Chapter 3.

Second Order Linear PDEs

3.1 Introduction
The general class of second order linear PDEs are of the form:

a(x, y)u xx + b(x, y)u xy + c(x, y)uyy

+ d(x, y)u x + e(x, y)uy + f (x, y)u = g(x, y). (3.1)

The three PDEs that lie at the cornerstone of applied mathematics are: the
heat equation, the wave equation and Laplace’s equation,i.e.

(i) ut = u xx , the heat equation

(ii) utt = u xx , the wave equation
(iii) u xx + uyy = 0, Laplace’s equation

or, using the same independent variables, x and y

(i) u xx − uy = 0, the heat equation (3.3a)

(ii) u xx − uyy = 0, the wave equation (3.3b)
(iii) u xx + uyy = 0. Laplace’s equation (3.3c)

Analogous to characterizing quadratic equations

ax2 + bxy + cy2 + dx + ey + f = 0,

as either hyperbolic, parabolic or elliptic determined by

b2 − 4ac > 0, hyperbolic,

b2 − 4ac = 0, parabolic,
b2 − 4ac < 0, elliptic,
2 Chapter 3. Linear Second Order Equations

we do the same for PDEs. So, for the heat equation a = 1, b = 0, c = 0

so b2 − 4ac = 0 and so the heat equation is parabolic. Similarly, the wave
equation is hyperbolic and Laplace’s equation is elliptic. This leads to a
natural question. Is it possible to transform one PDE to another where the
new PDE is simpler? Namely, under a change of variable

r = r(x, y), s = s(x, y),

can we transform to one of the following canonical forms:

urr − uss + l.o.t.s. = 0, hyperbolic, (3.5a)

uss + l.o.t.s. = 0, parabolic, (3.5b)
urr + uss + l.o.t.s. = 0, elliptic, (3.5c)

where the term “l.o.t.s” stands for lower order terms. For example, con-
sider the PDE
2u xx − 2u xy + 5uyy = 0. (3.6)

This equation is elliptic since the elliptic b2 − 4ac = 4 − 40 = −36 < 0. If

we introduce new coordinates,

r = 2x + y, s = x − y,

then by a change of variable using the chain rule

u xx = urr r2x + 2urs r x s x + uss s2x + ur r xx + us s xx ,

u xy = urr r x ry + urs (r x sy + ry s x ) + uss s x sy + ur r xy + us s xy ,
uyy = urr ry2 + 2urs ry sy + uss s2y + ur ryy + us syy ,

gives

u xx = 4urr + 4urs + uss ,

u xy = 2urr − urs − uss ,
uyy = urr − 2urs + uss .
 3.1. Introduction 3

Under (3.7), equation (3.6) becomes

urr + uss = 0,

which is Laplace’s equation (also elliptic). Before we consider transforma-

tions for PDEs in general, it is important to determine whether the equa-
tion type could change under transformation. Consider the general class
of PDEs
au xx + bu xy + cuyy = 0 (3.7)

where a, b, and c are functions of x and y and noting that we have sup-
pressed the lower terms as they will not affect the type. Under a change of
variable (x, y) → (r, s) with the change of variable formulas (3.7) gives
³ ´
a urr r2x + 2urs r x s x + uss s2x + ur r xx + us s xx
¡ ¢
+ b urr r x ry + urs (r x sy + ry s x ) + uss s x sy + ur r xy + us s xy (3.8)
³ ´
+ c uyy + urr ry2 + 2urs ry sy + uss s2y + ur ryy + us syy = 0

Rearranging (3.8), and again neglecting lower order terms, gives

(ar2x + br x ry + cry2 )urr + (2ar x s x + b(r x sy + ry s x ) + 2cry sy )urs (3.9)

+ (as2x + bs x sy + cs2y )uss = 0.

Setting

A = ar2x + br x ry + cr2y ,
B = 2ar x s x + b(r x sy + ry s x ) + 2cry sy , (3.10)
C = as2x + bs x sy + cs2y ,

gives (again suppressing lower order terms)

Aurr + Burs + Cuss = 0,

whose type is given by

¡ ¢2
B2 − 4AC = (b2 − 4ac) r x sy − ry s x ,
4 Chapter 3. Linear Second Order Equations

from which we deduce that

b2 − 4ac > 0, ⇒ B2 − 4AC > 0,

b2 − 4ac = 0, ⇒ B2 − 4AC = 0,
b2 − 4ac < 0, ⇒ B2 − 4AC < 0,

giving that the equation type is unchanged under transformation. We now

consider transformations to canonical form. As there are three types of
canonical forms, hyperbolic, parabolic and elliptic, we will deal with each
type separately.

