tutorial: non-linear dynamics - leonardo pacciani-mori

Tutorial: non-linear dynamics

Winter School on Quantitative Systems Biology:Quantitative Approaches in Ecosystem Ecology

Leonardo Pacciani-Mori (University of Padua, Italy)November 30th and December 1st, 2020

Structure of the tutorial

Introduction: why is non-linearity interesting?

→ Examples of non-linearity in ecology

Stream plots

→ How to get a sense of the behavior of a non-linear system without solving it

Stability in non-linear systems

→ Lyapunov functions

→ Spectral analysis

Examples

Important

The aim of these tutorials is also to encourage discussion: please don’t hesitate to interrupt meand ask questions if you have them!!!

Stream plots

Examples

Important

Stream plots

Examples

Important

Stream plots

Examples

Important

Stream plots

Examples

Important

Stream plots

Examples

Important

Introduction

We would like to solve non-linear differential equations (nLDE), i.e.:

x = f (x) (1)

where f (x) is a non-linear function. For example:

x = sinx (2a)

x2 = xy y =xy

xx = −y y = x2 (2c)

Why are nLDE important?

Every interesting and/or non-trivial phenomenon is described by nLDE.

Introduction

x = f (x) (1)

where f (x) is a non-linear function.

For example:

x = sinx (2a)

x2 = xy y =xy

xx = −y y = x2 (2c)

Introduction

x = f (x) (1)

x = sinx (2a)

x2 = xy y =xy

xx = −y y = x2 (2c)

Introduction

x = f (x) (1)

x = sinx (2a)

x2 = xy y =xy

xx = −y y = x2 (2c)

IntroductionExamples

1 Pendulum equation:

θ = −g

`sinθ (3)

2 Navier-Stokes equations (→ fluiddynamics). For an incompressible fluid:

∂~u∂t

+ (~u · ~∇)~u − ν∇2~u = −~∇w+ ~g (4)

3 Logistic growth equation:

x = rx(1− x

4 Lotka-Volterra equations:

x = αx − βxy (6a)

y = δxy −γy (6b)

where:x > 0→ prey’s population, y > 0→predator’s population; α > 0→ prey’sgrowth rate, β > 0→ predation rate, δ > 0→ predator’s growth rate, γ > 0→predator’s death rate.If written as ~z = f (~z), with ~z = (x,y), f (~z) isnon-linear.

θ = −g

`sinθ (3)

∂~u∂t

+ (~u · ~∇)~u − ν∇2~u = −~∇w+ ~g (4)

x = rx(1− x

θ = −g

`sinθ (3)

∂~u∂t

+ (~u · ~∇)~u − ν∇2~u = −~∇w+ ~g (4)

x = rx(1− x

θ = −g

`sinθ (3)

∂~u∂t

+ (~u · ~∇)~u − ν∇2~u = −~∇w+ ~g (4)

x = rx(1− x

θ = −g

`sinθ (3)

∂~u∂t

+ (~u · ~∇)~u − ν∇2~u = −~∇w+ ~g (4)

x = rx(1− x

Equilibria

nLDE are (almost) never solvable analytically(there is no theorem that can help us in general)

QuestionCan we obtain some information on anon-linear system without solving itanalytically?

One of the most interesting properties of dynamical systems are their equilibria:

x = f (x) ⇒ x∗ is an equilibrium if f (x∗) = 0 . (7)

“Informal” definitions

An equilibrium x∗ is said to be stable if any given solution x(t) with x0 = x(0) “closeenough” to x∗ will always remain “close” to x∗

An equilibrium x∗ is said to be unstable if there are solutions x(t) with x0 = x(0) “closeenough” to x∗ which go “away” from x∗

Question (revisited)

Can we obtain some information on a non-linear system’s equilibria (and their stability)without solving it analytically?

There are several ways we can do so. A very simple yet useful one is drawing stream plots, i.e.drawing trajectories of the system in the state space.

Equilibria

Stream plotsLet’s see how to do this in a particular case:

x = x3 − x2 − 2x (8)

f (x) = x3 − x2 − 2x = x(x − 2)(x+ 1) (9a)

Therefore, the equilibria are:

x∗ = −1 x∗ = 0 x∗ = 2 (9b)

When f (x) > 0, x > 0 so x increases, and vicev-ersa x decreases when f (x) < 0.

We can get a sense of which equilibria are stableand which are unstable:

x∗ = 0→ stable; x∗ = −1, x∗ = 2→ unstable.

x = x3 − x2 − 2x (8)

f (x) = x3 − x2 − 2x = x(x − 2)(x+ 1) (9a)

x∗ = −1 x∗ = 0 x∗ = 2 (9b)

x = x3 − x2 − 2x (8)

f (x)f (x) = x3 − x2 − 2x = x(x − 2)(x+ 1) (9a)

x∗ = −1 x∗ = 0 x∗ = 2 (9b)

x = x3 − x2 − 2x (8)

0 2−1

f (x) = x3 − x2 − 2x = x(x − 2)(x+ 1) (9a)

x∗ = −1 x∗ = 0 x∗ = 2 (9b)

x = x3 − x2 − 2x (8)

0 2−1

f (x) = x3 − x2 − 2x = x(x − 2)(x+ 1) (9a)

x∗ = −1 x∗ = 0 x∗ = 2 (9b)

x = x3 − x2 − 2x (8)

0 2−1

f (x) = x3 − x2 − 2x = x(x − 2)(x+ 1) (9a)

x∗ = −1 x∗ = 0 x∗ = 2 (9b)

x = x3 − x2 − 2x (8)

0 2−1

f (x) = x3 − x2 − 2x = x(x − 2)(x+ 1) (9a)

x∗ = −1 x∗ = 0 x∗ = 2 (9b)

Stream plots

Let’s see another example:

x = cosx (10)

f (x) = cosx (11a)

x∗ = ±(2k + 1)π2

k = 0,1,2, . . . (11b)

k even and x∗ > 0, or k odd and x∗ < 0→ stable;k odd and x∗ > 0, or k even and x∗ < 0→ unstable

Stream plots

x = cosx (10)

f (x) = cosx (11a)

x∗ = ±(2k + 1)π2

k = 0,1,2, . . . (11b)

Stream plots

x = cosx (10)

f (x)f (x) = cosx (11a)

x∗ = ±(2k + 1)π2

k = 0,1,2, . . . (11b)

Stream plots

x = cosx (10)

−π2

32π−3

f (x) = cosx (11a)

x∗ = ±(2k + 1)π2

k = 0,1,2, . . . (11b)

Stream plots

x = cosx (10)

−π2

32π−3

f (x) = cosx (11a)

x∗ = ±(2k + 1)π2

k = 0,1,2, . . . (11b)

Stream plots

x = cosx (10)

−π2

32π−3

f (x) = cosx (11a)

x∗ = ±(2k + 1)π2

k = 0,1,2, . . . (11b)

Stream plots

Let’s see an ecologically relevant case wherewe can compare the analytic solution with thestream plot, the logistic equation:

x = rx(1− x

We can solve the equation analytically:

x(t) =Kx0

x0 + (K − x0)e−rt(13)

Properties

x(t)t→∞−−−−→ K

x0 = 0⇒ x(t) = 0 or x0 = K ⇒ x(t) = K

Therefore, the behavior of the analytic solu-tion is the following:

The maximal population that the system cansustain is K (→ carrying capacity).

