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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
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 2 / 41
Some Examples
A Coupled Spring-Mass System
m1d2x
dt2= −k1x1 + k2(y − x),
m2d2y
dt2= −k2(y − x).
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 3 / 41
Some Examples
A Mixing Problem
dA1
dt= − r21
V1A1 +
r12
V2A2 + cr ,
dA2
dt=
r21
V1A1 −
r12 + r
V2A2.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 4 / 41
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 5 / 41
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 7 / 41
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 .
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 8 / 41
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 9 / 41
Elimination technique
Example
Solve the first order linear system
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 .
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 10 / 41
Elimination technique
Example
Solve the first order linear system
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 .
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 11 / 41
Exercise
EX. Solve the first order linear system
x ′1 = 4x1 + 2x2,
x ′2 = −x1 + x2
by the elimination method.Sol.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 12 / 41
Table of Contents
1 Systems of Linear Equations
2 Higher Order Linear Systems
3 Vector Formulation
4 Constant Matrix Vector DE
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 13 / 41
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 14 / 41
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 15 / 41
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)
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 16 / 41
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 17 / 41
Exercise
EX. Convert the ODE
y ′′′ + t2y ′ − ety = t.
Sol.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 18 / 41
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 19 / 41
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 20 / 41
Exercise
EX. Convert the system of equations
x ′ = ty + cos t,
y ′′ = x ′ − x + et .
Sol.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 21 / 41
Table of Contents
1 Systems of Linear Equations
2 Higher Order Linear Systems
3 Vector Formulation
4 Constant Matrix Vector DE
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 22 / 41
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)
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 23 / 41
Vector Differential Equations
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
].
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 24 / 41
Vector Differential Equations
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
.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 25 / 41
Vector Differential Equations
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)
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 26 / 41
Vector Differential Equations
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 27 / 41
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 28 / 41
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 29 / 41
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 30 / 41
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 31 / 41
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 32 / 41
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 33 / 41
Vector DE: Homogeneous
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 34 / 41
Vector DE: Homogeneous
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 35 / 41
Vector DE: Homogeneous
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
]. So ~x2 is a solution.
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
].
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 36 / 41
Vector DE: Homogeneous
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
]. So ~x1 is a solution.
~x ′2 =
[e2t(3 + 2t)−(2t + 1)e2t
],
[3 1−1 0
]~x2 =
[e2t(3 + 2t)−(2t + 1)e2t
]. So ~x2 is a solution.
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 37 / 41
Table of Contents
1 Systems of Linear Equations
2 Higher Order Linear Systems
3 Vector Formulation
4 Constant Matrix Vector DE
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 38 / 41
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.
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 39 / 41
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
].
S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 41 / 41
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