Lecture Notes for Lecture 17 of Dr. Z.’s Dynamical Systems in Biology

These notes are based on parts of Chapters 4,5, and 6 of Leah Edelstein-Keshet’s classic “Mathematical Models in Biology”. A 1-dimensional continuous dynamical system It is a differential equation, where the unknowns function x(t) of times satisfies the differential equation

\frac{dx}{dt} = F (x(t), t)

whree F is some function of two variables. If F only depends on x(t) (i.e. only depends on time via x(t)) it is called autonomous. We wiil only consider this case, so the format is

\frac{dx}{dt} = F (x(t)) .

Often you write it as

\frac{dx}{dt} = F (x) ,

where it is understood that x is a function of t. For any initial condition x(t_{0}) = x_{0} x(t) will evolve as time t goes to infinity. Often it would go to infinity, but sometimes it would tend to some steady steady-state, that in the continuous case is more often called steady equilibrium. Def. A number x_{0} is called an equilibrium if it solves the (algebraic) equation F (x_{0}) = 0. Problem 17.1: Find all the equilibrium points of the 1D continuous dynamical system

\frac{dx}{dt} = (x(t) − 1)(x(t) − 2)(x(t) − 3)(x(t) − 4)(x(t) − 5) .

Sol. of 17.1: We have to solve

(x − 1)(x − 2)(x − 3)(x − 4)(x − 5) = 0 .

whose solutions are x = 1,x = 2,x = 3,x = 4,x = 5. Ans. to 17.1: The set of equilibrium points of this differential equation is {1, 2, 3, 4, 5}. If the quantity (e.g. population of some species) is exactly at an equilibrium point, then it stays there for ever after, so the solution of the initial valule differential equation 1 with

• initial condition x(0) = 1 is x(t) = 1 • initial condition x(0) = 2 is x(t) = 2

• initial condition x(0) = 3 is x(t) = 3 etc. But for some eqilibrium points of you just move a timy bit you would go very far away. These are unstable equilibrium points. For other equilibrium points, if you start at a nearby value (and often pretty far), in the long run it would be attracted to it. How to decide whether an equilibrium point is stable? If x_{0} is an equilibrium point of x′(t) = F (x(t)), in other words F (x_{0}), look at F′(x_{0}) Stability Criterion x_{0} is stable if F′(x_{0}) < 0. It is unstable if F′(x_{0}) > 0. It is semi-stable if F′(x_{0}) = 0. Problem 17.2: Find all the stable equilibrium points of the 1D continuous dynamical system

\frac{dx}{dt} = (x(t) − 1)(x(t) − 2)(x(t) − 3)(x(t) − 4)(x(t) − 5) .

Sol. of 17.2: By the product rule (PLEASE do not expand it!) F′(x) = (x−2)(x−3)(x−4)(x−5)+(x−1)(x−3)(x−4)(x−5)+(x−1)(x−2)(x−4)(x−5)+(x−1)(x−2)(x−3)(x−5)+(x−1)(x−2)(x−3)(x−4)

please do not expand. We know from 17.1 that the candidates are {1, 2, 3, 4, 5}. When x = 1, we have F′(1) = (1 − 2)(1 − 3)(1 − 4)(1 − 5) = 24 since this is positive, we know that x = 1 is unstable. When x = 2, we have F′(2) = (2 − 1)(2 − 3)(3 − 4)(2 − 5) = −6 since this is negative, we know that x = 2 is stable. When x = 3, we have F′(3) = (3 − 1)(3 − 2)(3 − 4)(3 − 5) = 2 since this is positive, we know that x = 3 is unstable. When x = 4, we have F′(4) = (4 − 1)(4 − 2)(4 − 3)(4 − 5) = −6 since this is negative, we know that x = 4 is stable. When x = 5, we have F′(5) = (5 − 1)(5 − 2)(5 − 3)(5 − 4) = 24 since this is positive, we know that x = 3 is unstable. Ans. to 17.2: The set of stable equilibria is {2, 4}.

Dynamical Systems in Severl Variables (species) A k-dimensional continuous dynamical system (we only do autonomous ones) is a system of k differential equation, with k unknown functions of time (e.g. the relateive occurrence of each species, concentration of nutrients, ratio of infeced,) of the form

\frac{dx_{1}}{dt} = F_{1}(x_{1}(t), …, x_{k}(t)) ,

\frac{dx_{2}}{dt} = F_{2}(x_{1}(t), …, x_{k})(t) ,

…

\frac{dx_{k}}{dt} = F_{k}(x_{1}(t), …, x_{k})(t) ,

where F_{1}, ..., F_{k} are k multi-variable functions of k variables. The underlying transformation is

F : R^{k} → R^{k} ,

given by

F(x_{1}, …, x_{k}) = (F_{1}(x_{1}, … x_{k}), …, F_{k}(x_{1}, … x_{k})) .

How to find the Equilibrium points? Solve the system of k algebraic (or tanscendental) equations x = F(x). In other words

x_{1} = F_{1}(x_{1}, …, x_{k}) , x_{2} = F_{2}(x_{1}, …, x_{k}) , … x_{k} = F_{k}(x_{1}, …, x_{k}) .

