Lecture 3 Deterministic Dynamics

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Deterministic Dynamical Systems

Basic Model

Lecture 3 Deterministic Dynamics

A. Banerji

Department of Economics

February 5, 2015

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A. Banerji

Basic Model

Outline

Deterministic Dynamical Systems Basic Model

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Basic Model

Definition

A dynamical system is a pair(S,h), whereS = (S, ρ)is a metric space andh:S→S.

Thet^thiterate of a pointx ∈S,h^t(x) =h(h(h...(x))) (wherehis applied tox ttimes). By convention, h⁰(x) =x.

Definition

The trajectory ofx ∈Sunderhis the infinite sequence (h^t(x))^∞_t=0

The idea is that fort=0,1,2, .., the dynamical system describes the evolution of a state according to the equationx_t+1=h(x_t).

Example

S= [0,1].h(x) =0.25+0.5x.

See JS Figs 4.2, 4.3 for illustrations of trajectories.

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Basic Model

Stability

A fixed pointx^∗ ofhis also called stationary or

equilibrium point of the dynamical system(S,h)as its trajectory stays there.

Definition

The stable set of a fixed pointx^∗ is the set of all pointsx whose trajectories converge tox^∗.x^∗is locally stable, or an attractor, if its stable set contains an open ball that containsx^∗.

Remark: You may have seen a more conventional definition of stability in an earlier course; the definition here corresponds to local asymptotic stability. Above example is stronger than local stability.

Definition

A dynamical system(S,h)is globally stable if 1. hhas a unique fixed pointx^∗, and

2. h^t(x)→x^∗ast → ∞, for allx ∈S.

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Basic Model

Note: So ifhis a uniformly strict contraction, then by Banach’s theorem,(S,h)is globally stable.

JS 4.1.1. Ifx^∗is a fixed point to which every trajectory converges, thenx^∗ is the unique fixed point.

This is trivial. Another pointx can’t be a fixed point because its trajectory doesn’t stay there; it tends tox^∗. JS 4.1.2. Ify ∈Sis s.t. h^t(x)→y for somex ∈S, and if his continuous aty, theny is a fixed point ofh.

Indeed, sinceh^t(x)→y, by continuity

h^t+1(x) =h(h^t(x))→h(y). Since(h^t⁺¹(x))is a subsequence of the convergent sequence(h^t(x)), it shares the same limit: soh(y) =y.

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Basic Model

Exercises

JS 4.1.3. Ifhis continuous onSand invariant onA⊂S (i.e. h:A→A), thenhis invariant on the closure ofA.

Proof.

Lety ∈cl(A). Want to showh(y)∈cl(A).

Sincey ∈cl(A), there exists a sequence(x_t)in A converging toy. By continuity ofh,h(xt)→h(y).

Sincehis invariant onA, the sequence(h(x_t))lies in A and hence incl(A); sincecl(A)is closed, the limith(y)is in the cl(A).

For a pointx very close toy,h(x)∈A, andh(y)is close toh(x)by continuity ofh. Note. JS has a

one-dimensional example of a globally stable dynamical system. In the example,(S, ρ) = ((0,∞),| |)is a

complete metric space andh(x) =2+p

(x)is a

contraction. So it has a unique fixed point. Moreover, by the proof of Banach’s theorem, all trajectories converge to the fixed point. So the dynamical system(S,h)is globally stable.

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Basic Model

JS 4.1.4. LetS= (<,| |)andh(x) =ax+b. Then, for x ∈Sandt ∈N,

h^t(x) =a^tx +bPt−1 i=0aⁱ.

(S,h)is globally stable if|a|<1.

The expression is true fort=1. Now suppose it’s true for t−1, for allx. Then

h^t(x) =h(h^t⁻¹(x)) =a(a^t⁻¹x+bPt−2

i=0aⁱ) +b

=a^tx+bPt−1 i=0aⁱ.

Suppose|a|<1. Lettingt → ∞,h^t(x)→b/(1−a). This is true for allx ∈S, and moreover note thatb/(1−a)is the unique fixed point ofh. So(S,h)is globally stable.

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Basic Model

Exercises

JS 4.1.6. Prove that(S,h)in 4.1.5. is stable using a more widely applicable method.

Proof.

his a uniformly strict contraction. Indeed,

|h(x)−h(y)|=|a(x−y)| ≤ |a||x−y|

So if|a|<1,his a contraction with modulus|a|.

By Banach’s theorem, there is a unique fixed pointx^∗, and by the proof of Banach, for allx ∈S,h^t(x)→x^∗. So (S,h)is globally stable.

The next, growth model exercise has a similar easy visual intuition. We do a proof suggested by JS, although an easier proof seems obvious.

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Basic Model

JS 4.1.7.S = (0,∞),ρ(x,y) =|log(x)−log(y)|. Then (S, ρ)is a complete metric space. Supposek_t₊₁=sAk_t^α, wherek is capital,s >0 is the savings rate,A>0 a productivity parameter andα∈(0,1). Convert this into a dynamical system on(S, ρ)and use Banach’s theorem to prove stability. (The idea behind the log metric is to useα as the contraction modulus).

