Mat 598 Monotone Dynamical Systems in BiologyINSTRUCTOR: Hal L. Smith, PSA 631; halsmith@asu.edu,URL: http://math.la.asu.edu/∼halsmith

TIME 9:15-10:30 TTh LOCATION ED 250 LINE # 87049

COURSE DESCRIPTION: A monotone dynamical system is just a dy-namical system, arising from an ode, delay equation, or pde, for which acomparison principle holds so that “bigger” initial states give rise to big-ger future states in some sense. Monotone systems of odes are often called“cooperative systems” and their time-reversals are called “competitive sys-tems”. Such systems have SIMPLER dynamics than general ode systems:for cooperative systems the typical solution converges to equilibrium-periodicorbits and chaotic invariant sets may exist but their domains of attractionare insignificant; a Poincare-Bendixson theory holds for THREE dimensionalcompetitive systems. Moreover, these systems arise naturally in biologicalsettings such as population dynamics, gene regulatory networks, epidemiol-ogy, and physiology. Even when systems are not monotone, they can some-times be imbedded in larger systems that are monotone from which usefulinformation may be derived. Cyclic feedback systems, related to monotonesystems, often arise in neural networks or gene regulatory networks. Suchsystems will also be considered in the course.

I intend to present the basic results of the theory (proofs are given inreferences below). The main focus will be the application of the theory tosystems of ordinary, delay, and partial differential equations, such as:

1. 3-species-Lotka-Volterra Competition Model

2. Competition between 2 bacterial strains in spatially heterogenous bio-reactors

3. Testosterone Dynamics in vivo

4. Virus Dynamics in vivo

5. The Repressilator: Two genes whose protein products repress eachothers transcription

6. S → E → I → R Epidemic Model

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7. Chemical Reaction Networks

8. 2-sex S → I → S Sexually Transmitted Disease model

9. Antibiotic treatment of a bacterial infection

10. Microbial growth in a gradostat

REQUIREMENTS: Homework and presentation of homework problems orpresentation of a project.

REFERENCES:

• Monotone Dynamical Systems, an Introduction to the Theory of Com-petitive and Cooperative Systems, Hal Smith, American MathematicalSociety, Mathematical Surveys and Monographs 1995.

• Monotone Dynamical Systems, M. Hirsch and Hal Smith, Handbookof Differential Equations , Ordinary Differential Equations ( volume2), eds. A.Canada,P.Drabek,A.Fonda, Elsevier, 239-357, 2005. (on myweb page under publications)

• SIAM Review Article, on my web page below syllabus.

• Differential Inequality appendix to ”Theory of the Chemostat”, on myweb site.

• Monotone systems, a mini-review, with M.W. Hirsch, in Positive Sys-tems. Proceedings of the First Multidisciplinary Symposium on Posi-tive Systems (POSTA 2003). Lecture Notes on Control and Informa-tion Sciences vol. 294, Springer Verlag, Heidelberg, 2003.

• The Many Proofs and Applications of Perron’s Theorem, SIAM Review42 (2000) p 487-498.

• Nonnegative Matrices in the Mathematical Sciences, Berman & Plem-mons, Academic Press, 1979.

Topics

• Perron-Frobenius Theory

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– irreducible matrix,

– incidence graph of matrix

– A quasi-positive ⇒ eAt ≥ 0, t ≥ 0

– dominant eigenvalue-positive eigenvector

– Linear Compartmental Models-Mean Residence Time

• Partial order relations and Cones

• Monotonicity Properties of Maps

• Kamke-Muller Theorem for ODEs

• Competitive & Cooperative Systems of ODEs

– Planar Periodic Competitive & Cooperative Systems of ODEs

– Generic convergence to equilibrium for Cooperative Systems

– Poincare-Bendixson Theory for 3D competitive & cooperative sys-tem

– The Carrying Simplex and Invariant Manifolds

• Monotone Cyclic Feedback Systems

• Competitive & Cooperative Systems of Delay Equations

• Reaction Diffusion Equations-Traveling Waves

3

1 Partial Orders & Monotone Maps

We want a partial order relation on a normed vector space X. A partialorder, compatible with vector space operations and the metric, must satisfy:

(1) x ≤ x

(2) x ≤ y & y ≤ x ⇒ x = y

(3) x ≤ y, y ≤ z ⇒ x ≤ z

(4) x ≤ y, u ≤ v ⇒ x + u ≤ y + v

(5) x ≤ y, a ∈ R+ ⇒ ax ≤ zy

(6) xn ≤ yn, xn → x, yn → y ⇒ x ≤ y

Of course, these should hold for all values of the indicated variables. Notethat (4) implies: x ≤ y ⇒ −y ≤ −x.

The set of positive vectors

C = x ∈ X : x ≥ 0

is often called a cone. We can see from the above that C has the followingproperties:

(a) 0 ∈ C

(b) C + C = C

(c) R+C = C

(d) C is closed.