3.2 Canonical Forms

If we introduce the change of coordinates

r = r(x, y), s = s(x, y), (3.11)

the derivatives change according to:

First Order
u x = ur r x + u s s x , u y = ur r y + u s s y , (3.12)

Second Order

u xx = urr r2x + 2urs r x s x + uss s2x + ur r xx + us s xx ,

u xy = urr r x ry + urs (r x sy + ry s x ) + uss s x sy + ur r xy + us s xy , (3.13)
uyy = urr ry2 + 2urs ry sy + uss s2y + ur ryy + us syy ,

If we substitute (3.12) and (3.13) into the general linear equation (3.1) and
re-arrange we obtain
¡ ¢
(ar2x + br x ry + cry2 )urr + 2ar x s x + b(r x sy + ry s x ) + 2cry sy urs
+ (as2x + bs x sy + cs2y )uss + l.o.t.s. = 0. (3.14)

Our goal now is to target a given canonical form and solve a set of equa-
tions for the new coordinates r and s.
 3.2. Canonical Forms 5

3.2.1 Parabolic Canonical Form

Comparing (3.14) with the parabolic canonical form (3.5b) leads to choos-
ing
ar2x + br x ry + cr2y = 0, (3.15a)
2ar x s x + b(r x sy + ry s x ) + 2cry sy = 0, (3.15b)

b2
Since in the parabolic case b2 − 4ac = 0, then substituting c = 4a we find
both equations of (3.15) are satisfied if

2ar x + bry = 0. (3.16)

with the choice of s(x, y) arbitrary. The following examples demonstrate.

Example 1.
Consider
u xx + 6u xy + 9uyy = 0. (3.17)

Here, a = 1, b = 6 and c = 9 showing that b2 − 4ac = 0, so the PDE is

parabolic. Solving
r x + 3ry = 0,

gives
r = f (3x − y).

As we wish to find new coordinates as to transform the original equation

to canonical form, we choose

r = 3x − y, s = y.

Calculating second derivatives

u xx = 9urr , u xy = −3urr + 3urs , uyy = urr − 2urs + uss . (3.18)

Substituting (3.18) into (3.17) gives

uss = 0!†
† Not to be confused with factorial (!).
6 Chapter 3. Linear Second Order Equations

Solving gives
u = f (r)s + g(r).

where f and g are arbitrary functions. In terms of the original variables,

we obtain the solution

u = y f (3x − y) + g(3x − y).

Example 2.
Consider
x2 u xx − 4xyu xy + 4y2 uyy + xu x = 0. (3.19)

Here, a = x2 , b = −4xy and c = 4y2 showing that b2 − 4ac = 0, so the PDE

is parabolic. Solving
x2 r x − 2xyry = 0,

or
xr x − 2yry = 0,

gives
r = f (x2 y).

As we wish to find new coordinates, i.e. r and s, we choose simple

r = x2 y, s = y.

Calculating first derivatives gives

u x = 2xyur . (3.20)

Calculating second derivatives

u xx = 4x2 y2 urr + 2yur , (3.21a)

u xy = 2x3 yurr + 2xyurs + 2xur , (3.21b)
uyy = x4 urr + 2x2 urs + uss . (3.21c)

Substituting (3.20) and (3.21) into (3.19) gives

4y2 uss − 4x2 yur = 0.

 3.2. Canonical Forms 7

or, in terms of the new variables, r and s,

r
uss − ur = 0. (3.22)
s2
An interesting question is whether different choices of the arbitrary func-
tion f and the variable s would lead to a different canonical forms. For
example, suppose we chose

r = 2 ln x + ln y, s = ln y,

we would obtain
uss − ur − us = 0, (3.23)

a constant coefficient parabolic equation, whereas, choosing

r = 2 ln x + ln y, s = 2 ln x,

we would obtain
uss − ur = 0, (3.24)

the heat equation.