Stream plots

x = rx(1− x

x(t) =Kx0

x0 + (K − x0)e−rt(13)

Properties

x(t)t→∞−−−−→ K

x0 = 0⇒ x(t) = 0 or x0 = K ⇒ x(t) = K

Stream plots

x = rx(1− x

x(t) =Kx0

x0 + (K − x0)e−rt(13)

Properties

x(t)t→∞−−−−→ K

x0 = 0⇒ x(t) = 0 or x0 = K ⇒ x(t) = K

Stream plots

x = rx(1− x

x(t) =Kx0

x0 + (K − x0)e−rt(13)

Properties

x(t)t→∞−−−−→ K

x0 = 0⇒ x(t) = 0 or x0 = K ⇒ x(t) = K

Stream plots

x = rx(1− x

x(t) =Kx0

x0 + (K − x0)e−rt(13)

Properties

x(t)t→∞−−−−→ K

x0 = 0⇒ x(t) = 0 or x0 = K ⇒ x(t) = K

tspace

x0 = K

Stream plots

x = rx(1− x

x(t) =Kx0

x0 + (K − x0)e−rt(13)

Properties

x(t)t→∞−−−−→ K

x0 = 0⇒ x(t) = 0 or x0 = K ⇒ x(t) = K

tspace

x0 = K

x0 = 0

Stream plots

x = rx(1− x

x(t) =Kx0

x0 + (K − x0)e−rt(13)

Properties

x(t)t→∞−−−−→ K

x0 = 0⇒ x(t) = 0 or x0 = K ⇒ x(t) = K

tspace

x0 = Kx0 = K

x0 = 0

Stream plots

x = rx(1− x

x(t) =Kx0

x0 + (K − x0)e−rt(13)

Properties

x(t)t→∞−−−−→ K

x0 = 0⇒ x(t) = 0 or x0 = K ⇒ x(t) = K

tspace

x0 = Kx0 = Kx0 < K

x0 = 0

Stream plots

x = rx(1− x

x(t) =Kx0

x0 + (K − x0)e−rt(13)

Properties

x(t)t→∞−−−−→ K

x0 = 0⇒ x(t) = 0 or x0 = K ⇒ x(t) = K

tspace

x0 = Kx0 = Kx0 < K

x0 > K

x0 = 0

Stream plots

x = rx(1− x

x(t) =Kx0

x0 + (K − x0)e−rt(13)

Properties

x(t)t→∞−−−−→ K

x0 = 0⇒ x(t) = 0 or x0 = K ⇒ x(t) = K

tspace

x0 = Kx0 = Kx0 < K

x0 > K

x0 = 0

Stream plotsCan we study the behavior of the system without the analytical solution?

x = rx(1− x

f (x) = rx(1− x

)= rx − r

Kx2 (15a)

x∗ = 0 x∗ = K (15b)

x∗ = 0→ unstable; x∗ = K → stable.

ConclusionWe have indeed recovered the same behavior!

x = rx(1− x

f (x) = rx(1− x

)= rx − r

Kx2 (15a)

x∗ = 0 x∗ = K (15b)

x = rx(1− x

f (x) = rx(1− x

)= rx − r

Kx2 (15a)

x∗ = 0 x∗ = K (15b)

x = rx(1− x

f (x) = rx(1− x

)= rx − r

Kx2 (15a)

x∗ = 0 x∗ = K (15b)

x = rx(1− x

f (x) = rx(1− x

)= rx − r

Kx2 (15a)

x∗ = 0 x∗ = K (15b)

x = rx(1− x

f (x) = rx(1− x

)= rx − r

Kx2 (15a)

x∗ = 0 x∗ = K (15b)

x = rx(1− x

f (x) = rx(1− x

)= rx − r

Kx2 (15a)

x∗ = 0 x∗ = K (15b)

Stream plotsLet’s apply the same principle to the Lotka-Volterra equations:(

(αx − βxyδxy −γy

x, y > 0α,β,γ,δ > 0 (16)

The equilibria are:(x∗

) (x∗

(γ/δα/β

We can see immediately that:

x0 = 0⇒(xy

(0−γy

), y0 = 0⇒

)(18a)

x > 0 ⇒ x(α − βy) > 0 ⇒ y < α/β (18b)

y > 0 ⇒ y(δx −γ) > 0 ⇒ x > γ/δ (18c)

The trajectories oscillate around theequilibrium (γ/δ,α/β)!

x, y > 0α,β,γ,δ > 0 (16)

) (x∗

(γ/δα/β

x0 = 0⇒(xy

(0−γy

), y0 = 0⇒

)(18a)

x > 0 ⇒ x(α − βy) > 0 ⇒ y < α/β (18b)

y > 0 ⇒ y(δx −γ) > 0 ⇒ x > γ/δ (18c)

(γ/δα/β

x, y > 0α,β,γ,δ > 0 (16)

) (x∗

(γ/δα/β

x0 = 0⇒(xy

(0−γy

), y0 = 0⇒

)(18a)

x > 0 ⇒ x(α − βy) > 0 ⇒ y < α/β (18b)

y > 0 ⇒ y(δx −γ) > 0 ⇒ x > γ/δ (18c)

(γ/δα/β

x, y > 0α,β,γ,δ > 0 (16)

) (x∗

(γ/δα/β

x0 = 0⇒(xy

(0−γy

), y0 = 0⇒

)(18a)

x > 0 ⇒ x(α − βy) > 0 ⇒ y < α/β (18b)

y > 0 ⇒ y(δx −γ) > 0 ⇒ x > γ/δ (18c)

(γ/δα/β

x, y > 0α,β,γ,δ > 0 (16)

) (x∗

(γ/δα/β

x0 = 0⇒(xy

(0−γy

), y0 = 0⇒

)(18a)

x > 0 ⇒ x(α − βy) > 0 ⇒ y < α/β (18b)

y > 0 ⇒ y(δx −γ) > 0 ⇒ x > γ/δ (18c)

(γ/δα/β

x, y > 0α,β,γ,δ > 0 (16)

) (x∗

(γ/δα/β

x0 = 0⇒(xy

(0−γy

), y0 = 0⇒

)(18a)

x > 0 ⇒ x(α − βy) > 0 ⇒ y < α/β (18b)

y > 0 ⇒ y(δx −γ) > 0 ⇒ x > γ/δ (18c)

(γ/δα/β

x > 0 x > 0

x < 0 x < 0

x, y > 0α,β,γ,δ > 0 (16)

) (x∗

(γ/δα/β

x0 = 0⇒(xy

(0−γy

), y0 = 0⇒

)(18a)

x > 0 ⇒ x(α − βy) > 0 ⇒ y < α/β (18b)

y > 0 ⇒ y(δx −γ) > 0 ⇒ x > γ/δ (18c)

(γ/δα/β

x > 0 x > 0

x < 0 x < 0

x, y > 0α,β,γ,δ > 0 (16)

) (x∗

(γ/δα/β

x0 = 0⇒(xy

(0−γy

), y0 = 0⇒

)(18a)

x > 0 ⇒ x(α − βy) > 0 ⇒ y < α/β (18b)

y > 0 ⇒ y(δx −γ) > 0 ⇒ x > γ/δ (18c)

(γ/δα/β

x > 0y < 0

x > 0y > 0

x < 0y < 0

x < 0y > 0

x, y > 0α,β,γ,δ > 0 (16)

) (x∗

(γ/δα/β

x0 = 0⇒(xy

(0−γy

), y0 = 0⇒

)(18a)

x > 0 ⇒ x(α − βy) > 0 ⇒ y < α/β (18b)

y > 0 ⇒ y(δx −γ) > 0 ⇒ x > γ/δ (18c)

(γ/δα/β

x > 0y < 0

x > 0y > 0

x < 0y < 0

x < 0y > 0

x, y > 0α,β,γ,δ > 0 (16)

) (x∗

(γ/δα/β

x0 = 0⇒(xy

(0−γy

), y0 = 0⇒

)(18a)

x > 0 ⇒ x(α − βy) > 0 ⇒ y < α/β (18b)

y > 0 ⇒ y(δx −γ) > 0 ⇒ x > γ/δ (18c)

(γ/δα/β

x > 0y < 0

x > 0y > 0

x < 0y < 0

x < 0y > 0

Stream plots are a very useful tool, particularly to get a sense of the system

Stream plots don’t give certain information about the stability of the equilibria

How can we study the stability of equilibria without solving a non-linear system?