(e) C ∩ (−C) = 0Most importantly:

x ≤ y ⇔ y − x ∈ C (1)

If set C has properties (a)-(e) and we define ≤ by (1), then ≤ satisfies (1)-(6).There is a one-to-one relation between partial order relations on X satisfying(1)-(6) and cones C satisfying (a)-(e).

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Example 1: X = Rn, C = Rn+, and x ≤ y ⇔ xi ≤ yi, 1 ≤ i ≤ n.

Example 2: X = Rn, given 1 ≤ k ≤ n

Ck = x : xi ≥ 0, 1 ≤ i ≤ k, xi ≤ 0, k + 1 ≤ i ≤ n

sox ≤Ck

y ⇔ xi ≤ yi, 1 ≤ i ≤ k, xi ≥ yi, k + 1 ≤ i ≤ n.

Example 3: X = C([0, 1],R), C = f ∈ X : f(t) ≥ 0, 0 ≤ t ≤ 1, andf ≤ g ⇔ f(t) ≤ g(t), 0 ≤ t ≤ 1.

A map F : (X,≤X) → (Y,≤Y ) is monotone if

x1 ≤X x2 ⇒ F (x1) ≤Y F (x2)

Since our interest is dynamics, we will be primarily interested in maps froma space into itself.

Since X is a normed linear space L(X) = T : X → X : T is continuous and linearis also a normed linear space. If X is ordered by cone C, define

L(X)+ = T ∈ L(X) : T (C) ⊂ C

A linear operator T in L(X)+ takes positive vectors to positive vectors. Wecall T a positive linear operator and write T ≥ 0 (see below).

If X = Rn and C = Rn+ then L(X) consists of the n×n matrices and the

positive linear operators L(X)+ consist of the nonnegative matrices.A linear operator generates a monotone map if and only if it is positive.

Proposition 1. T ∈ L(X) defines a monotone map x → Tx if and only ifT ∈ L(X)+.

Proof. If x ≤ y we want to show that Tx ≤ Ty, or equivalently, 0 ≤ Ty −Tx = T (y − x). Since y − x ≥ 0 and T ∈ L(X)+, this last inequalityholds.

It is interesting and useful that L(X)+ is typically a cone in L(X).

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Proposition 2. L(X)+ satisfies all the properties (a)-(e) of being a cone inL(X) except possible (e). Property (e) holds if X = C − C.

Proof. It is easy to check that the 0 operator belongs to L(X), that sumsof positive operators are positive operators, and the positive multiples ofpositive operators are positive operators. If Tn is a sequence of positiveoperators and Tn → T , then for any x ≥ 0 we have Tx = limn Tnx. SinceTnx ∈ C and C is closed, we see that Tx ≥ 0. If T belongs to both L(X)+

and −L(X)+ and x ≥ 0, then Tx ≥ 0 and −Tx ≥ 0 so Tx ≤ 0 ≤ Tx.Consequently, Tx = 0, and since x ≥ 0 is arbitrary, we have shown thatTC = 0. To show T = 0 we must show TX = 0. If every vector can bewritten as the difference of two positive vectors, X = C −C, this is obvious.You should be able to see it also holds if C − C is dense in X.

In X = R2 with the usual order generated by C = R2+ we have

(−3, 2) = (0, 2)− (3, 0)

and, obviously (0, 2), (3, 0) ∈ C. This can clearly be generalized:

Rn = Rn+ − Rn

+

Lemma 1. If T : [0, 1] → L(X)+ is continuous then∫ 1

0T (t)dt ∈ L(X)+.

Proof. The integral is a limit as N →∞ of partial sums

SN =N∑

n=1

T (n/N)1

N

Since each SN ∈ L(X)+ and L(X)+ is closed, the result follows.

Here is the main result. Just like in ordinary calculus, a function ismonotone (increasing) if its derivative is positive. We just need to interpret”positive derivative” as ”positive operator”.

Theorem 1. Let F : X → X be continuously differentiable on the normedlinear space X ordered by cone C. If for every x ∈ X, F ′(x) ∈ L(X)+, thenF is monotone with respect to ≤C.

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Proof. Use the above Lemma, the definitions, and

F (y)− F (x) =

∫ 1

0

d

dtF (ty + (1− t)x)dt =

∫ 1

0

F ′(ty + (1− t)x)dt (y − x)

We remark that it is not necessary for the above proof that F is definedon all X. It suffices for F to be defined on D ⊂ X and D to have theconvexity property that ty + (1 − t)x ∈ D for 0 ≤ t ≤ 1 whenever x, y ∈ Dand x ≤ y.

