Linear Algebra and Differential Equations

Linear Algebra and Differential Equations

Sujin Khomrutai, Ph.D.

Department of Math & Computer Science Chulalongkorn University

Lecture 7

Table of Contents

1 Nonhomogeneous Vector DE

2 Dynamical Systems

Nonhomogeneous Vector DE

We study

~

x⁰ =A(t)~x+f~(t),

where A(t) isn×n matrix and ~f is ann-vector function.

Theorem

Assume A(t), ~f are continuous.

Then ~x⁰ =A(t)~x+f~(t) has a general solution

~

x(t) =C₁~x₁(t) +· · ·+C_n~x_n(t) +X(t),

where~x1, . . . , ~xn are fundamental solutions to the homogeneous equation

~x⁰ =A(t)~x,

and X(t) is a particular solution to the nonhomogeneous vector DE.

Nonhomogeneous Vector DE

Theorem (Variation of parameters method) Let Φ(t) be an n×n matrix value function given by

Φ(t) =

~

x1(t) · · · ~xn(t) , where~x1, . . . , ~xn are linearly independent solutions of

~

x⁰ =A(t)~x.

Then a particular solution to ~x⁰ =A(t)~x+~f(t) is X~(t) = Φ(t)

Z

Φ(s)⁻¹f~(s)ds.

Note. Φ(t) is called a fundamental matrixfor the vector DE.

Nonhomogeneous Vector DE

Remark. In practice, it is easier calculateX~(t) as follows.

Step 1. Find~x₁, . . . , ~x_n, then form Φ(t).

Step 2. Solve the linear algebraic equation

Φ(t)~v⁰(t) =f~(t) ⇒ get ~v⁰(t) = p⁰(t)

q⁰(t)

.

One may use Cramer’s rule/Gaussian elimination method.

Step 3. Integrate~v⁰(t) to get

~ v(t) =

Z

~

v⁰(t)dt = p(t)

q(t)

.

Step 4. A particular solution is

X~(t) = Φ(t)~v(t).

Nonhomogeneous Vector DE

Example

Find a particular solution of ~x⁰ = 1 2

4 3

~ x+

12e^3t 18e^2t

. Given that

~ x₁(t) =

−e^−t e^−t

, ~x₂(t) = e^5t

2e^5t

. Sol. We get Φ(t) =

−e^−t e^5t e^−t 2e^5t

. We solve Φ(t)~v⁰ =f~(t), i.e.

−e^−t e^5t e^−t 2e^5t

p⁰(t) q⁰(t)

= 12e^3t

18e^2t

, det

−e^−t e^5t e^−t 2e^5t

=−3e^4t By Cramer’s rule

p⁰(t) = det

12e^3t e^5t 18e^2t 2e^5t

−3e^4t =−8e^4t+ 6e^3t ∴p(t) =−2e^4t+ 2e^3t

Nonhomogeneous Vector DE

Next,

q⁰(t) = det

−e^−t 12e^3t e^−t 18e^2t

−3e^4t = 6e^−3t+ 4e^−2t

∴q(t) =−2e^−3t−2e^−2t From the calculations, we obtain

X~(t) = Φ(t) p(t)

q(t)

=

−4e^2t

−2e^2t−6e^3t

as a particular solution to the vector DE.

Nonhomogeneous Vector DE

Example

Find a particular solution of ~x⁰ =

3 2

−2 −1

~x+ −3e^t

6te^t

. Sol.

Exercise

EX.Find a particular solution to the vector DE~x⁰ =

2 −1

−1 2

~x+ 0

4e^t

. Sol.

Table of Contents

1 Nonhomogeneous Vector DE

2 Dynamical Systems

Autonomous Systems

We have developed theory to solve linear systems d

dt~x(t) =A~x(t) +~f(t)

explicitly, where Ais a constant matrix. This is impossible to do when Ais non-constant, or the system is nonlinear.

1 qualitative technique

2 numerical technique

We discuss qualitative technique for the system of 2 functions.

Consider the finding of x(t),y(t) satisfying d

dtx =F(x,y,t), d

dty =G(x,y,t).

