2.3.2
Parabolic
Since ∆
∗
= 0, B
2
− 4AC = 0 and thus
B =
±2
√
A
√
C.
(2.3.2.1)
Clearly we cannot arrange for both A
∗
and C
∗
to be zero, since the characteristic equation
(2.3.1.8) can have only one solution. That means that parabolic equations have only one
characteristic curve. Suppose we choose the solution φ
1
(x, y) of (2.3.1.8)
dy
dx
=
B
2A
(2.3.2.2)
to define
ξ = φ
1
(x, y).
(2.3.2.3)
Therefore A
∗
= 0.
Using (2.3.2.1) we can show that
0 = A
∗
= Aξ
2
x
+ Bξ
x
ξ
y
+ Cξ
2
y
= Aξ
2
x
+ 2
√
A
√
Cξ
x
ξ
y
+ Cξ
2
y
=
√
Aξ
x
+
√
Cξ
y
2
(2.3.2.4)
It is also easy to see that
B
∗
= 2Aξ
x
η
x
+ B(ξ
x
η
y
+ ξ
y
η
x
) + 2Cξ
y
η
y
= 2(
√
Aξ
x
+
√
Cξ
y
)(
√
Aη
x
+
√
Cη
y
)
= 0
22
The last step is a result of (2.3.2.4). Therefore A
∗
= B
∗
= 0. To obtain the canonical form
we must choose a function η(x, y). This can be taken judiciously as long as we ensure that
the Jacobian is not zero.
The canonical form is then
C
∗
u
ηη
= H
∗
and after dividing by C
∗
(which cannot be zero) we have
u
ηη
=
H
∗
C
∗
.
(2.3.2.5)
If we choose η = φ
1
(x, y) instead of (2.3.2.3), we will have C
∗
= 0. In this case B
∗
= 0
because the last factor
√
Aη
x
+
√
Cη
y
is zero. The canonical form in this case is
u
ξξ
=
H
∗
A
∗
(2.3.2.6)
Example
x
2
u
xx
− 2xyu
xy
+ y
2
u
yy
= e
x
(2.3.2.7)
A = x
2
B =
−2xy
C = y
2
∆ = (
−2xy)
2
− 4 · x
2
· y
2
= 4x
2
y
2
− 4x
2
y
2
= 0.
Thus the equation is parabolic for all x, y. The characteristic equation (2.3.2.2) is
dy
dx
=
−2xy
2x
2
=
−
y
x
.
(2.3.2.8)
Solve
dy
y
=
−
dx
x
ln y + ln x = C
In figure 7 we sketch the family of characteristics for (2.3.2.7). Note that since the problem
is parabolic, there is ONLY one family.
Therefore we can take ξ to be this family
ξ = ln y + ln x
(2.3.2.9)
and η is arbitrary as long as J
= 0. We take
η = x.
(2.3.2.10)
23
that is α and β are the real and imaginary parts of ξ. Clearly η is the complex conjugate
of ξ since the coefficients of the characteristic equation are real. If we use these functions
α(x, y) and β(x, y) we get an equation for which
B
∗∗
= 0,
A
∗∗
= C
∗∗
.
(2.3.3.3)
To show that (2.3.3.3) is correct, recall that our choice of ξ, η led to A
∗
= C
∗
= 0. These
are
A
∗
= (Aα
2
x
+Bα
x
α
y
+Cα
2
y
)
−(Aβ
2
x
+Bβ
x
β
y
+Cβ
2
y
)+i[2Aα
x
β
x
+B(α
x
β
y
+α
y
β
x
)+2Cα
y
β
y
] = 0
C
∗
= (Aα
2
x
+Bα
x
α
y
+Cα
2
y
)
−(Aβ
2
x
+Bβ
x
β
y
+Cβ
2
y
)
−i[2Aα
x
β
x
+B(α
x
β
y
+α
y
β
x
)+2Cα
y
β
y
] = 0
Note the similarity of the terms in each bracket to those in (2.3.1.12)-(2.3.1.14)
A
∗
= (A
∗∗
− C
∗∗
) + iB
∗∗
= 0
C
∗
= (A
∗∗
− C
∗∗
)
− iB
∗∗
= 0
where the double starred coefficients are given as in (2.3.1.12)-(2.3.1.14) except that α, β
replace ξ, η correspondingly. These last equations can be satisfied if and only if (2.3.3.3) is
satisfied.
Therefore
A
∗∗
u
αα
+ A
∗∗
u
ββ
= H
∗∗
(α, β, u, u
α
, u
β
)
and the canonical form is
u
αα
+ u
ββ
=
H
∗∗
A
∗∗
.
(2.3.3.4)
Example
e
x
u
xx
+ e
y
u
yy
= u
(2.3.3.5)
A = e
x
B = 0
C = e
y
∆ = 0
2
− 4e
x
e
y
< 0,
for all x, y
The characteristic equation
dy
dx
=
0
±
√
−4e
x
e
y
2e
x
=
±2i
√
e
x
e
y
2e
x
=
±i
e
y
e
x
dy
e
y/2
=
±i
dx
e
x/2
.