Example: X = R2 with the competitive order generated by the fourth (south-east) quadrant Q = (x, y)T : x ≥ 0, y ≤ 0. Recall u ≤Q v means v issoutheast of u. Check that L(X)+ consists of matrices of the following type

(+ −− +

)

since (+ −− +

)(+−

)=

(+−

)

Consider the mapping F : R2+ → R2

+ given by

F (x, y) = (r1x

1 + x + c12y,

r2y

1 + c21x + y)

where ri, cij > 0. It is easy to verify that its Jacobian matrix has the requisitesign structure for (x, y) ∈ R2

+. Therefore F is monotone with respect to the“southeast ordering”.

2 Graph theoretic issues related to Compet-

itive and Cooperative systems

The test for a C1 system x′ = f(x), whose jacobian is sign stable and signsymmetric on a convex domain, to be cooperative with respect to some partialorder ≤m, where m ∈ 0, 1n, and ≤m is defined by

x ≤m y ⇔ (−1)mixi ≤ (−1)miyi, 1 ≤ i ≤ n

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requires solving the equations

mi + mj = sij, mod 2, i < j (2)

for m ∈ 0, 1n where

sij =1 ∂fi

∂xj(x) +

∂fj

∂xi(x) < 0, some x

0 ∂fi

∂xj(x) +

∂fj

∂xi(x) > 0, some x

Note that the number of equations is exactly the number e of edges in thesigned influence graph for the system x′ = f(x), with (−1)sij being the signon the edge joining vertex i to vertex j. When fi is independent of xj andvice-versa, we have no edge for vertex i to vertex j, i 6= j.

If we rewrite (2) in matrix-vector format as

Gm = s (3)

then G is an e × n matrix whose rows g1, g2, · · · , ge characterize the edgesof the graph since each contains precisely two ones in the components corre-sponding to the vertices connected by that edge. We can identify the graphwith the matrix. This is not the usual influence matrix used to codify thegraph which is an n × n and contains a one in the i − j-th entry if an edgeconnects vertex i and vertex j.

As an example, consider the influence graph below

x4 x3

x1 x2

+

+

−−³³³³³³³³³

−

8

The corresponding equations are

m1 + m2 = 0

m1 + m4 = 1

m2 + m3 = 1

m2 + m4 = 1

m3 + m4 = 0

and the matrix-vector form of the equations is

1 1 0 01 0 0 10 1 1 00 1 0 10 0 1 1

m = s

where m = (m1, m2,m3,m4)T and s = (0, 1, 1, 1, 0)T . We may readily solve

these equations to get m = (0, 0, 1, 1) or m = (1, 1, 0, 0). Consequently, thecone is given by K = x : x1, x2 ≥ 0, x3, x4 ≤ 0.

We will always assume our influence graph is connected (path betweenany two vertices). Disconnected influence graphs correspond to disconnectedsystems; for example if there are two connected components in the influencegraph then our system has the form

x′ = f(x), y′ = g(y)

In this case, we deal with each subsystem separately.We begin by finding the kernel of G which we view as finding all partial

orders ≤m that are preserved when the system has only positive edges: s = 0.The answer is that only the standard order ≤ corresponding to m = 0 or itsopposite work.

Lemma 2. If the graph is connected then the null space of G is N(G) = 0, 1where 0 is the n-vector of zeros and 1 is the n-vector of ones.

Proof. Obviously 0, 1 ⊂ N(G). Suppose Gm = 0 and there is a propersubset A of 1, 2, · · · , n such that mi = 0, i ∈ A and mi = 1, i ∈ Ac. Butfrom the equations (2) we see that there cannot be an edge from a vertex inA to one in Ac since such an edge joining j ∈ A to k ∈ Ac yields the equationmj + mk = 0 + 1 6= 0. We have a contradiction to the connectedness of theinfluence graph.

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If h ∈ 0, 1e then q = hT G =∑e

i=1 higi may be viewed formally as theunion of edges gi for those i for which hi = 1. As such, q is a union of paths.It vanishes if and only if each vertex appears an even number of times in eachpath. But a path in which each vertex appears an even number of times is aloop. This means that q is a disjoint union of loops. This is the key idea inthe following.

Lemma 3. Let h ∈ 0, 1e. Then hT G = 0 if and only if the formal unionq =

∑ei=1 higi is a disjoint union of loops in the influence graph.

Our main result is the following.

Proposition 3. (2) has a solution m ∈ 0, 1n if and only if

hT s = 0, ∀h 3 hT G = 0

Equivalently, if and only if each loop has an even number of − edges.

Proof. The standard solvability criteria for Gm = s apply: vector s mustbe orthogonal to N(GT ). Since each h for which hT G = 0 corresponds to aformal union of loops in the graph and hT s counts the number of − edges inthe loop, the parity assertion follows.

Proposition 3 says our system is cooperative if and only if every loop hasan even number of negative edges. Our example apparently derives from acooperative system.

To test for a competitive system, we see if the time-reversed system x′ =−f(x) is cooperative. As the influence graph of this system is related tothe influence graph of x′ = f(x) by having all signs flipped, we must focuson the positive edges of the influence graph of x′ = f(x) as they becomethe negative edges of the influence graph of x′ = −f(x). Every loop of theinfluence graph of x′ = f(x) must have an even number of + edges for thesystem to be competitive relative to some order relation ≤m.

