Created by Claudia Neuhauser
Worksheet 8: Nonlinear Systems of Differential Equations
Nonlinear Systems of Differential Equations—Consumer-Resource
Models
Nonlinear, autonomous systems of ordinary differential equations are of the form
dx1
 f1 x1 , x2 ,, xn 
dt
dx2
 f 2 x1 , x2 ,, xn 
dt

dxn
 f n x1 , x2 ,, xn 
dt
where each of the functions fi on the right-hand side are real-valued functions in n
variables. Most of the time, we will restrict the analysis to systems of two variables. We
will focus on equilibria and stability.
Equilibria and Stability
Consider the system of two autonomous differential equations
(1)
dx
 f ( x, y )
dt
dy
 g ( x, y )
dt
The first step is to find the equations of the zero isoclines, which are defined as the set of
points that satisfy
0  f ( x, y )
0  g ( x, y )
Each equation results in a curve in the x-y space. Point equilibria occur where the two
isoclines intersect (Figure 1). A point equilibrium ( xˆ , yˆ ) of (1) therefore simultaneously
satisfies the two equations
f ( xˆ , yˆ )  0 and g ( xˆ , yˆ )  0
We will call point equilibria simply “equilibria.”
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Worksheet 8: Nonlinear Systems of Differential Equations
y
f ( x, y )  0
Equilibrium
g ( x, y )  0
x
Figure 1: Zero isoclines corresponding to the two differential equations. Equilibria occur where the
isoclines intersect.
The analytical approach to stability relies on analyzing the effects of small perturbations.
We say that the equilibrium ( xˆ , yˆ ) is locally stable if the system returns to ( xˆ , yˆ ) after a
small perturbation, and unstable otherwise. Mathematically, this can be analyzed through
linearizing the right-hand side of each of the two differential equations in (1) about the
equilibrium.
The system (1) can be written in matrix form
d  x(t )   f ( x, y )

dt  y (t )  g ( x, y ) 
where the right-hand side is a vector-valued function that maps a point in R2 (the twodimensional real plane) into a point in R2. To linearize a vector-valued function, we need
to linearize each component separately. Linearizing a function of two variables about a
specific point means to find the tangent plane at this point (this, of course, is not always
possible). The equation of a tangent plane of f(x,y) about ( xˆ , yˆ ) is given by
 ( x, y )  f ( xˆ, yˆ ) 
f ( xˆ , yˆ )
f ( xˆ , yˆ )
( x  xˆ ) 
( y  yˆ )
x
y
 f ( x, y ) 
We thus find for the linearization of the vector-valued function 

 g ( x, y ) 
 f ( xˆ, yˆ )
 ( x, y )   f ( xˆ, yˆ )   x
  ( x, y )    g ( xˆ, yˆ )    g ( xˆ, yˆ )

 
 
 x

2
f ( xˆ, yˆ ) 
y   x  xˆ 

g ( xˆ, yˆ )   y  yˆ 
y 
Worksheet 8: Nonlinear Systems of Differential Equations
Now, considering   x  xˆ and   y  yˆ as the perturbations, then with f ( xˆ , yˆ )  0
and g ( xˆ , yˆ )  0 , we find
(2)
 d   f ( xˆ , yˆ )
 dt   x
 d    g ( xˆ, yˆ )
  
 dt   x
f ( xˆ , yˆ ) 
y   

g ( xˆ, yˆ )   
y 
The matrix
 f ( x, y )
 x
J ( x, y )  
 g ( x, y )
 x
f ( x, y ) 
y 

