Discrete Dynamical Systems 1jfrabajante.weebly.com/uploads/1/1/5/5/11551779/lecture_2.pdf ·...

Discrete Dynamical Systems

Discrete Dynamical Systems 1

J. F. Rabajante

IMSP, UPLB

1st Sem, AY 2012-2013

Rabajante MATH 191: Special Topics

Example: Bank Account

Suppose you have an initial deposit of x(0) = x0. The annualinterest rate is r . What is the corresponding mathematicalmodel?

Compounded annualy:x(k + 1) = (1 + r)x(k) — this is called a recursive/iterativerelationSolution is x(k) = (1 + r)kx0.

Compounded m-times in a year:x(k + 1) = (1 + r

m )mx(k)Solution is x(k) = (1 + r

m )mkx0.

Compounded continuously (not anymore discrete):x(t) =

(limm→∞(1 + r

m )m)t x0 or dxdt = rx

Solution is x(t) = x0ert .

Compounded annualy:x(k + 1) = (1 + r)x(k) — this is called a recursive/iterativerelation

Solution is x(k) = (1 + r)kx0.

m )mkx0.

(limm→∞(1 + r

m )mkx0.

(limm→∞(1 + r

m )mx(k)

Solution is x(k) = (1 + rm )mkx0.

(limm→∞(1 + r

m )mkx0.

(limm→∞(1 + r

m )mkx0.

(limm→∞(1 + r

m )mkx0.

(limm→∞(1 + r

Solution is x(t) = x0ert .Rabajante MATH 191: Special Topics

Example: Biology

How many rabbits? (by Leonardo of Pisa)

Fibonacci sequence 1,1,2,3,5,8,13,...

x(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1

We will show later that Fibonacci sequence is related to thegolden/divine ratio φ = 1.618....

Example: Biology

x(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1

Example: Biology

x(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1

Example: Biology

x(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1

Difference equations∆x = x(k + 1)− x(k) ;x(k + m) = f (x(k + m − 1), x(k + m − 2), ..., x(k))

Let’s discuss- linear and nonlinear- homogeneous and nonhomogeneous- autonomous and nonautonomous- order of a discrete dynamical system

Let’s discuss- linear and nonlinear

- homogeneous and nonhomogeneous- autonomous and nonautonomous- order of a discrete dynamical system

Let’s discuss- linear and nonlinear- homogeneous and nonhomogeneous

- autonomous and nonautonomous- order of a discrete dynamical system

Let’s discuss- linear and nonlinear- homogeneous and nonhomogeneous- autonomous and nonautonomous

- order of a discrete dynamical system

Let’s discuss- linear and nonlinear- homogeneous and nonhomogeneous- autonomous and nonautonomous- order of a discrete dynamical system

Let us find the analytic solution to the IVPx(k + 1) = ax(k) + b, x(0) = x0.

x(1) = ax(0) + b

x(2) = ax(1) + b= a(ax(0) + b) + b= a2x(0) + ab + b= a2x(0) + b(a + 1)

x(3) = ax(2) + b= a(a2x(0) + b(a + 1)) + b= a3x(0) + ab(a + 1) + b= a3x(0) + b(a2 + a) + b= a3x(0) + b(a2 + a + 1)

In general, x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).

x(1) = ax(0) + b

x(2) = ax(1) + b

= a(ax(0) + b) + b= a2x(0) + ab + b= a2x(0) + b(a + 1)

x(1) = ax(0) + b

x(2) = ax(1) + b= a(ax(0) + b) + b

= a2x(0) + ab + b= a2x(0) + b(a + 1)

x(1) = ax(0) + b

x(2) = ax(1) + b= a(ax(0) + b) + b= a2x(0) + ab + b

= a2x(0) + b(a + 1)

x(1) = ax(0) + b

x(3) = ax(2) + b

= a(a2x(0) + b(a + 1)) + b= a3x(0) + ab(a + 1) + b= a3x(0) + b(a2 + a) + b= a3x(0) + b(a2 + a + 1)

x(1) = ax(0) + b

x(3) = ax(2) + b= a(a2x(0) + b(a + 1)) + b

= a3x(0) + ab(a + 1) + b= a3x(0) + b(a2 + a) + b= a3x(0) + b(a2 + a + 1)

x(1) = ax(0) + b

x(3) = ax(2) + b= a(a2x(0) + b(a + 1)) + b= a3x(0) + ab(a + 1) + b

= a3x(0) + b(a2 + a) + b= a3x(0) + b(a2 + a + 1)

x(1) = ax(0) + b

x(3) = ax(2) + b= a(a2x(0) + b(a + 1)) + b= a3x(0) + ab(a + 1) + b= a3x(0) + b(a2 + a) + b

= a3x(0) + b(a2 + a + 1)

x(1) = ax(0) + b

Continuation... x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).

But note that an−1 + an−2 + ...+ a + 1 is a geometric series withratio a.

So an−1 + an−2 + ...+ a + 1 = an−1a−1 if a 6= 1, and

an−1 + an−2 + ...+ a + 1 = n if a = 1.

Therefore, we have the following theorem.