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3 Homework Problems

Exercise 1. Let A ≥ 0 be a square matrix and denote by a(q)ij the ij-th entry

of Aq where q = 1, 2, · · · . Show a(q)ij > 0 if and only if there is a directed path

with q edges from vertex j to vertex i in G(A).

Exercise 2. Let A be a quasipositive square matrix with s(A) < 0. Showthat

−A−1 =

∫ ∞

0

eAtdt > 0

Exercise 3. Let A be a square matrix and G(A) its directed graph. Vertex iis equivalent to vertex j if either i = j or there is a directed path from i toj AND a directed path from j to i. This defines an equivalence relation onthe vertices: the equivalence class of vertex i is denoted as usual by [i]-thecollection of all vertices that one can get to from i and get back from. Nowlets define a new graph with vertices these equivalence classes. A directededge goes from class [i] to class [j] if there is a directed edge in G(A) fromSOME member of [i] to SOME member of [j]. Observe that this graph hasNO LOOPS!! It also has at least one vertex with no edges to it! I think it iscalled a TREE. If A is irreducible then this graph is trivial-only one vertex!This construction tells us how to permute the standard basis vectors so thatthe new matrix obtained from A is in ”canonical form” with irreducible sub-matrices down the main diagonal. Try this out on

0 0 1 00 0 1 11 0 0 01 1 0 0

This construction is related to Conley’s Fundamental Theorem of DynamicalSystems. See p 404 of “Dynamical Systems”, 2nd ed., Clark Robinson.

Exercise 4. Consider n tanks, each containing growth medium of volume V ,in a linear array with nearest neighbor two-way flows between vessels:

¿ © ¿ © ¿ · · · ¿ © ¿

The circles represent the tanks and the arrows denote the flow between tanksand between the first and last tanks and the environment. Each arrow rep-resents a flow of rate F in units of volume per unit time (e.g., gallons per

11

minute) of growth medium. The material of interest here is a nutrient sup-plied at concentration N0 (weight per unit volume) in the inflow from theexternal environment to the left-most tank. The inflow from the environ-ment to the right-most vessel contains only growth medium with no nutrient.Each tank is assumed to be well-mixed so its contents are homogeneously dis-tributed. The outflow from the two end-tanks to the environment containsjust the well-mixed contents of each vessel. We want to determine equationsfor

Ni(t) = concentration of nutrient in tank i at time t

(a) Write down the linear system of equations for Ni in case n = 2. Showthat it is a linear compartmental system with inputs. Calculate s(A)and determine the Mean Residence Time for nutrient in the gradostat.Find the steady state values of Ni. Why is this called a gradostat?

(b) Can you repeat (1) for the general case n ≥ 2?

Exercise 5. Particularly in pharmacology, it is common to administer drugsin a periodic fashion: take 2 tablets every 6 hours. Therefore it is natural toconsider compartmental systems

x′ = Ax + r(t)

where r(t + T ) = r(t) is periodic of period T and, of course, nonnegativer(t) ≥ 0. Suppose that A is a compartmental matrix as described in thepdf on my web site with s(A) < 0. Show that there is a unique T -periodicsolution x(t) ≥ 0. Hint: by the variation of constants formula

x(t) = eAtx(0) +

∫ t

0

eA(t−s)r(s)ds.

We want to find initial data x(0) so that x(0) = x(T ), which will guaranteethat the corresponding solution is T -periodic. Therefore we seek a solutionx(0) of the equation

x(0) = eAT x(0) +

∫ T

0

eA(T−s)r(s)ds

Show that there exists a unique solution x(0) ≥ 0 and provide a formula forit.

12

Exercise 6. Consider the discrete-time Juvenile-Adult population model

Jn+1 =rAn

b + An

An+1 = pAn + qJn

where 0 ≤ p, q < 1, 0 < q are survival probabilities and r, b > 0 have to dowith adult fecundity. Find all fixed points and determine their stability. Inthe case p = 0, find all period-two points and determine their stability. Hint:They are fixed points of P 2 where P is the map defined by the right side.

A 2× 2 matrix A has |λ| < 1 for all λ ∈ σ(A) if and only if

|tr (A)| < 1 + det A < 2

and |λ| > 1 for some eigenvalue if any one of the three inequalities

tr (A) > 1 + det A, tr (A) < −1− det A, det A > 1

holds.When q = 0 and qr > b, the fixed point (J,A) = (0, 0) is unstable and

(J∗, A∗) = (rq − b)(1, q−1) is asymptotically stable. In addition, there isthe period-2 orbit (J∗, 0), (0, A∗) where J∗, A∗ are the components of thepositive fixed point. The period-2 orbit is of saddle type since the eigenvaluesof the jacobian of the second iterate map are rq/b > 1 and b/rq < 1. Thisperiodic orbit attracts all orbits starting at points (J,A) 6= (0, 0) where eitherJ = 0 or A = 0. You can easily check this. All orbits starting at points (J,A)with J > 0 and A > 0 converge to the positive fixed point!