There are two possible approaches:

Using Lyapunov functions

Spectral analysis

Some formal definitions

Let x∗ be an equilibrium for the system x = f (x), with x ∈Rn and f :D ⊂Rn→R

n is continuous.

Definition: stable equilibrium

The equilibrium x∗ is said to be stable if, for every neighborhood A of x∗ there is a neighborhood B ⊂ Asuch that the solutions starting from points in B will always remain inside A. In other words, x∗ isstable if every solution starting “close” to x∗ always remains “close” to x∗. Graphically:

If x∗ is stable not only ∀t ≥ 0, but∀t ∈ R, x∗ is said to be stable at alltimes.

If x∗ is stable and limt→∞ x(t) = x∗, x∗ issaid to be asymptotically stable.

n is continuous.

The equilibrium x∗ is said to be stable if, for every neighborhood A of x∗ there is a neighborhood B ⊂ Asuch that the solutions starting from points in B will always remain inside A.

In other words, x∗ isstable if every solution starting “close” to x∗ always remains “close” to x∗. Graphically:

n is continuous.

The equilibrium x∗ is said to be stable if, for every neighborhood A of x∗ there is a neighborhood B ⊂ Asuch that the solutions starting from points in B will always remain inside A. In other words, x∗ isstable if every solution starting “close” to x∗ always remains “close” to x∗.

Graphically:

n is continuous.

Definition: unstable equilibrium

The equilibrium x∗ is said to be unstable if it is not stable, i.e. if it exists one neighborhood A of x∗

such that for any choice of B ⊂ A there is always (at least) one initial condition x(0) ∈ B such that itssolution goes out of A.In other words, x∗ is unstable if there are solutions starting “close” to x∗ that eventually go “faraway” from x∗. Graphically:

n is continuous.

such that for any choice of B ⊂ A there is always (at least) one initial condition x(0) ∈ B such that itssolution goes out of A.

In other words, x∗ is unstable if there are solutions starting “close” to x∗ that eventually go “faraway” from x∗. Graphically:

n is continuous.

such that for any choice of B ⊂ A there is always (at least) one initial condition x(0) ∈ B such that itssolution goes out of A.In other words, x∗ is unstable if there are solutions starting “close” to x∗ that eventually go “faraway” from x∗.

Graphically:

n is continuous.

such that for any choice of B ⊂ A there is always (at least) one initial condition x(0) ∈ B such that itssolution goes out of A.In other words, x∗ is unstable if there are solutions starting “close” to x∗ that eventually go “faraway” from x∗. Graphically:

Lyapunov functions

The principle behind this approach is the following:

Assume we have a system x = f (x) and x∗ isan equilibrium

Assume we can define a functionW : V →R (with V neighborhood of x∗)that is differentiable and has a relativeminimum in x∗

Let’s consider a solution x(t) of the system,and calculateW (x(t)). Then:→ ifW (x(t)) decreases, i.e. W < 0, x(t) will

move towards x∗, so x∗ is asymptoticallystable

→ ifW (x(t)) increases, i.e. W > 0, then x(t)will move away from x∗, and so x∗ isunstable

Lyapunov functions

Let’s consider a solution x(t) of the system,and calculateW (x(t)).

Then:→ ifW (x(t)) decreases, i.e. W < 0, x(t) will

Lyapunov functions

Lyapunov functionsLyapunov’s second theorem

Let x∗ be an equilibrium of the system x = f (x), V a neighborhood of x∗ andW : V →R adifferentiable function with a local minimum in x∗.

1 If W (x(t)) = 0⇒ x∗ is stable at all times

2 If W (x(t)) ≤ 0⇒ x∗ is stable

3 If W (x(t)) < 0⇒ x∗ is asymptotically stable

4 If W (x(t)) > 0⇒ x∗ is unstable

A functionW with the aforementioned properties is called Lyapunov function for x∗.

Pros and cons

This method can give a lot of information on equilibria

It only gives sufficient and not necessary conditions for (in)stability: an equilibrium can be(un)stable without having a Lyapunov function

It is not always easy to find Lyapunov functions (exception: conserved quantities)

Let x∗ be an equilibrium of the system x = f (x), V a neighborhood of x∗ andW : V →R adifferentiable function with a local minimum in x∗. Then:

Pros and cons

Lyapunov functionsExample

Let’s see if we can find a Lyapunov functions for the Lotka-Volterra equations.

= αx − βxydy

dt= δxy −γy (19)

Dividing the first equation by the second:

=αx − βxyδxy −γy

=xy·α − βyδx −γ

⇒ dxδx −γx

= dyα − βyy

⇒ dx(δ −

(αy− β

Integrating on both sides:

δx −γ lnx+ const. = α lny − βy + const. ⇒ W := δx −γ lnx+ βy −α lny = const. (21)

Important

There is no “general recipe” to find a system’s Lyapunov function!

= αx − βxydy

⇒ dxδx −γx

= dyα − βyy

⇒ dx(δ −

(αy− β

Important

= αx − βxydy

⇒ dxδx −γx

= dyα − βyy

⇒ dx(δ −

(αy− β

Important

= αx − βxydy

⇒ dxδx −γx

= dyα − βyy

⇒ dx(δ −

(αy− β

Important

= αx − βxydy

⇒ dxδx −γx

= dyα − βyy

⇒ dx(δ −

(αy− β

δx −γ lnx+ const. = α lny − βy + const.

⇒ W := δx −γ lnx+ βy −α lny = const. (21)

Important

= αx − βxydy

⇒ dxδx −γx

= dyα − βyy

⇒ dx(δ −

(αy− β

Important

= αx − βxydy

⇒ dxδx −γx

= dyα − βyy

⇒ dx(δ −

(αy− β

Important

We have found the functionW = δx −γ lnx+ βy −α lny.

Let’s find its extrema:(∂xW∂yW

(δ −γ/xβ −α/y

)= 0 ⇔

x = γ/δy = α/β

⇒(γ/δα/β

)is the minimum ofW (22)

NoticeThe partial derivatives ofW diverge for (x,y) = (0,0)!

→ The equilibrium of the Lotka-Volterra equations is a (global) minimum forW .Let’s see howW behaves on the trajectories of the system:

W = δx −γ xx

+ βy −αy

L-V eq= 0 (23)

ConclusionThe equilibrium (γ/δ,α/β) of the Lotka-Volterra equations is stable at all times.We can’t say anything on the equilibrium (0,0).

We have found the functionW = δx −γ lnx+ βy −α lny. Let’s find its extrema:(∂xW∂yW

x = γ/δy = α/β

⇒(γ/δα/β

W = δx −γ xx

+ βy −αy

L-V eq= 0 (23)

)= 0 ⇔

x = γ/δy = α/β

⇒(γ/δα/β

W = δx −γ xx

+ βy −αy

L-V eq= 0 (23)

)= 0 ⇔

x = γ/δy = α/β

⇒(γ/δα/β

W = δx −γ xx

+ βy −αy

L-V eq= 0 (23)

)= 0 ⇔

x = γ/δy = α/β

⇒(γ/δα/β

→ The equilibrium of the Lotka-Volterra equations is a (global) minimum forW .

Let’s see howW behaves on the trajectories of the system:

W = δx −γ xx

+ βy −αy

L-V eq= 0 (23)

)= 0 ⇔

x = γ/δy = α/β

⇒(γ/δα/β

W = δx −γ xx

+ βy −αy

L-V eq= 0 (23)

)= 0 ⇔

x = γ/δy = α/β

⇒(γ/δα/β

W = δx −γ xx

+ βy −αy

L-V eq=

0 (23)

)= 0 ⇔

x = γ/δy = α/β

⇒(γ/δα/β

W = δx −γ xx

+ βy −αy

L-V eq= 0

)= 0 ⇔

x = γ/δy = α/β

⇒(γ/δα/β

W = δx −γ xx

+ βy −αy

L-V eq= 0 (23)

Let’s take a look at the trajectories of the Lotka-Volterra equations:

0 1 2 3 4 5 6x (preys)

Trajectories of the Lotka-Volterra equations

0 5 10 15 20 25 30Time

Solution of the Lotka-Volterra equationx (preys)y (predators)

x0 = y0 = 3

Simulations with α = 2/3, β = 1/3, γ = 2, δ = 1.

