faculty of mathematics and natural sciences

Delay Differential Equations

Bachelor Project Mathematics

June 2016

Student: C.R. Terpstra

First supervisor: Dr. A.E. Sterk

Second supervisor: Prof.dr. A.J. van der Schaft

Abstract

This thesis introduces the concept of differential equations with delay, the so-called delay differential equations (DDEs). These differential equations dependon past history, and are therefore used in many models because they are morerealistic than models independent of past history.

In this thesis, simple cases and linear systems of DDEs with a single delaywill be discussed. We will look at proofs of existence and uniqueness, numericaland analytic solutions, and the stability of the steady-state solutions. Thisthesis ends with a discussion of the delayed logistic equation.

i

Contents

1 Introduction 1

2 Simple Cases 22.1 Simplest DDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2.1.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.1.2 Existence and Uniqueness . . . . . . . . . . . . . . . . . . 42.1.3 General Equation . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 Scalar DDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2.1 General Equation . . . . . . . . . . . . . . . . . . . . . . . 82.2.2 Existence and Uniqueness . . . . . . . . . . . . . . . . . . 9

3 Linear Systems 103.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.2 Linear DDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3.2.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2.2 General Equation . . . . . . . . . . . . . . . . . . . . . . . 13

3.3 General Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

4 Application 164.1 Delayed Logistic Equation . . . . . . . . . . . . . . . . . . . . . . 16

A Matlab Codes 19A.1 Chapter 2: Simple Cases . . . . . . . . . . . . . . . . . . . . . . . 19A.2 Chapter 4: Application . . . . . . . . . . . . . . . . . . . . . . . . 20

Bibliography 22

ii

Chapter 1

Introduction

Systems of ordinary or partial differential equations are independent of previousstates or rates. Systems of differential equations that are dependent on previousstates are called systems of delay differential equations (DDEs). In this thesis,these differential equations, also referred to as retarded functional differentialequations (RFDEs), will be analyzed.

Models including delay differential equations exist, among other things, in bi-ology, economics, and mechanics. An example of a DDE in the biology fieldis the Mackey-Glass equation for the density of certain blood cells. The delayin this differential equation is given by the time between initiation of cellularproduction in the bone marrow and release of mature cells into the blood.

Delay differential equations were introduced to create more realistic models sincemany processes depend on past history. A limitation of DDEs is that they onlydepend on past states and not past rates. Differential equations dependent onpast rates are the so-called neutral delay differential equations (NDDEs), butthese differential equations will not be discussed in this thesis. Furthermore,DDEs dependent on multiple past states will also not be discussed in this thesis.

In the next chapter, Chapter 2, simple cases of DDEs will be analyzed. InChapter 3, linear systems with delay will be discussed. For these systems and thesimple cases, we will look at analytic and numerical solutions. Furthermore, wewill discuss the stability of the steady-state solutions. For the simple cases, wewill also look at proofs of existence and uniqueness of solutions. This thesis endswith a discussion of an application: the delayed logistic equation, in Chapter 4.

1

Chapter 2

Simple Cases

2.1 Simplest DDE

2.1.1 Example

The simplest example of a DDE is given by

x′(t) = −x(t− τ) for t ≥ 0, (2.1)

where τ > 0 is called the delay. Suppose the initial condition for (2.1) is givenby

x(t) = 1 for t ∈ [−τ, 0]. (2.2)

Following the procedure called the method of steps described in [5, p.13-14], thesolution x(t) for t ∈ [(n−1)τ, nτ ], n ∈ N, can be determined in the following way.

For t ∈ [0, τ ], it follows that t− τ ∈ [−τ, 0]. Therefore,

x′(t) = −x(t− τ) = −1.

From this, we can conclude that

x(t) = x(0) +

∫ t

0

(−1) ds = 1− t, t ∈ [0, τ ]. (2.3)

Similarly, we can show that

x′(t) = −x(t− τ) = −[1− (t− τ)], t ∈ [τ, 2τ ].

Therefore,

x(t) = x(τ) +

∫ t

τ

−[1− (s− τ)] ds

= 1− τ + [−s+1

2(s− τ)2]s=ts=τ

= 1− t+1

2(t− τ)2, t ∈ [τ, 2τ ]. (2.4)

2

By induction, it can be proven that

x(t) = 1 +

n∑k=1

(−1)k[t− (k − 1)τ ]k

k!, t ∈ [(n− 1)τ, nτ ], n ∈ N. (2.5)

The solution x(t) is unique. This will be proven in the next section.

Let f(a−) and f(a+) denote lims→a− f(s) and lims→a+ f(s) for a functionf and constant a, respectively. From the method of steps done above, it canimmediately be seen that x′(0−) = 0 and x′(0+) = −1. Therefore, x is notdifferentiable at t = 0. In fact, x(n−1) is not differentiable at t = (n − 1)τ forall n ∈ N, since x(n)((n− 1)τ−) = 0 and x(n)((n− 1)τ+) = (−1)n.

0 1 2 3 4

t

0

0.5

1

x(t

)

τ = 0.5

0 2 4 6 8 10 12 14

t

-0.5

0

0.5

1

x(t

)τ = 1

0 5 10 15 20

t

-1

-0.5

0

0.5

1

x(t

)

τ = 1.5

0 2 4 6 8 10 12 14 16

t

-2

0

2

4

x(t

)

τ = 2

Numerical Solution Analytic Solution

Figure 2.1: Numerical and analytic solutions of (2.1) with initial condition (2.2)for different values of τ .