Autonomous Systems

Definition

IfF,G does not depend explicitly on t, i.e.

d

dtx =F(x,y), d

dty =G(x,y), the system is called an autonomous system.

Otherwise, it is called non-autonomous.

1 x(t),y(t) are velocities of two particles.

2 (x(t),y(t)) are moving in the phase plane, i.e. thexy-plane.

3 The curve of (x(t),y(t)) for allt is called thetrajectory.

4 All trajectories (varying initial conditions) form thephase portrait.

Autonomous Systems

Equilibriums

Definition

For an autonomous dynamical system d

dtx =F(x,y), d

dty =G(x,y), an equilibrium point is a point (x₀,y₀) such that

F(x₀,y₀) = 0, G(x₀,y₀) = 0.

If (x0,y0) is an equilibrium point, then

x(t) =x0, y(t) =y0

is a solution called an equilibrium solution.

Equilibriums

Example

Find all equilibrium points of the dynamical system x⁰=x(y−1), y⁰ =y(x+ 1).

Sol.

Equilibriums

Example

Find all equilibrium points of the dynamical system x⁰ =x(2x+y), y⁰=y(x−2y+ 4).

Sol.

Exercise

EX.Find all equilibrium points of the dynamical system x⁰ =x(x²+y²−1), y⁰ = 2y(xy −1).

Sol.

Classifications of Equilibrium Points

Consider a linear system d

dtx=ax+by, d

dty =cx+dy. Taking ~x(t) =

x(t) y(t)

, the linear system can be written as d

dt~x =A~x, A= a b

c d

. We assume throughout that

detA6= 0.

Theorem

(x₀,y₀) = (0,0)is the only equilibrium point of the system.

Two eigenvalues λ₁, λ₂ of A are non-zero.

Stability of Equilibrium Point

Definition

For a linear system, an equilibrium point (x0,y0) is said to be

1 stableif every solution stays bounded as t → ∞.

2 unstable if it is not stable.

3 asymptotically stable if every solution converges to (x₀,y₀) as t → ∞. The equilibrium in this case is called anattractor.

1 If all eigenvaluesλ_j <0, or Reλ_j <0, then asymptotically stable.

2 If one eigenvalueλ_j >0 or Reλ_j >0, then unstable.

3 If all eigenvaluesλj ≤0 or Reλj ≤0, then stable.

Classifications of Equilibrium Points

There are 3 cases for λ1, λ2:

1 Distinct real eigenvalues

Eigenvectors ~v₁, ~v₂ and fundamental solutions

~

x1(t) =e^λ1t~v1, ~x2(t) =e^λ2t~v2.

2 Complex Eigenvalues a±bi

Complex Eigenvector ~v =~u+iw~ and fundamental solutions

~x1(t) =e^at(cosbtu~−sinbtw~), ~x2(t) =e^at(sinbtu~+ cosbtw~).

3 Repeated Eigenvaluesλ₁ =λ₂.

Distinct Real Eigenvalues

General solution is

~

x(t) =C1e^λ1t~v1+C2e^λ2t~v2. If~x(0) is parallel to ~vj (C2= 0 or C1= 0) then

~

x(t) =C_je^λjt~v_j. For λ_j >0, trajectory is pointing away from (0,0),

Distinct Real Eigenvalues

For λj <0, trajectory pointing towards (0,0),

Stable Node

Theorem (Stable Node) Ifλ₂ < λ₁ <0 then

1 For t small,~x(t)k~v₂(t)

2 For t large,~x(t)k~v1(t)

3 lim

t→∞~x(t) = 0

0

So the trajectory starts parallel to ~v2 but eventually parallel to~v1 and approaches (0,0)at t =∞.

The equlibrium point (0,0)is called astable node. It is asymptotically stable.

Stable Node

Unstable Node

Theorem (Unstable Node) If0< λ₁ < λ₂

1 For t small,~x(t)k~v₁(t)

2 For t large,~x(t)k~v2(t)

3 lim

t→∞~x(t) unbounded

4 lim

t→−∞~x(t) = 0

0

So the trajectory starts parallel to ~v1 but eventually parallel to~v2 and unbounded at t =∞.

