5.1
Second-Order linear PDE
Consider a second-order linear PDE
L[u] = auxx + 2buxy + cuyy + dux + euy + f u = g,
(x, y) ∈ U
(5.1)
for an unknown function u of two variables x and y. The functions a, b and
c are assumed to be of class C 1 and satisfying a2 + b2 + c2 6= 0. The operator
L0 [u] := auxx + 2buxy + cuyy
consisting of the second order terms of L is called the principal part of L.
Many of the fundamental properties of the solutions of (5.1) are determined
by the sign of the discriminant of L. The discriminant ∆(L)(x, y) is defined
by
b a
∆(L)(x, y) = det
= b2 − ac
c b
where a, b and c are evaluated at the point (x, y).
We are interested in how the PDE is transformed under changes of coordinates. We consider a C 1 - map F (x, y) = (ξ(x, y), η(x, y)) whose Jacobian
satisfies
ξx ξy
6= 0
detJ(x, y) = det
ηx ηy
at each point of (x0 , y0 ) ∈ U . The inverse function theorem implies that near
the point (x0 , y0 ) the map F has an inverse F −1 (ξ, η) = (x(ξ, η), y(ξ, η)).
The inverse is of class C 1 . Now, assuming that u is a solution of (5.1),
define w(ξ, η) = u(x(ξ, η), y(ξ, η)). Then u(x, y) = w(ξ(x, y), η(x, y)) and,
by the chain rule,
ux = wξ ξx + wη ηx
uy = wξ ξy + wη ηy
uxx = wξξ (ξx )2 + 2wξη ξx ηx + wηη (ηx )2 + wξ ξxx + wη ηxx
uyy = wξξ (ξy )2 + 2wξη ξy ηy + wηη (ηy )2 + wξ ξyy + wη ηyy
uxy = uyx = wξξ ξx ξy + wξη (ξx ηy + ηx ξy ) + wηη ξx ηy + wξ ξxy + wη ηxy
Substituting into (5.1), we find that
e
L[w]
= Awξξ + 2Bwξη + Cwηη + Dwξ + Ewη + F w = G
32
(5.2)
e0 [w] = Awξξ + 2Bwξη + Cwηη
with the coefficients of the principal part L
given by
A(ξ, η) = aξx2 + 2bξx ξy + cξy2
B(ξ, η) = aξx ηx + b(ξx ηy + ξy ηx ) + cξy ηy
C(ξ, η) =
aηx2
+ 2bηx ηy +
(5.3)
cηy2
Observe that
t
A B
ξ ξ
a b
ξ η
= x y ·
· x x ,
B C
ηy ηy
b c
ξy ηy
where t denotes the transpose of the matrix. Recalling that the determinant
of the product of matrices is equal to the product of the determinants of
matrices and that the determinant of a transpose of a matrix is equal to the
determinant of a matrix, we get
a b
A B
· (J(x, y))2 .
(5.4)
= det
det
b c
B A
This shows that the discriminant of L has the same sign as the discriminant
of the transformed equation and so it is an invariant of the change of coordinates. Consequently, we can classify equations (5.1) according to the sign
of the discriminant.
Definition 5.1. The equation (5.1) is called
• hyperbolic at (x, y) if ∆(L)(x, y) > 0.
• parabolic at (x, y) if ∆(L)(x, y) = 0.
• elliptic at (x, y) if ∆(L)(x, y) < 0.
Example 5.2.
tion.
(i) The wave equation utt − uxx = 0 is a hyperbolic equa-
(ii) The heat equation ut − uxx = 0 is a parabolic equation.
(iii) The Laplace equation uxx = uyy = 0 is an elliptic equation.
Example 5.3. Consider the Tricomi equation
yuxx + uyy = 0.
(5.5)
Here a = y, b = 0, c = 1 and d = e = f = g = 0. Its discriminant is equal to
b a
0 y
det
= det
= −y.
c b
1 0
33
Hence the equation (5.5) is hyperbolic for y < 0, parabolic when y = 0, and
elliptic for y > 0.
