Linear Algebra and Differential Equationspioneer.netserv.chula.ac.th/~ksujin/Slide(216)5.pdf ·...

Linear Algebra and Differential Equations

Sujin Khomrutai, Ph.D.

Department of Math & Computer ScienceChulalongkorn University

Lecture 4

S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 1 / 41

Table of Contents

1 Systems of Linear Equations

2 Higher Order Linear Systems

3 Vector Formulation

4 Constant Matrix Vector DE


Some Examples

A Coupled Spring-Mass System

m1d2x

dt2= −k1x1 + k2(y − x),

m2d2y

dt2= −k2(y − x).


Some Examples

A Mixing Problem

dA1

dt= − r21

V1A1 +

r12

V2A2 + cr ,

dA2

dt=

r21

V1A1 −

r12 + r

V2A2.


First Order Linear Systems

Definition (First order linear systems)

A first order linear system consists of unknown functions

x1(t), x2(t), . . . , xn(t) (n ≥ 2)

that satisfy the following differential equations

x ′1 = a11(t)x1 + a12(t)x2 + · · ·+ a1n(t)xn + f1(t),

x ′2 = a21(t)x1 + a22(t)x2 + · · ·+ a2n(t)xn + f2(t),

...

x ′n = an1(t)x1 + an2(t)x2 + · · ·+ ann(t)xn + fn(t).

• f1 = f2 = · · · = fn = 0 ⇒ homogeneous.

• fi 6= 0 for some i ⇒ nonhomogeneous.


First Order Linear Systems

Example

For the first order linear system

x ′1 = x1 − 8x2, x ′2 = −x1 + 3x2

show thatx1(t) = −2e5t + 4e−t , x2(t) = e5t + e−t

is a solution to the linear system.

Sol. We calculate

x ′1 = −10e5t − 4e−t , x1 − 8x2 = −10e5t − 4e−t ,

x ′2 = 5e5t − e−t , −x1 + 3x2 = 5e5t − e−t ,

so x ′1 = x1 − 8x2 and x ′2 = −x1 + 3x2.

Thus (x1(t), x2(t)) is a solution of the system.S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 6 / 41

Uncoupled systems

EX. First order linear system

x ′1 = 2x1,

x ′2 = −x2

is called uncoupled because each equation can be solved separately.

• x ′1 = 2x1 ⇒ x1(t) = C1e2t .

• x ′2 = −x2 ⇒ x2(t) = C2et .

In fact, if there is an equation, says

x ′i = aiixi

that can be solved separately, the function xi can be removed. So thereare fewer number of unknown functions.


Uncoupled systems

Example

Solve the first order linear system

x ′1 = 2x1 − x2,

x ′2 = x2.

Sol. The second equation can be solved to get

x2 = C2et .

Pluggin into the first equation

x ′1 = 2x1 − C2et .

Solving this ODE we get

x1(t) = C1e−2t + C2e

t .


Elimination technique

For a constant coefficients first order linear system, i.e.

x ′1 = a11x1 + a12x2 + · · ·+ a1nxn + f1(t),

x ′2 = a21x1 + a22x2 + · · ·+ a2nxn + f2(t),

...

x ′n = an1x1 + an2x2 + · · ·+ annxn + fn(t),

where aij are constants, one can employ the elimination technique tosolve the system.

EX. (n = 2, coupled system) From

x ′1 = ax1 + bx2,

x ′2 = cx1 + dx2,

we eliminate x2 from the first equation:

x2 =1

b(x ′1 − ax1) ⇒ sub. into 2’nd eqn.



Example


x ′1 = x1 + 2x2, x ′2 = 2x1 − 2x2.

Sol. Eliminate x2 from 1’st eqn:

x2 =1

2(x ′1 − x1).

Substitute into 2’nd eqn:

1

2(x ′′1 − x ′1) = 2x1 − 2

1

2(x ′1 − x1), x ′′1 + x ′1 − 6x1 = 0.

Solving the last eqn:

x1(t) = C1e−3t + C2e

2t ,

x2(t) = −2C1e−2t +

C2

2e2t .



Example


x ′1 = 2x1, x ′2 = x2 − x3, x ′3 = x2 + x3.

Sol. The first equation can be solved directly: x1(t) = C1e2t .

