6

Question 1

The solution to the difference equation
x_t+1 = ax_t
is

Answer 1

x_t = a^tx₀, where x₀ is the first term of the sequence. (We assume that the sequence is x₀, x₁, x_{2 …} rather than x₁, x₂,…

Question 2

Solving systems of difference equations by CHANGE OF VARIABLE. Steps: Given system X_t+1 = AX_t

Answer 2

1. Let X_t = PZ_t (or, equivalently, Z_t = P^-1X_t)
2. Z_t+1 = P^-1X_t+1

Z_t+1 = P^-1AX_t

Z_t+1 = P^-1APZ_t

Z_t+1 = DZ_t ​

3. Find P
4. Write out Z_t+1 = DZ_t ​ and solve to give u_t, v_t and w^t using fact that the solution to the difference equation x_t+1 = ax_t is x_t = a^tx₀, where x₀ is the first term of the sequence

5 Write out X_t = PZ_t in matrix form

6​ Write out X₀ = PZ₀ and solve using row operations to find u₀, v₀ and W₀

7 Plug these back into step 5 (still in matrix form)

8 State x_t, y_t and z_t

Question 3

Solving systems of difference equations by MATRIX POWERS. Steps given system X_t+1 = AX_t

Answer 3

1. If X_t+1 = AX_t, then X_t = A^tX₀
2. FInd P

3 P^-1AP = D = diag(1, 5, 13)

so A = PDP^-1 and A^t = PD^tP^-1

4. Find P^-1
5. X_t = A^tX₀ = PD^tP^-1X₀ - fill in and work out

Question 4

Malthus equation

Answer 4

ẏ = ry where ẏ = dy/dt
with solution y(t) = e^rty(0).

Question 5

Steps to solve a linear system of differential equations,

ẏ = Ay

where A is an n x n matrix

Answer 5

If D is diagonal, this is easy to solve. For, suppose
D = diag(λ₁, λ₂, …, λ_n) then the system is precisely
ẏ₁ = λ₁y1, ẏ₂ = λ₂y₂, …, ẏ_n = λ_ny_n;
and so
y₁ = y₁(0)e^λ₁t, …, y_n = y_n(0)e^λ_nt

1. Suppose that A can be diagonalised. Then P^-1AP = D, with P = (v₁ … v_n); D = diag(λ₁, λ₂, …, λ_n);
   where λ_i are the eigenvalues and v_i the corresponding eigenvectors.

1b. Find P and D
2. Let z = P^-1y (or, equivalently, y = Pz).

so ẏ = Pż since P has constant entries.

Since ẏ = Ay, and Pż = ẏ, then Pż = Ay = APz

so ż = P^-1APz = Dz

3. ż = Dz in matrix form to write out z₁ and z₂ in z₁ = e^λ₁tz₁(0) form
4. Then write out in matrix form y = Pz to state ys in terms of zs
5. Write out in matrix form z₀ = P^-1y₀ to state z₁(0),… z_n(0) in terms of y_i(0)s
6. Finally, state y_i(t)s in terms of y_i(0)s

Question 6

The vector y* ∈ Rⁿ is a steady state of the system ẏ = F(y) if

Answer 6

F(y*) = 0

If a system has initial condition y(0)=y* then it satisfies y(t)=y* for all t. So y* is a constant solution of the system.

Question 7

A steady state y* is an asymptotically stable equilibrium if

Answer 7

every solution y(t) that starts
near y* converges to y* as t → ∞,

i.e. if there is some ε > 0 such that

if ẏ = F(y) and ∥y(0) - y*∥ < ε then y(t) → y*

Question 8

A can be diagonalised if it has

Answer 8

• n distinct eigenvalues
• n linearly independent eigenvectors (can have repeated eigenvalues)

Question 9

If the n x n matrix A has n distinct eigenvalues λ₁, λ₂, …, λ_n and corresponding eigenvectors v₁,…, v_n then the system ẏ = Ay has general solution

Answer 9

y(t) = c₁e^λ₁tv₁ + c₂e^λ₂tv₂ + … + c_ne^λ_ntv_n

where constants c_i are the initial values z_i(0) and

ż = P^-1APz = Dz.

Note: if ż = Dz then z = (c₁e^λ₁t … c_ne^λ_nt)^T

also y = Pz = (v₁ … v_n)(c₁e^λ₁t … c_ne^λ_nt)^T = c₁e^λ₁tv₁ + c₂e^λ₂tv₂ + … + c_ne^λ_ntv_n

Question 10

Suppose A has n distinct negative real eigenvalues. Then the only steady state of the system ẏ = Ay is

Answer 10

y* = 0, and this is asymptotically stable.

This is because:

• if all the λ_i are neg, then each term in

y(t) = c₁e^λ₁tv₁ + c₂e^λ₂tv₂ + … + c_ne^λ_ntv_n

will tend to zero as t → ∞

• if the matrix has non-zero eigenvalues and can be
  diagonalised then the only solution to Ay = 0 is y = 0; i.e., the only steady state is y* = 0 and, since y(t) → 0 as t → ∞, y* is asymptotically stable.