Exercise 7. Consider two competing species in an seasonally changing en-vironment

x′1 = x1[e(t)− a(t)x1 − b(t)x2]

x′2 = x2[f(t)− c(t)x1 − d(t)x2]

where a, b, c, d, e, f are T -periodic, continuous functions: a(t) = a(t + T ) forall t. Assume that a, b, c, d ≥ 0 for all t making this a “competitive system”.Let P : R2

+ → R2+ be the “period map” defined by P (x(0)) = x(T ).

(a) Show that P is a competitive map in the sense that x(0) ≤C y(0) impliesthat P (x(0)) ≤C P (y(0)) where ≤C is the “southeast ordering”.

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(b) Show that P satisfies det P ′(x) > 0, x ∈ R2+ and that P is injective.

Recall Abel’s formula (Thm 2.3 Brauer& Nohel)! Therefore P satisfies(H+) from the notes.

(c) Show that if∫ T

0e(t)dt 6= 0 and a(t) is not everywhere zero then there

can be at most one T -periodic solution (x1(t), 0) satisfying x1(t) ≥ 0and x1 not identically zero. Hint: let y(t) = 1/x1(t).

Recall that P (x0) = x(T, x0) so P ′(y) = ∂x∂x0

(T, y). We know that Φ(t) =∂x∂x0

(t, y) is the 2× 2 fundamental matrix for

z′ =∂F

∂x(t, x(t, y))z

satisfying Φ(0) = I. Here, F (t, x) denotes the right hand side of our compet-ing species system, x = (x1, x2)

T . By Abel’s formula

det Φ(t) = det Φ(0) exp(

∫ t

0

traceA(s)ds)

where A(s) = ∂F∂x

(s, x(s, y)). In any case, det Φ(T ) = det P ′(y) > 0.

Exercise 8. A model of bacteria growing in a two-vessel gradostat on anutrient S is given by the following equations (which have been scaled tosimplify them) in which Si denotes scaled nutrient concentration in vessel iand ui denotes scaled bacterial density in vessel i, i = 1, 2.

S ′1 = 1− 2S1 + S2 − f(S1)u1

S ′2 = S1 − 2S2 − f(S2)u2

u′1 = u1f(S1)− 2u1 + u2

u′2 = u2f(S2)− 2u2 + u1

The specific growth rate of the organism is

f(S) =mS

a + S

The leading 1 in the equation for S ′1 signifies fresh nutrient entering vesselone of the gradostat at scaled concentration one.

This system has an equilibrium (S1, S2, u1, u2) = (2/3, 1/3, 0, 0) called the“washout equilibrium” because all bacteria are washed out.

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(1) Show that R4+ is positively invariant for the system: solutions that start

nonnegative stay nonnegative in the future.

(2) Show that if S(0) ≤ (2/3, 1/3), then S(t) ≤ (2/3, 1/3) for t ≥ 0.

(3) Show that every solution with S(0) ≤ (2/3, 1/3) converges to the washoutequilibrium if the Jacobian matrix there has all eigenvalues with nega-tive real part.

For part (3), we assume the 4 × 4 variational equation at the washoutstate z′ = Az, where

A =

−2 1 f ′(2/3) 01 −2 0 f ′(1/3)0 0 f(2/3)− 2 10 0 1 f(1/3)− 2

has all eigenvalues with negative real part. Note that this holds preciselywhen the 2× 2 block B in the lower right has negative eigenvalues since theupper left 2 × 2 block has negative eigenvalues. Now, by part (2), since ifS(0) ≤ (2/3, 1/3), then S(t) ≤ (2/3, 1/3) for t ≥ 0 we have

u′1 ≤ u1f(2/3)− 2u1 + u2

u′2 ≤ u2f(1/3)− 2u2 + u1

oru′ ≤ Bu

If u satisfiesu′ = Bu, u(0) = u(0)

then we may compare: 0 ≤ u(t) ≤ u(t), t ≥ 0. But B has negative eigenval-ues so u(t) → 0 as t → ∞. It follows that u(t) → 0 too since it is squeezedbetween 0 and u(t)! It is now not hard to see that S(t) → (2/3, 1/3) att →∞. One way is to notice that v1 = S1 + u1 and v2 = S2 + u2 satisfy

v′ =( −2 1

1 −2

)v +

(10

)

and thus v(t) → (2/3, 1/3) as t →∞.

15

Exercise 9. Perhaps the simplest model of antibiotic treatment of a bacterialinfection is the following system

B′ = rB(1−B/K)− cAB

A′ = d(A0 − A)− cAB

where B denotes bacteria (cells/ml) and A denotes antibiotic (gms/ml). Bac-teria grow logistically in the absence of antibiotic: r,K > 0. Antibiotic issupplied at concentration S0, d > 0 is removal rate, and antibiotic kills atrate proportional to its concentration (c > 0).

