MAT-35006: Dynamical Systems and Chaos Part II: Biology...

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MAT-35006: Dynamical Systems and ChaosPart II: Biology Applications

Lecture 3: 2D Linear Systems

Ilya PotapovMathematics Department, TUT

Room [email protected]

Ilya Potapov: TD325, [email protected] Dynamical Systems in Biology. Lecture 3: 2D Linear Systems

Linear systems

Linear systems has one solution that is steady state.

It is possible using the linear algebra apparatus to analyze“type” of the steady state.

General form of a linear system and algebraic system ofequations corresponding to the steady state:

dx

dt= ax + by

dy

dt= cx + dy

SteadyState

⇒

ax + by = 0

cx + dy = 0

In order to determine the type of the steady state we need tosolve the so called characteristic equation:∣∣∣∣a− λ b

c d − λ

∣∣∣∣ = 0


Type of the steady state

Determinant (it is the name)∣∣∣∣a− λ bc d − λ

∣∣∣∣ = 0

expands to:

(a− λ)(d − λ)− bc = 0⇒ λ2 − (a + d)λ+ ad − bc = 0

The classical quadratic equation with regard to λ.

The solution to the equation is:

λ1,2 =a + d ±

√(a + d)2 − 4(ad − bc)

2


Physical meaning of λ1,2 parameters

λ1 and λ2 form the exponents of the corresponding solutionsof the original ODE system

dx

dt= ax + by

dy

dt= cx + dy

in the form of: x(t) = A · eλ1t and y(t) = B · eλ2t

So (as we saw in the Malthus model) the solutions x(t) andy(t) tend to either 0 or ∞. In other words: lim

t→∞x(t) = 0 or

limt→∞

x(t) =∞ (the same for y(t)).

Which case is realized depends on the real part of the λ1,2values. NOTE: the λ1,2 can be complex conjugates.


λ1, λ2 — both negative

Reλ1,2 < 0⇒ solutions x(t) and y(t) tend to zero. For example,a = −10, b = 2, c = 2, d = −5 ⇒λ1,2 =

−15±√

225−4·(50−4)2 = −7.5±

√412 (√

41 ≈ 6.4)

−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

X

Y


λ1, λ2 — both positive

Reλ1,2 > 0⇒ solutions x(t) and y(t) tend to ∞. For example,a = 10, b = 2, c = 2, d = 5 ⇒λ1,2 =

15±√

225−4·(50−4)2 = 7.5±

√412 (√

41 ≈ 6.4)

−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

X

Y


λ1, λ2 — different signs

Reλ1 > 0,Reλ2 < 0⇒ solution x(t) tends to ∞ whereas y(t)tends to zero. For example, a = 1, b = 1, c = 1, d = −1 ⇒λ1,2 =

0±√

02−4·(−1−1)2 = ±

√2

−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

X

Y


One λ equals zero

Equations ax +by = 0 and cx +dy = 0 become linearly dependent.Geometrically this means the lines in 2D coordinate space becomecollinear and the solution to the system is one line correspondingto both equations. ⇒ There are infinite number of solutions.For example, a = 1, b = 1, c = 1, d = 1 ⇒λ1,2 =

2±√

22−4·(1−1))2 = 0 ∨ 2

−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

X

Y


λ1, λ2 — complex conjugates

λ1,2 is of the form a± bi . For example, a = −1, b = 1, c = −4,

d = −1 ⇒ λ1,2 =−2±√

(−2)2−4·(1−(−4))2 = −1± 2i

−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

X

Y


λ1, λ2 — complex conjugates. Reλ1,2 = 0

Special case of focus point, so called center ⇒ oscillatory regime.For example, a = 1, b = 1, c = −4, d = −1 ⇒λ1,2 =

0±√

02−4·(−1−(−4))2 = ±

√3i

−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

X

Y


Example

Linear chemical equation:

k1→ X

k2→ Y

k3→

Corresponding ODE system:dX

dt= k1 − k2X

dY

dt= k2X − k3Y

Characteristic equation:∣∣∣∣−k2 − λ 0k2 −k3 − λ

∣∣∣∣ = 0⇒ λ2 + (k2 + k3)λ+ k2k3 = 0

⇒ λ1,2 =−(k2 + k3)±

√(k2 − k3)2

2show !

= −k2 ∨ −k3


Law of Mass Action

When a reaction happens, the concentrations of the reactants andproducts are changing. The concentrations of the reactants aredecreasing and products are increasing. The law of mass action(Waage and Guldberg, 1864) states that the rate of a chemicalreaction is proportional to the product of concentrations ofthe reacting molecules.For the reaction

aA + bBk−→ dD + f F,

the reaction rate is

v = −1

a

d[A]

dt= −1

b

d[B]

dt=

1

d

d[D]

dt=

1

f

d[F]

dt= k[A]a[B]b.


Deterministic modeling of chemical reactions

Degradation:

Ak−→ ∅

Rate of the reaction is:

v = −d [A]

dt= k[A] ,

where [A] is the concentration of A.Transformation:

Ak−→ B

Rate of the reaction is:

v = −d [A]

dt=

d [B]

dt= k[A]

More complex example:

2Ak−→ B + C

v = −1

2

d [A]

dt=

d [B]

dt=

d [C ]

dt= k[A]2


Reversible reaction

2A + Bk1−−⇀↽−−k−1

C

means that we have two reactions2A + Bk1−→ C

Ck−1−−→ 2A + B

,

where A, B, and C are chemical species. k1 is the forward and k−1the backward rate constant. The reaction rates for forward andbackward reaction are

v1 = k1[A]2[B] and v−1 = k−1[C].

The reaction rate for the reversible reaction is

v1,−1 = v1 − v−1 = k1[A]2[B]− k−1[C].


The reaction rates for each chemical species are

d[A]

dt= −2v1 + 2v−1 = −2v1,−1,

d[B]

dt= −v1 + v−1 = −v1,−1, and

d[C]

dt= v1 − v−1 = v1,−1.

The numbers before reaction rates are called stoichiometriccoefficients.In steady state, d[A]

dt = d[B]dt = d[C]

dt = 0 which means that v1 = v−1

and [A]2[B][C] = k−1

k1= Kd , where Kd is the equilibrium constant.


Stoichiometric Matrix

One way to present a model is

d[Ci ]

dt=

∑j∈{Jto ,Jout}

sijvj ,

where Jto and Jout are the set of indices of reactions leading to thechemical species Ci (Ci is the product) and out of the chemicalspecies Ci (Ci is the reactant), respectively. sij are thestoichiometric coefficients and vj is the reaction rate for thereaction j . One can write in matrix format:

dx

dt= Sv,

where vector x describes the concentrations of each chemicalspecies, S is the stoichiometric matrix, and vector v describes thedeterministic reaction rates for each reaction.


For example, the reversible reaction 2A + Bk1−−⇀↽−−k−1

C can be

presented as

dx

dt=

dAdtdBdtdCdt

= Sv =

−2 2−1 11 −1

[ v1v−1

],

where we can easily see that in the forward reaction, 2 moles of Aand 1 mole of B are forming 1 mole of C, and in the backwardreaction, 1 mole of C is forming 2 moles of A and 1 mole of B.And we get the same result as before:

d[A]

dt= −2v1 + 2v−1,

d[B]

dt= −v1 + v−1, and

d[C]

dt= v1 − v−1.


Summary

Linear 2D systems are easy to analyze analytically.

Knowing the characteristic equation solutions λ1,2 one canpredict the behavior with great details.

Molecular systems with chemical transformations can bewritten in the “language” of ODE’s by applying the Law ofMass Action.


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