3.2.2 Hyperbolic Canonical Form

In order to obtain the canonical form for the hyperbolic type, i.e.

urr − uss + l.o.t.s. = 0, (3.25)

it is necessary to choose
³ ´
ar2x + br x ry + cry2 = − as2x + bs x sy + cs2y ,
2ar x s x + b(r x sy + ry s x ) + 2cry sy = 0. (3.26)

The problem is that this system is still a very hard problem to solve (both
PDEs are nonlinear and coupled!). Therefore, we introduce a modified hy-
perbolic form that is much easier to work with.
8 Chapter 3. Linear Second Order Equations

3.2.3 Modified Hyperbolic Form

The modified hyperbolic canonical form is defined as

urs + l.o.t.s. = 0, (3.27)

noting that a = 0, b = 1 and c = 0 and that b2 − 4ac > 0 still! In order to

target the modified hyperbolic form, it is now necessary to choose

ar2x + br x ry + cry2 = 0, (3.28a)

as2x + bs x sy + cs2y = 0. (3.28b)

If we re-write (3.28a) and (3.28b) as follows

µ ¶2
rx rx
a + 2b + c = 0, (3.29a)
ry ry
µ ¶2
sx sx
a + 2b + c = 0, (3.29b)
sy sy

rx sx
then we can solve equations (3.29a) and (3.29b) separately for ry and sy .
This leads to two first order linear PDEs for r and s. The solutions of these
then gives rise to the correct canonical variables. The following examples
demonstrate.

Example 3.
Consider
u xx − 5u xy + 6uyy = 0 (3.30)

Here, a = 1, b = −5 and c = 6 showing that b2 − 4ac = 1 > 0, so the PDE

is hyperbolic. Thus, (3.57a) becomes

r2x − 5r x ry + 6ry2 = 0, s2x − 5s x ry + 6s2y = 0,

and factoring gives

¡ ¢¡ ¢ ¡ ¢¡ ¢
r x − 2ry r x − 3ry = 0, s x − 2sy s x − 3sy = 0,
 3.2. Canonical Forms 9

from which we choose

r x − 2ry = 0, s x − 3sy = 0,

giving rise to solutions

r = f (2x + y), s = g(3x + y).

As we wish to find new co-ordinates as to transform the original equation

to canonical form, we choose

r = 2x + y, s = 3x + y.

Calculating second derivatives

u xx = 4urr + 12urs + 9uss ,

u xy = 2urr + 5urs + 3uss , (3.31)
uyy = urr + 2urs + uss .

Substituting (3.31) into (3.30) gives

urs = 0.

Solving gives
u = f (r) + g(s).

where f and g are arbitrary functions. In terms of the original variables,

we obtain the solution

u = f (2x + y) + g(3x + y).

Example 4.
Consider
xu xx − (x + y)u xy + yuyy = 0. (3.32)

Here, a = x, b = −(x + y) and c = y showing that b2 − 4ac = (x − y)2 > 0,

so the PDE is hyperbolic. Solving

xr2x − (x + y)r x ry + yr2y = 0,

10 Chapter 3. Linear Second Order Equations

or, upon factoring

¡ ¢¡ ¢
xr x − yry r x − ry = 0.

As s satisfies the same equation, we choose the first factor for r and the
second for s
xr x − yry = 0, s x − sy = 0. (3.33)

Upon solving (3.33), we obtain

r = f (xy), s = g(x + y).

As we wish to find new co-ordinates, i.e. r and s, we choose simple

r = xy, s = x + y.

Calculating first derivatives gives

u x = yur + us , uy = xur + us . (3.34)

Calculating second derivatives

u xx = y2 urr + 2yurs + uss , (3.35a)

u xy = xyurr + (x + y)urs + uss + ur , (3.35b)
uyy = x2 urr + 2xurs + uss . (3.35c)

Substituting (3.34) and (3.35) into (3.32) gives

³ ´
2
4xy − (x + y) urs − (x + y)ur = 0,

or, in terms of the new variables, r and s,

s
urs + ur = 0.
s2 − 4r

3.2.4 Regular Hyperbolic Form

We now wish to transform a given hyperbolic PDE to its regular canonical
form
urr − uss + l.o.t.s. = 0. (3.36)
 3.2. Canonical Forms 11

First, let us consider the following example.

x2 u xx − y2 uyy = 0. (3.37)

If we were to transform to modified canonical form, we would solve

xr x − yry = 0, xs x + ysy = 0,

which gives
r = f (xy), s = g(x/y).

As we wish to find new co-ordinates, i.e. r and s, we choose simple

r = xy, s = x/y.

In doing so, the original PDE then becomes

1
urs − us = 0. (3.38)
2r

However, if we choose

r = ln x + ln y, s = ln x − ln y,

then the original PDE becomes

urs − us = 0, (3.39)

which is clearly an easier PDE. However, if we introduce new coordinates

α and β such that
r+s r−s
α= , β= ,
2 2
noting that derivatives transform

1 1 1 1 1 1
ur = uα + u β , us = uα − u β urs = uαα − u ββ , (3.40)
2 2 2 2 4 4

and the PDE (3.39) becomes

uαα − u ββ − 2uα + 2u β = 0,
12 Chapter 3. Linear Second Order Equations

a PDE in regular hyperbolic form. Thus, combining the variables r and s

and α and β gives directly

α = ln x, β = ln y.

