Difference equations v3 - riemann.math.klte.hu

Difference equations v3

Laszlo Losonczi

University of Debrecen, Faculty of Economics and Business Administration

Laszlo Losonczi (DE) Difference equations v3 1 / 26

3.1 Difference equations, existence and uniqueness theorem

Denote by N0 the set of nonnegative integers i.e. N0 = N ∪ {0}. Letf : N0 × R→ R be a given function then the equation

y(n + 1) = f (n, y(n)) (n ∈ N0)

is called an explicite difference equation of first order.

It is clear that if y(0) is known/given then the equation uniquelydetermines all elements of the (unknown) sequence y(n) (n ∈ N0).

Existence and Uniqueness TheoremFor given f : N0 × R→ R and a(0) ∈ R there exist exactly onesequence y(n) (n ∈ N0) which satisfies the equation

y(n + 1) = f (n, y(n)) (n ∈ N0) and the initial condition y(0) = a(0).


3.2 A simple first order equation

Consider the following equation

y(n + 1) + py(n) = f (n) (n ∈ N0)

where p 6= 0 is a constant and f : N0 → R is given.

Using the equation for n = 0,1,2 . . . we get

y(1) = −py(0) + f (0)y(2) = −py(1) + f (1) = −p(−py(0) + f (0)) + f (1)

= (−p)2y(0)− pf (0) + f (1)y(3) = −py(2) + f (2) = −p((−p)2y(0)− pf (0) + f (1)) + f (2)

= (−p)3y(0) + (−p)2f (0)− pf (1) + f (2)

By induction we can show that

y(n) = (−p)ny(0) +n−1∑k=0

(−p)n−1−k f (k) (n ∈ N0)

is the general solution of our equation where y(0) is arbitrary.Laszlo Losonczi (DE) Difference equations v3 3 / 26

3.3 A Multiplier-Accelerator Model of Growth

Let Y (n) denote the national income, I(n) total investment, and S(n)total saving all at time n. Suppose that savings are proportional tonational income, and that investment is proportional to the change inincome during the time from n to n + 1. Then we have the followingsystem

S(n) = αY (n)I(n + 1) = β (Y (n + 1)− Y (n))

S(n) = I(n).

The last equation is the familiar equilibrium condition that savingequals investment in each time. Here α, β are positive constants, andwe assume that β > α > 0. Find Y (n) if Y (0) is given!


3.4 A Multiplier-Accelerator Model of Growth: the solution

Substituting the third and then the first equation into the second we get

I(n + 1) = S(n + 1) = αY (n + 1) = β (Y (n + 1)− Y (n))

orY (n + 1)− β

β − αY (n) = Y (n + 1)−

(1 +

α

β − α

)Y (n) = 0

Using the solution of the previous section with −p = 1 + αβ−α , f (n) = 0

we get that

Y (n) =

(1 +

α

β − α

)n

Y (0) (n ∈ N0).

This gives constant rate of growth of 100αβ−α% at each time, or

αβ−α = Y (n+1)−Y (n)

Y (n) being the relative growth.


3.5 Cobweb model

If the function f in our difference equation of the section 3.1 dependsonly on the second variable, i.e. the equation has the form

y(n + 1) = f (y(n)) (n ∈ N0) and y(0) = a(0)

then the equation is called autonomous. Such equations can bestudied by help of the so called cobweb model:

first we draw the graph of the function f , and draw the line y = x ,connect the points (y(0), y(0)) = (a(0),a(0)) and(y(0), f (y(0))) = (y(0), y(1)) by a line segment then connect thelatter point with the point (y(1), y(1)) by a line segment,repeat the process starting with the point (y(1), y(1)) andcontinue similarly.

In the above algorithm drawing the line segments parallel to the y axisup to the x axis we get (at crossing points on the x axis) the elementsof the sequence y(0), y(1), . . . . In the next example in our figure theseline segments were not drawn only the elements of the sequence arelabeled on the x axis.