See Figure 2.1 for plots of numerical and analytic solutions of the initialvalue problem (2.1), (2.2), for some values of τ . The analytic solution in thisfigure is given by (2.5). The Matlab codes used for computing the plots in thischapter can be found in Appendix A.1.

Using the package DDE23 in Matlab, plots can be made with on the hor-izontal axis the time t and on the vertical axis the (numerical) solution x(t)for different values of τ . Comparing both solutions, we can conclude that thenumerical solution approximates the analytic solution quite well. Furthermore,the accuracy increases as t increases.

From Figure 2.1, we can also conclude that the solution x(t) for τ = 0.5 isstarting to oscillate. The solutions oscillate around the steady-state solution of(2.1) given by x ≡ 0. Also, the oscillations increase as τ increases. Accordingto the plots in this figure, at least until τ = 1.5, x = 0 is stable. When τ = 2,

3

x = 0 is no longer a stable solution. Later in this chapter, it will be proven thatx(t) oscillates for values of τ > 1/e ≈ 0.37, and x = 0 is stable when τ < π/2and unstable when τ > π/2 ≈ 1.57.

2.1.2 Existence and Uniqueness

Let τ ≥ 0 be a constant in J = [ξ, ξ+ a], where ξ ≥ 0, and a > 0. The equation

x′(t) = f(t, x(t− τ)) for t ∈ J (2.6)

is called a delay differential equation, where τ > 0 is called the delay. An initialcondition for (2.6) is given by

x(t) = φ(t) for t ∈ J− = [ξ − τ, ξ], (2.7)

where φ is a given continuous function.

Theorem 2.1. We consider the initial value problem (2.6), (2.7), where f iscontinuous in the strip S = J × R, φ is continuous in J−, and τ > 0 is aconstant in J . Then, there exists exactly one solution.

The proof of this theorem is based on the proof of Theorem 7.V. in [6], andthe method of steps used in the previous section.

Proof. Let

x′n(t) = f(t, xn(t− τ)) for t ∈ [ξ + (n− 1)τ, ξ + nτ ],

where n ∈ N.

For t ∈ [ξ, ξ + τ ], it follows that t− τ ∈ [ξ − τ, ξ]. Therefore,

x′1(t) = f(t, x1(t− τ)) = f(t, φ(t− τ)), and

x1(ξ) = φ(ξ).

From this, we can conclude that

x1(t) = x1(ξ) +

∫ t

ξ

f(s, x1(s− τ)) ds

= φ(ξ) +

∫ t

ξ

f(s, φ(s− τ)) ds. (2.8)

Hence, x1 is uniquely defined for t ∈ [ξ, ξ + τ ]. Similarly, we can show that

x′2(t) = f(t, x2(t− τ)) = f(t, x1(t− τ)), t ∈ [ξ + τ, ξ + 2τ ], and

x2(ξ + τ) = x1(ξ + τ).

Therefore,

x2(t) = x2(ξ + τ) +

∫ t

ξ+τ

f(s, x2(s− τ)) ds

= x1(ξ + τ) +

∫ t

ξ+τ

f(s, x1(s− τ)) ds. (2.9)

4

So x2 is uniquely defined for t ∈ [ξ + τ, ξ + 2τ ]. We can conclude that xnis defined by xn−1 for all n ∈ N. Therefore, xn is uniquely defined for t ∈[ξ + (n− 1)τ, ξ + nτ ], n ∈ N. From this, we can conclude that x, defined by

x(t) =

φ(t) for t ∈ J− = [ξ − τ, ξ],

φ(ξ) +

∫ t

ξ

f(s, x(s− τ)) ds for t ∈ J = [ξ, ξ + a],(2.10)

is well-defined for all values t ∈ J− ∪ J = [ξ − τ, ξ + a]. Furthermore, x(t) is asolution of the initial value problem (2.6), (2.7). From the definition of x(t), wecan conclude that x(t) is the only solution. �

2.1.3 General Equation

The general equation of the simplest DDE is given by

x′(t) = −αx(t− τ), (2.11)

where α is a constant, and τ > 0 is the delay. The case α > 0 corresponds tonegative feedback, and the case α < 0 corresponds to positive feedback.

By scaling this DDE, the number of parameters can be reduced, which makessolving easier. We will determine a DDE with U(µ) = x(t), where the delay isthe constant 1. For the scaling, we set µ := ηt, η > 0. Then,

dU

dµ=

dx

η dt= −αη−1x(t− τ) = −αη−1U(ηt− ητ) = −αη−1U(µ− ητ).

If we set η := 1/τ and β := ατ , then

dU

dµ= −βU(µ− 1). (2.12)

We introduce the following linear operator, defined on the differentiable func-tions:

L(U) =dU

dµ+ βU(µ− 1).

We seek (complex) values of λ such that U(µ) = eλµ is a solution of (2.12).Filling this expression in for L(U), we get

L(eλµ) = λeλµ + βeλ(µ−1) = eλµ[λ+ βe−λ]. (2.13)

We want to solve the characteristic equation

h(λ) ≡ λ+ βe−λ = 0. (2.14)

Then, L(eλµ) is the zero function, hence U(µ) = eλµ is a solution of (2.12).To solve (2.14), we set λ := x + iy. Considering real and imaginary parts,

we get the systemx = −βe−x cos(y)

y = βe−x sin(y).(2.15)

We call λ ∈ C a root of (2.12) of order l, where l ≥ 1, if

h(λ) = h′(λ) = h′′(λ) = · · · = h(l−1)(λ) = 0, h(l)(λ) 6= 0.