In fact, one can show that if

r+s r−s
α= , β= ,
2 2

where r and s satisfies (3.28a) and (3.28b) then α and β satisfies

³ ´
aα2x + bα x αy + cα2y = − aβ2x + bβ x β y + cβ2y , (3.41a)
2aα x β x + b(α x β y + αy β x ) + 2cαy β y = 0. (3.41b)

which is (3.58a) with r and s replaces with α and β. This give a convenient
way to go directly to the coordinates that lead to the regular hyperbolic
form. We note that
r±s
α, β = , (3.42)
2
so we can essentially consider

ar2x + br x ry + cry2 = 0, (3.43a)

as2x + bs x sy + cs2y = 0. (3.43b)

but instead of factoring, treat each as a quadratic equation in r x /ry or s x /sy

and solve according. We demonstrate with an example.

Example 5.
Consider
8u xx − 6u xy + uyy = 0. (3.44)

The corresponding equations for r and s are

8r2x − 6r x ry + r2y = 0, (3.45a)

8s2x − 6s x sy + s2y = 0, (3.45b)
 3.2. Canonical Forms 13

but as they are identical it suffices to only consider one. Dividing (3.57a)
by ry2 gives
µ ¶2
rx rx
8 −6 + 1 = 0.
ry ry
Solving by the quadratic formula gives
rx 6±2
= ,
ry 16
or
8r x − (3 ± 1)ry = 0.

The method of characteristics gives

dx dy
=− ; dr = 0.
8 3±1
which gives
r = f ((3 ± 1)x + 8y) ,

which we choose
r = 3x + 8y ± x,

which leads to the chose

r = 3x + 8y, s = x,

Under this transformation, the original equation (3.44) becomes

urr − uss = 0,

the desired canonical form.

Example 6.
Consider

xy3 u xx − x2 y2 u xy − 2x3 yuyy − y2 u x + 2x2 uy = 0. (3.46)

The corresponding equations for r and s are

xy3 r2x − x2 y2 r x ry − 2x3 yr2y = 0, (3.47a)

xy3 s2x − x2 y2 s x ry − 2x3 ys2y = 0, (3.47b)
14 Chapter 3. Linear Second Order Equations

and choosing the first gives

µ ¶2
2 rx rx
y − xy − 2x2 = 0.
ry ry
Solving by the quadratic formula gives
rx (1 ± 3)x
= ,
ry 2y
or
2yr x − (1 ± 3) xry = 0.

Solving gives
³ ´
2 2 2
r = f x + 2y ± 3x .

If we choose f to be simple and split according to the ± gives

r = x2 + 2y2 , s = 3x2 ,

Under this transformation, the original equation (3.46) becomes

urr − uss = 0,

the desired canonical form.

3.2.5 Elliptic Canonical Form

In order to obtain the canonical form for the elliptic type, i.e.

urr + uss + l.o.t.s. = 0,

it is necessary to choose
³ ´
ar2x + br x ry + cry2 = as2x + bs x sy + cs2y ,
2ar x s x + b(r x sy + ry s x ) + 2cry sy = 0. (3.48)

The problem, like the regular hyperbolic type, is still difficult to solve.
However, we find that if we let †
α+β α−β
r= , s= (3.49)
2 2i
† Please note the switch in the variables r and s and α and β.
 3.2. Canonical Forms 15

where α and β satisfy

aα2x + bα x αy + cα2y = 0, (3.50a)

aβ2x + bβ x β y + cβ2y = 0, (3.50b)

then (3.48) is satisfied. This is much like the connection between modi-
fied and regular hyperbolic canonical form. As solving (3.50a) gives rise
to complex roots, the formulas (3.49) will take real and complex parts of
the solved αandβ equations as new variables. The next few examples will
illustrate.

Example 7.
Consider
u xx − 4u xy + 5uyy = 0. (3.51)

The corresponding equations for r and s are

r2x − 4r x ry + 5r2y = 0, (3.52a)

s2x − 4s x sy + 5s2y = 0, (3.52b)

but as they are identical it suffices to only consider one. Dividing (3.52a)
by ry2 gives
µ ¶2
rx rx
−4 + 5 = 0.
ry ry
Solving by the quadratic formula gives
rx
= 2 ± i,
ry
or
r x − (2 ± i)ry = 0.