Question 11

Taylor’s Theorem

Answer 11

Let F : Rⁿ → Rⁿ be continuously differentiable
and let y* ∈ Rⁿ. Then for h ∈ Rⁿ,
F(y* + h) = F(y*) + DF(y*)h + R(h).

DF(y*) is the Jacobian evaluated at y*, and R(h) has the property that R(h)/∥h∥ → 0 as h → 0.
Loosely speaking, if each entry of h is small, then
F(y* + h) ≈ F(y*) + DF(y*)h;

Question 12

Let y* be a steady state solution of the system ẏ = F(y) (where F : Rⁿ → Rⁿ is continuously differentiable).

Let DF(**y\***) denote the Jacobian matrix
evaluated at y\*. If DF(**y\***) has n negative real eigenvalues then **y\*** is

Answer 12

asymptotically stable

Question 13

Now suppose that we have two species of animal and that they compete with each other (for food, for instance). Denote the corresponding populations at time t by y₁(t) and y₂(t). We assume that, in the absence of the other, the population of either species
would exhibit logistic growth, as above. But given that they compete with each other, we assume that the presence of each has a negative effect on the growth rate of the other. That is, we assume that for some positive numbers a₁, a₂, b₁, b₂, c₁, c₂,
ẏ₁/y₁ = a₁ - b₁y₁ - c₁y₂
ẏ₁/y₂ = a₂ - b₂y₂ - c₂y_1-

What is the coupled system of difference equations?

Answer 13

ẏ₁ = a₁y₁ - b₁y₁² - c₁y₂y₁

ẏ₁ = a₂y₂ - b₂y₂² - c₂y₁y₂

Question 14

A quadratic form in n x 2 variables, e.g.

(q(x₁, x₂, x₃) = 5x₁² + 10x₂² + 2x₃²+ 4x₁x₂ + 2x₁x₃ + 2x₂x₃) variables is of the form

Answer 14

x^TAx

A = (5 2 1)

(2 10 1)

(1 1 2)

Nb: Diagonal entries of A are the coefficients of the squared variables and the other entries are half of the coefficients)

Question 15

Suppose that q(x) is a quadratic form. Then
q(x) is positive definite if

Answer 15

q(x) ≥ 0 for all x, and q(x) = 0 only when x = 0, the
zero-vector

Question 16

Suppose that q(x) is a quadratic form. Then
q(x) is positive semi-definite if

Answer 16

q(x) ≥ 0 for all x

Question 17

Suppose that q(x) is a quadratic form. Then
q(x) is negative definite if

Answer 17

q(x) ≤ 0 for all x, and q(x) = 0 only when x = 0, the
zero-vector

Question 18

Suppose that q(x) is a quadratic form. Then
q(x) is negative semi-definite if

Answer 18

q(x) ≤ 0 for all x

Question 19

Suppose that q(x) is a quadratic form. Then
q(x) is indefinite if

Answer 19

it is neither positive definite, nor positive semi-definite, nor negative definite, nor negative semi-definite; in other words, if there are x₁, x₂ such that q(x₁) < 0 and q(x₂) > 0.

Question 20

Suppose that the quadratic form q(x) has matrix representation
q(x) = x^TAx. Then:
If all eigenvalues of A are positive, then q is

Answer 20

positive definite

Question 21

Suppose that the quadratic form q(x) has matrix representation
q(x) = xTAx. Then:
If all eigenvalues of A are non-negative, then q is

Answer 21

positive semi-definite

Question 22

Suppose that the quadratic form q(x) has matrix representation
q(x) = xTAx. Then:
If all eigenvalues of A are negative, then q is

Answer 22

negative definite

Question 23

Suppose that the quadratic form q(x) has matrix representation
q(x) = xTAx. Then:
If all eigenvalues of A are non-´positive, then q is

Answer 23

negative semi-definite

Question 24

Suppose that the quadratic form q(x) has matrix representation
q(x) = xTAx. Then:
If some eigenvalues of A are negative, and some are positive, then q is

Answer 24

indefinite

Question 25

Suppose that A is an n x n matrix and that A₁, A₂, …, A_n are its leading principal submatrices. Then: A is positive definite if and only if

Answer 25

all its leading principal subdeterminants are positive

all its eigenvalues are positive

Question 26

Suppose that A is an n x n matrix and that A1, A2, …, An are its leading principal submatrices. Then: A is negative definite if and only if

Answer 26

• its negative, -A, is positive definite.
• its leading principal subdeterminants alternate in sign, with the first negative, as follows:

IA₁I < 0, IA2I > 0, IA₃I < 0, …, IA_nI > 0.
It should be noted that there is no test quite this simple to check whether a matrix is positive or negative semi-definite.

A matrix has all its eigenvalues negative if and only if the principal subdeterminants are alternately negative and positive.