g ( x, y ) 
y 
is called the Jacobi matrix.
The system (2) is a linear system of two equations, and we can use the results from linear
systems of two differential equations to determine the stability of the equilibria. Namely,
 The equilibrium is a node if both eigenvalues of the Jacobian evaluated at the
equilibrium are real, distinct, nonzero, and are of the same sign. The node is
locally stable if the eigenvalues are negative, and unstable if the eigenvalues
are positive.
 The equilibrium is a saddle if both eigenvalues of the Jacobian evaluated at
the equilibrium are real and nonzero but have opposite signs. A saddle is
unstable.
 The equilibrium is a spiral if both eigenvalues of the Jacobian evaluated at the
equilibrium are complex conjugates with nonzero real parts. The spiral is
locally stable if the real parts of the eigenvalues are negative, and unstable if
the real parts of the eigenvalues are positive.
Task 1:
Show that (3,2) is an equilibrium of
dx
 5  x  xy  2 y
dt
dy
 xy  3 y
dt
and determine its stability.
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Worksheet 8: Nonlinear Systems of Differential Equations
Task 2:
Consider the following set of two differential equations:
dx1
 x1 10  2 x1  x2 
dt
dx2
 x2 10  x1  2 x2 
dt
(a) Graph the zero isoclines into the vector field (see handout in class).
The vector field was created in Matlab using
>> [x,y] = meshgrid(-2:.1:4,-2:0.1:4);
>> M=x.*(10-2*x-y);
>> N=y.*(10-x-2*y);
>> quiver(x,y,M,N,2)
>> grid on
Note that the multiplication is “.*” since we are multiplying componentwise.
 10 10 
,  is an equilibrium, and determine its stability using the analytical
 3 3
(b) Show that 
approach.
Task 3:
Find all equilibria of
dx1
 x1  x2
dt
dx2
 x1 x2  x2
dt
and use the analytical approach to determine the stability of each equilibrium. Graph the
zero isoclines to confirm that you found all equilibria.
Euler Method for Systems of Differential Equations
To numerically solve a system of differential equations, we can use the Euler method.
This is the simplest approximation scheme. It has disadvantages, like slow convergence
and instabilities. There are far superior numerical schemes (like the Runge-Kutta
method), but the Euler scheme suffices for our purposes here.
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Worksheet 8: Nonlinear Systems of Differential Equations
The Euler scheme is directly based on the definition of the derivative. Namely,
dx
x(t  h)  x(t )
 lim
dt h0
h
To approximate the system of differential equations
dx
 f ( x, y )
dt
dy
 g ( x, y )
dt
we approximate the left-hand side using the definition of the derivative
x(t  h)  x(t )
 f  x(t ), y (t ) 
h
y (t  h)  y (t )
 g x(t ), y (t ) 
h
and then rewrite this as
x(t  h)  x(t )  hf  x(t ), y (t ) 
y (t  h)  y (t )  hg  x(t ), y (t ) 
which can be coded up in a spreadsheet. The time step h needs to be small enough so that
the approximation is sufficiently good but not too small so that rounding errors
accumulate too quickly.
Consumer-Resource Models
Model 1: Lotka-Volterra Predator-Prey Model
There are a large number of models that deal with consumer and resource interactions.
The oldest such model is the Lotka-Volterra model, which describes the interaction
between a predator and its prey. If x denotes the abundance of the prey and y the
abundance of the predator, then the model is given by the following set of differential
equations:
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Worksheet 8: Nonlinear Systems of Differential Equations
(3)
dx
 bx  cxy
dt
dP
 ecxy  dy
dt
where the parameters b, c, e, and d on the right-hand side are positive constants. The
parameter b is the prey birth rate, c is the consumption rate, e is the conversion fraction of
prey into new predators, and d is the predator death rate. The equilibria can be obtained
algebraically by setting the right-hand sides equal to 0 and solving for x and y. We find
 d b
the trivial equilibrium (0,0) and a nontrivial equilibrium  xˆ , yˆ    ,  . The
 ec c 
corresponding Jacobi matrix is given by
 cx 
b  cy
J ( x, y )  

 ecy ecx  d 
If we evaluate this at the trivial equilibrium (0,0), we find
b 0 
J (0,0)  

0  d 
Therefore, the two eigenvalues are 1  b and 2  d . Both eigenvalues are real.
Since one eigenvalue is positive and the other is negative, we conclude that the trivial
equilibrium is a saddle and thus unstable.
 d b
If we evaluate the Jacobi matrix at the nontrivial equilibrium  xˆ , yˆ    ,  , we
 ec c 
find