Eliminate x3 from 2’nd eqn: x3 = −x ′2 + x2.

Substitute into 3’rd eqn:

− x ′′2 + x ′2 = x2 − x ′2 + x2,

x ′′2 − 2x ′2 + x2 = 0.

Solving the last eqn:

x2(t) = (C2 + C3t)et ,

x3(t) = −C3et .


Exercise

EX. Solve the first order linear system

x ′1 = 4x1 + 2x2,

x ′2 = −x1 + x2

by the elimination method.Sol.


Table of Contents






Higher order linear systems

Recall. m’th order linear ODE is

y (m) + pm(t)y (m−1) + · · ·+ p2(t)y ′ + p1y = f (t).

Definition (Higher order linear systems)

A higher order linear system consists of unknown functions

y1(t), . . . , yn(t)

and equations

y(m1)1 =

∑linear terms,

...

y(mn)n =

∑linear terms

Highest derivative of y1 is m1, y2 is m2, and so on.


Conversion to a first order linear system

We show that

1 m’th order linear ODEs and

2 higher order linear systems

can always be rewritten into first order linear systems!

So it suffices to develop theory and techniques for first order linearsystems.

In the rest of this chapter we will concentrate on first order linear systems.


Conversion: ODE

Theorem (m’th order linear ODEs)

For m’th order ODE

y (m) + pm(t)y (m−1) + · · ·+ p2(t)y ′ + p1(t)y = f (t),

we introduce m functions

x1 = y , x2 = y ′, . . . , xm = y (m−1),

then we get an equivalent first order linear system

x ′1 = x2,

x ′2 = x3,

...

x ′m = −p1(t)x1 − p2x2 − · · · − pm(t)xn + f (t)


Conversio: ODE

Example

Convert the ODEy ′′ + 4ety ′ − 9t2y = 7t2

into an equivalent first order linear system.

Sol. Let x1 = y , x2 = y ′. Then we get

x ′1 = x2,

x ′2 = 9t2x1 − 4etx2 + 7t2,

where in the last equation we substitute x1 = y , x2 = y ′, and x ′2 = y ′′:

x ′2 + 4etx2 − 9t2x1 = 7t2

then solving for x ′2 to get the desired equation.


Exercise

EX. Convert the ODE

y ′′′ + t2y ′ − ety = t.

Sol.


Conversion: System

For a higher linear system

y(m1)1 =

∑linear terms,

...

y(mn)n =

∑linear terms

we introduce m1 functions from y1:

x11 = y1, x12 = y ′1, . . . , x1m1 = y(m1−1)1

m2 functions from y2:

x21 = y2, x22 = y ′2, . . . , x2m2 = y(m2−1)2

and so forth. Then we construct the equation for x ′ij as before.


Conversion: System

Example

Rewrite the linear system

x ′′ = 4y + et ,

y ′′ = −t2x ′ + sin t.

Sol. We introduce functions

x1 = x , x2 = x ′, x3 = y , x4 = y ′.

Then we get the first order linear system

x ′1 = x2,

x ′2 = 4x3 + et ,

x ′3 = x4,

x ′4 = −t2x2 + sin t

in the unknown functions x1, x2, x3, x4.


Exercise

EX. Convert the system of equations

x ′ = ty + cos t,

y ′′ = x ′ − x + et .

Sol.


Table of Contents






Vector Differential Equations

Definition (Vector differential equation)

We can express

x ′1 = a11(t)x1 + · · ·+ a1nxn + f1(t),

...

x ′n = an1(t)x1 + · · ·+ ann(t)xn + fn(t)

in matrix-vector notation, called n-vector differential equation as

~x ′ = A(t)~x + ~f (t), where

~x =

x1...xn

, A(t) =

a11(t) · · · a1n(t)...

. . ....

an1(t) · · · ann(t)

, ~f (t) =

f1(t)...

fn(t)



Example

Express system

x ′1 = 2x1 − 3x2 + 1

x ′2 = −x1 + 5x2 + et

into 2-vector differential equation.

Sol. We have

~x ′(t) =

[2 −3−1 5

]~x(t) +

[1et

].



Example

Express system

x ′1 = cos tx1 + x2 − sin tx3

x ′2 = sin tx1 − x2 + cos tx3

x ′3 = 4x2 + 5

into 3-vector differential equation.