5

Lemma 2.1. µjeλµ, j = 0, 1, . . . , k are solutions of (2.12) if and only if λ is aroot of order at least k + 1 of h.

The proof of this lemma is taken from [5, p.17].

Proof. Differentiating (2.13) k times with respect to λ and using that this kth-derivative commutes with L we find, by Leibniz’ rule for the derivative of aproduct, that

L(µkeλµ) = (∂

∂λ)k[eλµh(λ)] = eλµ[

k∑j=0

Ckj h(j)(λ)µk−j ]

where Ckj = k!/j!(k − j)! are the binomial coefficients. The result follows im-mediately from this observation. �

As h is an analytic function of the complex variable λ it has the followingelementary properties. See Appendix A in [5].

(A) The set of roots can have no accumulation point in C; therefore, for eachR > 0, the set of roots satisfying |λ| ≤ R is finite. It follows that the set ofroots is a countable (possible finite) set.

(B) If the set of roots is infinite, denoted by {λn}∞n=1, then |λn| → ∞. Because|β|e−<(λn) = |λn|, it follows that <(λn) → −∞. Consequently, for eacha ∈ R, <(λ) ≥ a for at most finitely many roots.

(C) If λ is a root, then it is a root of finite order.

(D) If λ is a root, so is its conjugate λ̄.

Lemma 2.2. The following hold.

1. If β < 0, then there is exactly one real root and it is positive.

2. If 0 < β < 1/e, then there are exactly two real roots x1 < x2, both negative.x1 → −∞ and x2 → 0 as β → 0.

3. If β = 1/e, then there is a single real root of order two, namely λ = −1.

4. If β > 1/e, then there are no real roots.

Proof. For (1): h is an increasing function that crosses the x-axis, hence thereis exactly one real root. Because β < 0, it follows that λ = −βe−λ > 0.

For (2): h has a minimum below the x-axis, hence there are exactly two realroots. Because β > 0, it follows that λ = −βe−λ < 0. The last assertion followsfrom the following:

β = −λeλ → 0 if and only if λ→ 0 or λ→ −∞.

For (3): h has a minimum on the x-axis, hence there is exactly one real root.Filling β = 1/e in for h gives the root λ = −1. Furthermore, h′(−1) = 0 andh′′(−1) = 1, hence the order of λ = −1 is two.

For (4): h has a minimum above the x-axis. �

The next result summarizes important information concerning the roots inthe case β > 0.

6

Proposition 2.2. The following hold for (2.14).

1. If 0 < β < π/2, then there exists δ > 0 such that <(λ) ≤ −δ for all roots.

2. If β = π/2, then λ = ±iπ/2 are roots of order one.

3. If β > π/2, then there are roots λ = x± iy with x > 0, y ∈ (π/2, π).

The proof of this proposition is taken from [5, p.18-20].

Proof. Because β > 0, if there is a root x + iy with x ≥ 0 and y > 0 of (2.15),then cos(y) ≤ 0 < sin(y). So y ∈ S ≡

⋃∞n=0{[π/2, π) + 2nπ}. Furthermore,

sin(y)

y=ex

β

must hold. As

d

dy

sin(y)

y=y cos(y)− sin(y)

y2< 0, y ∈ S

and sin(y)/y = 2/π when y = π/2, we conclude that sin(y)/y ≤ 2/π for y ∈ S.Therefore,

1

β≤ ex

β=

sin(y)

y≤ 2

π.

So it follows that β ≥ π/2. Thus, if β < π/2, then <(λ) < 0 for every root λ.This and the last assertion in (B) above proves (1).

The second assertion follows immediately from computation.Let’s now turn to the final assertion (3). If we write our root as λ = reiθ,

then (2.14) becomes

r[cos(θ − π) + i sin(θ − π)] = βe−x[cos(−y) + i sin(−y)].

Equivalently,r = βe−x and θ − π = −y + 2kπ

for some integer k. Let’s search for a root in the first quadrant on the raythrough the origin making angle θ ∈ (0, π/2) with the positive x-axis. Then,taking k = 0, y(θ) = π − θ > 0 and so x(θ) > 0 is determined by trigonometrybecause tan(θ) = y/x.

We claim that:x(θ) = (π − θ) cot(θ)

y(θ) = π − θ, 0 < θ < π/2

β(θ) =π − θsin(θ)

ex(θ)(2.16)

is a one-parameter family of solutions of (2.14) satisfying x > 0 and π/2 < y < π.Clearly, x(θ), y(θ), β(θ) depend continuously on θ ∈ (0, π/2). Also,

x(θ)→ +∞, y(θ)→ π, β(θ)→ +∞, θ → 0

andx(θ)→ 0, y(θ)→ π/2, β(θ)→ π/2, θ → π/2.

Inasmuch as β(θ) is strictly decreasing on (0, π/2) it follows that the range of βis (π/2,∞). �

7

From Theorem 4.3 in [5], we can conclude that the stability of the steady-state solution of a delay differential equation is determined in the same wayas for an ordinary differential equation. So the steady-state solution of (2.11)given by x ≡ 0 is asymptotically stable if <(λ) < 0 for all roots λ of (2.14),and this solution is unstable if there is a root λ with positive real part. Usingthis result, the following corollary follows immediately from Proposition 2.2 andLemma 2.2.

