Math155: Calculus for Biological Scientists Analysis of Discrete-Time …shipman/math155/files/Ek... · 2012. 8. 26. · Analysis of Discrete-Time Dynamical Systems David Eklund Cobwebbing:

Math155:Calculus forBiologicalScientistsAnalysis of

Discrete-TimeDynamicalSystems

David Eklund

Math155: Calculus for Biological ScientistsAnalysis of Discrete-Time Dynamical Systems

David Eklund

Colorado State University

August 26, 2012

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David Eklund

Cobwebbing: A graphical solution technique

Consider a discrete-time dynamical system with updatingfunction f :

mt+1 = f(mt).

For example, the graph of f might look like this:

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The diagonal is the line below defined by mt+1 = mt:

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Note that the point (m0,m1) lies on the graph of f ,m1 = f(m0). For example, if m0 = 3.0:

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Cobwebbing step 1: find (m0,m1).

For m0 = 3.0:

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Cobwebbing step 1: find (m0,m1). For m0 = 3.0:

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Step 2: move horizontally to the diagonal, ending at (m1,m1).

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Step 3: move vertically to (m1,m2) on the graph of f .

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Move horizontally to the diagonal again: move to (m2,m2).

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And so on...move vertically to (m2,m3).

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Move horizontally to (m3,m3).

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Move vertically to (m3,m4).

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Move horizontally to (m4,m4)

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We approach a point where the graph of f intersects thediagonal, that is where f(mt) = mt. This is called anequilibrium.

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• That a physical system is at an equilibrium means that therelevant properties remain constant over time.

• In the present context this makes sense: if mt is anequilibrium then f(mt) = mt and therefore

mt+1 = f(mt) = mt,

hence mt+1 = mt and the measured quantity does notchange!

• Physical systems often tend toward some equilibrium overtime.

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What happens if we start at m0 = 1.0?

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Cobweb as before. First move horizontally to the diagonal.

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Move vertically to the graph of f .

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Move horizontally to the diagonal.

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Move horizontally to the diagonal.

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This time we approach a different equilibrium, namely theorigin.

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If we start at m0 = 6.0, we approach the same equilibrium asbefore, when we started at m0 = 3.0.

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If we start at m0 = 6.0, we approach the same equilibrium asbefore. Zooming in:

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The equilibrium in the middle is special. If we start there wenever leave, but if we merely start close to it then we moveaway from it!

The middle equilibrium is called unstable, and the other twoare called stable.

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Examples

Example 1: Consider the medication dynamics from a previouslecture. Let Mt denote the concentration (in mg/l) ofmedication in a persons bloodstream and suppose thisconcentration is measured every day and that the updatingfunction is Mt+1 = 0.5Mt + 1.0.

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Examples

We cobweb from M0 = 1.0:

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Examples


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Examples


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Examples


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Examples


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Examples


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Examples


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Examples

Now we cobweb from M0 = 3.0:

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Examples


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Examples


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Examples


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Examples


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Examples


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Examples


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Examples

In both cases we approach the equilibrium 2.0! Recall that wehave seen this before.

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Examples

Example 2: Now consider the a tree that grows 1.0 meter peryear, let ht denote the tree height (in meters), which ismeasured every year. The updating function is ht+1 = ht +1.0.

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Examples


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Examples


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Examples


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Examples


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Examples


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Examples

The sequence keeps growing! No equilibrium is approached.

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Finding equilibria

Let f be the updating function of a dynamical system:mt+1 = f(mt). An equilibrium is a point m

∗ such that

f(m∗) = m∗.

For some f we can compute the equilibria explicitly.

Example

Consider the medication dynamics from before:

Mt+1 = 0.5Mt + 1.0

If M∗ is an equilibrium, then 0.5M∗ + 1.0 = M∗.

This implies that 1.0 = 0.5M∗, and hence M∗ = 2.0.

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Finding equilibria

Example

Now consider the growing tree from before:

ht+1 = ht + 1.0

If h∗ is an equilibrium, then h∗ = h∗ + 1.0. But this equationhas no solutions and hence there is no equilibrium.

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Finding equilibria

Example

We will find the equilibria of the dynamical systemxt+1 =

cxtxt+1

, where c is some number with c 6= 0.

If x∗ is an equilibrium then x∗ = cx∗

x∗+1 .This implies that x∗(x∗ + 1) = cx∗ and hencex∗(x∗ + 1− c) = 0.Therefore x∗ = 0 or x∗ + 1− c = 0, which gives x∗ = 0 orx∗ = c− 1.These points are really equilibria, as long as x∗ + 1 6= 0, whichis true since c 6= 0.Note that if c = 1, then c− 1 = 0 and hence the equilibriacoincide in this case.Conclusion: When c 6= 1 there are two equilibria, x∗ = 0 andx∗ = c− 1. When c = 1 there is only one equilibrium, x∗ = 0.

