2.1
Linearisation
In this section you will learn to

understand the technique of linearisation

linearise a nonlinear system using a change of variable

linearise a nonlinear system using Taylor’s series.
2.1.1
The technique
The technique of linearisation involves approximating a complicated
system of equations with a simpler linear system. We hope to gain insight
into the behaviour of the nonlinear system through an analysis of the
behaviour of its linearisation. We hope that the nonlinear system will
behave locally like its linearisation, at least in a qualitative sense.
In general, the linearisation of a system of equations about an equilibrium
point can be achieved by changing variables so that the equilibrium point
is transformed to the origin. Points in the original system close to the
equilibrium point will correspond to points close to the origin in the new
system. Thus we are only concerned with values of the new variables
close to zero and under certain conditions the nonlinear terms can be
neglected. The equations that result are linear and are the linearisation of
the original system. In the examples that you will meet the nonlinear
terms will be polynomial and are small enough to be neglected.
Consider the nonlinear system
x  f ( x , y )
y  g ( x , y )
with equilibrium point (p,q).
The transformation
u x p
v  yq
transforms the equilibrium point
bp, qgto the origin
Differentiating gives
u  x
v  y
Substituting for x and y in the original equations gives
u  au  bv  F (u, v )
v  cu  dv  G (u, v )
where F(u,v) and G(u,v) consist only of nonlinear terms.
Then the linear system
u  au  bv
v  cu  dv
is said to be the linearisation of the nonlinear system provided that
F ( u, v )
u v
2
2
 0, and
G ( u, v )
u v
2
2
 0, as u 2  v 2  0
bg bg
These last conditions ensure that the nonlinear terms F u, v and G u, v
are negligible in comparison to u and v as the equilibrium point is
approached.
bg
In the examples that you will meet the nonlinear terms F u, v and
bg
G u, v will be polynomials with all terms of degree two or higher.
Let
bg
F u, v  au 2  buv  cv 2 
Then
lim
F au  bv  cv
lim G
b g b gHu  v
u v
u v
Fau buv  cv IJ
 lim G 
b g b gHu
K
u
v
 lim b
au  bv  cv g
b gb g
b g
F u, v
bx , y gb0,0g u 2  v 2
2
x , y  0,0
2
2
2
2
2
x , y  0,0
2
2
2
2
2
2
2
2
x , y  0,0
0
bg
Therefore if F u, v consists of nonlinear polynomial terms it can be
neglected for small values of u and v.
Worked Example 1
Consider the nonlinear system
x  f ( x , y )  2 x  y
bg
y  g x , y   y  x 2
To find the equilibrium point.
2x+y=0
y + x2 = 0
The solutions of these equations are x = 0 or 2,
y = 0 or 4
bg bg
Therefore the equilibrium points are 0,0 and 2,4
IJ
K