The method of characteristics gives

dx dy
=− , dr = 0.
1 2±i
which gives
r = f (2x + y ± ix) ,
16 Chapter 3. Linear Second Order Equations

which we choose
r = 2x + y ± x,

which leads to the chose

r = 2x + y, s = x,

Under this transformation, the original equation (3.51) becomes

urr + uss = 0,

the desired canonical form.

Example 8.
Consider
³ ´2 ³ ´³ ´ ³ ´2 ³ ´
2 1 + x2 u xx − 2 1 + x2 1 + y2 u xy + 1 + y2 uyy + 4x 1 + x2 u x = 0.
(3.53)
The corresponding equations for r and s are
³ ´2 ³ ´³ ´ ³ ´2
2 2 2 2 2
2 1+x rx − 2 1 + x 1 + y r x ry + 1 + y r2y = 0, (3.54a)
³ ´2 ³ ´³ ´ ³ ´2
2 2 2 2 2
2 1+x sx − 2 1 + x 1 + y s x sy + 1 + y s2y = 0, (3.54b)

but as they are identical it suffices to only consider one. Solving by the
quadratic formula gives
¡ ¢
rx (2 ± i) 1 + y2
= ,
ry 1 + x2
or
³ ´ ³ ´
2 1 + x2 r x − (1 ± i) 1 + y2 ry = 0.

The method of characteristics gives the solution as

³ ´
−1 −1 −1
r = f tan x + 2 tan y ± tan x ,

which we choose

r = tan−1 x + 2 tan−1 y ± tan−1 x,

 3.2. Canonical Forms 17

which leads to the choice

r = tan−1 x + 2 tan−1 y, s = tan−1 x,

Under this transformation, the original equation (3.53) becomes

urr + uss − 2yur = 0,

and upon using the original transformation gives

r−s
urr + uss − 2 tan ur = 0,
2
the desired canonical form.

Exercises
1. Determine the type of the following second order PDEs

(i) x2 u xx − y2 uyy = u x + uy
(ii) u xx + 2u xy + uyy = 0
(iii) y2 u xx + 2yu xy − uyy = 0
(iv) u xy + u = u x + uy

2. Transform the following parabolic PDEs to canonical form. Find the

general solution if possible.

(i) u xx + 2u xy + uyy = 0
(ii) y2 u xx − 2xyu xy + x2 uyy = 0.
(iii) y2 u xx + 2xyu xy + x2 uyy − 2xu x = 0.

3. Reduce the following second order PDEs to modified hyperbolic canon-

ical form

(i) 2u xx − 3u xy + uyy = u x + uy ,
(ii) x2 u xx − 3xyu xy + 2y2 uyy = 0.
18 Chapter 3. Linear Second Order Equations

4. Reduce the following second order PDEs to canonical form

(i) 4u xx − 8u xy + 3uyy = 0,
(ii) 4u xx + 4u xy + 5uyy = 1,
(iii) x2 u xx + y2 uyy = 1,
(iv) u xx − (1 + y2 )2 uyy = 0.

5. If a linear second order PDE

a(x, y)u xx + 2b(x, y)u xy + c(x, y)uyy + lots = 0.

is hyperbolic, then it is possible to transform to a modified hyperbolic

canonical form
urs + l.o.t.s. = 0,

by choosing r and s such that they satisfy

ar2x + 2br x ry + cry2 = 0,

as2x + 2bs x sy + cs2y = 0.

5(i) Show that by introducing new variables α and β such that

α = r + s, β = r − s,

then the following equations are satisfied

³ ´
aα2x + 2bα x αy + cα2y = − aβ2x + 2bβ x β y + cβ2y ,
aα x β x + b(α x β y + ry β x ) + cαy β y = 0.

This then leads to the hyperbolic canonical form

uαα − u ββ + l.o.t.s. = 0.

5(ii). If, instead of the α and β introduced in 5(i), we introduce α and β

such that
α = c11 r + c12 s, β = c21 r + c22 s,
 3.2. Canonical Forms 19

where c11 , c12 , c21 and c22 are constant and r and s satisfy the above first
order PDEs, namely

ar2x + 2br x ry + cry2 = 0,

as2x + 2bs x sy + cs2y = 0,

find conditions on c11 , c12 , c21 and c22 such that

³ ´
aα2x + 2bα x αy + cα2y = − aβ2x + 2bβ x β y + cβ2y ,
aα x β x + b(α x β y + ry β x ) + cαy β y = 0.

are still satisfied.