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David Eklund

Finding equilibria

Example


cxtxt+1

, where c is some number with c 6= 0.If x∗ is an equilibrium then x∗ = cx

∗

x∗+1 .

This implies that x∗(x∗ + 1) = cx∗ and hencex∗(x∗ + 1− c) = 0.Therefore x∗ = 0 or x∗ + 1− c = 0, which gives x∗ = 0 orx∗ = c− 1.These points are really equilibria, as long as x∗ + 1 6= 0, whichis true since c 6= 0.Note that if c = 1, then c− 1 = 0 and hence the equilibriacoincide in this case.Conclusion: When c 6= 1 there are two equilibria, x∗ = 0 andx∗ = c− 1. When c = 1 there is only one equilibrium, x∗ = 0.

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David Eklund

Finding equilibria

Example


cxtxt+1


∗

x∗+1 .This implies that x∗(x∗ + 1) = cx∗ and hencex∗(x∗ + 1− c) = 0.

Therefore x∗ = 0 or x∗ + 1− c = 0, which gives x∗ = 0 orx∗ = c− 1.These points are really equilibria, as long as x∗ + 1 6= 0, whichis true since c 6= 0.Note that if c = 1, then c− 1 = 0 and hence the equilibriacoincide in this case.Conclusion: When c 6= 1 there are two equilibria, x∗ = 0 andx∗ = c− 1. When c = 1 there is only one equilibrium, x∗ = 0.

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David Eklund

Finding equilibria

Example


cxtxt+1


∗

x∗+1 .This implies that x∗(x∗ + 1) = cx∗ and hencex∗(x∗ + 1− c) = 0.Therefore x∗ = 0 or x∗ + 1− c = 0, which gives x∗ = 0 orx∗ = c− 1.

These points are really equilibria, as long as x∗ + 1 6= 0, whichis true since c 6= 0.Note that if c = 1, then c− 1 = 0 and hence the equilibriacoincide in this case.Conclusion: When c 6= 1 there are two equilibria, x∗ = 0 andx∗ = c− 1. When c = 1 there is only one equilibrium, x∗ = 0.

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David Eklund

Finding equilibria

Example


cxtxt+1


∗

x∗+1 .This implies that x∗(x∗ + 1) = cx∗ and hencex∗(x∗ + 1− c) = 0.Therefore x∗ = 0 or x∗ + 1− c = 0, which gives x∗ = 0 orx∗ = c− 1.These points are really equilibria, as long as x∗ + 1 6= 0, whichis true since c 6= 0.

Note that if c = 1, then c− 1 = 0 and hence the equilibriacoincide in this case.Conclusion: When c 6= 1 there are two equilibria, x∗ = 0 andx∗ = c− 1. When c = 1 there is only one equilibrium, x∗ = 0.

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David Eklund

Finding equilibria

Example


cxtxt+1


∗

x∗+1 .This implies that x∗(x∗ + 1) = cx∗ and hencex∗(x∗ + 1− c) = 0.Therefore x∗ = 0 or x∗ + 1− c = 0, which gives x∗ = 0 orx∗ = c− 1.These points are really equilibria, as long as x∗ + 1 6= 0, whichis true since c 6= 0.Note that if c = 1, then c− 1 = 0 and hence the equilibriacoincide in this case.

Conclusion: When c 6= 1 there are two equilibria, x∗ = 0 andx∗ = c− 1. When c = 1 there is only one equilibrium, x∗ = 0.

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David Eklund

Finding equilibria

Example


cxtxt+1


∗

x∗+1 .This implies that x∗(x∗ + 1) = cx∗ and hencex∗(x∗ + 1− c) = 0.Therefore x∗ = 0 or x∗ + 1− c = 0, which gives x∗ = 0 orx∗ = c− 1.These points are really equilibria, as long as x∗ + 1 6= 0, whichis true since c 6= 0.Note that if c = 1, then c− 1 = 0 and hence the equilibriacoincide in this case.Conclusion: When c 6= 1 there are two equilibria, x∗ = 0 andx∗ = c− 1. When c = 1 there is only one equilibrium, x∗ = 0.

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Math155: Calculus for Biological Scientists Analysis of Discrete-Time …shipman/math155/files/Ek... · 2012. 8. 26. · Analysis of Discrete-Time Dynamical Systems David Eklund Cobwebbing: