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Lecture Notes 6: Dynamic Equations

Part A: First-Order Dierence Equations

in One Variable

Peter J. Hammond

latest revision 2017 September 23rd, plus minor correction typeset fromdiffEqSlides18A.tex University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 1 of 54

Lecture Outline

Introduction: Dierence vs. Dierential Equations

First-Order Dierence Equations

First-Order Linear Dierence Equations: Introduction

General First-Order Linear Equation

Particular, General, and Complementary Solutions

Explicit Solution as a Sum

Constant and Undetermined Coecients

Stationary States and Stability for Linear First-Order Equations Local Stability of Nonlinear First-Order Equations University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 2 of 54

Walking as a Simple Dierence Equation

What is the dierence

between dierence and dierential equations?

It is relatively common to indicate by:

a subscript a d iscretetime function lik em7!xm; parentheses a continuou stime function l iket7!x(t).

Walking on two feet can be modelled as a

discrete time p rocess with time domain T=f0;1;2;:::g=Z+=f0g [N that counts the number of completed steps.

Aftermsteps, the respective positions`;r2R2

of the left and right feet on the ground can be described by the two functionsT3m7!(`m;rm). University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 3 of 54

Walking as a More Complicated Dierence Equation

Athletics rules limit a walking step to be no longer than a stride. So a walking process that starts with the left foot might be described by the two coupled equations m=( (rm1) ifmis odd m1ifmis evenandrm=( (`m1) ifmis even r m1ifmis odd form= 0;1;2;:::. Or, if the length and direction of each pace are aected by the length and direction of its predecessor, by m=( (rm1;`m2) ifmis odd m1ifmis even andrm=( (`m1;rm2) ifmis even r m1ifmis odd form= 0;1;2;:::. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 4 of 54

Walking as a Dierential Equation

Newtonian physics implies that a walker's centre of mass must be a continuous function of time , described by a 3-vector valued mappingR+3t7!(x(t);y(t);z(t))2R3.

The time domain is thereforeT:=R+.

The same will be true for the position of, for instance, the extreme end of the walker's left big toe.

Newtonian physics requires that the

acceleration

3-vecto r

described by the second derivative d2dt2(x(t);y(t);z(t))2R3 should be well dened for allt. The biology of survival requires it to be bounded. Actually, the motion becomes seriously uncomfortable unless the acceleration (or deceleration) is continuous | as my driving instructor taught me more than 50 years ago! University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 5 of 54

Lecture Outline

Introduction: Dierence vs. Dierential Equations

First-Order Dierence Equations

First-Order Linear Dierence Equations: Introduction

General First-Order Linear Equation

Particular, General, and Complementary Solutions

Explicit Solution as a Sum

Constant and Undetermined Coecients

Stationary States and Stability for Linear First-Order Equations Local Stability of Nonlinear First-Order Equations University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 6 of 54

Basic Denition

LetT=Z+3t7!xt2Xdescribe a discrete time process,

withX=R(orX=Rm) as thestate space . Its dierence at time tis dened as xt:=xt+1xt

A standard

rst-o rderdierence equation tak esthe fo rm x t+1xt= xt=dt(xt) where eachdt:X!X, or equivalently,

TX3(t;x)7!dt(x)

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 7 of 54

Equivalent Recurrence Relations

Obviously, the dierence equationxt+1xt= xt=dt(xt)

is equivalent to the recurrence relation xt+1=rt(xt) whereTX3(t;x)7!rt(x) =x+dt(x), or equivalently,dt(x) =rt(x)x.

Thus dierence equations and recurrence relations

are entirely equivalent

We follow standard mathematical practice

in using the notation for recurrence relations, even when discussing dierence equations.

We may write \dierence equation"

even when considering a recurrence relation. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 8 of 54

Existence of Solutions

Example

Consider the dierence equationxt=px

t11 withx0= 5.

Evidentlyx1=p51 = 2, thenx2=p21 = 1,

and nextx3=p11 = 0, leavingx4=p01 undened as a real number.

