider
2
y
φ xy
2
x
φ = 0
on
R
2
. Then
a
=
xy, b
= 0
, c
= 1. So
b
2
ac
=
xy
. So the type is elliptic if
xy < 0, hyperbolic if xy > 0, and parabolic if xy = 0.
In the regions where it is hyperbolic, we find
b ±
b
2
ac
a
= ±
1
xy
.
Hence the two characteristics are given by
dy
dx
= ±
1
xy
.
This has a solution
1
3
y
3/2
± x
1/2
= c.
We now let
u =
1
3
y
3/2
+ x
1/2
,
v =
1
3
y
3/2
x
1/2
.
Then the equation becomes
2
φ
u∂v
+ lower order terms = 0.
Example. Consider the wave equation
2
φ
t
2
c
2
2
φ
x
2
= 0
on
R
1,1
. Then the equation is hyperbolic everywhere, and the characteristic
curves are
x ± ct
= const. Let’s look for a solution to the wave equation that
obeys
φ(x, 0) = f(x),
t
φ(x, 0) = g(x).
Now put u = x ct, v = x + ct. Then the wave equation becomes
2
u∂v
= 0.
So the general solution to this is
φ(x, t) = G(u) + H(v) = G(x ct) + H(x + ct).
The initial conditions now fix these functions
f(x) = G(x) + H(x), g(x) = cG
0
(x) + cH
0
(x).
Solving these, we find
φ(x, t) =
1
2
[f(x ct) + f (x + ct)] +
1
2c
Z
x+ct
xct
g(y) dy.
This is d’Alembert’s solution to the 1 + 1 dimensional wave equation.
Note that the value of
φ
at any point (
x, t
) is completely determined by
f, g
in the interval [
x ct, x
+
ct
]. This is known as the (past) domain of dependence
of the solution at (
x, t
), written
D
(
x, t
). Similarly, at any time
t
, the initial
data at (
x
0
,
0) only affects the solution within the region
x
0
ct x x
0
+
ct
.
This is the range of influence of data at x
0
, written D
+
(x
0
).
p
D
(p)
B
S
D
+
(S)
We see that disturbances in the wave equation propagate with speed c.