Sol. We have

~x ′(t) =

cos t 1 − sin tsin t −1 cos t

0 4 0

~x(t) +

005

.



Let Rn denote the set of all n-vectors:

Rn =

a1

...an

: a1, . . . , an ∈ R

.

Definition

Vn(I ) denotes the set of all n-vector functions ~x : I → Rn.

Theorem

Vn(I ) is a vector space.

Note.

a~x(t) + b~y(t) =

ax1(t) + by1(t)...

axn(t) + byn(t)



Example (Linearly dependence)

Determine whether the 2-vector functions

~x1(t) =

[et

2et

], ~x2(t) =

[3 sin tcos t

]are linearly independent?

Sol. Note that ~x1, ~x2 ∈ Vn(R). Suppose there are C1,C2 such that

C1~x1(t) + C2~x2(t) = ~0 for all t ∈ R.

This must be true at t = 0:

C1

[12

]+ C2

[01

]=

[00

]⇒

[1 02 1

] [C1

C2

]=

[00

]Since det

[1 02 1

]= 1 6= 0, C1 = C2 = 0. So ~x1, ~x2 are linearly independent.


Wronskian

Definition

For ~x1, . . . , ~xn ∈ Vn(I ), we define

W (~x1, . . . , ~xn)(t) = det[~x1(t), . . . , ~xn(t)] (n-row matrix).

It is called the Wronskian of ~x1, . . . , ~xn.

Theorem

If the Wronskian

W (~x1, . . . , ~xn)(t0) 6= 0 for some t0 ∈ I ,

then ~x1, . . . , ~xn are linearly independent.

If W (~x1, . . . , ~xn)(t) = 0 for all t ∈ I , then ~x1, . . . , ~xn are linearly dependent.


Wronskian

Example

Determine whether the 3-vector functions

~x1(t) =

tt2

−t3

, ~x2(t) =

2t3t2

0

, ~x3(t) =

−t03t3

are linearly independent?

Sol. We haveW (~x1, . . . , ~xn)(t) = 0 for all t ∈ R,

so these vectors are linearly dependent.


Exercise

EX. Determine whether the 3-vector functions

~x1(t) =

t + 1t − 1

2t

, ~x2(t) =

et

e2t

e3t

, ~x3(t) =

1sin tcos t

are linearly independent?

Sol.


IVP of Vector DE

Definition

A problem of n-vector DE{~x ′ = A(t)~x + ~f (t)

~x(t0) = ~α = [α1, . . . , αn]T

is called an initial value problem (or IVP).

• t0 is called an initial time, ~α ∈ Rn is called an initial data.

Theorem (Existence and uniqueness)

If A(t), ~f (t) are continuous on an interval I and t0 ∈ I , then IVP

~x ′ = A(t)~x + ~f (t), ~x(t0) = ~α,

has a unique solution defined on I , for each given initial data ~α.


IVP of Vector DE

Example

Let A(t) be continuous n × n-matrix function on an interval I and t0 ∈ I .Solve the IVP

~x ′ = A(t)~x , ~x(t0) = ~0.

Sol. (Trial) Observe that the zero vector function

~x(t) = 0

satisfies the given IVP.

(Apply theorem) By the existence and uniqueness theorem, we concludethat the only solution to IVP is

~x = ~0.


Vector DE: Homogeneous

We investigate homogeneous n-vector DE

~x ′ = A(t)~x . (H)

Theorem (Superposition)

If ~x1(t), . . . , ~xn(t) solve (H) then

~x = C1~x1 + · · ·+ Cn~xn

also solves the n-vector DE for arbitrary constants C1, . . . ,Cn.

Corollary

Sn(I ) = {~x : I → Rn solves (H)} is a vector space.



Assume ~x1, . . . , ~xn solve (H). Does ~x1, . . . , ~xn constitute general solution,i.e. any solution ~x of

~x ′ = A(t)x , ~x(t0) = ~α,

can be expressed as~x = C1~x1 + · · ·+ Cn~xn

for a certain choice of C1, . . . ,Cn.