Next we shall show that we can find changes of coordinates in which the
(5.1) takes a simple form.
• Hyperbolic equations.
Suppose that equation (5.1) is hyperbolic on the domain U . This means
that b2 − ac > 0 at each point of U . We shall show that in this we can
choose (x, y) 7→ (ξ(x, y), η(x, y)) so that
A(ξ, η) = aξx2 + 2bξx ξy + cξy2 = 0
(5.6)
C(ξ, η) = aηx2 + 2bηx ηy + cηy2 = 0.
(5.7)
Under such a change of coordinates and dividing by 2B the hyperbolic equation (5.1) takes its canonical form
e
L[w]
= wξη + ℓ[w] = G,
(5.8)
where ℓ is a first-order linear operator and G is function.
Note that if a and c are equal to 0, then the equation (5.1) is already in its
canonical form (just divide by 2b). Hence without loss of generality we may
assume that a 6= 0. Note also that the equation (5.7) for η is the same as
(5.6) for ξ. Hence it suffices to consider only one of the equations, say (5.6).
It can be written as a product
# "
#
"
√
√
−b + b2 − ac
−b − b2 − ac
ξy · ξx −
ξy = 0
a ξx −
a
a
and so, we need to solve the following two linear equations
ξx − µ1 ξy = 0
(5.9)
ξx − µ2 ξy = 0,
(5.10)
and
where we have abbreviated
√
−b − b2 − ac
µ1 =
a
and µ2 =
−b −
√
b2 − ac
.
a
Note that µ1 and µ2 are the real solutions of the equation
aµ2 + 2bµ + c = 0.
34
(5.11)
In order to obtain a nonsingular map (x, y) 7→ (ξ(x, y), η(x, y)), we
choose ξ to be the solution of (5.9) and η to be the solution of (5.10).
To solve (5.9), we use the method of characteristics (except that we don’t
specify the initial condition). The characteristic equations are
dx
= 1,
dt
dy
= −µ1 ,
dt
dz
= 0.
dt
The last equation says that the solution ξ is constant along each of the characteristics (x(t), y(t)). In view of the first two equations, the characteristics
can be obtain as curves y = y(x) solving
dy
=
dx
dy
dt
dx
dt
= −µ1 ,
(5.12)
and then the solution ξ is constant at points (x, y(x)). Similarly, one solves
(5.10) to obtain η.
In summary, to choose ξ and η one solves the (5.11) to obtain two real
roots µ1 and µ2 . Then, denoting by
f (x, y) = C1
and
g(x, y) = C2
the solutions of characteristics equations
dy
= −µ1 (x, y
dx
and
dy
= −µ2 (x, y),
dx
(5.13)
the variables ξ and η are defined by
ξ(x, y) = f (x, y),
η(x, y) = g(x, y).
The solutions of both equations in (5.13) are called the two families of characteristics of (5.1).
Example 5.4. Consider
yuxx + uyy = 0
In the region where y < 0, the equation is hyperbolic. Solving yµ2 + 1 = 0,
one finds two real solutions
µ1 = −
1
(−y)1/2
and µ2 =
We look for two real families of characteristics,
dy
1
=0
−
dx (−y)1/2
and
35
1
(−y)1/2
dy
dx
+ µ1 = 0 and
dy
1
= 0.
+
dx (−y)1/2
dy
dx
+ µ2 = 0,
The solutions of the equations are
2
(−y)3/2 + x = C1
3
2
− (−y)3/2 + x = C2 .
3
and
Therefore, we set
2
ξ = (−y)3/2 + x
3
and
2
η = − (−y)3/2 + x.
3
The derivatives of ξ and η are,
ξx = 1,
ξy = −(−y)1/2 ,
ηx = 1,
ηy = (−y)1/2 .
