MODULE 3: SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS 7

Lecture 2 Canonical Forms or Normal Forms

By a suitable change of the independent variables we shall show that any equation of the
form
Auxx + Buxy + Cuyy + Dux + Euy + F u + G = 0, (1)

where A, B, C, D, E, F and G are functions of the variables x and y, can be reduced to a

canonical form or normal form. The transformed equation assumes a simple form so that
the subsequent analysis of solving the equation will be become easy.

Consider the transformation of the indpendent variables from (x, y) to (ξ, η) given by

ξ = ξ(x, y), η = η(x, y). (2)

Here, the functions ξ and η are continuously differentiable and the Jacobian
 
∂(ξ, η)  ξx ξy 
J= =  = (ξx ηy − ξy ηx ) ̸= 0 (3)
∂(x, y)  ηx ηy 

in the domain where (1) holds.

Using chain rule, we notice that

ux = uξ ξx + uη ηx
uy = uξ ξy + uη ηy
uxx = uξξ ξx2 + 2uξη ξx ηx + uηη ηx2 + uξ ξxx + uη ηxx
uxy = uξξ ξx ξy + uξη (ξx ηy + ξy ηx ) + uηη ηx ηy + uξ ξxy + uη ηxy
uyy = uξξ ξy2 + 2uξη ξy ηy + uηη ηy2 + uξ ξyy + uη ηyy

Substituting these expression into (1), we obtain

Ā(ξx , ξy )uξξ + B̄(ξx , ξy ; ηx , ηy )uξη + C̄(ηx , ηy )uηη = F (ξ, η, u(ξ, η), uξ (ξ, η), uη (ξ, η)), (4)

where

Ā(ξx , ξy ) = Aξx2 + Bξx ξy + Cξy2

B̄(ξx , ξy ; ηx , ηy ) = 2Aξx ηx + B(ξx ηy + ξy ηx ) + 2Cξy ηy
C̄(ηx , ηy ) = Aηx2 + Bηx ηy + Cηy2 .

An easy calculation shows that

B̄ 2 − 4ĀC̄ = (ξx ηy − ξy ηx )2 (B 2 − 4AC). (5)

MODULE 3: SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS 8

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:
Case I: B 2 − 4AC > 0 (Hyperbolic type)
Case II: B2 − 4AC = 0 (Parabolic type)
Case III: B2 − 4AC < 0 (Elliptic type)
Case I: Note that B 2 − 4AC > 0 implies the equation Aα2 + Bα + C = 0 has two real
and distinct roots, say λ1 and λ2 . Now, choose ξ and η such that
∂ξ ∂ξ ∂η ∂η
= λ1 and = λ2 . (6)
∂x ∂y ∂x ∂y
Then the coefficients of uξξ and uηη will be zero because

Ā = Aξx2 + Bξx ξy + Cξy2 = (Aλ21 + Bλ1 + C)ξy2 = 0,

C̄ = Aηx2 + Bηx ηy + Cηy2 = (Aλ22 + Bλ2 + C)ηy2 = 0.

Thus, (5) reduces to

B̄ 2 = (B 2 − AC)(ξx ηy − ξy ηx )2 > 0
as B 2 −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
+ λi (x, y) = 0, i = 1, 2. (7)
dx
Let the family of curves determined by the solution of (7) for i = 1 and i = 2 be

f1 (x, y) = c1 and f2 (x, y) = c2 , (8)

respectively. These family of curves are called characteristics curves of PDE (5). With
this choice, divide (4) throughout by B̄ (as B̄ > 0) and use (7)-(8) to obtain
∂2u
= ϕ(ξ, η, u, uξ , uη ), (9)
∂ξ∂η
which is the canonical form of hyperbolic equation.
EXAMPLE 1. Reduce the equation uxx = x2 uyy to its canonical form.

Solution. Comparing with (1) we find that A = 1, B = 0, C = −x2 .

The roots of the equations Aα2 + Bα + C = 0 i.e., α2 + x2 = 0 are given by λi = ±x.