Corollary 2.3. The following hold for (2.11).

1. If α < 0, then x = 0 is unstable.

2. If 0 < ατ < π/2, then x = 0 is asymptotically stable.

3. If ατ = π/2, then x = sin(πµ/2) and x = cos(πµ/2) are solutions.

4. If ατ > π/2, then x = 0 is unstable.

The next result summarizes important information concerning oscillations.

Theorem 2.4. For every real α and τ > 0, the following are equivalent.

1. Every solution of (2.11) is oscillatory.

2. ατ > 1/e.

Proof. For every real λ, U(µ) = eλµ is either a monotonic or constant function.Therefore, x(t) = U(µ) is not oscillatory for real λ. So x(t) is oscillatory if andonly if λ is a complex (not real) root. By Lemma 2.2, λ is a complex (not real)root if and only if β = ατ > 1/e. This proves the theorem. �

2.2 Scalar DDE

2.2.1 General Equation

The homogeneous equation of the scalar DDE is given by

x′(t) = ax(t) + bx(t− τ), (2.17)

where a, b are constants, and τ > 0 is the delay. The nonhomogeneous equationwill be discussed in the next chapter.

We seek a nontrivial solution of (2.17) of the form

x(t) = eλtc, c 6= 0,

where λ is complex and c is a constant.The characteristic equation we want to solve is given by

h(λ) ≡ λ− a− be−λτ = 0. (2.18)

This expression can be simplified by multiplying the right-hand side by τ andsetting

z := λτ, α := aτ, β := bτ.

We geth(z) ≡ z − α− βe−z = 0. (2.19)

8

Solving this characteristic equation can be done in a similar way as thecharacteristic equation for the simplest equation in the previous section.

We can conclude the following about the stability of the steady-state solutiongiven by x ≡ 0.

Theorem 2.5. The following hold for (2.17).

1. If a+ b > 0, then x = 0 is unstable.

2. If a+ b < 0 and b ≥ a, then x = 0 is asymptotically stable.

3. If a+ b < 0 and b < a, then there exists τ∗ > 0 such that x = 0 is asymptot-ically stable for 0 < τ < τ∗ and unstable for τ > τ∗.

The proof of this theorem can be found in [5, p.53-54].

2.2.2 Existence and Uniqueness

Let τ > 0 be a constant in J = [ξ, ξ+a], where ξ ≥ 0, and a > 0. Let’s considerthe delay differential equation

x′(t) = f(t, x(t), x(t− τ)) for t ∈ J (2.20)

with a single delay τ > 0. Assume that f(t, x, y) and fx(t, x, y) are continuouson R3. An initial condition for (2.20) is given by

x(t) = φ(t) for t ∈ J− = [ξ − τ, ξ], (2.21)

where φ is a given continuous function on R.Equation (2.20) can be solved by the method of steps as follows. For t ∈

[ξ, ξ + τ ], x(t) must satisfy the initial value problem for the ODE:

x′1(t) = f(t, x1(t), φ(t− τ)),

x1(ξ) = φ(ξ).

As g(t, x1) ≡ f(t, x1, φ(t − τ)) and gx1(t, x1) are continuous, a unique solutionof this ODE is guaranteed by standard existence results from ODE theory.

For t ∈ [ξ + τ, ξ + 2τ ], x(t) must satisfy the initial value problem for theODE:

x′2(t) = f(t, x2(t), x1(t− τ)),

x2(ξ + τ) = x1(ξ + τ).

Again, standard existence results for such problems guarantee the existenceof a unique solution. We may repeat this process to extend the solution to[ξ + 2τ, ξ + 3τ ], and so on.

Following this procedure, a unique solution of the initial value problem(2.20), (2.21) can be determined.

9

Chapter 3

Linear Systems

3.1 Preliminaries

A DDE with a single delay is of the form

x′(t) = f(t, x(t), x(t− τ)) for t ≥ ξ, (3.1)

where f is a given continuous function, ξ ≥ 0 is a constant, and τ > 0 is thedelay. An initial condition for (3.1) is given by

x(t) = φ(t) for t ∈ [ξ − τ, ξ], (3.2)

where φ is a given continuous function.We are interested in determining the state of the system (3.1), (3.2) at time

t ≥ ξ ≥ 0. The state of a system at time t ≥ 0 includes all information neededto determine the state of the system at future times s ≥ t. Therefore, the stateof the system (3.1), (3.2) at time t ≥ ξ includes x(η) for all η ∈ [t− τ, t]. Hence,we conclude that this state, which we denote by xt, is given by

xt(θ) := x(t+ θ) for − τ ≤ θ ≤ 0. (3.3)

In this chapter, we will look at linear systems with delay. We will discusslinear DDEs of the form

x′(t) = L(xt) for t ≥ ξ, (3.4)

where ξ ≥ 0 is a constant, xt is defined as (3.3), and L is the map C → Cnwhere C = C([−τ, 0],Cn). An initial condition for (3.4) is of the form

xξ = φ, (3.5)

where xξ is defined as (3.3), and φ is a given continuous function in C.The map L is linear if it satisfies

L(aφ+ bψ) = aL(φ) + bL(ψ), φ, ψ ∈ C, a, b ∈ C.