With u(x, y) = v(ξ(x, y), η(x, y), one gets
ux = vξ + vη
uy = −(−y)1/2 vξ + (−y)1/2 vη
uxx = vξξ + 2vξη + vηη
1
uyy = −yvξξ + 2yvξη − yvηη + (−y)−1/2 [vξ − vη ].
2
Substituting into the equation, one obtains
0 = yuxx + uyy
1
1
−1/2
−3/2
[vξ − vη ] = 4y vξη − (−y)
(vξ − vη ) .
= 4yvξη + (−y)
2
8
Since ξ − η = 43 (−y)3/2 , one concludes that
vξη −
1
(vξ − vη ) = 0.
6(ξ − η)
• Parabolic equations.
Suppose that (5.1) is parabolic on the domain U . Hence b2 − ac = 0 at
each point of U . As before assume that a 6= 0 on U . We find a map
(x, y) 7→ (ξx, y), η(x, y)) so that B(ξ, η) = A(ξ, η) = 0. It suffices to make
A = 0 since 0 = B 2 − AC = B 2 implies that B(ξ, η) = 0. Under such
a change of coordinates the parabolic equation (5.1) can be brought to its
canonical form
e
L[w]
= wξξ + ℓ[w] = G(ξ, η)
where ℓ is a first-order linear operator and G is function.
36
To do this we look for (x, y) 7→ ξ(x, y) so that
A(ξ, η) = aξx2 + 2bξx ξy + cξy2 = 0.
Since b2 = ac, we have
b
b2
aξx2 + 2bξx ξy + cξy2 = a ξx2 + 2 ξx ξy + 2 ξy2 = a (ξx − µξy )2 ,
a
a
where µ = − ab is the double root of
aµ2 + 2bµ + c = 0.
So, we look for the solution ξ of the first-order linear equation,
ξx − µξy = 0.
The solution ξ is constant along each characteristic which is determined by
the equation
b
dy
= −µ = .
(5.14)
dx
a
For the map (x, y) 7→ η(x, y) we can take any map so that ξx ξ − ξy ξx 6= 0.
In summary, in the parabolic case to choose ξ and η one solves the (5.11)
to obtain double root µ = −b/a. Then, denoting by
f (x, y) = C
the solution of characteristics equation
dy
= −µ
dx
(5.15)
the variable ξ is defined by
ξ(x, y) = f (x, y)
and the variable η is chosen so that ξx ξ − ξy ξx 6= 0.
Example 5.5. Reduce the following equation to its canonical form and then
find the general solution,
x2 uxx − 2xyuxy + y 2 uyy + xux + yuy = 0
for x > 0.
37
The discriminant is equal to
−xy x2
=0
det
y 2 −xy
so that the equation is parabolic. The quadratic equation x2 µ2 −2xyµ+y 2 =
0 has exactly one solution, µ = xy . Next we look for characteristics. These
are solutions of the equation
dy
= µ,
dx
i.e.,
dy
y
=− .
dx
x
The family of solutions is given by xy = C. Therefore, we define ξ(x, y) = xy
and we take as the second independent variable η(x, y) = x. Then
ξx = y,
ξy = x,
ηx = 1,
ηy = 0.
The Jacobian of the map (x, y) 7→ (ξ(x, y), η(x, y)) is nonzero. Let v(ξ, η) =
u(x(ξ, η), y(ξ, η)), that is, u(x, y) = v(ξ(x, y), η(x, y)). Using the chain rule,
ux = yvξ + vη
uy = xvξ
uxx = y 2 vξξ + 2yvξη + vηη
uxy = xyvξξ + xvξη + vξ
uyy = x2 vξξ .
Substituting into the equation, one gets
x2 vηη + xvη = 0,
so that, using x = η,
1
vηη + vη = 0.