Let’s take a look at the trajectories of the Lotka-Volterra equations:

0 1 2 3 4 5 6x (preys)

Trajectories of the Lotka-Volterra equations

0 5 10 15 20 25 30Time

Solution of the Lotka-Volterra equationx (preys)y (predators)

x0 = y0 = 3

Simulations with α = 2/3, β = 1/3, γ = 2, δ = 1.

Spectral analysis

The simplest approach that can be taken to study the stability of a non-linear system islinearizing it around its equilibria.

The underlying principle is that a non-linear system can be approximated by a linear one, if weare close enough to an equilibrium x∗:

x = f (x) = f (x∗)︸︷︷︸=0

+J(x − x∗) +H(x − x∗)2 +O(‖x − x∗‖3) ⇒ x ∼ J(x − x∗) (24)

where J is the Jacobian matrix of f computed in x∗:

Jij =∂if

∂xj(x∗) (f : Ω ⊂R

n→Rn) (25)

The eigenvalues of the jacobian matrix give us valuable information on the stability of x∗.

Spectral analysis

x = f (x) = f (x∗)︸︷︷︸=0

+J(x − x∗) +H(x − x∗)2 +O(‖x − x∗‖3) ⇒ x ∼ J(x − x∗) (24)

Jij =∂if

∂xj(x∗) (f : Ω ⊂R

n→Rn) (25)

Spectral analysis

x = f (x) = f (x∗)︸︷︷︸=0

+J(x − x∗) +H(x − x∗)2 +O(‖x − x∗‖3) ⇒ x ∼ J(x − x∗) (24)

Jij =∂if

∂xj(x∗) (f : Ω ⊂R

n→Rn) (25)

Spectral analysis

x = f (x) = f (x∗)︸︷︷︸=0

+J(x − x∗) +H(x − x∗)2 +O(‖x − x∗‖3)

⇒ x ∼ J(x − x∗) (24)

Jij =∂if

∂xj(x∗) (f : Ω ⊂R

n→Rn) (25)

Spectral analysis

x = f (x) = f (x∗)︸︷︷︸=0

+J(x − x∗) +H(x − x∗)2 +O(‖x − x∗‖3) ⇒ x ∼ J(x − x∗) (24)

Jij =∂if

∂xj(x∗) (f : Ω ⊂R

n→Rn) (25)

Spectral analysis

x = f (x) = f (x∗)︸︷︷︸=0

+J(x − x∗) +H(x − x∗)2 +O(‖x − x∗‖3) ⇒ x ∼ J(x − x∗) (24)

Jij =∂if

∂xj(x∗) (f : Ω ⊂R

n→Rn) (25)

Spectral analysis

x = f (x) = f (x∗)︸︷︷︸=0

+J(x − x∗) +H(x − x∗)2 +O(‖x − x∗‖3) ⇒ x ∼ J(x − x∗) (24)

Jij =∂if

∂xj(x∗) (f : Ω ⊂R

n→Rn) (25)

Spectral analysisLyapunov’s first theorem

Let x∗ be an equilibrium of the non-linear system x = f (x), x ∈Rn. Then:

1 All eigenvalues of J have negative real part⇒ x∗ is asymptotically stable2 At least one eigenvalue of J has positive real part⇒ x∗ is unstable

Pros and cons

This approach can be used in any non-linear system

The stability properties of a non-linear system are the same of its linearization

If at least one of the eigenvalues has null real part (i.e., it’s 0 or purely imaginary) and allothers have negative real part, we can’t say anything on x∗ (→ higher orders)

Marginal stability

An equilibrium is said to be marginally stable if it is neither asymptotically stable nor unstable.3 All the eigenvalues of J are purely imaginary⇒ x∗ is marginally stable

Pros and cons

Marginal stability

1 All eigenvalues of J have negative real part⇒ x∗ is asymptotically stable

2 At least one eigenvalue of J has positive real part⇒ x∗ is unstable

Pros and cons

Marginal stability

Pros and cons

Marginal stability

Pros and cons

Marginal stability

Pros and cons

Marginal stability

Pros and cons

Marginal stability

Pros and cons

Marginal stability

Pros and cons

Marginal stability

Pros and cons

Marginal stability

An equilibrium is said to be marginally stable if it is neither asymptotically stable nor unstable.

3 All the eigenvalues of J are purely imaginary⇒ x∗ is marginally stable

Pros and cons

Marginal stability

Spectral analysisExamples

Let’s see a couple of general examples.

x = x3 − x2 − 2x (26)

0 2−1

f (x) = x3 − x2 − 2x (27a)

In this case the Jacobian is simply the derivative:

f ′(x) = 3x2 − 2x − 2 (27b)

and computing it in the three equilibria:

f ′(−1) = 3 f ′(0) = −2 f ′(2) = 6 (27c)

Therefore:x∗ = −1,x∗ = 2→ unstablex∗ = 0→ asymptotically stable

x = x3 − x2 − 2x (26)

0 2−1

f (x) = x3 − x2 − 2x (27a)

f ′(x) = 3x2 − 2x − 2 (27b)

f ′(−1) = 3 f ′(0) = −2 f ′(2) = 6 (27c)

x = x3 − x2 − 2x (26)

0 2−1

f (x) = x3 − x2 − 2x (27a)

f ′(x) = 3x2 − 2x − 2 (27b)

f ′(−1) = 3 f ′(0) = −2 f ′(2) = 6 (27c)

x = x3 − x2 − 2x (26)

0 2−1

f (x) = x3 − x2 − 2x (27a)

f ′(x) = 3x2 − 2x − 2 (27b)

f ′(−1) = 3 f ′(0) = −2 f ′(2) = 6 (27c)

x = x3 − x2 − 2x (26)

0 2−1

f (x) = x3 − x2 − 2x (27a)

f ′(x) = 3x2 − 2x − 2 (27b)

f ′(−1) = 3 f ′(0) = −2 f ′(2) = 6 (27c)

x = x3 − x2 − 2x (26)

0 2−1

f (x) = x3 − x2 − 2x (27a)

f ′(x) = 3x2 − 2x − 2 (27b)

f ′(−1) = 3 f ′(0) = −2 f ′(2) = 6 (27c)

x = −x y = ky3 with k > 0 (28)

(−xky3

)(29a)

The only equilibrium is (x,y) = (0,0), and the ja-cobian matrix is:

J =(−1 00 3ky2

=(−1 00 0

)(29b)

Therefore, the eigenvalues are −1 and 0 → wecan’t say anything about the equilibrium!

What if we draw the stream plot?

The equilibrium is unstable.In this case (0,0) is called saddle point.

