Ordinary Di erential Equations. Session 12roquesol/Math_308...Session 12. Systems of First Order...

Systems of First Order Linear Equations ISystems of First Order Linear Equations II

Ordinary Differential Equations. Session 12

Dr. Marco A Roque Sol

04/09/2019

Dr. Marco A Roque Sol Ordinary Differential Equations. Session 12


Basic Theory of Systems of First Order Linear EquationsHomogeneous Linear Systems with Constant Coefficients

Basic Theory of Systems of First Order LinearEquations

The general theory of a system of n first order linear equations

x ′1 = p11x1 + p12x2 + . . .+ p1nxn + g1(t)x ′2 = p21x1 + p22x2 + . . .+ p2nxn + g2(t)...

...x ′n = pn1x1 + pn2x2 + . . .+ pnnxn + gn(t)

or

X′ = P(t)X + g(t)

closely parallels that of a single linear equation of nth order.





we assume that P and g are continuous on some intervalα < t < β; that is, each of the scalar functionsp11, ..., pnn, g1, ..., gn is continuous there.

Theorem 7.4

If the vector functions x(1) and x(2) are solutions of thehomogeneus system ( g(t) = 0 ) then the linear combinationc1x(1) + c2x(2) is also a solution for any constants c1 and c2.

This is the principle of superposition





By repeated application of Theorem, we can conclude that ifx(1), ..., x(k) are solutions of the homogeneous system, then

c1x(1) + ...+ ckx(k)

is also a solution for any constants c1, ..., ck .

Theorem 7.5

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





If the constants c1, ..., cn are thought of as arbitrary, then theabove equation includes all solutions of the system, and it iscustomary to call it the general solution.

Any set of solutions x(1), ..., x(n) of the homogeneus system that islinearly independent at each point in the interval α < t < β is saidto be a fundamental set of solutions for that interval.

Theorem 7.6

If x(1), ..., x(n) are solutions of the homogeneus system on theinterval α < t < β, then in this interval W [x(1), ..., x(n)] given by





W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

(n)1

x(1)2 x

(2)2 . . . x

(n)2

...... . . .

...

x(1)n x

(2)n . . . x

(n)n

∣∣∣∣∣∣∣∣∣∣either is identically zero or else never vanishes.To prove this theorem is necessary to establish that

dW

dt= [p11 + p22 + ...+ pnn]W

Hence

W (t) = ce∫[p11+p22+...+pnn]dt





Theorem 7.7

Let x(1), ..., x(n) be the solutions of the homogeneus system thatsatisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively, where t0 is any point inα < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

Then, x(1), ..., x(n) form a fundamental set of solutions of thehomogeneous system.

Finally in the case that the solution is complex-valued, we have thefollowing result.





Theorem 7.8

Consider the homogeneous system

X′ = P(t)X

where each element of P is a real-valued continuous function. Ifx = u(t) + i v(t) is a complex-valued solution, then its real partu(t) and its imaginary part v(t) are also solutions of this equation.




Homogeneous Linear Systems with ConstantCoefficients

We will concentrate most of our attention on systems ofhomogeneous linear equations with constant coefficients

x′ = Ax

where A is a constant n × n matrix. Unless stated otherwise, wewill assume further that all the elements of A are real (rather thancomplex) numbers.

The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.





A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot some solution curves, or trajectories. Aplot that shows a representative sample of trajectories for a givensystem is called a phase portrait .





Now, for the system

x′ = Ax

we look for solutions of the form

x = veλt

where the expon entλ and the vector v are to be determined.Substituting x in the system gives

λveλt = Aveλt

(A− λI) v = 0





Thus, to solve the system of differential equations, we must solvethe above system of algebraic equations. That is, we need to findthe eigenvalues and eigenvectors of the matrix A.

