V Stability analysis
Consider the following two coupled differential equations:
x& = f ( x, y )
y& = g ( x, y )
[V.1]
The nullclines are defined as:
x& = 0 → f ( xo , y o ) = 0
y& = 0 → g ( xo , y o ) = 0
[V.2]
in order to solve [V.2] we linearize around the fixed point (xo,yo):
~
x ≡ x−x
o
~
y ≡ y − yo
[V.3]
If f(x,y) and g(x,y) are approximated by a first order Taylor expansion, [V.2] can be
written as:
∂f
∂f
x& ≈ ~
x
+~
y
≡ a~
x + b~
y
∂x ( xo , y o )
∂y ( xo , y o )
∂g
y& ≈ ~
x
∂x
∂g
+~
y
≡ c~
x + d~
y
∂
y
( xo , yo )
( xo , yo )
[V.4]
or in matrix notation:
r
r
X& = AX
a b 
A=

c d 
r  x& 
X& =  
 y& 
r ~
x
X =  ~
y
[V.5]
The matrix A is characterized by its trace and the determinant:
τ = trace( A) = a + d
[V.6]
∆ = det( A) = ad − bc
Let’s try to find a solution of the convenient form:
v
v
v
v& = λv = Av
24
[V.7]
7.81/8.591/9.531 Systems Biology – A. van Oudenaarden – MIT– September 2004
This vector is called the eigenvector, λ is the corresponding eigenvalue. [V.7] can be
solved by:
a − λ
det 
 c
b 
=0
d − λ 
[V.8]
leading to:
λ1 =
λ2 =
τ + τ 2 − 4∆
2
[V.9]
τ − τ 2 − 4∆
2
or
∆ = λ1λ2
[V.10]
τ = λ1 + λ2
For a stable fixed point both λ1 and λ2 should be negative. Therefore a stable fixed point
is characterized by:
∆>0
[V.11]
τ <0
Now let us use the evaluate the stability of the toggle switch (Chapter IV) as an example:
α1
−u
1+ vβ
α
v& = g(u,v) = 2 γ − v
1+ u
u& = f(u,v) =
[V.12]
The fixed points are:
α1
1+ vβ
α
v= 2 γ
1+ u
u=
[V.13]
The matrix A is given by:
25
7.81/8.591/9.531 Systems Biology – A. van Oudenaarden – MIT– September 2004

 −1
A=
γ −1
 − α 2 γu
 (1 + u γ ) 2
− α1β vβ−1 
(1 + vβ ) 2 

−1 

[V.14]
The trace of this matrix is always negative so the only requirement for stability is that the
determinant of this matrix is larger than zero. Let us focus on the conditions for which the
determinant equals zero. This would define the boundary in parameter space that
separates the bistable from monostable region. Setting the determinant to zero gives:
α1βvβ−1 α 2 γu γ −1
=1
(1 + vβ ) 2 (1 + u γ ) 2
[V.15]
Using the conditions for the fixed points [V.13] gives:
βγvβ+1u γ +1 = α1α 2
[V.16]
To be able to solve [V.16] we have to make some assumptions. We assume that we are
working with strong promoters and therefore α1 and α2 are large and that expression ratio
between the ON and OFF state is large. In the case where u >> v , the fixed points are
approximated by:
u ≈ α1
v≈
α2
α1γ
[V.17]
Condition [V.16] now becomes:
1
βγα β2α1−βγ = 1 ⇔ log α1 ≈ log α 2
γ
[V.18]
Consistent with Fig. 2c-d in the toggle switch paper. The other boundary is obtained by
realizing that the system is symmetry. By replacing α1 by α2 and γ by β, the other
boundary is found:
log α1 ≈ β log α 2
26
[V.19]
7.81/8.591/9.531 Systems Biology – A. van Oudenaarden – MIT– September 2004