Therefore
ξ =
−2e
−y/2
− 2ie
−x/2
η =
−2e
−y/2
+ 2ie
−x/2
25
The real and imaginary parts are:
α =
−2e
−y/2
(2.3.3.6)
β =
−2e
−x/2
.
(2.3.3.7)
Evaluate all necessary partial derivatives of α, β
α
x
= 0,
α
y
= e
−y/2
,
α
xx
= 0,
α
xy
= 0,
α
yy
=
−
1
2
e
−y/2
β
x
= e
−x/2
,
β
y
= 0,
β
xx
=
−
1
2
e
−x/2
,
β
xy
= 0,
β
yy
= 0
Now, instead of using both transformations, we recall that (2.3.1.12)-(2.3.1.18) are valid with
α, β instead of ξ, η. Thus
A
∗
= e
x
· 0 + 0 + e
y
e
−y/2
2
= 1
B
∗
= 0 + 0 + 0 = 0
as can be expected
C
∗
= e
x
e
−x/2
2
+ 0 + 0 = 1
as can be expected
D
∗
= 0 + 0 + e
y
−
1
2
e
−y/2
=
−
1
2
e
y/2
E
∗
= e
x
−
1
2
e
−x/2
+ 0 + 0 =
−
1
2
e
x/2
F
∗
=
−1
H
∗
=
−D
∗
u
α
− E
∗
u
β
− F
∗
u =
1
2
e
y/2
u
α
+
1
2
e
x/2
u
β
+ u.
Thus
u
αα
+ u
ββ
=
1
2
e
y/2
u
α
+
1
2
e
x/2
u
β
+ u.
Using (2.3.3.6)-(2.3.3.7) we have
e
x/2
=
−
2
β
e
y/2
=
−
2
α
and therefore the canonical form is
u
αα
+ u
ββ
=
−
1
α
u
α
−
1
β
u
β
+ u.
(2.3.3.8)
26
Problems
1.
Find the characteristic equation, characteristic curves and obtain a canonical form for
each
a.
x u
xx
+ u
yy
= x
2
b.
u
xx
+ u
xy
− xu
yy
= 0
(x
≤ 0, all y)
c.
x
2
u
xx
+ 2xyu
xy
+ y
2
u
yy
+ xyu
x
+ y
2
u
y
= 0
d.
u
xx
+ xu
yy
= 0
e.
u
xx
+ y
2
u
yy
= y
f.
sin
2
xu
xx
+ sin 2xu
xy
+ cos
2
xu
yy
= x
2. Use Maple to plot the families of characteristic curves for each of the above.
27
2.4
Equations with Constant Coefficients
In this case the discriminant is constant and thus the type of the equation is the same
everywhere in the domain. The characteristic equation is easy to integrate.
2.4.1
Hyperbolic
The characteristic equation is
dy
dx
=
B
±
√
∆
2A
.
(2.4.1.1)
Thus
dy =
B
±
√
∆
2A
dx
and integration yields two families of straight lines
ξ = y
−
B +
√
∆
2A
x
(2.4.1.2)
η = y
−
B
−
√
∆
2A
x.
(2.4.1.3)
Notice that if A = 0 then (2.4.1.1) is not valid. In this case we recall that (2.4.1.2) is
Bζ
x
ζ
y
+ Cζ
2
y
= 0
(2.4.1.4)
If we divide by ζ
2
y
as before we get
B
ζ
x
ζ
y
+ C = 0
(2.4.1.5)
which is only linear and thus we get only one characteristic family. To overcome this difficulty
we divide (2.4.1.4) by ζ
2
x
to get
B
ζ
y
ζ
x
+ C
ζ
y
ζ
x
2
= 0
(2.4.1.6)
which is quadratic. Now
ζ
y
ζ
x
=
−
dx
dy
and so
dx
dy
=
B
±
√
B
2
− 4 · 0 · C
2C
=
B
± B
2C
or
dx
dy
= 0,
dx
dy
=
B
C
.
(2.4.1.7)
The transformation is then
ξ = x,
(2.4.1.8)
η = x
−
B
C
y.
(2.4.1.9)
The canonical form is similar to (2.3.1.14).
28
2.4.2
Parabolic
The only solution of (2.4.1.1) is
dy
dx
=
B
2A
.
Thus
ξ = y
−
B
2A
x.
(2.4.2.1)
Again η is chosen judiciously but in such a way that the Jacobian of the transformation is
not zero.
Can A be zero in this case? In the parabolic case A = 0 implies B = 0 (since ∆ = B
2
−4·0·C
must be zero.) Therefore the original equation is
Cu
yy
+ Du
x
+ Eu
y
+ F u = G
which is already in canonical form
u
yy
=
−
D
C
u
x
−
E
C
u
y
−
F
C
u +
G
C
.
(2.4.2.2)
2.4.3
Elliptic
Now we have complex conjugate functions ξ, η
ξ = y
−
B + i
√
−∆
2A
x,
(2.4.3.1)
η = y
−
B
− i
√
−∆
2A
x.