What happens if k < 0 or k = 0?

x = −x y = ky3 with k > 0 (28)

(−xky3

)(29a)

J =(−1 00 3ky2

=(−1 00 0

)(29b)

x = −x y = ky3 with k > 0 (28)

(−xky3

)(29a)

J =(−1 00 3ky2

=(−1 00 0

)(29b)

x = −x y = ky3 with k > 0 (28)

(−xky3

)(29a)

J =(−1 00 3ky2

=(−1 00 0

)(29b)

x = −x y = ky3 with k > 0 (28)

(−xky3

)(29a)

J =(−1 00 3ky2

=(−1 00 0

)(29b)

x = −x y = ky3 with k > 0 (28)

(−xky3

)(29a)

J =(−1 00 3ky2

=(−1 00 0

)(29b)

x > 0y < 0

x < 0y < 0

x > 0y > 0

x < 0y > 0

x = −x y = ky3 with k > 0 (28)

(−xky3

)(29a)

J =(−1 00 3ky2

=(−1 00 0

)(29b)

x > 0y < 0

x < 0y < 0

x > 0y > 0

x < 0y > 0

x = −x y = ky3 with k > 0 (28)

(−xky3

)(29a)

J =(−1 00 3ky2

=(−1 00 0

)(29b)

x > 0y < 0

x < 0y < 0

x > 0y > 0

x < 0y > 0

x > 0y < 0

x < 0y < 0

x > 0y > 0

x < 0y > 0

x = −x y = ky3 with k > 0 (28)

(−xky3

)(29a)

J =(−1 00 3ky2

=(−1 00 0

)(29b)

x > 0y < 0

x < 0y < 0

x > 0y > 0

x < 0y > 0

x > 0y < 0

x < 0y < 0

x > 0y > 0

x < 0y > 0

x = −x y = ky3 with k > 0 (28)

(−xky3

)(29a)

J =(−1 00 3ky2

=(−1 00 0

)(29b)

-2 -1 0 1 2

x = −x y = ky3 with k > 0 (28)

(−xky3

)(29a)

J =(−1 00 3ky2

=(−1 00 0

)(29b)

-2 -1 0 1 2

x = −x y = ky3 with k > 0 (28)

(−xky3

)(29a)

J =(−1 00 3ky2

=(−1 00 0

)(29b)

-2 -1 0 1 2

Let’s study both the logistic equation and the Lotka-Volterra system with the spectral analysis.

Logistic equation:

x = f (x) = rx(1− x

)⇒ f ′(x) = r − 2

rKx (30)

Therefore: f ′(K) = −r < 0⇒ x∗ = K is asymptotically stable, f ′(0) = r > 0⇒ x∗ = 0 is unstable.Lotka-Volterra equations:(

)⇒ J =

(α − βy −βxδy δx −γ

)|x=0,y=0

=(α 00 −γ

)is unstable (31)

Let’s look at the other equilibrium: J =(α − βy −βxδy δx −γ

)|x=γ/δ,y=α/β

0 −βγ/δδα/β 0

Eigenvalues: ±i√γα→ this equilibrium is marginally stable.

Let’s study both the logistic equation and the Lotka-Volterra system with the spectral analysis.Logistic equation:

x = f (x) = rx(1− x

⇒ f ′(x) = r − 2rKx (30)

)⇒ J =

)|x=0,y=0

=(α 00 −γ

)is unstable (31)

)|x=γ/δ,y=α/β

x = f (x) = rx(1− x

)⇒ f ′(x) = r − 2

rKx (30)

)⇒ J =

)|x=0,y=0

=(α 00 −γ

)is unstable (31)

)|x=γ/δ,y=α/β

x = f (x) = rx(1− x

)⇒ f ′(x) = r − 2

rKx (30)

Therefore: f ′(K) = −r < 0⇒ x∗ = K is asymptotically stable,

f ′(0) = r > 0⇒ x∗ = 0 is unstable.Lotka-Volterra equations:(

)⇒ J =

)|x=0,y=0

=(α 00 −γ

)is unstable (31)

)|x=γ/δ,y=α/β

x = f (x) = rx(1− x

)⇒ f ′(x) = r − 2

rKx (30)

Therefore: f ′(K) = −r < 0⇒ x∗ = K is asymptotically stable, f ′(0) = r > 0⇒ x∗ = 0 is unstable.

Lotka-Volterra equations:(xy

)⇒ J =

)|x=0,y=0

=(α 00 −γ

)is unstable (31)

)|x=γ/δ,y=α/β

x = f (x) = rx(1− x

)⇒ f ′(x) = r − 2

rKx (30)

Therefore: f ′(K) = −r < 0⇒ x∗ = K is asymptotically stable, f ′(0) = r > 0⇒ x∗ = 0 is unstable.Lotka-Volterra equations:

)⇒ J =

)|x=0,y=0

=(α 00 −γ

)is unstable (31)

)|x=γ/δ,y=α/β

x = f (x) = rx(1− x

)⇒ f ′(x) = r − 2

rKx (30)

⇒ J =(α − βy −βxδy δx −γ

)|x=0,y=0

=(α 00 −γ

)is unstable (31)

)|x=γ/δ,y=α/β

x = f (x) = rx(1− x

)⇒ f ′(x) = r − 2

rKx (30)

)⇒ J =

)|x=0,y=0

=(α 00 −γ

)is unstable (31)

)|x=γ/δ,y=α/β

x = f (x) = rx(1− x

)⇒ f ′(x) = r − 2

rKx (30)

)⇒ J =

)|x=0,y=0

=(α 00 −γ

⇒(00

)is unstable (31)

)|x=γ/δ,y=α/β

x = f (x) = rx(1− x

)⇒ f ′(x) = r − 2

rKx (30)

)⇒ J =

)|x=0,y=0

=(α 00 −γ

)is unstable (31)

)|x=γ/δ,y=α/β

x = f (x) = rx(1− x

)⇒ f ′(x) = r − 2

rKx (30)

)⇒ J =

)|x=0,y=0

=(α 00 −γ

)is unstable (31)

Let’s look at the other equilibrium:

J =(α − βy −βxδy δx −γ

)|x=γ/δ,y=α/β

x = f (x) = rx(1− x

)⇒ f ′(x) = r − 2

rKx (30)

)⇒ J =

)|x=0,y=0

=(α 00 −γ

)is unstable (31)

)|x=γ/δ,y=α/β

x = f (x) = rx(1− x

)⇒ f ′(x) = r − 2

rKx (30)

)⇒ J =

)|x=0,y=0

=(α 00 −γ

)is unstable (31)

)|x=γ/δ,y=α/β

x = f (x) = rx(1− x

)⇒ f ′(x) = r − 2

rKx (30)

)⇒ J =

)|x=0,y=0

=(α 00 −γ

)is unstable (31)

)|x=γ/δ,y=α/β

Eigenvalues: ±i√γα

→ this equilibrium is marginally stable.

x = f (x) = rx(1− x

)⇒ f ′(x) = r − 2

rKx (30)

)⇒ J =

)|x=0,y=0

=(α 00 −γ

)is unstable (31)

)|x=γ/δ,y=α/β

Another exampleLet’s see some other interesting examples.

x = −x3 + x = −x(x2 − 1) = f (x) (33)

The equilibria are x∗ = 0 and x∗ = ±1.

The stream plot already suggests that x∗ = ±1 arestable, while x∗ = 0 is unstable. Using spectralanalysis:

f ′(x) = −3x2 + 1 (34a)

f ′(0) = 1 f ′(±1) = −2 (34b)

Therefore, x∗ = ±1 are indeed stable, while x∗ =0 is unstable. This is an example of bistability.

x = −x3 + x = −x(x2 − 1) = f (x) (33)

f ′(x) = −3x2 + 1 (34a)

f ′(0) = 1 f ′(±1) = −2 (34b)

x = −x3 + x = −x(x2 − 1) = f (x) (33)

0 1−1

f ′(x) = −3x2 + 1 (34a)

f ′(0) = 1 f ′(±1) = −2 (34b)

x = −x3 + x = −x(x2 − 1) = f (x) (33)

0 1−1

f ′(x) = −3x2 + 1 (34a)

f ′(0) = 1 f ′(±1) = −2 (34b)

x = −x3 + x = −x(x2 − 1) = f (x) (33)

0 1−1

f ′(x) = −3x2 + 1 (34a)

f ′(0) = 1 f ′(±1) = −2 (34b)

x = −x3 + x = −x(x2 − 1) = f (x) (33)

0 1−1

f ′(x) = −3x2 + 1 (34a)

f ′(0) = 1 f ′(±1) = −2 (34b)

x = −x3 + x = −x(x2 − 1) = f (x) (33)

0 1−1

f ′(x) = −3x2 + 1 (34a)

f ′(0) = 1 f ′(±1) = −2 (34b)

Therefore, x∗ = ±1 are indeed stable, while x∗ =0 is unstable.