(a) Show that the system is competitive in the first quadrant.

(b) Show that solutions are bounded. Therefore, all solutions converge toequilibrium since the system is competitive.

(c) Find all equilibria and describe the global behavior of all solutions.

The trivial equilibrium (B,A) = (0, A0) is locally asymptotically stable ifcA0 > r and unstable when cA0 < r. Positive equilibria (B,A) must satisfy0 < B < K and 0 < A < A0. Letting x = B/K ∈ (0, 1) we find that xsatisfies p(x) = x2 + ( d

cK− 1)x + d

K(A0

r− 1

c) = 0.

Case I: cA0 − r < 0. In this case p(0) < 0 < p(1) and there is a uniqueroot x ∈ (0, 1). Therefore there is a unique positive equilibrium (0, A0) ¿K

(B∗, A∗) in this case (here ≤K is the southeast order). Using the equilibriumconditions, the Jacobian at (B∗, A∗) is

( −rB∗/K −cB∗

−cB∗ −dA∗

)

which has negative trace and positive determinant. Thus (B∗, A∗) is asymp-totically stable. It attracts all solutions not on the A-axis.

Case II: cA0− r > 0. In this case p(0), p(1) > 0 so we expect zero,one ortwo roots of p(x) = 0 in (0, 1). p′(x) = 0 gives x0 = (1− d

cK)/2 which belongs

to (0, 1) only if d < cK. If d ≥ cK or if d < cK and p(x0) > 0 then thereare no other equilibria. Note that p(x0) = d

K(A0

r− 1

c) − 1

4(1 − d

cK)2. In this

case, (0, A0) is globally asymptotically stable since it is the only equilibriumand every solution converges to equilibrium.

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If d < cK and dK

(A0

r− 1

c)− 1

4(1− d

cK)2 < 0 (in addition to cA0 − r > 0)

then there are two positive steady states (B1, A1), (B2, A2) where

(0, A0) ¿K (B1, A1) ¿K (B2, A2)

(0, A0) and (B2, A2) are asymptotically stable and (B1, A1) is a saddle pointwhose stable manifold forms the separatrix boundary between the basins ofattraction of the two attractors. This case is biologically interesting becausea small inoculum of bacteria (initial data near (0, A0)) results in eradicationof the infection while a larger inoculum of bacteria will end up at (B2, A2)with the infection not contained!

Exercise 10. Decide whether each of the following is competitive, coopera-tive, or neither. Cooperative (competitive) should be interpreted in the broadsense that there is some ordering, generated by an orthant, that is preservedin the future (past). Find the appropriate ordering in each case. Unlessotherwise specified, the domain of each system is the positive cone.

(1) A model of two genes which interact with each other:

x′1 = β1(y1 − x1)

y′1 = α1f1(x2)− y1

x′2 = β2(y2 − x2)

y′2 = α2f2(x1)− y2

where fi > 0 satisfy f ′1 6= 0, f ′2 6= 0. xi = mRNA, yi = protein productof gene i. Consider the different cases depending on the signs of theinteraction terms f ′i .

(2) S → E → I → R epidemic model.

S ′ = µN − µS − βIS/N

E ′ = βIS/N − (µ + γ)E

I ′ = γE − (µ + α)I

Here the domain is (S, E, I) : S, E, I ≥ 0, S + E + I ≤ N.(3) Virus dynamics model:

T ′ = δ − αT − kV T

(T ∗)′ = −βT ∗ + kV T

V ′ = −γV + NβT ∗ − kV T

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Here, V denotes virus density, T denotes target cell density, and T ∗

denotes infected target cell density in the blood.

(4) A two-sex, two strain STD model boils down to the following equations:

˙Imi = −σm

i Imi + am

i (pm − Im1 − Im

2 )Ifi

˙Ifi = −σf

i Ifi + af

i (pf − If

1 − If2 )Im

i , i = 1, 2

Ifi denotes the females infected with strain i of the disease. The domain

is (Im1 , Im

2 , If1 , If

2 ) ≥ 0 : Ix1 + Ix

2 ≤ px, x = m, f.

Exercise 11. Let x0 be an equilibrium of the cooperative system x′ = f(x), x ∈D where f ∈ C1(D) and D ⊂ Rn is open and convex. For simplicity, letsassume the cooperativity is relative to the usual ordering since we can alwayschange variables so that this is so.

(a) Show that x ∈ D : x ≥ x0 and x ∈ D : x ≤ x0 are positivelyinvariant.

(b) If s = s(Df(x0)) > 0 and Df(x0)v = sv where v À 0, which holdsfor example if Df(x0) is irreducible, show that f(x0 + rv) À 0 for allsufficiently small r > 0. This means, by our convergence theorem, thatx(t, x0 + rv) → p for some equilibrium p À x0 provided this solutionhas compact orbit closure in D. A similar conclusion f(x0 − rv) ¿ 0holds implying that x(t, x0 − rv) → q ¿ x0. Hint: Taylor series withremainder!