3.6 Cobweb model: example

Our figure shows the cobweb model for the equationy(n + 1) =

√4− y(n) (n ∈ N0) with initial value y(0) = 0. It is clear

that y(1) = 2, y(2) =√

2 ≈ 1,414, y(3) =√

4−√

2 ≈ 1,608. Thefigure shows the graphs of the functions f (x) =

√4− x and h(x) = x .


3.7 Second order difference equations, existence and uniqueness

theorem

Let f : N0 × R× R→ R be a given function then the equation

y(n + 2) = f (n, y(n + 1), y(n)) (n ∈ N0)

is called an explicit difference equation of second order.

It is clear that if y(0), y(1) are given then the equation uniquelydetermines the sequence (y(n)) (n ∈ N0).

Existence and Uniqueness TheoremFor a given function f : N0 × R× R→ R and given initial valuesa(0),a(1) ∈ R there exists exactly one sequence (y(n)) (n ∈ N0)satisfying the difference equation

y(n + 2) = f (n, y(n + 1), y(n)) (n ∈ N0)

and the initial conditions y(0)=a(0),y(1)=a(1).Laszlo Losonczi (DE) Difference equations v3 8 / 26

3.8 Linear difference equations

For the sake of simplicity we formulate the results for second orderequations. Similar results hold for equations of higher order.Linear difference equationsLet p,q, f : N0 → R be given functions then the equation

y(n + 2) + p(n)y(n + 1) + q(n)y(n) = f (n) (n ∈ N0) (1)

is called a linear difference equation of second order. The functionsp,q are the coefficients of (y(n + 1), y(n) respectively), f is called thefree term of the equation. We suppose that the coefficient q is not zero(sequence)otherwise the order of the equation would be less than 2.(1) is called homogeneous if f (n) = 0 (n ∈ N0), otherwise we speakof inhomogeneous equation. The homogeneous equation obtainedfrom (1) by substituting f (n) = 0 is called the associatedhomogeneous equation to (1).

For given a(0),a(1) ∈ R there is exactly one sequence(y(n)), (n ∈ N0) which satisfies the equation (1) and the initialconditions y(0) = a(0), y(1) = a(1).


3.9 General solution of linear homogeneous difference equations

The general form of a linear homogeneous difference equation is

y(n + 2) + p(n)y(n + 1) + q(n)y(n) = 0 (n ∈ N0). (2)

where p,q 6= 0 are given sequences.

It is easy to prove that if ha y1(n), y2(n) (n ∈ N0) are solutions of theabove equation then their linear combinationy(n) = C1y1(n) + C2y2(n) (n ∈ N0) with arbitrary constant coefficientsC1,C2 is also a solution, moreover all solutions can be written in thisform, provided that the solutions y1(n), y2(n) (n ∈ N0) are linearlyindependent (that is their linear combinationC1y1(n) + C2y2(n) = 0 (n ∈ N0) only if C1 = C2 = 0. This means thenone of y1), y2 is a constant multiple of the other. We also say thaty(n) = C1y1(n) + C2y2(n) (n ∈ N0) is the general solution of theequation.


3.10 Linear independence of solutions, Casorati determinant

The next theorem gives a necessary and sufficient condition for thelinear independence of two solutions of homogeneous differenceequations.

One can prove that the solutions y1(n), y2(n) (n ∈ N0) of thehomogeneous equation are linearly independent if and only if thedeterminant

Cy1,y2(n) :=

∣∣∣∣ y1(n) y2(n)y1(n + 1) y2(n + 1)

∣∣∣∣ 6= 0 for some n0 ∈ N0.

Moreover, if Cy1,y2(n0) 6= 0 for some n0 ∈ N0 then Cy1,y2(n) 6= 0 for alln ∈ N0.

Cy1,y2(n) is called the Casorati determinant of the solutionsy1(n), y2(n). Usually we take n0 = 0 in the above theorem.