Proof.

(i)ρis obviously a metric. Now, let(x_n)be a Cauchy sequence in(S, ρ), and let(zn) = (logxn). Let >0.

Then there existsMs.t. n,m≥Mimplies|z_n−z_m|<

(since(x_n)is Cauchy in(S, ρ)). So(z_n)is Cauchy in (<,| |), and by completeness of<, converges to some y ∈ <. By continuity of the exponential function, x_n =e^zⁿ →e^y ∈S. So(S, ρ)is complete.

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Basic Model

Exercises

Proof.

(ii) Leth(k) =sAk_t^α.h:S→S. Consider the dynamical system(S,h).

ρ(h(x),h(y)) =|log(sAx^α)−log(sAy^α)|

≤ |α||logx−logy|=|α|ρ(x,y).

Since|α|<1,his a uniformly strict contraction with modulus|α|. So it has a unique fixed pointx^∗ and by the proof of Banach’s theorem, all sequences(h^t(x))

converge tox^∗. So(S,h)is globally stable.

Draw Picture.

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Basic Model

JS 4.1.8. Suppose(S, ρ)is a complete metric space and h:S→S a uniformly strict contraction. IfA⊂Sis nonempty, closed and invariant underh, then the fixed pointx^∗ lies inA.

Proof.

Take somex ∈Aand consider the sequence

(xn) = (hⁿ(x)). SinceAis invariant underh,(xn)⊂A;

sincehis a uniformly strict contraction,(x_n)is Cauchy (recall Proof of Banach’s theorem); sinceSis complete, xn →x^∗ ∈S. And by the proof of Banach’s theorem,x^∗is the unique fixed point ofh. SinceAis closed,x^∗ ∈A.

The next result is used in the discussion on Markov Chains.

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Basic Model

Some Results

Lemma

Let(S,h)be a dynamical system. If h is nonexpansive and(S,h^N)is globally stable for some N ∈N, then(S,h) is globally stable.

Proof.

h^Nhas a unique fixed pointx^∗, and for allx ∈S, h^kN(x)→x^∗ask → ∞. Let >0, and choosek large enough thatρ(h^kN(h(x^∗)),x^∗)< . (puttingx =h(x^∗)as the initial state). Sinceh^kN(x^∗) =x^∗, we have

ρ(h(x^∗),x^∗) =ρ(h(h^kN(x^∗)),x^∗) =ρ(h^kN(h(x^∗)),x^∗)< . Since this is true for all >0,h(x^∗) =x^∗;x^∗ is a fixed point ofh.

Now letx ∈Sand >0, andk large enough that

ρ(h^kN(x),x^∗)< . Applyinghseveral times toh^kN(x)and using the nonexpansiveness inequality, forn≥kN we have

ρ(hⁿ(x),x^∗) =ρ(h^n−kN(h^kN(x)),x^∗)≤ρ(h^kN(x),x^∗)< . So(S,h)is globally stable.

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Basic Model

Lemma

Suppose(S,f)is a dynamical system s.t. S⊂ <is open and f is C¹. If x^∗is a fixed point s.t. |f⁰(x^∗)|<1, then x^∗ is locally stable.

Proof.

Step 1. Since|f⁰(x^∗)|<1, by continuity off⁰ |f⁰(x)|<1 for allx in some interval aroundx^∗. In painful detail, G=f⁰⁻¹((−1,1))is open since (−1,1)is open andf⁰ is continuous. Sincef⁰(x^∗)∈(−1,1),x^∗∈Gand hence an interior point ofG, soB(x^∗, )⊂G, for some. So there exista,bs.t.x^∗− <a<x^∗<b<x^∗+. So,[a,b]⊂G.

Since[a,b]is compact andf⁰ is continuous, there exists z ∈[a,b]s.t.f⁰(z) =sup_x∈[a,b]|f⁰(x)|=λ <1.

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Basic Model

Proof contd.

Proof.

Step 2. f is a uniformly strict contraction on some closed interval containingx^∗. Indeed, letx,y ∈[a,b]and let x ≥y, WLOG. Thenf(x) =f(y) +Rx

y f⁰(u)du, so f(x)−f(y)≤λ(x−y)(asf⁰(u)≤ |f⁰(u)| ≤λ). On the other hand,f⁰(u)≥ −λ, sof(x)−f(y)≥ −λ(x−y).

So,|f(x)−f(y)| ≤λ|x −y|.

Now supposef⁰(x^∗)>0 (or<0). By continuity, the same inequality holds on some[a⁰,b⁰]⊂[a,b]that containsx^∗ in its interior. Supposex^∗ <z <b⁰. We have

f(z)−f(x^∗)≤λ(z−x^∗). Sincef(x^∗) =x^∗, this means f(z)−x^∗ ≤λ(z−x^∗)sof(z)<z. Also, sincef⁰ >0 in this interval,f(z)>f(x^∗) =x^∗. Thusx^∗<f(z)<z <b⁰. We can extend this to other cases and conclude that f : [a⁰,b⁰]→[a⁰,b⁰]. By Banach’s theorem,x^∗is the unique fixed point and for allx ∈[a⁰,b⁰],f^t(x)→x^∗ as t → ∞, sox^∗is an attractor.

