Canonical Forms Or Normal Forms Math Worksheet Page 2
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MODULE 3: SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS
The equation (5) shows that the transformation of the independent variables does not
modify the type of PDE.
We shall determine ξ and η so that (4) takes the simplest possible form. We now
consider the following cases:
2
Case I: B
4AC > 0
(Hyperbolic type)
2
Case II: B
4AC = 0
(Parabolic type)
2
Case III: B
4AC < 0
(Elliptic type)
2
2
Case I: Note that B
4AC > 0 implies the equation Aα
+ Bα + C = 0 has two real
and distinct roots, say λ
and λ
. Now, choose ξ and η such that
1
2
∂ξ
∂ξ
∂η
∂η
= λ
and
= λ
.
(6)
1
2
∂x
∂y
∂x
∂y
Then the coefficients of u
and u
will be zero because
ξξ
ηη
¯ A = Aξ
2
2
2
2
+ Bξ
ξ
+ Cξ
= (Aλ
+ Bλ
+ C)ξ
= 0,
x
y
1
x
y
1
y
¯ C = Aη
2
2
2
2
+ Bη
η
+ Cη
= (Aλ
+ Bλ
+ C)η
= 0.
x
y
2
x
y
2
y
Thus, (5) reduces to
¯ B
2
2
2
= (B
AC)(ξ
η
ξ
η
)
> 0
x
y
y
x
2
as B
4AC > 0. Note that (6) is a first-order linear PDE in ξ and η whose characteristics
curves are satisfy the first-order ODEs
dy
+ λ
(x, y) = 0, i = 1, 2.
(7)
i
dx
Let the family of curves determined by the solution of (7) for i = 1 and i = 2 be
f
(x, y) = c
and f
(x, y) = c
,
(8)
1
1
2
2
respectively. These family of curves are called characteristics curves of PDE (5). With
this choice, divide (4) throughout by ¯ B (as ¯ B > 0) and use (7)-(8) to obtain
2
∂
u
= ϕ(ξ, η, u, u
, u
),
(9)
ξ
η
∂ξ∂η
which is the canonical form of hyperbolic equation.
2
E
1. Reduce the equation u
= x
u
to its canonical form.
XAMPLE
xx
yy
2
Solution. Comparing with (1) we find that A = 1, B = 0, C =
x
.
= ±x.
2
2
2
The roots of the equations Aα
+ Bα + C = 0 i.e., α
+ x
= 0 are given by λ
i
The differential equations for the family of characteristics curves are
dy
± x = 0.
dx
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