How to find the Stable Equilibrium points? Recall that for discrete dynamical systems, we find the Jacobian of the underlying transformation, a certain matrix of functions. Then we find all the steady-states, and for each and every one of them we plug into the Jacobian, getting a matrix of numbers. Then we look at the eigenvalues if all of them have absolute value less than 1 then we know it is stable. For the continuous case we do the same except that we have a different criterion. An equilibrium point x_{0} is stable if and only if all the eigenvalues of J(x_{0}) have negative real part. If they have non-negative real parts and some of them are purely imaginary (i.e. the real part is 0) then it is a semi-stable equilibrium point. Why is true? Again using multi-variable Taylor we get that, near the equilbrium point x_{0},

F(x − x_{0} = J(x_{0})(x − x_{0})plusT INY .

A linear system

x′(t) = Lx(t) ,

where L is a numerical matrix, 0 is obviously an equilibrium. By diagonalizing

L = UDU^{−1} ,

where D is a diagonal matrix consisting of the eigenvalues, By a change of variable y = U^{−1}x we have the system

y′(t) = Dy(t) ,

spelling it out

y′_{1}(t) = λ_{1}y_{1}(t) ,
y′_{2}(t) = λ_{2}y_{2}(t) ,
…
y′_{k}(t) = λ_{k}y_{k}(t) ,

Hence

y_{1}(t) = e^{λ_{1} t} , y_{k}(t) = e^{λ_{k} t} ,

If the real part of any of the λ_{i}-s is positive the correpsonding y_{i}(t) would blow up as t gets larger and larger. Problem 17.3: Find all equilibrium points of the continuous dynamical system

\frac{dx}{dt} = 2 − \frac{5x}{1 + 4y}

\frac{dy}{dt} = \frac{5}{9} − \frac{5y}{1 + 4x}

Solution pf 17.3 We have to solve

2 − \frac{5x}{1 + 4y} = 0 , \frac{5}{9} − \frac{5y}{1 + 4x} = 0 .

Cross-multiplying and simplifying we get

5x − 8y = 2 , −4x + 9y = 5 ,

whose only solution is (x, y) = (2, 1). Answer to 17.3: The only equilibrium point of the system is (2, 1). Problem 17.4: Find all stable equilibrium points of the continuous dynamical system

\frac{dx}{dt} = 2 − \frac{5x}{1 + 4y}

\frac{dy}{dt} = \frac{5}{9} − \frac{5y}{1 + 4x}

The underlying transformation is

(x, y) → (2 − \frac{5x}{1 + 4y} , \frac{5}{9} − \frac{5y}{1 + 4x}) .

The Jacobian is

J(x, y) = [ \begin{matrix}−\frac{5}{1+4y} & \frac{20x}{(1+4y)^{2}} \\ \frac{20y}{(1+4x)^{2}} & −frac51 + 4x \end{matrix} ] .

From 17.3 we know that the only equilibrium point is (2, 1). Plugging it in we get

J(2, 1) = [ \begin{matrix}−\frac{5}{1+4·1} & \frac{20·2}{(1+4·1)^{2}} \\ \frac{20·1}{(1+4·2)^{2}} & −frac51 + 4 · 1 \end{matrix} ] .

[ \begin{matrix}−1 & \frac{8}{5} \\ \frac{20}{81} & −\frac{5}{9} \end{matrix} ]

The characteristic equation is

(−1 − λ)(−\frac{5}{9} − λ) − \frac{8}{5} · \frac{20}{81} = 0 .

Simplifying:

λ^{2} + \frac{14}{9}λ + \frac{13}{81} = 0 .

Factorizing:

(λ + \frac{1}{9})(λ + \frac{13}{9}) = 0 .

So the two eigenvalues are −\frac{1}{9} and −\frac{13}{9}. Both are negative real numbers (and hence they have negative real parts, the imaginary part being 0, and we found out that (2, 1) is a stable equilibrium point.

Ans. to 17.4: The only stable equiibrium point is (2, 1).

Two Important Dynamical Systems in Mathematical Biology

The Chemostat: If N is the bacterial population and C is the nutrient, and a_{1}, a_{2} are two parameters that depend on the particular situation (determined experimentally) then the system is

\frac{dN}{dt} = a_{1} · \frac{CN}{C + 1} − N

\frac{dC}{dt} = −fracCNC + 1 − C + a_{2}

The equilibrium points are

(0, a_{2}) , (\frac{a_{1}(a_{2}a_{1} − a_{2} − 1)}{a_{1} − 1}, \frac{1}{a_{1} − 1}) .

(0, a_{2}) is not stable, but the other one often is.

The (continuous) SIRS model, due to Kermack and MacKendrick with population N = S + I + R, where S is susceptible but not infected, I is infected, and R = N − I − S is removed. there are biological parameters β, γ, ν and we also consider N as a parameter. The variables are S and I

\frac{dS}{dt} = −βS I + γ(N − S − I)

\frac{dI}{dt} = βS I − νI

The equilibrium points are

(N, 0) , (\frac{ν}{β} , γ \frac{N − \frac{ν}{β}}{ν + γ} .

In the first one no one is infected (but everyone is susceptible), the second one can only happen if it is positive i.e.

N − \frac{ν}{β} > 0 ,

i.e.

\frac{Nβ}{ν} > 1 .

R_{0} = \frac{Nβ}{ν}

is a famous number called the infectious contact number. It is the cutoff . If it is less than 1 no one is infectedm but if it is larger than 1 some pople will be infected.