In the next section, an example of the following linear DDE will be discussed:

x′(t) = Ax(t) +Bx(t− τ), (3.6)

10

where A,B are n × n matrices, and τ > 0 is the delay. This equation can berewritten in the form of (3.4) by defining the map L as

L(y) = Ay(0) +By(−τ),

because then, using (3.3),

L(xt) = Axt(0) +Bxt(−τ) = Ax(t) +Bx(t− τ),

which is precisely (3.6).We introduce the Laplace transform:

F (s) :=

∫ ∞0

e−stf(t) dt, (3.7)

where f(t) is a function on [0,∞). The domain of F (s) consists of all the valuesfor which the integral in (3.7) exists. The Laplace transform of f is denoted byboth F and L{f}. We let f(t) be an exponentially bounded function of orderα; that is,

|f(t)| ≤Meαt for all t ≥ Tfor some positive constants M,T . Then, if f(t) is also piecewise continuous on[0,∞), then L{f}(s) exists for s > α.

The key property of the Laplace transform that we exploit is that the trans-form of a convolution is the product of the transforms. If f(t) and g(t) arepiecewise continuous on [0,∞), then their convolution, denoted by f ∗ g, isdefined by

(f ∗ g)(t) :=

∫ t

0

f(s)g(t− s) ds =

∫ t

0

f(t− s)g(s) ds. (3.8)

The Laplace transform satisfies

L{f ∗ g}(s) = F (s)G(s),

where F = L{f} and G = L{g}.

3.2 Linear DDE

3.2.1 Example

In this section, we will discuss the following homogeneous linear DDE:

x′(t) = Ax(t) +Bx(t− τ) for t ≥ 0, (3.9)

where A =

[0 1−2 3

], B =

[0 11 0

], and τ > 0 is the delay. The initial condition

for (3.9) is given by

φ(t) := x(t) =

[11

]for t ∈ [−τ, 0]. (3.10)

We will start with determining the eigenvalues and the corresponding eigen-vectors of matrix A. The characteristic equation is given by

h(λ) ≡ det(A− λI) = λ2 − 3λ+ 2 = 0. (3.11)

11

Therefore, the eigenvalues of A are λ1 = 1 and λ2 = 2.From

(A− λ1I)v1 =

[−1 1−2 2

]v1 =

[00

],

it follows that an eigenvector corresponding to λ1 is given by v1 =

[11

]. From

(A− λ2I)v2 =

[−2 1−2 1

]v2 =

[00

],

it follows that an eigenvector corresponding to λ2 is given by v2 =

[12

].

Therefore, we can conclude that the solution of

x′(t) = Ax(t)

is given by

xh(t) = c1et

[11

]+ c2e

2t

[12

], (3.12)

where c1, c2 are constants.For t ∈ [0, τ ],

x′(t) = Ax(t) +Bφ(t− τ)

=

[0 1−2 3

]x(t) +

[0 11 0

] [11

]=

[0 1−2 3

]x(t) +

[11

]. (3.13)

A particular solution of (3.13) is of the form

xp(t) =

[d1d2

],

where d1, d2 are constants.Substituting xp(t) for x(t) in (3.13) gives

x′p(t) =

[00

]=

[0 1−2 3

] [d1d2

]+

[11

]=

[d2 + 1

−d1 + 3d2 + 1

].

Therefore, d1 = −2 and d2 = −1. So the particular solution of (3.13) is givenby

xp(t) =

[−2−1

]= −

[21

]. (3.14)

Hence, the general solution of (3.13) is of the form

x(t) = xh(t) + xp(t) = c1et

[11

]+ c2e

2t

[12

]−[21

],

where c1, c2 are constants. Because x(0) = φ(0), it follows that

x(0) =

[11

]= c1

[11

]+ c2

[12

]−[21

]=

[c1 + c2 − 2c1 + 2c2 − 1

].

12

Therefore, c1 = 4 and c2 = −1. So the general solution of (3.9), (3.10), fort ∈ [0, τ ], is given by

x(t) = xh(t) + xp(t) = 4et[11

]− e2t

[12

]−[21

]. (3.15)

In a similar way, general solutions of the initial value problem (3.9), (3.10)defined on other intervals of the form [(n− 1)τ, nτ ], n ∈ N, can be determined.

3.2.2 General Equation

The general equation of the nonhomogeneous linear DDE is given by

x′(t) = Ax(t) +Bx(t− τ) + f(t) for t ≥ 0, (3.16)

where A,B are n× n matrices, τ > 0 is the delay, and f is a given continuousfunction. The initial condition for (3.16) is of the form

x(t) = φ(t) for t ∈ [−τ, 0], (3.17)

where φ is a given continuous function.Applying the Laplace transform to the left-hand side of (3.16), we obtain

L{x′}(s) =

∫ ∞0

e−stx′(t) dt

= [e−stx(t)]t=∞t=0 + s

∫ ∞0

e−stx(t) dt

= sX(s)− φ(0).