3.11 Linear inhomogeneous equations: structure of the solution

Consider a linear inhomogeneous difference equation

y(n + 2) + p(n)y(n + 1) + q(n)y(n) = f (n) (n ∈ N0).

where q(n) 6= 0 for some n ∈ N0. It is easy to show that the generalsolution this inhomogeneous equation is of the form

y(n) = C1y1(n) + C2y2(n) + y(n) (n ∈ N0)

where yh(n) = C1y1(n) + C2y2(n) is the general solution of theassociated homogeneous equation (i.e. y1(n), y2(n) are linearlyindependent solutions of the associated homogeneous equation,C1,C2 are arbitrary constants) and y(n) is a solution (called aparticular solution) of our inhomogeneous equation.


3.12 Linear homogeneous difference equations with constant

coefficients

If in (1) p,q 6= 0 are constant functions (sequences) andf (n) = 0 (n ∈ N0), then we get a linear homogeneous differenceequations with constant coefficients:

y(n + 2) + py(n + 1) + qy(n) = 0 (n ∈ N0). (3)

Parallel to (3) consider the algebraic equation of second degree

λ2 + pλ+ q = 0, (4)

called the characteristic equation of the difference equation (3).Multiplying (4) by λn we get that

λn+2 + pλn+1 + qλn = 0.

This shows that y(n) = λn is a solution of (3) if and only if λ is asolution of the characteristic equation (4). The solutions depend onthe discriminant D = p2 − 4q of the equation (4).


3.13 The method of solution

If p2 − 4q > 0 then the characteristic equation has two distinctreal roots λ1 6= λ2, and then

y1(n) = λn1, y2(n) = λn

2 (n ∈ N0)

are solutions of equation (3), namely substituting λ = λi (i = 1,2)into (4), and multiplying the resulting equation by λn

i we get that

λn+2i + pλn+1

i + qλni = 0 (i = 1,2; n ∈ N0).

They are linearly independent as their Casorati determinant at thepoint 0

Cy1,y2(0) =

∣∣∣∣ 1 1λ1 λ2

∣∣∣∣ = λ2 − λ1 6= 0.

Therefore the general solution of equation (3) is

y(n) = C1λn1 + C2λ

n2 (n ∈ N0)

where C1,C2 are arbitrary constants.



If p2 − 4q = 0 then the characteristic equation has one(double) real root λ1 = λ2 = −p

2 , and then we can easily showthat

y1(n) = λn1, y2(n) = nλn

1 (n ∈ N0)

are solutions of equation (3). λ1 = −p2 6= 0 as if

−p2 = 0,p2 − 4q = 0 would imply that p = q = 0, which

contradicts to the assumption q 6= 0. They are linearlyindependent as their Casorati determinant at the point 0

Cy1,y2(0) =

∣∣∣∣ 1 0λ1 λ1

∣∣∣∣ = λ1 6= 0.

The general solution of (3) is

y(n) = C1λn1 + C2nλn

1 (n ∈ N0)

where C1,C2 are arbitrary constants.



If p2 − 4q < 0 then the characteristic equation has two distinctconjugate complex roots λ1,2 = α± iβ where

α = −p2 , β =

√4q−p2

2 6= 0, or in trigonometric formλ1,2 = r(cosϑ± i sinϑ) with (r =

√α2 + β2, cosϑ = α

r ).Then we have λn

1,2 = rn(cos(ϑn)± i sin(ϑn)). The real andimaginary parts of this power are solutions of (3), they are linearlyindependent as their Casorati determinant at the point 0

Cy1,y2(0) =

∣∣∣∣ 1 0r cosϑ r sinϑ

∣∣∣∣ = r sinϑ = rβ

r= β 6= 0.

Hence the general solution is

y(n) = C1rn cos(ϑn) + C2rn sin(ϑn) (n ∈ N0)

where C1,C2 are arbitrary constants.The above solution can be rewritten as

y(n) = Arn cos(ϑn + ω) or y(n) = Arn sin(ϑn + ω) (n ∈ N0)

where A, ω are arbitrary constants.Laszlo Losonczi (DE) Difference equations v3 16 / 26

3.14 The method of undetermined coefficients

Decomposition of the solution of the inhomogeneous equationIf y i(n) are solutions of the equationsy(n + 2) + p(n)y(n + 1) + q(n)y(n) = fi(n) (n ∈ N0, i = 1, . . . , k).

then y(n) = y1(n) + · · ·+ yk (n) is a solution of the equationy(n + 2) + p(n)y(n + 1) + q(n)y(n) = f1(n) + · · ·+ fk (n) (n ∈ N0).