If we assume that A is a real-valued matrix, then we must considerthe following possibilities for the eigenvalues of A:

1. All eigenvalues are real and different from each other.

2. Some eigenvalues occur in complex conjugate pairs.

3. Some eigenvalues, either real or complex, are repeated. If the neigenvalues are all real and different, as in the





Example 7.14

Consider the system

x′ = Ax =

(1 14 1

)x

Solution

Let’s find the eigenvalues of the matrix A





|A− λI| =

∣∣∣∣1− λ 14 1− λ

∣∣∣∣ = 0

(1− λ)2 − 4 = 0 =⇒

(λ2 − 2λ− 3 = (λ− 3)(λ+ 1) = 0 =⇒





λ1 = 3, λ2 = −1

If λ1 = 3, then the system reduces to the single equation

−2v1 + v2 = 0, =⇒ v2 = 2v1

and a corresponding eigenvector is

v(1) =

(12

)





Similarly, corresponding to λ2 = −1, we find that a correspondingeigenvector is

v(2) =

(1

− 2

)The corresponding solutions of the differential equation are

x(1) =

(12

)e3t ; x(2) =

(1

− 2

)e−t





The Wronskian of these solutions is

W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = −4e2t 6= 0

Hence the solutions x(1) and x(2) form a fundamental set, and thegeneral solution of the system is

x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t



Complex EigenvaluesFundamental MatriceseAt Matrix and Diagonalization

Complex Eigenvalues

In this section we consider again a system of n linear homogeneousequations with constant coefficients

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

In the case, λ is complex, we have complex eigenvalues andeigenvectors always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues

are two complex-conjugate eigenvalues and eigenvectors de lamatrix a, then

X±(t) = e(µ±i ν)t (a± i b)

are complex-valued solutions, but taking in account that

e(µ±i ν)t = eµt (cos(νt)± i sin(νt))

and the principle of superposition, then we have that

X1(t) =1

2

(X+(t) + X−(t)

)X2(t) =

1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues

are two (real) solutions !!!

X1(t) = eµt (acos(νt)− bsin(νt))

X2(t) = eµt (acos(νt) + bsin(νt))




Complex Eigenvalues

Example 7.17

Solve the following ODE

x′ = Ax =

3 1 10 2 10 −1 2

x

Solution





Complex Eigenvalues

|A− λI| =

∣∣∣∣∣∣3− λ 1 1

0 2− λ 10 −1 2− λ

∣∣∣∣∣∣ = 0

(3− λ)

∣∣∣∣2− λ 1−1 2− λ

∣∣∣∣− (1)

∣∣∣∣0 10 2− λ

∣∣∣∣+ (1)

∣∣∣∣1 2− λ1 1

∣∣∣∣ =⇒

(3− λ)(λ2 − 4λ+ 6) = 0 =⇒




Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

=

0 1 10 −1 10 −1 −1

v1v2v3

=

0 1 10 0 20 0 0

v1v2v3

=

000




Complex Eigenvalues


v(1) =

100

If λ2 = 2 + i , then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

=




Complex Eigenvalues

3− (2 + i) 1 10 2− (2 + i) 10 −1 2− (2 + i)

v1v2v3

=

1− i 1 10 −i 10 −1 −i

v1v2v3

=

1− i 1 10 −i 10 0 0

v1v2v3

=

1− i 0 1− i0 −i 10 0 0

v1v2v3

=

000




Complex Eigenvalues


v(2) =

10−1

+ i

010

The corresponding solutions of the differential equation are

x(1) =

100

e3t ; x(2) = e2t

10−1

cos(t)−

010

sin(t)

x(3) = e2t

10−1

cos(t) +

010

sin(t)




Complex Eigenvalues


W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣e3t e2tcos(t) e2tsin(t)0 −e2tsin(t) e2tcos(t)0 −e2tcos(t) −e2tsin(t)

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣1 cos(t) sin(t)0 −sin(t) cos(t)0 −cos(t) −sin(t)

∣∣∣∣∣∣ =

e3te2te2t(sin2(t) + cos2(t)

)= e7t 6= 0




Complex Eigenvalues

Hence the solutions x(1), x(2) and x(3) form a fundamental set, andthe general solution of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

100

e3t + c2

10−1

e2tcos(t)−

010

e2tsin(t)

+

c3

10−1

e2tcos(t) +

010

e2tsin(t)




Complex Eigenvalues

X =

x1x2x3

=

c1e3t + e2t(c2cos(t) + c3sin(t))

0 e2t(−c2sin(t) + c3cos(t))0 −e2t(c2cos(t) + c3sin(t))

