4.3Inhomogeneous Di erence Equations

In a completely analogous way to the ODE case, one proves that every solution to the inhomogeneous linear di erence equation (4.1) is of the form

(yn)homo + (yn)part

where (yn)homo is a solution to the homogeneous equation (4.1) with fn = 0 and (yn)part is a particular solution to the inhomogeneous equation (4.1).

4.4Exercises

#1. Degenerate Roots

Consider the constant coe cient di erence equation (4.2) but now assume the two roots λ1,2 are equal. Show that

nλn1 is a second linearly independent solution to (4.2).

√

#2. Rational Approximations to 2

Solve the di erence equation

xn+1 = 2xn + xn−1, n ≥ 1

with initial conditions x0 = 0 and x1 = 1 that corresponds to the sequence 0, 1, 2, 5, 12, 29,. . . . Show that

provide us with very good approximations to the square root of two.2

#3. Catalan numbers

Many times nonlinear recurrence relations arise. For example, Catalan numbers Tn satisfy the nonlinear recurrence relation

n−1

X

Tn = TkTn−1−k, n = 1, 2, . . .

k=0

where T0 := 1. Define

∞

X

T (z) = Tnzn.

n=0

2The square root of two is irrational.

Chapter 5

Matrix Di erential Equations

Linear systems are almost the only large class of di erential equations for which there exists a definitive theory. This theory is essentially a branch of linear algebra, and allows us to solve all autonomous linear equations.

V.A. Arnold, Ordinary Di erential Equations [2]

5.1The Matrix Exponential

Let A be a n×n matrix with constant entries. In this chapter we study the matrix di erential equation

Remark: In an advanced course you will prove that this infinite series of matrices converges to a n × n matrix.

It is important to note that for matrices A and B, in general, exp(tA) exp(tB) 6= exp(tA + tB).

1We put the scalar factor t directly into the definition of the matrix exponential since it is in this form we will use the matrix exponential.

53

The last equality follows by factoring A out on the right instead of the left.

5.2.2Solution to Matrix ODE with Constant Coe cients

We now use (5.7) to prove

Theorem: Let

where x Rn and A is a n × n matrix with constant coe cients. Then every solution of (5.8) is of the form

for some constant vector x0 Rn.

Proof : (i) First we show that x(t) = etAx0 is a solution:

=AetAx0

=Ax(t).

(ii)We now show that every solution of (5.8) is of the form (5.9). Let y(t) by any solution to (5.8). Let

Δ(t) := e−tAy(t).

If we can show that Δ(t) is independent of t—that it is a constant vector which we call x0, then we are done since multiplying both sides by etA shows

etAx0 = etAΔ(t) = etAe−tAy(t) = y(t).

2In a complex analysis course you will prove that convergent complex power series can be di erentiated term-by-term and the resulting series has the same radius of convergence. Note there really is something to prove here since there is an interchange of two limits.

=0.

The next theorem relates the solution x(t) of (5.8) to the eigenvalues and eigenvectors of the matrix A (in the case A is diagonalizable).

Theorem: Let A be a diagonalizable matrix. Any solution to (5.8) can be written as

Proof : All solutions of (5.8) are of the form (5.9). Since A is diagonalizable, the eigenvectors of A can be used to form a basis: {ψ1, . . . , ψn}. Since this is a basis there exist constants c1, . . . , cn such that

x0 = c1ψ1 + c2ψ2 + · · · + cnψn.

(x0 is the constant vector appearing in (5.9).)

For any eigenvector ψ of A with eigenvalue λ we have

etAψ = etλψ.

(This can be proved by applying the infinite series (5.2) to the eigenvector ψ and noting Ak ψ = λk ψ for all positive integers k.) Thus

etAx0 = c1etAψ1 + · · · cnetAψn

=c1etλ1 ψ1 + · · · + cnetλN ψn.

Here are two immediate corollaries of this theorem:

1.If A is diagonalizable and has only real eigenvalues, then any solution x(t) of (5.1) will have no oscillations.

2.If A is diagonalizable and the real part of every eigenvalue is negative, then

x(t) → 0 (zero vector), as t → +∞ To see this recall that if λ = σ + iµ (σ and µ both real), then

eλt = eσteiµt.

If σ < 0, eσt → 0 as t → +∞. Now apply preceding theorem.