Theorem (Homogeneous: General solutions)

If ~x1, . . . , ~xn are solutions of ~x ′ = A(t)~x and

W (~x1, . . . , ~xn)(t0) 6= 0

for some t0 ∈ I , then~x = C1~x1 + · · ·+ Cn~xn

is a general solution. Such ~x1, . . . , ~xn are called fundamental solutions.



Theorem (Dimension)

The set Sn(I ) of solutions to

~x ′ = A(t)~x

is an n-dimensional vector space.

Proof. Let~e1 = [1, 0, . . . , 0]T , . . . , ~en = [0, . . . , 0, 1]T .

Solve the IVP~x ′ = A(t)~x , ~x(t0) = ~ej

for j = 1, 2, . . . , n, we obtain solutions

~x1, . . . , ~xn

corresponding to ~e1, . . . , ~en, respectively. Note that the Wronskian at t0 is1. So we are done.



Example

Show that ~x1 =

[sin 3tcos 3t

], ~x2 =

[− cos 3t

sin 3t

]are fundamental solutions of

~x ′ =

[0 3−3 0

]~x .

Sol. ~x ′1 =

[3 cos 3t−3 sin 3t

],

[0 3−3 0

]~x1 =

[3 cos 3t−3 sin 3t

]. So ~x1 is a solution.

Next, ~x ′2 =

[3 sin 3t3 cos 3t

]and

[0 3−3 0

]~x2 =

[3 sin 3t3 cos 3t


W (~x1, ~x2)(0) = det

[0 −11 0

]= 1. So ~x1, ~x2 are fundamental solutions and

general solution is ~x = C1~x1 + C2~x2 =

[C1 sin 3t − C2 cos 3tC1 cos 3t + C2 sin 3t

].



Example

Show that ~x1 =

[e2t

−e2t

], ~x2 =

[e2t(1 + t)−te2t

]are fundamental solutions of

~x ′ =

[3 1−1 0

]~x .

Sol. ~x ′1 =

[2e2t

−2e2t

],

[3 1−1 0

]~x1 =

[2e2t

−2e2t


~x ′2 =

[e2t(3 + 2t)−(2t + 1)e2t

],

[3 1−1 0

]~x2 =

[e2t(3 + 2t)−(2t + 1)e2t


W (~x1, ~x2)(0) = det

[1 1−1 0

]= 1. So ~x1, ~x2 are fundamental solutions and

general solution is ~x = C1~x1 + C2~x2.


Table of Contents






A is Constant

Theorem

Let A be a constant (n × n)-matrix. If λ ∈ R is an eigenvalue of A witheigenvector ~v then

~x(t) = eλt~v

is a solution of~x ′ = A~x .

Proof. We calculate

~x ′(t) = (eλt)′~v = λeλt~v .

Also, since λ is an eigenvalue of A with eigenvector ~v , A~v = λ~v . So

A~x = eλtA~v = λeλt~v .

So ~x ′ = A~x as claim.


A is Constant

Example

Find the general solution to ~x ′ =

[2 1−3 −2

]~x .

Sol. Let A =

[2 1−3 −2

]. We solve the eigenvalue problem A~v = λ~v .

det(A− λI ) = det

[2− λ 1−3 −2− λ

]= 0 ⇒ (2− λ)(−2− λ) + 3 = 0

λ1 = −1, λ2 = 1.

λ1 = −1: We solve (A− λ1I )~v = ~0.[3 1−3 −1

] [rs

]=

[00

]⇒ 3r + s = 0.

Taking r = 1, s = −3 we get ~v1 = [1,−3]T .S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 40 / 41

A is Constant

λ2 = 1: We solve (A− λ2I )~v = ~0.[1 1−3 −3

] [rs

]=

[00

]⇒ r + s = 0.

Taking r = 1, s = −1 we get ~v2 = [1,−1]T .By theorem, we get solutions

~x1(t) = e−t[

1−3

], ~x2(t) = et

[1−1

].

Wronskian

W (~x1, ~x2)(0) =

[1 1−3 −1

]= 2

So ~x1, ~x2 are fundamental solutions and a general solution is

~x = C1~x1 + C2~x2

[C1e

−t + C2et

−3C1e−t − C2e

t

].


Linear Algebra and Differential Equationspioneer.netserv.chula.ac.th/~ksujin/Slide(216)5.pdf ·...

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