Theorem 1 The analytic solution to x(k + 1) = ax(k) + b,x(0) = x0 is

x(k) = akx0 + b ak−1a−1 if a 6= 1 (geometric series)

x(k) = akx0 + nb if a = 1 (arithmetic series).

an−1 + an−2 + ...+ a + 1 = n if a = 1.

Example: Amortization

The discrete IVP is x(k + 1) = (1 + r)x(k)− p, x(0) = L.

To do: Find the analytic solution. Then solve for p such thatx(n) = 0 where n is the last payment time.

Now, consider the following ax(k + 2) + bx(k + 1) + cx(k) = 0,a 6= 0.

Let x(k) = λk 6= 0.So we have aλk+2 + bλk+1 + cλk = 0.Factoring out, λk (aλ2 + bλ+ c) = 0.

The equation aλ2 + bλ+ c = 0 is a characteristic equation andthe solutions to the characteristic equation are calledeigenvalues.

Let x(k) = λk 6= 0.

So we have aλk+2 + bλk+1 + cλk = 0.Factoring out, λk (aλ2 + bλ+ c) = 0.

Let x(k) = λk 6= 0.So we have aλk+2 + bλk+1 + cλk = 0.

Factoring out, λk (aλ2 + bλ+ c) = 0.

Theorem 2 The general solution toax(k + 2) + bx(k + 1) + cx(k) = 0, a 6= 0 is

For real, distinct eigenvalues λ1 and λ2:x(k) = d1λ

k1 + d2λ

For repeated real eigenvalue λ: x(k) = d1λk + d2kλk .

For complex eigenvalues λ1,λ2= re±iθ:x(k) = r k (d1 cos(kθ) + d2 sin(kθ)).

Recall: α + βi = r cos(θ) + ir sin(θ) = reiθ.r =

√α2 + β2, θ = tan−1

Example: Try to graph the solution of2x(k + 2)− 2x(k + 1) + x(k) = 0.

k1 + d2λ

√α2 + β2, θ = tan−1

k1 + d2λ

√α2 + β2, θ = tan−1

k1 + d2λ

√α2 + β2, θ = tan−1

k1 + d2λ

√α2 + β2, θ = tan−1

k1 + d2λ

√α2 + β2, θ = tan−1

Example: Recall the Fibonacci sequencex(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1.

x(k + 2)− x(k + 1)− x(k) = 0.

Eigenvalues: 1±√

General solution is: x(k) = d1

(1+√

)k+ d2

(1−√

Specific solution is: x(k) = 5+√

(1+√

)k+ 5−

(1−√

Furthermore, get the ratio limk→∞x(k+1)

x(k) , the ratio is equal to Φ

(golden ratio).

Example: Recall the Fibonacci sequencex(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1.x(k + 2)− x(k + 1)− x(k) = 0.

Eigenvalues: 1±√

(1+√

)k+ d2

(1−√

(1+√

)k+ 5−

(1−√

(golden ratio).

Eigenvalues: 1±√

(1+√

)k+ d2

(1−√

(1+√

)k+ 5−

(1−√

(golden ratio).

Eigenvalues: 1±√

(1+√

)k+ d2

(1−√

(1+√

)k+ 5−

(1−√

(golden ratio).

Eigenvalues: 1±√

(1+√

)k+ d2

(1−√

(1+√

)k+ 5−

(1−√

(golden ratio).

Eigenvalues: 1±√

(1+√

)k+ d2

(1−√

(1+√

)k+ 5−

(1−√

x(k) ,

the ratio is equal to Φ

(golden ratio).

Eigenvalues: 1±√

(1+√

)k+ d2

(1−√

(1+√

)k+ 5−

(1−√

(golden ratio).

Exercise: Prove that the golden mean satisfies Φ = 1 + 1Φ . Also

prove that the golden ratio can also be rewritten as a continuedfraction: 1 + 1

1+ 11+...

Equilibria and Stability: First Discussion

Consider x(k + 1) = f (x(k)). Applying the function f m-times,x(k + 1) = f (f (f (...f (x(0))))) = f n(x(0)).

Definition 1 An equilibrium or fixed point is a value x∗ for whichx∗ = f (x∗).

Example: Find the fixed point of x(k + 1) = ax(k) + b.Determine if the fixed point of x(k + 1) = 0.8x(k) + 2, x(0) = 1attract or repel. Determine if the fixed point ofx(k + 1) = −1.04x(k), x(0) = 1 attract or repel.

Example: Find the fixed point of x(k + 1) = ax(k) + b.

Determine if the fixed point of x(k + 1) = 0.8x(k) + 2, x(0) = 1attract or repel. Determine if the fixed point ofx(k + 1) = −1.04x(k), x(0) = 1 attract or repel.

Example: Find the fixed point of x(k + 1) = ax(k) + b.Determine if the fixed point of x(k + 1) = 0.8x(k) + 2, x(0) = 1attract or repel.

Determine if the fixed point ofx(k + 1) = −1.04x(k), x(0) = 1 attract or repel.

What is the fixed point of ax(k + 2) + bx(k + 1) + cx(k) = 0?

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