Method of undetermined coefficientsIf in (1) the coefficients are constants and f (n) is a linearcombination of terms of the form

an, nk , cosβn, sinβn (5)

(or products of such forms), then (1) has a particular solution whichis the linear combination of the functions

an, (1,n,n2, . . . ,nk ), (sinβn, cosβn), (cosβn, sinβn)

(or products of them), provided that none of the functions in (5) aresolutions of the associated homogeneous equation.



Method of undetermined coefficientsIn the table below we give the form of a solution of the inhomogeneousequation (trial solution) (provided that no term of y(n) is a solutionof the associated homogeneous equation):

f (n) y(n)

an A1an

nk Aknk + Ak−1nk−1 + · · ·+ A1n + A0nkan (Aknk + Ak−1nk−1 + · · ·+ A1n + A0)an

sinβn A1 sinβn + A2 cosβncosβn A1 sinβn + A2 cosβnan sinβn an(A1 sinβn + A2 cosβn)

an cosβn an(A1 sinβn + A2 cosβn)

annksinβn an(Aknk +. . .+A0)sinβn+an(Bknk +. . .+B1n+B0)cosβnannkcosβn an(Aknk +. . .+A0)sinβn+an(Bknk +. . .+B1n+B0)cosβn



The trial solution y(n) in the previous table does not change ifanywhere in the column of f (n) we replace nk by a polynomial ofdegree k .

Method of undetermined coefficientsIf some terms of the trial solution y(n) (in the previous table) aresolutions of the associated homogeneous equation then the(new), trial solution is nr y(n) where y(n) is the trial solution in theprevious table, r is the minimal exponent such that no term ofnr y(n) is solution of the associated homogeneous equation. Forsecond order equations r is 1 or 2.


3.14 Method of undetermined coefficients

The previous statements can be reformulated as follows. If the freeterm is of the form

f (n) = anPk (n) cos(βn) or f (n) = anPk (n) sin(βn),

where Pk (n) is a polynomial of degree k then the inhomogeneousequation has a solution of the form

y(n) = nr an(Aknk +. . .+A0)sin(βn)+nr an(Bknk +. . .+B0)cos(βn)

where r is the multiplicity of the root λ1,2 = a(cosβ ± i sinβ) in thecharacteristic equation (that is r = 0, if λ1,2 = a(cosβ ± i sinβ) is not asolution of the characteristic equation, r = 1, if λ1,2 = a(cosβ ± i sinβ)is a simple root of the characteristic equation, finally r = 2, ifλ1,2 = a(cosβ ± i sinβ) is a double root of the characteristic equation.


3.15 Examples for the method of undetermined coefficients

EXAMPLE 1. The characteristic equation ofy(n + 2)− 5y(n + 1) + 6y(n) = 0 (n ∈ N0) is λ2 − 5λ+ 6 = 0 whoseroots are λ1 = 2, λ2 = 3 hence its general solution isyh(n) = C12n + C23n.

EXAMPLE 2. The general solution of the inhomogeneous equationy(n + 2)− 5y(n + 1) + 6y(n) = 4n + n2 + 3 (n ∈ N0) isy(n) = C12n + C23n + y1(n) + y2(n), where y1(n) = A 4n is thesolution corresponding to the term 4n and y2(n) = Bn2 + Cn + D is thesolution corresponding to the term n2 + 3, where A,B,C,D areconstants (undetermined coefficients).