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Basic Model

Example

Growth model with “threshold" nonconvexities (Azariadis and Drazen (1990))

k_t₊₁=sA(k)k_t^α, whereα∈(0,1),

A(k) =

A₁ if0<k <k_b A₂ ifk_b≤k <∞

where 0<A₁<A₂. Letk_i^∗ be the unique solution to k =sA_ik^α,i =1,2. For the casek₁^∗<k_b<k₂^∗, the two fixed points are attractors, by the above lemma. History (initial condition?) determines which one a country ends up at, and so how rich it is. See JS Figure 4.4.

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Linear Affine Systems

Linear (affine) Dynamical System:

(<ⁿ,h), whereh:<ⁿ→ <ⁿis defined byh(x) =Ex+b, whereE is ann×nreal matrix andbis a vector in<ⁿ. A sufficient condition for an affine system to be globally stable is that for any one of thep-norms 1,2,∞,||Ex||_p be bounded by a number less than 1 for allx lying on the unit circle.

Letλp=sup{||Ex||_p:||x||_p =1}, wherep=1,2,or∞.

JS 4.1.11. For allx ∈ <ⁿ,||Ex||_p≤λp||x||_p. Indeed,(x/||x||_p)is on the unit circle, so

||E(x/||x||_p)||_p≤λp. Now just multiply both sides by

||x||_p.

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Basic Model

JS 4.1.12. Ifλp<1 forp=1 or 2 or∞, the affine system (<ⁿ,h)is globally stable.

Proof.

Letx,y ∈ <ⁿ.||h(x)−h(y)||_p=||E(x −y)||_p

≤λp||x−y||_pby the earlier result. So ifλp<1,his a uniformly strict contraction mapping with modulusλpon the complete metric space<ⁿ, so Banach’s theorem implies that a unique fixed point exists, to which all trajectories(h^t(x))^∞_t=0converge.

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Basic Model

Long and Plosser 1983

Modeling business cycles in multisector optimal growth models with iid production shocks. Solving their model, log output in 6 sectors follows the difference equation y_t₊₁=Ayt +b

whereyt is a 6×1 vector of log outputs andAis a 6×6 matrix of input-output elasticities. bis random (but we take it to be a constant). Long and Plosser estimateA and find that each row, and each column sums to < 1.

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JS 4.1.13. Long and Plosser’s dynamical system is stable (as the row sums are all less than 1).

Proof.

Let row sumi,α_i =Pn

j=1|a_ij|and letα=max_iα_i <1. Let x be on the unit circle according top=∞, i.e.

||x||_∞=max_i|x_i|=1.

Ax =x₁a_.1+...+xna.n. So mod of theith row ofAx

=|x₁a_i1+...+x_na_in| ≤ |x₁||a_i1|+...+|x_n||a_in|

≤1.|a_i1|+...+1.|a_in|=αi ≤α <1

So,||Ax||_∞< α <1, for allx on the unit circle. So λ∞=sup||Ax||_∞≤α <1. By JS 4.1.12.,(<ⁿ,h), with h(y) =Ay +bis globally stable, andy^∗ = (I−A)⁻¹bis the unique fixed point.

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Basic Model

Long and Plosser III

JS 4.1.14. LP’s system is stable as the column sums are less than 1.

Proof.

Letβ_j be the column sums of a square matrixBand let β =max_jβ_j <1. LetB= (b₁, ...,b_n). Then

||Bx||₁||Pn

i=1x_ib_i||₁≤Pn

i=1||x_ib_i||₁≤ |x₁|β₁+...+|x_n|β_n. Ifx is on the unit circle according top=1,P

i|x_i|=1, so the right hand is a convex combination of theβ_j’s and is therefore≤β <1. Soλ1=sup||Bx||₁≤β <1 and stability follows from JS 4.1.12.

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Definition

A dynamical system(S,h)is Lagrange stable if every trajectory is precompact. (i.e., every(h^t(x))has a limit point inS).

JS 4.1.15. IfSis bounded,(S,h)is Lagrange stable.

Bolzano-Weierstrass.(h^t(x))⊂S, so is bounded, as a set;(h^t(x))is a sequence in this bounded set, so has a convergent subsequence. IfSis complete, the limit of this subsequence lies inS.

JS 4.1.16. Example of Lagrange stable system for unboundedS.

LetS= (0,∞)andh(x) =0.5+0.5x.

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Monotone Systems

JS 4.1.17. Ifh:< → <is increasing (y ≥x implies h(y)≥h(x)), then every trajectory in(<,h)is monotone.

The idea is that in<, eitherx ≥h(x)orx ≤h(x).

Suppose the former. Then sincehis increasing, applying hto both sides we geth(x)≥h²(x), and so on.

JS 4.1.18. This is not true in(<ⁿ,h).

As a counterexample, take(<²,h)with

h(x¹,x²) = (x²,x¹), for all(x¹,x²)∈ <².his increasing.

But(h^t(2,1))^∞₀ is not monotone.