This is an example of bistability.

x = −x3 + x = −x(x2 − 1) = f (x) (33)

0 1−1

f ′(x) = −3x2 + 1 (34a)

f ′(0) = 1 f ′(±1) = −2 (34b)

Bistability

If we look at the system as a particle in a potential:

x = −V ′(x) with V (x) =x4

4− x

−V ′(x)

−x3 + x

0 1−1

The system has two alternative stable equilibria.

External perturbations can move the systemfrom one equilibrium to the other.

Bistability

x = −V ′(x) with V (x) =x4

4− x

⇒(xv

−V ′(x)

−x3 + x

0 1−1

Bistability

x = −V ′(x) with V (x) =x4

4− x

−V ′(x)

−x3 + x

0 1−1

Bistability

x = −V ′(x) with V (x) =x4

4− x

−V ′(x)

−x3 + x

0 1−1

Bistability

x = −V ′(x) with V (x) =x4

4− x

−V ′(x)

−x3 + x

0 1−1

Bistability

x = −V ′(x) with V (x) =x4

4− x

−V ′(x)

−x3 + x

0 1−1

Timescale separation

A technique that is often useful in studying non-linear systems is that of timescale separation.In abstract, assume a system is described by two sets of variables xi and yj :

xi = fi(~x,~y) yj = gj (~x,~y) with i = 1, . . . ,Nx j = 1, . . . ,Ny (36)

It often happens that the timescales of the two variables are very different→ we can distinguishbetween “slow” and “fast” variables.Let’s assume xi → slow, yj → fast⇒ yj reach their stationary value very quickly⇒ we assumeyj = 0. Therefore:

yj = gj (~x,~y) = 0 ⇒ ~y = g−1(~x) ⇒ xi = fi(~x,g−1(~x)) (37)

This allows us to write the system only with xi .

NoticeThe validity of this approach depends on the system considered. We need somephenomenological knowledge on the system to justify this separation of timescales.

Timescale separationA technique that is often useful in studying non-linear systems is that of timescale separation.

In abstract, assume a system is described by two sets of variables xi and yj :

Timescale separationA technique that is often useful in studying non-linear systems is that of timescale separation.In abstract, assume a system is described by two sets of variables xi and yj :

It often happens that the timescales of the two variables are very different

→ we can distinguishbetween “slow” and “fast” variables.Let’s assume xi → slow, yj → fast⇒ yj reach their stationary value very quickly⇒ we assumeyj = 0. Therefore:

It often happens that the timescales of the two variables are very different→ we can distinguishbetween “slow” and “fast” variables.

Let’s assume xi → slow, yj → fast⇒ yj reach their stationary value very quickly⇒ we assumeyj = 0. Therefore:

It often happens that the timescales of the two variables are very different→ we can distinguishbetween “slow” and “fast” variables.Let’s assume xi → slow, yj → fast

⇒ yj reach their stationary value very quickly⇒ we assumeyj = 0. Therefore:

It often happens that the timescales of the two variables are very different→ we can distinguishbetween “slow” and “fast” variables.Let’s assume xi → slow, yj → fast⇒ yj reach their stationary value very quickly

⇒ we assumeyj = 0. Therefore:

yj = gj (~x,~y) = 0

⇒ ~y = g−1(~x) ⇒ xi = fi(~x,g−1(~x)) (37)

yj = gj (~x,~y) = 0 ⇒ ~y = g−1(~x)

⇒ xi = fi(~x,g−1(~x)) (37)

Timescale separationExamples

An example where this is often done is chemical reactions. For example, Gierer-Meinhardt’sactivator-inhibitor equations:

x = a− by + cx2

yy = d · x2 − ey (38)

(x→ activator, y → inhibitor).

1 The activator is much faster than the inhibitor:

x = 0 ⇒ x2 =1cy(−a+ by) ⇒ y = −

+ e)y +

bdcy2 (39)

2 The inhibitor is much faster than the activator:

y = 0 ⇒ y =dex2 ⇒ x = a+

cde− bdex2 (40)

An example where this is often done is chemical reactions.

For example, Gierer-Meinhardt’sactivator-inhibitor equations:

x = a− by + cx2

yy = d · x2 − ey (38)

x = 0 ⇒ x2 =1cy(−a+ by) ⇒ y = −

+ e)y +

bdcy2 (39)

y = 0 ⇒ y =dex2 ⇒ x = a+

cde− bdex2 (40)

x = a− by + cx2

yy = d · x2 − ey (38)

x = 0 ⇒ x2 =1cy(−a+ by) ⇒ y = −

+ e)y +

bdcy2 (39)

y = 0 ⇒ y =dex2 ⇒ x = a+

cde− bdex2 (40)

x = a− by + cx2

yy = d · x2 − ey (38)

x = 0 ⇒ x2 =1cy(−a+ by) ⇒ y = −

+ e)y +

bdcy2 (39)

y = 0 ⇒ y =dex2 ⇒ x = a+

cde− bdex2 (40)

x = a− by + cx2

yy = d · x2 − ey (38)

⇒ x2 =1cy(−a+ by) ⇒ y = −

+ e)y +

bdcy2 (39)

y = 0 ⇒ y =dex2 ⇒ x = a+

cde− bdex2 (40)

x = a− by + cx2

yy = d · x2 − ey (38)

x = 0 ⇒ x2 =1cy(−a+ by)

⇒ y = −(adc

+ e)y +

bdcy2 (39)

y = 0 ⇒ y =dex2 ⇒ x = a+

cde− bdex2 (40)

x = a− by + cx2

yy = d · x2 − ey (38)

x = 0 ⇒ x2 =1cy(−a+ by) ⇒ y = −

+ e)y +

bdcy2 (39)

y = 0 ⇒ y =dex2 ⇒ x = a+

cde− bdex2 (40)

x = a− by + cx2

yy = d · x2 − ey (38)

x = 0 ⇒ x2 =1cy(−a+ by) ⇒ y = −

+ e)y +

bdcy2 (39)

y = 0 ⇒ y =dex2 ⇒ x = a+

cde− bdex2 (40)

x = a− by + cx2

yy = d · x2 − ey (38)

x = 0 ⇒ x2 =1cy(−a+ by) ⇒ y = −

+ e)y +

bdcy2 (39)

⇒ y =dex2 ⇒ x = a+

cde− bdex2 (40)

x = a− by + cx2

yy = d · x2 − ey (38)

x = 0 ⇒ x2 =1cy(−a+ by) ⇒ y = −

+ e)y +

bdcy2 (39)

y = 0 ⇒ y =dex2

⇒ x = a+cde− bdex2 (40)

x = a− by + cx2

yy = d · x2 − ey (38)

x = 0 ⇒ x2 =1cy(−a+ by) ⇒ y = −

+ e)y +

bdcy2 (39)

y = 0 ⇒ y =dex2 ⇒ x = a+

cde− bdex2 (40)

Let’s see an ecologically relavant example: MacArthur’s consumer-resource model.