Applying the Laplace transform to the right-hand side of (3.16), we obtain

sX(s)− φ(0) =

∫ ∞0

e−st[Ax(t) +Bx(t− τ) + f(t)] dt

= A

∫ ∞0

e−stx(t) dt+B

∫ ∞0

e−stx(t− τ) dt+

∫ ∞0

e−stf(t) dt

= AX(s) +B

[∫ τ

0

e−stφ(t− τ) dt+

∫ ∞τ

e−stx(t− τ) dt

]+ F (s)

= AX(s) +B

[∫ τ

0

e−stφ(t− τ) dt+

∫ ∞0

e−s(t+τ)x(t) dt

]+ F (s)

= AX(s) +B

[∫ τ

0

e−stφ(t− τ) dt+ e−sτX(s)

]+ F (s)

= [A+ e−sτB]X(s) +B

∫ τ

0

e−stφ(t− τ) dt+ F (s)

= [A+ e−sτB]X(s) +BΦ(s) + F (s),

where Φ = L{φ(· − τ)}, and where we have extended φ to [−τ,∞) by makingit zero for t > 0. This leads to

X(s) = K(s)[φ(0) +BΦ(s) + F (s)], (3.18)

13

whereK(s) = (sI −A− e−sτB)−1

is a matrix-valued function.In order to make use of the convolution result, we need to know the inverse

transform k of K. In view of the calculations above, we see that k is a solutionof (3.16), (3.17), with f = 0, for the initial data

ξ(θ) =

{I for θ = 0,0 for θ ∈ [−τ, 0).

(3.19)

In spite of the discontinuity of ξ at zero, the method of steps readily estab-lishes that the solution k exists for t ≥ 0. The matrix function k is called thefundamental matrix solution of (3.16), (3.17).

Lemma 3.1. We may express the solution of the initial value problem (3.16),(3.17) as

x(t) := x(t;φ, f) = x(t;φ, 0) + x(t; 0, f). (3.20)

Proof. For t ∈ [0, τ ], the following hold:

x′(t;φ, f) = Ax(t;φ, f) +Bφ(t− τ) + f(t);

x′(t;φ, 0) = Ax(t;φ, 0) +Bφ(t− τ);

x′(t; 0, f) = Ax(t; 0, f) + f(t).

So, for t ∈ [0, τ ],

x′(t;φ, 0) + x′(t; 0, f) = A[x(t;φ, 0) + x(t; 0, f)] +Bφ(t− τ) + f(t).

Hence,x(t) := x(t;φ, f) = x(t;φ, 0) + x(t; 0, f) for t ∈ [0, τ ]. (3.21)

We can conclude from (3.21) that (3.20) holds for all t ≥ 0, by uniqueness ofthe solutions x(t;φ, f), x(t;φ, 0), and x(t; 0, f) for all t ≥ 0. �

3.3 General Results

In this chapter, we are interested in solving the linear DDE:

x′(t) = L(xt), (3.4)

where L is the map C → Cn where C = C([−τ, 0],Cn), and where xt is definedas

xt(θ) := x(t+ θ) for − τ ≤ θ ≤ 0. (3.3)

We seek a nontrivial solution of (3.4) of the form

x(t) = eλtv, v 6= 0,

where λ and the elements of the vector v are complex. We will use the notationexpλ, where expλ(θ) = eλθ. Then,

xt(θ) = x(t+ θ) = eλ(t+θ)v = eλteλθv = eλt expλ(θ)v.

14

Therefore, xt = eλt(expλ)v.Using that L is a linear map and that eλt ∈ C for all λ, t; for x(t) = eλtv to

be a solution of (3.4), we must have

x′(t) = λeλtv = L(xt) = L(eλt(expλ)v) = eλtL((expλ)v).

Dividing both sides of this equation by eλt, we get

λv = L((expλ)v).

Let {e1, e2, . . . , en} be the standard basis for Cn. Then, we can writev =

∑j vjej . Using this notation, we get L((expλ)v) =

∑j vjL((expλ)ej),

by linearity of L and the fact that vj ∈ C for all j.Define Lλ to be the n× n matrix

Lλ = (L((expλ)e1) |L((expλ)e2) | · · · |L((expλ)en)) = (Li((expλ)ej)),

where Li(φ) is element i of L(φ). Then,

λv = L((expλ)v)

=∑j

vjL((expλ)ej)

= (L((expλ)e1) |L((expλ)e2) | · · · |L((expλ)en)) · (v1 v2 · · · vn)T

= Lλv.

Therefore, x(t) = eλtv is a nontrivial solution of (3.4) if and only if λ is asolution of the characteristic equation:

det(λI − Lλ) = 0, (3.22)

and the vector v belongs to the null space of λI − Lλ.Let’s return to the DDE discussed before:

x′(t) = Ax(t) +Bx(t− τ), (3.6)

where A,B are n× n matrices, and τ > 0 is the delay. Rewriting this DDE inthe form of (3.4), see Section 3.1, the map L is defined as

L(y) = Ay(0) +By(−τ).

For this DDE, the n× n matrix Lλ is given by

Lλ = (Li((expλ)ej))

= ((A expλ(0)ej +B expλ(−τ)ej)i)

= ((Aej + e−λτBej)i)

= (Ae1 + e−λτBe1 |Ae2 + e−λτBe2 | · · · |Aen + e−λτBen)

= (A+ e−λτB) (e1 e2 · · · en)

= (A+ e−λτB)I

= A+ e−λτB.

Therefore, the characteristic equation is given by

det(λI − Lλ) = det(λI −A− e−λτB) = 0. (3.23)

Solving (3.23) for λ and determining the vector v from the null space of λI −A− e−λτB, we can conclude that x(t) = eλtv is a nontrivial solution of (3.6).

15

Chapter 4

Application

4.1 Delayed Logistic Equation

The logistic equation is given by

x′(t) = ax(

1− x

K

), (4.1)

where a is the growth rate and K is the carrying capacity of the ecosystem.According to the logistic equation, the growth rate of a population is directlyproportional to the current population and the availability of resources in theecosystem.