bg
At the point 0,0
For the equilibrium point (0,0) no transformation is necessary as it is
already at the origin and the equations remain as
x.  2 x  y
y   y  x 2
Since the nonlinear terms are polynomials they can be neglected
Hence the linearisation equations are
x  2 x  y
y   y
bg
At the point 2,4
For the equilibrium point (2,4) use the substitution
x  u2
y  v4
to move the equilibrium point to the origin.
The equations become
u  2(u  2)  v  4  2u  v
v  (v  4)  (u  2) 2  4u  v  u2
Neglecting the nonlinear terms since they are polynomial gives the
linearisation
u  2u  v
v  4u  v
Examples 1
For each of the following systems
a) find the equilibrium points
b) state the transformation necessary to move the equilibrium point to
the origin
c) find the linearisation of the system
1) x  y
2)
y  x 2  y 2
y   x  x 3
2.1.2
x  y 2  3x  2
Linearisation by Taylor Series
Consider the nonlinear system
x  f ( x , y )
y  g ( x , y )
bg
with equilibrium point p, q
Any function which is differentiable can be written as a Taylor series as
follows
f ( x , y )  f ( p, q )  ( x  p)
f
x
 ( y  q)
( p ,q )
f
y
 F ( x, y)
( p ,q )
b g
where F ( x, y) consists of nonlinear polynomial terms in x  p and
by  qg.
Therefore
x  f ( x , y )  f ( p, q )  ( x  p)
y  g ( x , y )  g ( p, q )  ( x  p)
If we use the change of variable
f
x
 ( y  q)
( p ,q )
f
y
g
g
 ( y  q)
x ( p ,q )
y
 F ( x, y)
( p ,q )
 G( x, y)
( p ,q )
u x p
y  vq
this will transform the equilibrium point to the origin.
Differentiating gives
u  x
v  y
bg
Since p, q is an equilibrium
bgbg
f p, q  g p, q  0
bx  pgand by  qgare small
and the nonlinear terms F b
can be neglected.
x , y gand Gb
x, yg
and for points near to the equilibrium point
Substituting in the original equations gives
u  au  bv
v  cu  dv
where
a
f
f
g
g
,b 
,c 
,and d 
x
y
x
y
This can be written in the form
u OL
a bO
uO
L
L
M P
M
P
M
v QN
c dQ
vP
N
N
Q
where J=
a bO
L
M
c dP
N
Qis called the Jacobian matrix.
Here is a worked example for you to look at. There is a parallel Maple
version on the internet.
Worked Example 2
Consider the set of equations
bg
y  gb
x, yg
 y  x
x  f x , y  2 x  y
2
Find the equilibrium points
The equilibrium points are given by
2 x  y  0
 y  x2  0
Solving these equations gives x  0 or 2, y  0 or 4
Therefore the equilibrium points are (0,0) and (2,4)
Find the Jacobian matrix
f
 2 ,
x
f
 1,
y
J
bg
g
 2x ,
x
2 1 O
L
M
2 x 1P
N
Q
At the point 0,0 the Jacobian matrix is
g
 1 ,
y
J ( 0,0) 
2 1 O
L
M
N0 1P
Q
Therefore the linearisation at the point (0,0) is
u  2u  v
v  v
bg
At the point 2,4 the Jacobian matrix is
J ( 2 ,4 ) 
2 1 O
L
M
N4 1P
Q
Therefore the linearisation at (2,4) is
u  2u  v
v  4u  v
Examples 2
For the following examples find
a)
the equilibrium points
b)
the Jacobian matrix of the linearisations
c)
the linearisation at each equilibrium point.
1) x  y , y   x  x 3
2) x  y 2  3x  2 , y  x 2  y 2
3) x  x  y  e x , y  x  y 1
4) x   y  x  xy, y  x  y  y 2
5) x  x 2  e y , y  y(1  y)
6) x  sin( x  y ), y  y
7) x  x 2  sin y  1, y  sinh( x  1)
8) x   x  y y , y  y  x x
hint
Ans. 1
a) (0,0)
(  1,0)
d(x x )
2x
dx
b) For (0,0) no substitution is required
For (1,0) substitute x = u +1, y = v
For (-1,0) substitute x = u –1, y = v
c) At (0,0) linearisation is x  y, y  x
At (  1,0) linearisation is
u  v, v  2u
2 a) (1,1), (1,-1), (2,2), (2,-2)
b) For (1,1) substitute x = u + 1, y = v + 1
For(1,-1) substitute x = u + 1, y = v - 1
For (2,2) substitute x = u + 2, y = v + 2
For (2,-2) substitute x = u + 2, y = v – 2
c) At (1,1) lineaisation is u  3u  2v, v  2u  2v
At (1,-1) linearisation is u  3u  2v, v  2u  2v
At (2,2) linearisation is u  3u  4v, v  4u  4v
At (2,-2) linearisation is u  3u  4v, v  4u  4v
Answers:
1) a) (0,0)
b) J 
(  1,0)
0
1O
L
M
1  3x 0P
N
Q
2
c) At (0,0) linearisation is u  v, v  u
At (  1,0) linearisation is u  v , v  2u
2) a) (1,1), (1,-1), (2,2), (2,-2)
b) J 
3
L
M
2x
N
O
2 y P
Q
2y
c) At (1,1) lineaisation is u  3u  2v, v  2u  2v
At (1,-1) linearisation is u  3u  2v, v  2u  2v
At (2,2) linearisation is u  3u  4v, v  4u  4v
At (2,-2) linearisation is u  3u  4v, v  4u  4v
3) a) (0,-1)
b) J 
0 1O
L
M
1 1P
N
Q
c) Linearisation is u  v, v  u  v
4) a) (0,0)
J
a)
1 1O
L
M
1 1P
N
Q
c) Linearisation is u  u  v, v  u  v
5)

1
a) ( 1,0) , ( e 2 ,1)
L
2 x e O
M
N0 1  2 yP
Q
y
b) J 
c) Linearisation at ( 1,0) is u  2u  v, v  v

1
2
Linearisation at ( e ,1) is u  
6)
2
1
u  v, v  v
e
e
a) ( n ,0 ) for all integer n
b) J 
cos( x  y )
L
M
N0
cos( x  y )
1
O
P
Q
c) linearisation is u  u  v, v  v if n is even, u  u  v, v  v if n
is odd
7)
a) (1, n ) for all integer n
b) J 
2x
L
M
cosh( x  1)
N
cos y
0
O
P
Q
c) Linearisation is u  2u  v, v  u if n is even, u  2u  v, v  u if n
is odd
8)
a) (0,0), (1,1), (-1,-1)
b) J 
1
L
M
2 x
N
O
P
1 Q
2y
c) Linearisation at (0,0) is u  u, v  v
Linearisation at (1,1) is u  u  2v, v  2u  v
Linearisation at (-1,-1) is u  u  2v, v  2u  v