It describes a system of NS speciescompeting for NR resources

The resources are supplied with constantrates, and the species uptake the resources

No other interaction between the species ispresent

The equations of the model are:

nσ = nσ

NR∑i=1

viασiri(ci)− δσ

ci = si −NS∑σ=1

nσασiri(ci) (41)

where:ri(ci) =

cici +Ki

and σ = 1, . . . ,NS and i = 1, . . . ,NR (42)

How is the consumer-resource model built?

nσ = nσ

NR∑i=1

nσασiri(ci) (41)

where:ri(ci) =

cici +Ki

and σ = 1, . . . ,NS and i = 1, . . . ,NR (42)

nσ = nσ

NR∑i=1

nσασiri(ci) (41)

where:ri(ci) =

cici +Ki

and σ = 1, . . . ,NS and i = 1, . . . ,NR (42)

nσ = nσ

NR∑i=1

nσασiri(ci) (41)

where:ri(ci) =

cici +Ki

and σ = 1, . . . ,NS and i = 1, . . . ,NR (42)

nσ = nσ

NR∑i=1

nσασiri(ci) (41)

where:ri(ci) =

cici +Ki

and σ = 1, . . . ,NS and i = 1, . . . ,NR (42)

How is the consumer-resource model built?27

nσ = nσ

NR∑i=1

nσασiri(ci) (43)

It is often assumed that the resources ci are faster than the populations nσ (→ the moleculardynamics is faster than the population dynamics). Therefore:

ci = 0 ⇒ ri(ci) =si∑NS

σ=1nσασi⇒ nσ = nσ

NR∑i=1

viασisi∑NS

ρ=1nραρi− δσ

We can therefore describe the system using only the species’ populations.

nσ = nσ

NR∑i=1

nσασiri(ci) (43)

It is often assumed that the resources ci are faster than the populations nσ (→ the moleculardynamics is faster than the population dynamics).

Therefore:

NR∑i=1

viασisi∑NS

nσ = nσ

NR∑i=1

nσασiri(ci) (43)

ci = 0

⇒ ri(ci) =si∑NS

NR∑i=1

viασisi∑NS

nσ = nσ

NR∑i=1

nσασiri(ci) (43)

σ=1nσασi

⇒ nσ = nσ

NR∑i=1

viασisi∑NS

nσ = nσ

NR∑i=1

nσασiri(ci) (43)

NR∑i=1

viασisi∑NS

Let’s now consider the consumer-resource model with some slight changes:

nσ = nσ

NR∑i=1

viασici − δσ

ci = gici

(1− ci

)−NS∑σ=1

nσασici (45)

If we apply again the separation of timescales so that ci are the fast variables:

ci = 0 ⇒ ci =Qi

1− 1gi

NS∑σ=1

nσασi

Substituting in nσ and rearranging , we get:

nσ = nσλσ

1−NS∑ρ=1

βρσnρ

where: λσ =NR∑i=1

viασiQi − δσ βρσ =1λσ

NR∑i=1

viQigiασiαρi (47)

These are called generalized Lotka-Volterra equations.

nσ = nσ

NR∑i=1

viασici − δσ

ci = gici

(1− ci

)−NS∑σ=1

nσασici (45)

ci = 0 ⇒ ci =Qi

1− 1gi

NS∑σ=1

nσασi

nσ = nσλσ

1−NS∑ρ=1

βρσnρ

NR∑i=1

nσ = nσ

NR∑i=1

viασici − δσ

ci = gici

(1− ci

)−NS∑σ=1

nσασici (45)

ci = 0 ⇒ ci =Qi

1− 1gi

NS∑σ=1

nσασi

nσ = nσλσ

1−NS∑ρ=1

βρσnρ

NR∑i=1

nσ = nσ

NR∑i=1

viασici − δσ

ci = gici

(1− ci

)−NS∑σ=1

nσασici (45)

ci = 0 ⇒ ci =Qi

1− 1gi

NS∑σ=1

nσασi

nσ = nσλσ

1−NS∑ρ=1

βρσnρ

NR∑i=1

nσ = nσ

NR∑i=1

viασici − δσ

ci = gici

(1− ci

)−NS∑σ=1

nσασici (45)

ci = 0 ⇒ ci =Qi

1− 1gi

NS∑σ=1

nσασi

nσ = nσλσ

1−NS∑ρ=1

βρσnρ

NR∑i=1

That’s all!

Questions?

Backup slides

Solution of the logistic equationLet’s see how we can solve the logistic equation analytically:

x = rx(1− x

⇒ dxx (1− x/K)

= rdt ⇒ dxx

1− x/K= rdt (48)

lnx+ c1 − ln(1− x

)+ c2 = rt + c3 ⇒ ln

x1− x/K

= c+ rt ⇒ x1− x/K

= ec+rt = Cert (49)

We can determine the value of C by computing this expression at t = 0:

1− x0/K=

K − x0(50)

Therefore, with simple rearrangements:

x1− x/K

K − x0ert ⇒ x(t) =

x0 + (K − x0)e−rt(51)

Back to original slide

x = rx(1− x

)⇒ dx

x (1− x/K)= rdt

⇒ dxx

1− x/K= rdt (48)

)+ c2 = rt + c3 ⇒ ln

x1− x/K

= c+ rt ⇒ x1− x/K

= ec+rt = Cert (49)

1− x0/K=

K − x0(50)

x1− x/K

K − x0ert ⇒ x(t) =

x0 + (K − x0)e−rt(51)

x = rx(1− x

)⇒ dx

x (1− x/K)= rdt ⇒ dx

x+dx/K

1− x/K= rdt (48)

)+ c2 = rt + c3 ⇒ ln

x1− x/K

= c+ rt ⇒ x1− x/K

= ec+rt = Cert (49)

1− x0/K=

K − x0(50)

x1− x/K

K − x0ert ⇒ x(t) =

x0 + (K − x0)e−rt(51)

x = rx(1− x

)⇒ dx

x+dx/K

1− x/K= rdt (48)

)+ c2 = rt + c3

⇒ lnx

1− x/K= c+ rt ⇒ x

1− x/K= ec+rt = Cert (49)

1− x0/K=

K − x0(50)

x1− x/K

K − x0ert ⇒ x(t) =

x0 + (K − x0)e−rt(51)

x = rx(1− x

)⇒ dx

x+dx/K

1− x/K= rdt (48)

)+ c2 = rt + c3 ⇒ ln

x1− x/K

= c+ rt

⇒ x1− x/K

= ec+rt = Cert (49)

1− x0/K=

K − x0(50)

x1− x/K

K − x0ert ⇒ x(t) =

x0 + (K − x0)e−rt(51)

x = rx(1− x

)⇒ dx

x+dx/K

1− x/K= rdt (48)

)+ c2 = rt + c3 ⇒ ln

x1− x/K

= c+ rt ⇒ x1− x/K

= ec+rt = Cert (49)

1− x0/K=

K − x0(50)

x1− x/K

K − x0ert ⇒ x(t) =

x0 + (K − x0)e−rt(51)

x = rx(1− x

)⇒ dx

x+dx/K

1− x/K= rdt (48)

)+ c2 = rt + c3 ⇒ ln

x1− x/K

= c+ rt ⇒ x1− x/K

= ec+rt = Cert (49)

1− x0/K=

K − x0(50)

x1− x/K

K − x0ert ⇒ x(t) =

x0 + (K − x0)e−rt(51)

x = rx(1− x

)⇒ dx

x+dx/K

1− x/K= rdt (48)

)+ c2 = rt + c3 ⇒ ln

x1− x/K

= c+ rt ⇒ x1− x/K

= ec+rt = Cert (49)

1− x0/K=

K − x0(50)

x1− x/K

K − x0ert

⇒ x(t) =Kx0

x0 + (K − x0)e−rt(51)

x = rx(1− x

)⇒ dx

x+dx/K

1− x/K= rdt (48)

)+ c2 = rt + c3 ⇒ ln

x1− x/K

= c+ rt ⇒ x1− x/K

= ec+rt = Cert (49)

1− x0/K=

K − x0(50)

x1− x/K

K − x0ert ⇒ x(t) =

x0 + (K − x0)e−rt(51)

Example on spectral analysis

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

The jacobian matrix is always the same→→ it doesn’t depend on k.Let’s draw the stream plot for k < 0.The equilibrium is asymptotically stable!Let’s draw the stream plot for k = 0.All points with x = 0 are equilibria.

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

The jacobian matrix is always the same→→ it doesn’t depend on k.Let’s draw the stream plot for k < 0.The equilibrium is asymptotically stable!Let’s draw the stream plot for k = 0.All points with x = 0 are equilibria.

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

The jacobian matrix is always the same→→ it doesn’t depend on k.