The differential equations for the family of characteristics curves are
dy
± x = 0.
dx
MODULE 3: SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS 9

whose solutions are y + 12 x2 = c1 and y − 12 x2 = c2 . Choose

1 1
ξ = y + x2 , η = y − x2 .
2 2
An easy computation shows that

ux = uξ ξx + uη ηx ,
uxx = uξξ ξx2 + 2uξη ξx ηx + uηη ηx2 + uξ ξxx + uη ηxx
= uξξ x2 − 2uξη x2 + uηη x2 + uξ − uη ,
uyy = uξξ ξy2 + 2uξη ξy ηy + uηη ηy2 + uξ ξyy + uη ηyy ,
= uξξ + 2uξη + uηη .

Substituting these expression in the equation uxx = x2 uyy yields

4x2 uξη = (uξ − uη )

1
or 4(ξ − η)uξη = (uξ − uη )
4(ξ − η)
1
or uξη = (uξ − uη )
4(ξ − η)

which is the required canonical form.

CASE II: B 2 − 4AC = 0 =⇒ the equation Aα2 + Bα + C = 0 has two equal roots, say
dy
λ1 = λ2 = λ. Let f1 (x, y) = c1 be the solution of dx + λ(x, y) = 0. Take ξ = f1 (x, y) and
η to be the any function of x and y which is independent of ξ.

As before, Ā(ξx , ξy ) = 0 and hence from equation (5), we obtain B̄ = 0. Note that
C̄(ηx , ηy ) ̸= 0, otherwise η would be a function of ξ. Dividing (4) by C̄, the canonical form
of (2) is
uηη = ϕ(ξ, η, u, uξ , uη ). (10)

which is the canonical form of parabolic equation.

EXAMPLE 2. Reduce the equation uxx + 2uxy + uyy = 0 to canonical form.

Solution. In this case, A = 1, B = 2, C = 1. The equation α2 + 2α + 1 = 0 has equal

dy
roots λ = −1. The solution of dx − 1 = 0 is x − y = c1 Take ξ = x − y. Choose η = x + y.
proceed as in Example 1 to obtain uηη = 0 which is the canonical form of the given PDE.

CASE III: When B 2 − 4AC < 0, the roots of Aα2 + Bα + C = 0 are complex. Following
the procedure as in CASE I, we find that

uξη = ϕ1 (ξ, η, u, uξ , uη ). (11)

MODULE 3: SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS 10

The variables ξ, η are infact complex conjugates. To get a real canonical form use the
transformation
1 1
α = (ξ + η), β = (ξ − η)
2 2i
to obtain
1
uξη = (uαα + uββ ), (12)
4
which follows from the following calculation:
1 1
uξ = uα αξ + uβ βξ = uα + uβ
2 2i
1 1
uξη = (uαα αη + uαβ βη ) + (uβα αη + uββ βη )
2 2i
1
= (uαα + uββ ).
4
The desired canonical form is

uαα + uββ = ψ(α, β, u(α, β), uα (α, β), uβ (α, β)). (13)

EXAMPLE 3. Reduce the equation uxx + x2 uyy = 0 to canonical form.

Solution. In this case, A = 1, B = 0, C = x2 . The roots are λ1 = ix, λ2 = −ix.

Take ξ = iy + 12 x2 , η = −iy + 12 x2 . Then α = 21 x2 , β = y. The canonical form is
1
uαα + uββ = − uα .
2α

Practice Problems

1. Reduce the following equations to canonical/normal form:

(A) 2uxx − 4uxy + 2uyy + 3u = 0.

(B) uxx + yuyy = 0.
(C) uxy + ux + uy = 2x.

2. Show that the equation

uxx − 6uxy + 12uyy + 4ux − u = sin(xy)

is of elliptic type and obtain its canonical form.

3. Determine the regions where Tricomi’s equation uxx + xuyy = 0 is of elliptic,

parabolic, and hyperbolic types. Obtain its characteristics and its canonical form in
the hyperbolic region.