(c) Suppose that x0, y0 are two equilibria satisfying x0 ¿ y0 with s(Df(x)) >0, x = x0, y0. If the Jacobian is irreducible at these two points, showthat there must exist an equilibrium p with x0 ¿ p ¿ y0.

(d) If s = s(Df(x0)) < 0 and Df(x0)v = sv where v À 0, show that forall small r > 0 we have f(x0 + rv) ¿ 0 ¿ f(x0 − rv). Therefore, thesolution starting at any point of the box

[x0 + rv, x0 − rv] = y : x0 − rv ≤ y ≤ x0 + rv

converges to x0.

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Part (b) simply uses the definition of differentiability of f :

f(x0 + rv) = f(x0) + rDf(x0)v + o(r) = r[sv + O(r)] À 0

where the last inequality holds for all small r > 0 since all components of svare positive and O(r) → 0 as r → 0+.

Part (c) uses that x(t, x0 + rv1) (x(t, y0− rv2)) is increasing (decreasing)in all its components, where vi À 0 is the eigenvector associated with Df(x0)for i = 1 and Df(y0) for i = 2, and r > 0 is sufficiently small that x0+rv1 ¿y0 − rv2. Therefore:

x0 ¿ x(1, x0+rv1) ¿ x(t, x0+rv1) ¿ x(t, y0−rv2) ¿ x(1, y0−rv2) ¿ y0, t > 1

Since x(t, x0 +rv1) p and x(t, y0−rv2) q where p, q are (not necessarilydistinct) equilibria by the convergence criterion, we conclude that:

x0 ¿ p ≤ q ¿ y0.

Exercise 12. Consider the example graph in the section of these notes onGraph theoretic issues related to Competitive and Cooperative sys-tems above. Find the set of vectors h ∈ 0, 1e such that hT G = 0 and showthat these h correspond to loops in the graph as stated in Proposition 3 ofthat section. Find m such that the system preserves the order relation ≤m inforward time.

Exercise 13. The system

x′1 =1

1 + xn3

− γx1

x′2 = x1 − γx2

x′3 = x2 − γx3

arises in gene regulation (see my web page pdf on gene regulatory networks).Its domain is R3

+; γ > 0 and n > 0. Find conditions on γ and n for theexistence of a periodic orbit.

See the pdf file on “gene regulatory networks” for answers.

Exercise 14. You et al. (Nature 428 22Apr 2004) build a gene circuit tocontrol the density of E. coli. x1 denotes bacterial density, x3 denotes thedensity of a quorum sensing molecule emitted by these cells. As this molecule

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builds up it activates a ”killer” gene circuit for a protein x2 which promotesprogrammed cell death in the bacteria. The model system is given by:

x′1 = k1x1(1− x1)− dx1x2

x′2 = x3 − γ1x2

x′3 = x1 − γ2x3

where all parameters are positive and x ∈ R3+. Analyze the dynamics of this

system as completely as you can. Can it have a periodic orbit?Equilibria are 0 and p À 0. 0 is unstable and its stable manifold consists

of the x1 = 0 plane. R3+ is positively invariant and orbits are bounded by a

standard argument using differential inequalities. If D = x : xi > 0, i =1, 2, 3, then the Butler-McGehee lemma shows that every orbit that startsin D has compact closure in D. D contains only one equilibrium p. Oursystem is competitive and irreducible. The determinant of the jacobian Jat p satisfies det J = −k1γ1γ2 < 0 so J either has all eigenvalues withnegative real part or precisely two with positive real part and one negativeeigenvalue. We use Routh-Hurewicz test to show that the latter can occur.The characteristic polynomial λ3 +a1λ

2 +a2λ+a3 = 0 satisfies ai > 0 so thestability test fails if

[a1a2 − a3]/γ1γ2 =

(k2

1γ1γ2

k1γ1γ2 + d+ γ1 + γ2

)[(γ1 + γ2)

k21

k1γ1γ2 + d+ 1

]− k1

is negative. One can see that this holds, for example, if the γi are sufficientlysmall since the product of the first two terms in parentheses goes to zero withγi. In this case, Prop. 6.1 of the notes on competitive and cooperative systemsimplies that the omega limit set of every orbit starting off the one-dimensionalstable manifold of p is a periodic orbit.

Exercise 15. Species x competes with species y but species x has a refugewhere it can hide from species y. We assume there are two habitats, speciesy is restricted to habitat one but species two can wander between the twohabitats. Here is the model system where xi denotes species x in habitat i,same for y.

x′1 = r1x1(1− ax1 − by1) + δ(x2 − x1)

x′2 = r1x2(1− ex2) + δ(x1 − x2)

y′1 = r2y1(1− cx1 − dy1)

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Assume all parameters are positive. Is this a cooperative and irreduciblesystem? If so what is the appropriate order relation? Determine the behaviorof this systems as completely as you can. This should include the stability ofall steady states and identifying subsets of their basin of attraction. Thereare several cases but you may assume that the determinant of the jacobianat each equilibrium is not zero!