Let’s draw the stream plot for k < 0.The equilibrium is asymptotically stable!Let’s draw the stream plot for k = 0.All points with x = 0 are equilibria.

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

The jacobian matrix is always the same→→ it doesn’t depend on k.Let’s draw the stream plot for k < 0.

The equilibrium is asymptotically stable!Let’s draw the stream plot for k = 0.All points with x = 0 are equilibria.

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

x > 0y > 0

x < 0y < 0

x > 0y > 0

x < 0y > 0

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

x > 0y > 0

x < 0y < 0

x > 0y > 0

x < 0y > 0

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

x > 0y > 0

x < 0y < 0

x > 0y > 0

x < 0y > 0

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

x > 0y > 0

x < 0y < 0

x > 0y > 0

x < 0y > 0

x > 0y > 0

x < 0y > 0

x > 0y < 0

x < 0y < 0

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

x > 0y > 0

x < 0y < 0

x > 0y > 0

x < 0y > 0

x > 0y > 0

x < 0y > 0

x > 0y < 0

x < 0y < 0

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

The jacobian matrix is always the same→→ it doesn’t depend on k.Let’s draw the stream plot for k < 0.The equilibrium is asymptotically stable!

Let’s draw the stream plot for k = 0.All points with x = 0 are equilibria.

-2 -1 0 1 2

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

Let’s draw the stream plot for k < 0.The equilibrium is asymptotically stable!

Let’s draw the stream plot for k = 0.

All points with x = 0 are equilibria.

-2 -1 0 1 2

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

x > 0y > 0

x < 0y < 0

x > 0y > 0

x < 0y > 0

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

x > 0y > 0

x < 0y < 0

x > 0y > 0

x < 0y > 0

x > 0y = 0

x < 0y = 0

x > 0y = 0

x < 0y = 0

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

x > 0y > 0

x < 0y < 0

x > 0y > 0

x < 0y > 0

x > 0y = 0

x < 0y = 0

x > 0y = 0

x < 0y = 0

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

-2 -1 0 1 2

x = −x y = ky3 with k < 0 and k = 0 (52)

J =(−1 00 3ky2

=(−1 00 0

)(53a)

-2 -1 0 1 2

The consumer-resource model

Let’s see how the consuemr-resource model is built.

We start from a very general form:

nσ = nσ (growth−death) ci = supply−uptake (54)

We assume that:

each species σ uptakes resource i with a rate Jσi

this uptake is converted proportionally to a growth contribution, i.e. g(i)σ = viJσi

the total growth rate is the sum of all these contributions, i.e. gσ =∑NRi=1 g

the death rates δσ are constant

the resurces are supplied with constant rates siTherefore:

nσ = nσ

NR∑i=1

viJσi − δσ

Jσinσ (55)

Let’s see how the consuemr-resource model is built. We start from a very general form:

We assume that:

nσ = nσ

NR∑i=1

viJσi − δσ

Jσinσ (55)

We assume that:

nσ = nσ

NR∑i=1

viJσi − δσ

Jσinσ (55)

We assume that:

nσ = nσ

NR∑i=1

viJσi − δσ

Jσinσ (55)

We assume that:

nσ = nσ

NR∑i=1

viJσi − δσ

Jσinσ (55)

We assume that:

nσ = nσ

NR∑i=1

viJσi − δσ

Jσinσ (55)

We assume that:

the resurces are supplied with constant rates si

Therefore:

nσ = nσ

NR∑i=1

viJσi − δσ

Jσinσ (55)

We assume that:

nσ = nσ

NR∑i=1

viJσi − δσ

Jσinσ (55)

We need some assumption for the uptake rates Jσi .

We use:

Jσi = ασic

ci +Ki(56)

This way the uptake rate saturates at high resource concentrations. Therefore:

nσ = nσ

NR∑i=1

viασic

ci +Ki− δσ

nσασic

ci +Ki, (57)

Alternatively, we can assume Jσi = ασici , but we need a logistic term instead of si to limit theuptake rates:

nσ = nσ

NR∑i=1

viασici − δσ

ci = gici

(1− ci

)−NS∑σ=1

nσασici , (58)

We need some assumption for the uptake rates Jσi . We use:

Jσi = ασic

ci +Ki(56)

This way the uptake rate saturates at high resource concentrations.

Therefore:

nσ = nσ

NR∑i=1

viασic

ci +Ki− δσ

nσασic

ci +Ki, (57)

nσ = nσ

NR∑i=1

viασici − δσ

ci = gici

(1− ci

)−NS∑σ=1

nσασici , (58)

Jσi = ασic

ci +Ki(56)

nσ = nσ

NR∑i=1

viασic

ci +Ki− δσ

nσασic

ci +Ki, (57)

nσ = nσ

NR∑i=1

viασici − δσ

ci = gici

(1− ci

)−NS∑σ=1

nσασici , (58)

Jσi = ασic

ci +Ki(56)

nσ = nσ

NR∑i=1

viασic

ci +Ki− δσ

nσασic

ci +Ki, (57)

nσ = nσ

NR∑i=1

viασici − δσ

ci = gici

(1− ci

)−NS∑σ=1

nσασici , (58)

From C-R to generalized L-V

nσ = nσ

NR∑i=1

viασici − δσ

ci = gici

(1− ci

)−NS∑σ=1

nσασici

ci = 0 ⇒ ci =Qi

1− 1gi

NS∑σ=1

nσασi

Substituting in nσ :

NR∑i=1

viασiQi −NR∑i=1

viασiQigi

NS∑ρ=1

nραρi − δσ

NR∑i=1

viασiQi − δσ

1− 1∑NR

i=1 viασiQi − δσ

NS∑ρ=1

NS∑i=1

viQigiασiαρi

nσ = nσ

NR∑i=1

viασici − δσ

ci = gici

(1− ci

)−NS∑σ=1

nσασici

ci = 0 ⇒ ci =Qi

1− 1gi

NS∑σ=1

nσασi

Substituting in nσ :

NR∑i=1

viασiQi −NR∑i=1

viασiQigi

NS∑ρ=1

nραρi − δσ

NR∑i=1

viασiQi − δσ

1− 1∑NR

NS∑ρ=1

NS∑i=1

viQigiασiαρi

nσ = nσ

NR∑i=1

viασiQi − δσ

1− 1∑NR

NS∑ρ=1

NS∑i=1

viQigiασiαρi

λσ :=NR∑i=1

viασiQi − δσ ⇒ nσ = nσλσ

1− 1λσ

NS∑ρ=1

NS∑i=1

viQigiασiαρi

Finally:

βρσ :=1λσ

NS∑i=1

viQigiασiαρi ⇒ nσ = nσλσ

1−NS∑ρ=1

βρσnρ

nσ = nσ

NR∑i=1

viασiQi − δσ

1− 1∑NR

NS∑ρ=1

NS∑i=1

viQigiασiαρi

λσ :=NR∑i=1

viασiQi − δσ

⇒ nσ = nσλσ

1− 1λσ

NS∑ρ=1

NS∑i=1

viQigiασiαρi

Finally:

βρσ :=1λσ

NS∑i=1

1−NS∑ρ=1

βρσnρ

nσ = nσ

NR∑i=1

viασiQi − δσ

1− 1∑NR

NS∑ρ=1

NS∑i=1

viQigiασiαρi

λσ :=NR∑i=1

1− 1λσ

NS∑ρ=1

NS∑i=1

viQigiασiαρi

Finally:

βρσ :=1λσ

NS∑i=1

1−NS∑ρ=1

βρσnρ

nσ = nσ

NR∑i=1

viασiQi − δσ

1− 1∑NR

NS∑ρ=1

NS∑i=1

viQigiασiαρi

λσ :=NR∑i=1

1− 1λσ

NS∑ρ=1

NS∑i=1

viQigiασiαρi

Finally:

βρσ :=1λσ

NS∑i=1

1−NS∑ρ=1

βρσnρ

tutorial: non-linear dynamics - leonardo pacciani-mori

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