The domain of (t;x)7!px1 is limited to

D:=Z+[1;1).Generally, consider a mappingD3(t;x)7!rt(x) whose domain is restricted to a subsetDZ+X. For the dierence equationxt+1=rt(xt) to have a solution one must ensure that (t;x)2D=)(t+ 1;rt(xt))2Dfor allt2Z+ University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 9 of 54

Lecture Outline

Introduction: Dierence vs. Dierential Equations

First-Order Dierence Equations

First-Order Linear Dierence Equations: Introduction

General First-Order Linear Equation

Particular, General, and Complementary Solutions

Explicit Solution as a Sum

Constant and Undetermined Coecients

Stationary States and Stability for Linear First-Order Equations Local Stability of Nonlinear First-Order Equations University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 10 of 54

Application: Wealth Accumulation in Discrete Time

Consider a consumer who, in discrete timet= 0;1;2;:::: I starts each periodt with an amountwtof accumulated wealth; I receives incomeyt; I spends an amountet; I earns interest on the residual wealthwt+ytetat the ratert. The process of wealth accumulation is then described by any of the equivalent equations w t+1= (1 +rt)(wt+ytet) =t(wtxt) =t(wt+st) where, at each timet, I t:= 1 +rtis theinterest facto r; I xt=etytdenotesnet exp enditure; I st=ytet=xtdenotesnet saving . University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 11 of 54

Compound Interest

Dene the

comp oundinterest facto r R t:=Y t1 k=0(1 +rk) =Y t1 k=0k with the convention that the product of zero terms equals 1 | just as the sum of zero terms equals 0. This compound interest factor is the unique solution to the recurrence relationRt+1= (1 +rt)Rt that satises the initial conditionR0= 1.

In the special case whenrt=r(allt),

it reduces toRt= (1 +r)t=t. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 12 of 54

Present Discounted Value (PDV)

We transform the dierence equationwt+1=t(wtxt)

by using the compound interest factorRt=Qt1 k=0k in order to discount both future wealth and expenditure.

To do so, dene new variables!t;t

for the present discounted values (PDVs) of, respectively: 1. w ealthwtat timetas!t:= (1=Rt)wt; 2. net exp enditurextat timetast:= (1=Rt)xt.

With these new variables,

the wealth equationwt+1=t(wtxt) becomes R t+1!t+1=tRt(!tt)

ButRt+1=tRt, so eliminating this common factor

reduces the equation to!t+1=!tt, with the evident solution!t=!0Pt1 k=0kfork= 1;2;:::. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 13 of 54

Lecture Outline

Introduction: Dierence vs. Dierential Equations

First-Order Dierence Equations

First-Order Linear Dierence Equations: Introduction

General First-Order Linear Equation

Particular, General, and Complementary Solutions

Explicit Solution as a Sum

Constant and Undetermined Coecients

Stationary States and Stability for Linear First-Order Equations Local Stability of Nonlinear First-Order Equations University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 14 of 54

General First-Order Linear Equation

The general rst-order linear dierence equation

can be written in the form x tatxt1=ftfort= 1;2;:::;T for non-zero constantsat2Rand a forcing termN3t7!ft2R. When this equation holds fort= 1;2;:::;T, whereT6, this equation can be written in the following matrix form: 0 B

BBBBBB@a11 0:::0 0 0

0a21:::0 0 0

0 0a3:::0 0 0

0 0 0:::aT11 0

0 0 0:::0aT11

C

CCCCCCA0

B

BBBBBBBB@x

0 x 1 x 2... x T2 x T1 x T1 C

CCCCCCCCA=0

B

BBBBBB@f

1 f 2 f 3... f T1 f T1 C

CCCCCCA

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 15 of 54

Matrix Form

The matrix form of the dierence equation isCx=f, where:

1.Cis theT(T+ 1)co ecientmatrix whose elements a re

c st=8 :asift=s

1 ift=s+ 1

0 otherwise

fors= 1;2;:::;Tandt= 0;1;2;:::;T;

2.xis theT+ 1-dimensional column vector (xt)Tt=0of endogenous unknowns, to be determined;

3.fis theT-dimensional column vector (ft)Tt=1of exogenous shocks.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 16 of 54

Partitioned Matrix Form

The matrix equationCx=fcan be written in partitioned form as

U eTxT1

x T =f where:

1.Uis an upper triangularTTmatrix;

2.eT= (0;0;0;:::;0;1)>is theTth column vector

of the canonical basis of the vector spaceRT;

3.xT1denotes the column vector which is the transpose

of the rowT-vector (x0;x1;x2;:::;xT2;xT1).

In fact the matrixUsatises

(U;eT) = (diag(a1;a2;:::;aT);eT) + (0T1;ITT) Hence there areTindependent equations inT+ 1 unknowns, leaving one degree of freedom in the solution. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 17 of 54

An Initial Condition

Consider the dierence equationxtatxt1=ft,

orCx=fin matrix form. An initial condition sp eciesan exogenous value x0 for the valuex0at time 0.

This removes the only degree of freedom

in the system ofTequations inT+ 1 unknowns.

Consider the special case whenat= 1 for allt2N.