In this section, we will discuss the delayed logistic equation:

x′(t) = ax(t)[1− bx(t− τ)] for t ≥ 0, (4.2)

where a, b > 0 are constants with b ≡ 1/K, and τ > 0 is the delay. As initialcondition for (4.2), we consider

x(t) = 0.01 for t ∈ [−τ, 0]. (4.3)

We denote the solution of the initial value problem (4.2), (4.3) by x(t).We start with determining the equilibrium points of (4.2), which we denote

by x∗. Because x∗ is constant, it follows that x∗ = x(t) = x(t−τ). Substitutingthis result in (4.2), we get

ax∗[1− bx∗] = 0.

Therefore, the equilibrium points of (4.2) are given by x∗ = 0 and x∗ = 1/b.Following the procedure described in [4], we can linearize (4.2) in the follow-

ing way. Let p(t) be a small variation in the population such that higher powersof p may be neglected. Substituting x(t) ≡ x∗ + p(t) in (4.2) gives the DDE:

p′(t) = a[x∗ + p(t)][1− bx∗ − bp(t− τ)] for t ≥ 0. (4.4)

We will use the same initial condition for (4.4) as for (4.2):

p(t) = 0.01 for t ∈ [−τ, 0]. (4.5)

Substituting x∗ = 0 in (4.4) gives

p′(t) = ap(t)[1− bp(t− τ)].

16

Since we may neglect higher powers of p, we end up with the following ODE:

p′(t) = ap(t). (4.6)

Therefore,p(t) = ceat,

where c is a constant. Taking the initial condition (4.5) into account, the solutionof the initial value problem (4.4), (4.5), for x∗ = 0, is given by

p1(t) = 0.01eat. (4.7)

Since a > 0, it follows that x∗ = 0 is unstable.Substituting x∗ = 1/b in (4.4) gives

p′(t) = a[1/b+ p(t)][−bp(t− τ)].

Since we may neglect higher powers of p, we end up with the following DDE:

p′(t) = −ap(t− τ). (4.8)

This differential equation is already discussed in Section 2.1.3. The equilibriumpoint of (4.8) is given by p∗ = 0. Because p∗ 6= x∗ and we are interested inhaving p∗ = x∗, we will add a certain constant to (4.8) such that p∗ = x∗ holds.This constant has to be equal to a/b, because then, we have the following DDE:

p′(t) = −a[p(t− τ)− 1/b], (4.9)

where the equilibrium point is given by p∗ = x∗ = 1/b. Instead of (4.5), weconsider (4.7) as initial condition for (4.9). The reasoning behind this becomesclear by looking at the plots of Figure 4.1. We denote the solution of the initialvalue problem (4.9), (4.7) by p2(t).

Using the package DDE23 in Matlab, plots of the numerical solution x(t)for different values of τ, a, and b can be made. Because changing the value of bdoes not affect the shapes of the plots, we will only consider one value for b.

See Figure 4.1 for the plots of x(t), p1(t), and p2(t) for some values of τ, a, andb = 0.5. Because determining p2(t) analytically is quite difficult, the numericalsolution p2(t) is included in these plots. The Matlab codes used for computingthe plots in this chapter can be found in Appendix A.2.

17

0 5 10 15 20

t

0

1

2

x(t

)

τ = 0.5, a = 0.5

1/b

0 5 10

t

0

1

2

x(t

)

τ = 0.5, a = 1

1/b

0 5 10

t

0

1

2

x(t

)

τ = 0.5, a = 1.5

1/b

0 5 10 15 20

t

0

1

2

x(t

)

τ = 1, a = 0.5

1/b

0 5 10 15 20

t

0

1

2

3x(t

)τ = 1, a = 1

1/b

0 5 10 15

t

0

2

4

x(t

)

τ = 1, a = 1.5

1/b

0 5 10 15 20 25

t

0

1

2

x(t

)

τ = 1.5, a = 0.5

1/b

0 5 10 15 20

t

0

2

4

x(t

)

τ = 1.5, a = 1

1/b

0 5 10 15

t

-10

0

10

x(t

)

τ = 1.5, a = 1.5

1/b

x(t) p1(t) p2(t)

Figure 4.1: Plots of x(t), p1(t), and p2(t), for b = 0.5, and different values of aand τ .

From the plots in Figure 4.1, we can conclude that p1(t) and p2(t) approx-imate x(t) very well, except for the last case. For this case, aτ = (1.5)2 =2.25 > π/2. By Corollary 2.3, it follows that x∗ = 1/b is unstable. From theplot of this case, we can conclude that x(t) contains complex (not real) values,and p1(t) and p2(t) only contain real values. Therefore, x(t) cannot be approx-imated by p1(t) and p2(t). For the other cases, we can conclude from the plots,and Corollary 2.3, that x∗ = 1/b is stable.

Furthermore, we can conclude from these plots that the cases where aτhas the same value, the graphs look similar. For example, consider the casesτ = 0.5, a = 1, and τ = 1, a = 0.5. Then, the plots of these cases look the same,but the time intervals are different. The same holds for the other cases with thesame value of aτ .