Substituting y1(n) into the inhomogeneous equationy(n + 2)− 5y(n + 1) + 6y(n) = 4n and dividing by 4n, we get42A− 5 · 4A + 6A = 1 hence A = 1/2, y1(n) = 1/2 · 4n.Next, substituting y2(n) = Bn2 + Cn + D into the equationy(n + 2)− 5y(n + 1) + 6y(n) = n2 + 3 we get

B(n+2)2+C(n+2)+D−5(B(n+1)2+ C(n+1)+D

)+6(Bn2 + Cn + D

)= n2 + 3, or after rearranging the terms

2Bn2 + (−6B + 2C)n + (−B − 3C + 2D) = n2 + 3.

As n2,n,1 are linearly independent sequences from the comparison ofthe coefficients we get 2B = 1, −6B + 2C = 0, −B − 3C + 2D = 3.The solution of this system is B = 1/2,C = 3/2,D = 4, thus thegeneral solution of our inhomogeneous equation is

y(n) = C12n + C23n +12

4n +32

n2 +12

n + 4 (n ∈ N0).



EXAMPLE 3. The general solution of the inhomogeneous equationy(n + 2)− 5y(n + 1) + 6y(n) = 2n(n + 3) (n ∈ N0) isy(n) = C12n + C23n + y(n), where y(n) = 2n(An + B)n (n ∈ N0),A,B are undetermined coefficients. Substituting y(n) into the equationand calculating the coefficients we get that A = −1

4 ,B = −94 ,

y(n) = −142n(n + 9)n.


3.16 Problems

1. Find the general solutions of the equations

y(n + 2)− 6y(n + 1) + 8y(n) = 0, y(n + 2)+2y(n+1)+3y(n)=0,y(n + 2)− 8y(n + 1) + 16y(n) = 0, 3y(n + 2) + 2y(n) = 4n,y(n + 2) + 2y(n + 1) + (n) = 9 · 2n, y(n + 1) + 2y(n) = 5n2 + 3.

2. (a) A model due to Ball and Smolensky is based on the system

C(n + 1) = cY (n), K (n + 1) = σY (n),Y (n + 1) = C(n + 1) + K (n + 1)− K (n) (n ∈ N0)

where C(n), K (n), Y (n) denotes consumption, capital stock, netnational product respectively, c, σ are positive constants. Give aneconomic interpretation of the equations!(b) Derive a difference equations of second order for Y (n). Findnecessary and sufficient conditions for the solution of this equation tohave explosive oscillations (i.e. the solution has sin and/or cos termswith coefficients/amplitudes tending to∞ as n→∞).


3.17 Problems: Multiplier-Accelerator Growth Model

Multiplier-Accelerator Growth ModelLet Y (n) denote the national income, C(n) total consumption, and I(n)total investment in a country all at time n. Suppose that

Y (n) = C(n) + I(n),C(n + 1) = aY (n) + b,I(n + 1) = c (C(n + 1)− C(n)) , (n ∈ N0),

where a,b, c are positive constants.The first equation states that the national income is divided betweenconsumption and investment. The second equation expresses theassumption that consumption at the time n + 1 is a linear function ofthe national income in the previous time period. This is the ”multiplier”part of the model. Finally the third equation says that the investmentin the time period n + 1 is proportional to the change in consumptionfrom the previous period. So investment is only needed whenconsumption increases. This is the ”accelerator” part of the model.


3.17 Problems: Multiplier-Accelerator Growth Model

Multiplier-Accelerator Growth ModelThe combined ”multiplier-accelerator” model has been studied byseveral economists, notably by P.A. Samuelson. Substituting the firstequation into the second we getC(n + 1) = aY (n) + b = a (C(n) + I(n)) + b = aC(n) + aI(n) + b. Inorder to use the third equation we need to replace n by n + 1 here:C(n + 2) = aC(n + 1) + aI(n + 1) + b. Now we can use the thirdequation to substitute here I(n + 1) and obtain

C(n+2) = aC(n+1)+aI(n+1)+b = aC(n+1)+ac (C(n + 1)− C(n))+b,

orC(n + 2)− a(1 + c)C(n + 1) + acC(n) = b (n ∈ N0). (6)

(a) Find a particular solution of (6).(b) Find the characteristic equation of the associated homogeneousequation and determine when it has two different real roots, or adouble root, or two complex roots.


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