The equlibrium point (0,0)is called an unstable node.

Saddle Point

Theorem (Saddle Point) Ifλ2 <0< λ1 then

1 For t small,~x(t)k~v2

2 For t large,~x(t)k~v1

3 lim

t→∞~x(t) unbounded

4 lim

t→−∞~x(t) unbounded

The equlibrium point (0,0)is called asaddle point. It is unstable.

Saddle Point

Complex Eigenvalues

Next, assume the eigenvalues are

λ_1,2=a±ib (b 6= 0).

Then

~

x(t) =C1~x1(t) +C2~x2(t)

=e^at(C1(cosbt~u−sinbtw~) +C2(sinbt~u+ cosbtw~)

=e^atP~(t).

Observe that

P~

t+2π b

=P(t).~

Center

Theorem (Center)

Ifλ_1,2=±ib (a pure imaginary, i.e. a= 0) then~x(t) =P~(t) is periodic.

So the trajectory is a closed curve.

The equilibrium (0,0)is called acenter. It is stable.

Spiral Point

Theorem

Ifλ_1,2=a±bi where a6= 0,b6= 0 then

1 a<0, the trajectory spiral towards (0,0). So (0,0)is called a stable spiral point. It is asymptotically stable.

2 a>0, the trajectory spiral away from (0,0)and unbounded at t =∞. So (0,0)is called an unstable spiral point.

Repeated Eigenvalues

For the repeated eigenvalues case, there are two subcases

1 A=kI (k 6= 0). Then λ₁ =λ₂=k and we can choose eigenvectors

~ v₁ =

1 0

, ~v₂= 0

1

.

Then the solution

~

x(t) =e^kt(C1~v1+C2~v2).

~x(t)k(C1~v1+C2~v2). Since C1~v1+C2~v2 spans all 2-vectors, the phase portrait contains all straight lines passing through (0,0).

2 Ais not a multiple of I. In this case, we need a generalized eigenvector.

Proper Node

Theorem

If A=kI (k 6= 0) then

1 If k >0 the trajectory is pointing away from(0,0). So(0,0)is called an unstable proper node.

2 If k <0 the trajectory is pointing towards(0,0). So (0,0)is called a stable proper node. It is asymptotically stable.

Degenerate Node

Finally, assume Ais not a multiple of I. Let V~ be a generalized eigenvector andV~⁽¹⁾ = (A−λI)V~.

General solution

~x(t) =e^λt(C1V~⁽¹⁾+C2(V~ +tV~⁽¹⁾)) Theorem

Then

1 For t small,~x(t)kV~⁽¹⁾

2 For t large,~x(t)kV~⁽¹⁾.

Ifλ <0, the equilibrium(0,0)is called astable degenerate node. It is asymptotically stable.

Ifλ >0, the equilibrium(0,0)is called an unstable degenerate node.

Degenerate Node

Some Examples

Example

Characterize the equilibrium point for the system ~x⁰ = 1 3

1 −3

and sketch the phase portrait.

Sol.

Some Examples

Example

Characterize the equilibrium point for the system ~x⁰ =

0 2

−2 0

and sketch the phase portrait.

Sol.

Some Examples

Example

Characterize the equilibrium point for the system ~x⁰ = 1 0

3 1

and sketch the phase portrait.

Sol.

Exercise

EX.Characterize the equilibrium point for the system ~x⁰=

2 3

−1 2

and sketch the phase portrait.

Sol.

Exercise

EX.Characterize the equilibrium point for the system ~x⁰=

1 1

−9 −5

and sketch the phase portrait.

Sol.

More Examples

Example

(Fin 2017) Consider the initial value problem ~x⁰ =

α 1

−1 α

~x

1 Find eigenvalues and eigenvectors in terms of α. Also, write the general real value solutions.

2 Give the condition on α so that the equilibrium (0,0) is stable.

Sol.

More Examples

Example

(Fin 2017) Consider the system of linear equation~x⁰ =

−1 0

2 −1

~x+ 1

2

1 Find the equilibrium point ~x0 and determine its stability (asymptotically stable, stable, or unstable).

2 Find the general solution and sketch some trajectories in the phase plane.

Sol.