(2.4.3.2)
Therefore
α = y
−
B
2A
x,
(2.4.3.3)
β =
−
√
−∆
2A
x.
(2.4.3.4)
(Note that
−∆ > 0 and the radical yields a real number.) The canonical form is similar to
(2.3.3.4).
Example
u
tt
− c
2
u
xx
= 0
(wave equation)
(2.4.3.5)
A = 1
B = 0
C =
−c
2
∆ = 4c
2
> 0
(hyperbolic).
29
The characteristic equation is
dx
dt
2
− c
2
= 0
and the transformation is
ξ = x + ct,
(2.4.3.6)
η = x
− ct.
(2.4.3.7)
The canonical form can be obtained as in the previous examples
u
ξη
= 0.
(2.4.3.8)
This is exactly the example from Chapter 1 for which we had
u(ξ, η) = F (ξ) + G(η).
(2.4.3.9)
The solution in terms of x, t is then (use (2.4.3.6)-(2.4.3.7))
u(x, t) = F (x + ct) + G(x
− ct).
(2.4.3.10)
30
Problems
1. Find the characteristic equation, characteristic curves and obtain a canonical form for
a.
4u
xx
+ 5u
xy
+ u
yy
+ u
x
+ u
y
= 2
b.
u
xx
+ u
xy
+ u
yy
+ u
x
= 0
c.
3u
xx
+ 10u
xy
+ 3u
yy
= x + 1
d.
u
xx
+ 2u
xy
+ 3u
yy
+ 4u
x
+ 5u
y
+ u = e
x
e.
2u
xx
− 4u
xy
+ 2u
yy
+ 3u = 0
f.
u
xx
+ 5u
xy
+ 4u
yy
+ 7u
y
= sin x
2. Use Maple to plot the families of characteristic curves for each of the above.
31
2.5
Linear Systems
In general, linear systems can be written in the form:
∂u
∂t
+ A
∂u
∂x
+ B
∂u
∂y
+ r = 0
(2.5.1)
where u is a vector valued function of t, x, y.
The system is called hyperbolic at a point (t, x) if the eigenvalues of A are all real and
distinct. Similarly at a point (t, y) if the eigenvalues of B are real and distinct.
Example The system of equations
v
t
= cw
x
(2.5.2)
w
t
= cv
x
(2.5.3)
can be written in matrix form as
∂u
∂t
+ A
∂u
∂x
= 0
(2.5.4)
where
u =
v
w
(2.5.5)
and
A =
0
−c
−c 0
.
(2.5.6)
The eigenvalues of A are given by
λ
2
− c
2
= 0
(2.5.7)
or λ = c,
−c. Therefore the system is hyperbolic, which we knew in advance since the system
is the familiar wave equation.
Example The system of equations
u
x
= v
y
(2.5.8)
u
y
=
−v
x
(2.5.9)
can be written in matrix form
∂w
∂x
+ A
∂w
∂y
= 0
(2.5.10)
where
w =
u
v
(2.5.11)
and
A =
0
−1
1
0
.
(2.5.12)
The eigenvalues of A are given by
λ
2
+ 1 = 0
(2.5.13)
or λ = i,
−i. Therefore the system is elliptic. In fact, this system is the same as Laplace’s
equation.
32
2.6
General Solution
As we mentioned earlier, sometimes we can get the general solution of an equation by trans-
forming it to a canonical form. We have seen one example (namely the wave equation) in
the last section.
Example
x
2
u
xx
+ 2xyu
xy
+ y
2
u
yy
= 0.
(2.6.1)
Show that the canonical form is
u
ηη
= 0
for y
= 0
(2.6.2)
u
xx
= 0
for y = 0.
(2.6.3)
To solve (2.6.2) we integrate with respect to η twice (ξ is fixed) to get
u(ξ, η) = ηF (ξ) + G(ξ).
(2.6.4)
Since the transformation to canonical form is
ξ =
y
x
η = y
(arbitrary choice for η)
(2.6.5)
then
u(x, y) = yF
y
x
+ G
y
x
.
(2.6.6)
Example
Obtain the general solution for
4u
xx
+ 5u
xy
+ u
yy
+ u
x
+ u
y
= 2.
(2.6.7)
(This example is taken from Myint-U and Debnath (19 ).) There is a mistake in their solution
which we have corrected here. The transformation
ξ = y
− x,
η = y
−
x
4
,
(2.6.8)
leads to the canonical form
u
ξη
=
1
3
u
η
−
8
9
.
(2.6.9)
Let v = u
η
then (2.6.9) can be written as
v
ξ
=
1
3
v
−
8
9
(2.6.10)
which is a first order linear ODE (assuming η is fixed.) Therefore
v =
8
3
+ e
ξ/3
φ(η).
(2.6.11)
33
Now integrating with respect to η yields
u(ξ, η) =
8
3
η + G(η)e
ξ/3
+ F (ξ).
(2.6.12)
In terms of x, y the solution is
u(x, y) =
8
3
y
−
x
4
+ G
y
−
x
4
e
(y−x)/3
+ F (y
− x).
(2.6.13)
34
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