0
 d b 
J ,   
 ec c  ec b
 c
d
d

ec    0  

e
0  eb 0 

c
To compute the eigenvalues, we solve
d 

  d b

 d
0  
2
det  J  ,   I   det 
e  0   0     eb      bd  0
ec
c
e
 eb 0   



 



which yields the two complex conjugate eigenvalues 1  i bd and  2  i bd . The
real parts of these two eigenvalues are both equal to 0. The linear stability analysis is thus
inconclusive.
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Worksheet 8: Nonlinear Systems of Differential Equations
It turns out that the equilibrium is neutral and this system of equations exhibits
neutral oscillations. This will be explored numerically in the computer lab exercises.
Model 2: Density-dependent Growth of the Prey
It is possible to stabilize the predator-prey model by including density-dependent growth
of the prey in the form of logistic growth. This takes the form
(4)
dx
x

 bx1    cxy
dt
 K
dy
 ecxy  dy
dt
where all parameters are positive. In the absence of the predator, the prey dynamics
reduce to logistic growth in Equation (4). Namely, If we set y  0 , then
dx
 bx 1  x / K  .
dt
The system of equations (4) has the nontrivial equilibrium (0,0), which is always
unstable. In addition, it has the prey only equilibrium (K,0), which is locally stable
d
d
provided K 
. If K  , an additional nontrivial equilibrium in the first quadrant
ec
ec
d
appears, which is locally stable. If K  , the prey only equilibrium is unstable.
ec
Model 3: Saturating Functional Response
The functional response of a predator is defined as the per predator predation rate. In this
case, the functional response is simply cx, a linear function of prey density. In nature, one
often observes functional response curves that saturate with increasing prey density. This
happens, for instance, if predators need to spend time handling prey.
A common form for a saturating functional response is
(5)
dx
x
cx

 bx1   
y
dt
 K ax
dy
cx
e
y  dy
dt
ax
This model is known as Rosenzweig’s model. It has the trivial equilibrium (0,0) and the
ad
 K , there is an additional equilibrium in the
prey only equilibrium (K,0). If 0 
ec  d
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Worksheet 8: Nonlinear Systems of Differential Equations
first quadrant, which may be locally stable or unstable. The interesting property of this
model is that when increasing the carrying capacity K, the equilibrium becomes unstable.
This is known as Rosenzweig’s paradox of enrichment. (You can read more about this in
the attached paper.)
Task 4:
cx
for c=0.81 and a=0.25. The parameter c
ax
is called the saturation level since lim p ( x )  c , and a is the half-saturation constant since
Graph the functional response curve p ( x ) 
x 
p (a )  c / 2 . Label a and c on your graph.
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Worksheet 8: Nonlinear Systems of Differential Equations
Computer Lab (In Class)
In three separate worksheets (Tabs: “LV,” “LV with DD,” and “Rosenzweig”), we will
code up the three different predator-prey models using the Euler method. Use the
following parameters:
b  1.2, c  0.81, e  0.8, d  0.25, a  0.25
Use initial conditions x(0)  0.5 and y (0)  0.25 . As the time step, choose h=0.007.
Your worksheet for the first model (Tab “LV”) looks as follows:
A
1
2
3
4
5
6
7
8
9
B
b
K
1.2
Time
0
0.007
0.014
0.021
0.028
C
c
0.25
D
E
e
0.81
Producer
Consumer
0.5
0.503491
0.507007
0.510546
0.51411
0.25
0.25013
0.250263
0.250401
0.250542
d
0.8
F
h
0.25
0.007
G
x_0
0.5
H
y_0
0.25
Row 1 contains the parameter names and Row 2 the corresponding values. In A5, enter 0.
Cell A6 is “=A5+$F$2”. Drag this cell down so that the final time is about 32. In B5 and
C5, copy “=$G$2” and “=$H$2”. In B6, write “=B5+$F$2*($A$2*B5-$C$2*B5*C5)”
and in C6, write “=C5+$F$2*($D$2*$C$2*B5*C5-$E$2*C5)”. Drag both cells down to
about row 5000.
Investigate the behavior of the three models as the carrying capacity K changes. Simulate
the three models for K=0.1, 0.25, 0.75, 1.0, and 2.0, and describe what you see.
Literature Cited
Rosenzweig, M.L. (1971) Paradox of enrichment: destabilization of exploitation
ecosystems in ecological time. Science 171: 385-387.
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