Exercise 16. Species x1 competes with species x2 and species x3 competeswith species x2 but species x1 and x3 do not interact.

x′1 = r1x1(1− x1 − c12x2)

x′2 = r2x2(1− c21x1 − x2 − c23x3)

x′3 = r3x3(1− c32x2 − x3)

where ri, cij > 0. assume further that

cij < 1 and c21 + c23 < 1

Is this a cooperative and irreducible system? If so what is the appropriateorder relation? Determine the behavior of this systems as completely as youcan. This should include the stability of all steady states and identifyingsubsets of their basin of attraction. You may assume that the determinant ofthe jacobian at each equilibrium is not zero!

In this and the previous problem, use our results that if two equilibria arerelated but there are no other equilibria in between them, then all trajectoriesstarting at a point strictly between them converges to the stable equilibrium.

Exercise 17. A model of bacterial growth in the chemostat with delay betweennutrient uptake and conversion to biomass is given by:

S ′(t) = 1− S(t)− f(S(t))x(t)

x′(t) = e−rf(S(t− r))x(t− r)− x(t)

where f(S) = mS/(a + S) and x, S ≥ 0. Assume that m/(1 + a) > er andm, a, r > 0. The factor exp(−r) accounts for a suppression in growth ratedue to substrate stored inside cells that washes out of the culture vessel beforegiving rise to new growth.

(a) Show that solutions corresponding to nonnegative initial data remainnonnegative.

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(b) Show that u(t) = x(t) + e−rS(t− r) → e−r, t →∞.

(c) Using b., obtain a scalar delay equation satisfied by x(t), called thelimiting equation. Change variables to let y(t) = erx(t) and write theequation using this variable. Notice that there is a biological restrictionon the domain of this delay equation.

(d) Show that the limiting system generates a strongly monotone dynamicalsystem. Describe the behavior of its solutions as completely as you can.

Exercise 18. Recall that x0 ∈ X is a super-equilibrium (sub-equilibrium)for monotone semiflow Φ : X × [0,∞) → X if Φ(x0, t) ≥ x0, t ≥ 0(Φ(x0, t) ≤ x0, t ≥ 0). Monotonicity of Φ then implies that t → Φ(x0, t)is non-decreasing (non-increasing) and, if Φ(x0, t) : t ≥ 0 has compactclosure in X, that Φ(x0, t) p (Φ(x0, t) p) for some equilibrium p. Obvi-ously, every equilibrium is both a sub-equilibrium and a super-equilibrium.

A point x0 ∈ Rn is a super-equilibrium for x′ = f(x), where f satisfiesthe quasimonotone condition, if and only if f(x0) ≥ 0. This last condition isboth necessary and sufficient (by convergence criterion from notes)!

How do we identify a super-equilibrium for the delay differential systemsx′(t) = f(x(t), x(t − 1)) where ∂f

∂x(x, y) is quasi-positive and ∂f

∂y(x, y) ≥ 0?

Assume f : Rn ×Rn → Rn is continuously differentiable. Recall that a superequilibrium must be a state so it must by a φ ∈ C([−1, 0],Rn).

(a) Given x0 ∈ Rn, find sufficient conditions such that φ = x0, the functionidentically equal to x0, is a super-equilibrium and justify your conditions.Hint#1: try to reformulate the problem as one of positive invariance of acertain set; hint#2: consider the special case x0 = 0; hint#3: make yourconditions as weak as possible by using the above hypotheses on f .

(b) In problem 17 (d) show that 1 is a sub-equilibrium. Show that yt(1) : t ≥0 has compact closure in C([−r, 0],R+) and therefore yt(1) e for someequilibrium e. Determine e.

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Exercise 19. The initial boundary value problem for u(x, t):

ut = d∆u + f(u), x ∈ Ω, t > 0

0 =∂u

∂n(x), x ∈ ∂Ω

u(x, 0) = u0(x), x ∈ Ω

where u0 ∈ C(Ω,R), generates a monotone semiflow Φ(u0, t) = u(•, t). Here,n is the outer normal vector to the boundary of Ω ⊂ Rn, assumed to besmooth, and Ω is bounded.

(a) Let a ∈ R and a be the element of C(Ω,R) whose value at every x is a.Give conditions for a to be a super-equilibrium and justify your conditions.

(b) If f(u) = u(u−0.5)(1−u), find all constant equilibria, all stable equilibria,all constant sub-equilibria, and all constant super-equilibria. Determine thelargest set B ⊂ C(Ω,R) such that Φ(t, b) → 1 as t →∞ for each b ∈ B thatyou can.

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