PDF Lecture 3 Deterministic Dynamics - Delhi School of Economics

Deterministic Dynamical Systems

Basic Model

Lecture 3 Deterministic Dynamics

A. Banerji

Department of Economics

March 4, 2016

A. Banerji

Deterministic Dynamical Systems

Basic Model

Outline

Deterministic Dynamical Systems Basic Model

Deterministic Dynamical Systems

Basic Model

The idea is that fort=0,1,2, .., the dynamical system should describe the evolution of a state according to a nonlinear first-order difference equationx_t+1=h(x_t).

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

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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Deterministic Dynamical Systems

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^∗.

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.

Deterministic Dynamical Systems

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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Deterministic Dynamical Systems

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(x_t)→h(y).

Sincehis invariant onA, the sequence(h(xt))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.

Deterministic Dynamical Systems

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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Deterministic Dynamical Systems

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.

Deterministic Dynamical Systems

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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Deterministic Dynamical Systems

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.

Deterministic Dynamical Systems

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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Deterministic Dynamical Systems

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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.

Deterministic Dynamical Systems

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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(x^∗−δ,x^∗+δ). Fix[a,b]in this interval. Since[a,b]is compact andf⁰ is continuous, there existsz ∈[a,b]s.t.

f⁰(z) =sup_x∈[a,b]|f⁰(x)|=λ <1.

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Deterministic Dynamical Systems

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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.

Deterministic Dynamical Systems

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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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Deterministic Dynamical Systems

Basic Model

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.

Deterministic Dynamical Systems

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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Deterministic Dynamical Systems

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.

Deterministic Dynamical Systems

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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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Deterministic Dynamical Systems

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.

Deterministic Dynamical Systems

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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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Deterministic Dynamical Systems

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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.