The obvious unique solution ofxtxt1=ft

is then that eachxtis thefo rwardsum x t= x0+X t s=1fs of the initial state x0, and of thetexogenously specied succeeding dierencesfs(s= 1;2;:::;t). University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 18 of 54

A Terminal Condition

Alternatively, a

terminal condition for the dierence equationxtxt1=ft species an exogenous value xTfor the valuexT at the terminal time T.

It leads to a unique solution as a

backw ardsum x t= xTX Tt1 s=0fTs of the exogenously specied I terminal state xT; I preceding backward dierencesfTs (s= 0;1;:::;Tt1). University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 19 of 54

Lecture Outline

Introduction: Dierence vs. Dierential Equations

First-Order Dierence Equations

First-Order Linear Dierence Equations: Introduction

General First-Order Linear Equation

Particular, General, and Complementary Solutions

Explicit Solution as a Sum

Constant and Undetermined Coecients

Stationary States and Stability for Linear First-Order Equations Local Stability of Nonlinear First-Order Equations University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 20 of 54

Particular and General Solutions

We are interested in solving the systemCx=f

ofTequations inT+ 1 unknowns, whereCis aT(T+ 1) matrix. When the rank ofCisT, there is one degree of freedom.

The associated

homoge neousequation Cx=0 will have a one-dimensional space of solutionsxHt=xHt(2R).

Given any

pa rticularsolution xPtsatisfyingCxP=f for the particular time seriesfof forcing terms, the general solution xGtmust also satisfyCxG=f.

Simple subtraction leads toC(xGxP) =0, soxGxP=xH

for some solutionxHof the homogeneous equationCx=0.

Soxsolves the equationCx=f

i there exists a scalar2Rsuch thatx=xP+xH, which leads to the formulaxG=xP+xHfor the general solution. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 21 of 54

Complementary Solutions

Consider again the general rst-order linear equation which takes the inhomogeneous fo rmxtatxt1=ft.

The associated

hom ogeneousequation tak esthe fo rm x tatxt1= 0 (for allt2N) with a zero right-hand side.

The associated

c omplementarysolutions make up the one-dimensional linear subspaceL of solutions to this homogeneous equation. The spaceLconsists of functionsZ+3t7!xt2Rsatisfying x t=x0Y t s=1as(for allt2N) wherex0is an arbitrary scaling constant. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 22 of 54

From Particular to General Solutions

Consider again the

inhomogeneous equation x tatxt1=ft for a general RHSft.

The associated

hom ogeneousequation tak esthe fo rm x tatxt1= 0

LetxPtdenote apa rticularsolution ,

andxGtany alternativegeneral solution , of the inhomogeneous equation. Our assumptions imply that, for eacht= 1;2;:::, one has x

PtatxPt1=ft

xGtatxGt1=ft Subtracting the second equation from the rst implies that x

GtxPtat(xGt1xPt1) = 0

This shows thatxHt:=xGtxPtsolves the homogeneous equation. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 23 of 54

Characterizing the General Solution

Theorem

Consider the inhomogeneous equationxtatxt1=ft

with fo rcingterm ft.

Its general solutionxGtis the sumxPt+xHtof

I any particular solutionxPtof the inhomogeneous equation; I

the general complementary solutionxHtof the corresponding homogeneous equationxtatxt1= 0.University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 24 of 54

Linearity in the Forcing Term, I

Theorem

Suppose thatxPtandyPtare particular solutions

of the two respective dierence equations x tatxt1=dtandytatyt1=et

Then, for any scalarsand,

the equationztatzt1=dt+ethas as a particular solution the corresponding linear combinationzPt:=xPt+yPt.Proof.

Routine algebra.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 25 of 54

Linearity in the Forcing Term, II

Consider any equation of the formxtatxt1=ft

whereftis a linear combinationPn k=1kfktofnforcing termshfktink=1.

The theorem implies that a particular solution

is the corresponding linear combinationPn k=1kxPktof particular solutionshxPktink=1to the respectivenequationsxtatxt1=fkt(k= 1;2;:::;n). University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 26 of 54

Lecture Outline

Introduction: Dierence vs. Dierential Equations

First-Order Dierence Equations

First-Order Linear Dierence Equations: Introduction

General First-Order Linear Equation

Particular, General, and Complementary Solutions

Explicit Solution as a Sum

Constant and Undetermined Coecients

Stationary States and Stability for Linear First-Order Equations Local Stability of Nonlinear First-Order Equations University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 27 of 54

Solving the General Linear Equation

Consider a rst-order linear dierence equation

x t+1=atxt+ft for a p rocessT3t7!xt2R, where eachat6= 0quotesdbs_dbs29.pdfusesText_35

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