18

Appendix A

Matlab Codes

A.1 Chapter 2: Simple Cases

Figure 2.1

1 function [ dxdt ] = SCDDE1dde ( t , x , Z)2 % x ’ ( t ) = −x ( t−tau ) f o r t>=03 % tau>0 i s the d e l a y4 x lag = Z ( : , 1 ) ;5 dxdt = [−xlag ( 1 ) ] ;6 end

1 function [ s ] = SCDDE1hist ( t )2 % I n i t i a l data i s g i ven by3 % x ( t ) = 1 f o r −tau<=t<=04 s = [ 1 ] ;5 end

1 %Analy t i c S o l u t i o n2 n = . . . ;3 tau = . . . ;4 t0 = 0 ;5 tend = . . . ;6 t = linspace ( t0 , tend ) ;7 x = zeros ( length ( t ) ) ;8 for i =1:n9 for j =1: length ( t )

10 i f ( ( i −1)∗ tau <= t ( j ) ) && ( t ( j ) <= i ∗ tau )11 A = zeros ( i ) ;12 for k=1: i13 A( k ) = (−1)ˆk ∗ ( ( t ( j )−(k−1)∗ tau )ˆ k ) / . . .14 f a c t o r i a l ( k ) ;15 end16 x ( j ) = 1 + sum(A) ;17 end

19

18 end19 end2021 %Numerical S o l u t i o n22 s o l = dde23 (@SCDDE1dde , [ tau ] , @SCDDE1hist , [ t0 , tend ] ) ;2324 %Figure25 f igure26 plot ( s o l . x , s o l . y , ’ Color ’ , [ 2 5 5 , 1 2 8 , 0 ] . / 2 5 5 ) ; hold on27 plot ( t , x , ’−−k ’ ) ;

A.2 Chapter 4: Application

Figure 4.1

1 function [ dxdt ] = APPDDE41dde ( t , x , Z)2 % x ’ ( t ) = ax ( t )[1− bx ( t−tau ) ] f o r t>=03 % tau>0 i s the d e l a y4 a = . . . ;5 b = 0 . 5 ;6 x lag = Z ( : , 1 ) ;7 dxdt = a∗x(1)∗ [1−b∗ xlag ( 1 ) ] ;8 end

1 function [ s ] = APPDDE41hist ( t )2 % I n i t i a l data i s g i ven by3 % x ( t ) = 0.01 f o r −tau<=t<=04 s = [ 0 . 0 1 ] ;5 end

1 function [ dp2dt ] = APPDDE42dde ( t , p2 , Z)2 % p2 ’ ( t ) = −a [ p2 ( t−tau )−1/b ] f o r t>=t023 % tau>0 i s the d e l a y4 a = . . . ;5 b = 0 . 5 ;6 p2lag = Z ( : , 1 ) ;7 dp2dt = −a ∗ [ p2 lag (1)−1/b ] ;8 end

1 function [ s ] = APPDDE42hist ( t )2 % I n i t i a l data i s g i ven by3 % p1 ( t ) = 0.01 exp ( at ) f o r t02−tau<=t<=t024 a = . . . ;5 b = 0 . 5 ;6 s = zeros ( length ( t ) ) ;7 for i =1: length ( t )

20

8 s ( i ) = 0.01∗exp( a∗ t ( i ) ) ;9 end

10 end

1 a = . . . ;2 b = 0 . 5 ;3 tau = . . . ;4 t01 = 0 ;5 t02 = . . . ;6 tend = . . . ;78 % x ( t )9 s o l = dde23 (@APPDDE41dde , [ tau ] , @APPDDE41hist , [ t01 , tend ] ) ;

1011 % p1 ( t )12 t = linspace ( t01 , tend ) ;13 p1 = zeros ( length ( t ) ) ;14 p1 (1 ) = 0.01∗exp( a∗ t ( 1 ) ) ;15 for i =1: length ( t )−116 i f ( p1 ( i ) <= max( s o l . y ) )17 p1 ( i +1) = 0.01∗exp( a∗ t ( i +1)) ;18 else19 break20 end21 end2223 % p2 ( t )24 s o l 2 = dde23 (@APPDDE42dde , [ tau ] , @APPDDE42hist , [ t02 , tend ] ) ;2526 %Figure27 f igure28 plot ( s o l . x , s o l . y , ’ k ’ ) ; hold on29 plot ( t ( 1 : i ) , p1 ( 1 : i ) , ’ Color ’ , [ 0 , 1 5 3 , 7 6 ] . / 2 5 5 ) ; hold on30 plot ( s o l 2 . x , s o l 2 . y , ’ Color ’ , [ 2 5 5 , 1 2 8 , 0 ] . / 2 5 5 ) ; hold on31 plot ( s o l . x , 1/ b∗ ones ( length ( s o l . x ) ) , ’−−k ’ )32 text ( tend +0.01∗ tend , 2 . 0 1 , ’ 1/b ’ )

21

Bibliography

[1] Dimitri Breda, Stefano Maset, and Rossana Vermiglio. Stability of Lin-ear Delay Differential Equations: A Numerical Approach with MATLAB.Springer Science & Business Media, 2014.

[2] Jack K. Hale and Sjoerd M. Verduyn Lunel. Introduction to FunctionalDifferential Equations. Springer Science & Business Media, 1993.

[3] R. Kent Nagle, Edward B. Saff, and Arthur David Snider. Fundamentals ofDifferential Equations. Pearson Education, Inc., 2008.

[4] Milind M. Rao and K.L. Preetish. Stability and hopf bifurcation analysis ofthe delay logistic equation, 2012.

[5] Hal Smith. An Introduction to Delay Differential Equations with Applica-tions to the Life Sciences. Springer Science & Business Media, 2010.

[6] Wolfgang Walter. Ordinary Differential Equations. Springer Science & Busi-ness Media, 1998.

22