η
To solve this equation introduce the function w = vη . Then
1
wη = − w
η
which has the general solution w = η1 A(ξ). So,
vη =
1
A(ξ)
η
38
which after integration gives
v(ξ, η) = A(ξ) ln η + B(ξ).
Since ξ(x, y) = xy and η(x, y) = x and u(x, y) = v(ξ(x, y), η(x, y)), the
general solution of the equation has the form
u(x, y) = A(xy) ln x + B(xy)
where A, B are two arbitrary functions of class C 2 .
• Elliptic equations.
Suppose that (5.1) is elliptic on the domain U . Then b2 − ac < 0 at each
point of U . This time we look for the map (x, y) 7→ (xξ(x, y), η(x, y)) so
that
A(ξ, η) = aξx2 + 2bξx ξy + cξy2 = C(ξ, η) = aηx2 + 2bηx ηy + cηy2
B(ξ, η) = aξx ηx + b(ξx ηy + ξy ηx ) + cξy ηy = 0.
Under such a change of coordinates and dividing by A, the elliptic equation (5.1) can be brought to its canonical form
e
L[w]
= wξξ + wηη + ℓ[w] = G(ξ, η)
where ℓ is a first-order linear operator.
The above system consists of two nonlinear first-order equations. Subtracting C(ξ, η) form A(ξ, η) and multiplying B(ξ, η) by 2i leads to the
following system,
a(ξx2 − ηx2 ) + 2b(ξx ξy − ηx ηy ) + c(ξy2 − ηy2 ) = 0
aξx (2iηx ) + b(ξx (2iηy ) + ξy (2iηx )) + cξy (2iηy ) = 0.
Consequently, setting φ = ξ + iη, one finds that the above system is equivalent to
aφ2x + 2bφx φy + cφ2y = 0.
This can be written as a product,
a [φx − µ1 φy ] · [φx − µ2 φy ] = 0,
where we have abbreviated
√
−b − i ac − b2
µ1 =
a
and
39
√
−b − i ac − b2
µ2 =
.
a
These are complex roots of
aµ2 + 2bµ + c = 0.
Note that µ1 and µ2 are conjugated, i.e., µ1 = µ2 . As in the hyperbolic case
we solve the characteristics equations
dy
= −µ1
dx
and
dy
= −µ2 .
dx
(5.16)
This time the solutions are complex. If f (x, y) = C1 and g(x, y) = C2 are
complex solutions of (5.16), then
φ(x, y) = f (x, y)
Then set
ξ=
1
(φ + ψ)
2
and ψ(x, y) = g(x, y).
and η =
1
(φ − ψ).
2i
Example 5.6. Consider
yuxx + uyy = 0
In the region where y > 0, the equation is elliptic. Solving yµ2 + 1 = 0, one
finds two complex solutions
µ1 =
i
y 1/2
and
µ2 = −
i
y 1/2
dy
+µ1 = 0 and
We look for two complex families of characteristics, dx
0,
dy
dy
i
i
+ 1/2 = 0 and
− 1/2 = 0.
dx y
dx y
The solutions of the equations are
2 3/2
y + ix = C1
3
and
2 3/2
y − ix = C2 .
3
and
2
ψ = y 3/2 − x,
3
Therefore, we set
2
φ = y 3/2 + x
3
and then
1
2
ξ = (φ + ψ) = y 3/2
2
3
and
40
η=
1
(φ − ψ) = x.
2i
dy
dx +µ2
=
The derivatives of ξ and η are,
ξx = 0,
ξy = y 1/2 ,
ηx = 1,
With u(x, y) = v(ξ(x, y), η(x, y), one gets
ux = vη
uy = y 1/2 vξ
uxx = vηη
1
uyy = yvξξ + y −1/2 vξ .
2
Substituting into the equation, one obtains
vξξ + vηη +
1
vξ = 0.
2y 3/2
Finally, since ξ = 23 y 3/2 , the equation becomes
3
vξξ + vηη + vξ = 0.
ξ
41
ηy = 0.