Here is the direction field associated with the system

x ′1x ′2x ′3

=

3 1 10 2 10 −1 2

x1x2x3




Complex Eigenvalues




Complex Eigenvalues

Example 7.18

Solve the following ODE

X′ = AX =

(−1/2 1−1 −1/2

)X

Solution


|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

4= 0




Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

)=

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

)=(

−i 1−1 −i

)(v1v2

)=

(00

)




Complex Eigenvalues


v(1) =

(1i

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

)=




Complex Eigenvalues

(i 1−1 i

)(v1v2

)=


v(2) =

(1−i

)The corresponding solutions of the differential equation are

x(1) =

(1i

)e(−1/2+i)t ; x(2) =

(1

− i

)e(−1/2−i)t




Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose the realand imaginary parts of either x (1) or x (2). In fact,

x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)

)Hence a set of real-valued solutions of is

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues


W [x(1), x(2)](t) =

∣∣∣∣ e−t/2cos(t) e−t/2sin(t)

−e−t/2sin(t) e−t/2cos(t)

∣∣∣∣ =

e−t/2e−t/2∣∣∣∣ cos(t) sin(t)−sin(t) cos(t)

∣∣∣∣ = e−t 6= 0

Hence the solutions x(1), x(2) form a fundamental set,




Complex Eigenvalues

and the general solution of the system is

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

)Here is the direction field associated with the system(

x ′1x ′2

)=

(−1/2 1−1 −1/2

)(x1x2

)




Complex Eigenvalues




Fundamental Matrices

Let’s start with the system

x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t) form a fundamental set ofsolutions on some interval α < t < β. Then the matrix

Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.





For instance, the system

x′ =

(1 14 1

)x

has solutions

x(1)(t) =

(e3t

2e3t

); x(2)(t) =

(e−t

−2e−t

)which are linearly independent. Thus a fundamental matrix for thesystem is

Ψ(t) =

(e3t e−t

2e3t −2e−t

)





Using a fundamental matrix the general solution can be written as

x = Ψ(t)c; c = constant

and if we imposed initial conditions x(t0) = x0, where t0 is a givenpoint on α < t < β and x0 is given initial vector, we obtain

Ψ(t0)c = x0

c = Ψ−1(t0)x0

x = Ψ(t)Ψ−1(t0)x0

is the solution of the initial value problem

x′ = P(t)x; x(t0) = x0





Recall that each column of the fundamental matrix Ψ(t) is asolution of the ODE. It follows that Ψ(t) satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ

Sometimes it is convenient to make use of the specialfundamental matrix , denoted by Φ, such that the initialcondition

x(j) = e(j); e(j) =

0...1...0

j − th row





Thus Φ(t) has the property that

Φ(t0) =

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

= I

and the solution of the IVP is given by

x = Φ(t)Φ−1(t0)x0 = Φ(t)x0

in another words

Φ(t) = Ψ(t)Ψ−1(t0)





Thus, for instance the system

x′ =

(1 14 1

)x

subject to the different initial conditions

x(1)(0) =

(10

); x(2)(0) =

(01

)has the particular solutions equal to

x(t) =1

2

(e3t

2e3t

)+

1

2

(e−t

−2e−t

)





x(t) =1

4

(e3t

2e3t

)− 1

4

(e−t

−2e−t

)Hence

Φ(t) =

12e

3t + 12e−t 1

2e3t − 1

2e−t

e3t − e−t 12e

3t + 12e−t

OBS

Note that the elements of Φ(t) are more complicated than those ofthe fundamental matrix Ψ(t); however, it is now easy to determinethe solution corresponding to any set of initial conditions.




eAt Matrix and Diagonalization

The Matrix eAt

Recall that the solution of the initial value problem

x ′ = ax , x(0) = x0, a = constant

is given by

x(t) = x0eat

Now, consider the corresponding initial value problem for an n × nsystem

x′ = Ax, x(0) = x0

where A is a constant matrix.





Applying the results of already obtained, we can write its solutionas

x = Φ(t)x0

where Φ(0) = I. Thus, Φ(t), is playing the roll of eat . let’s seethis with more detail.

The scalar exponential function eat can be represented by thepower series

eat = 1 +∞∑n=1

antn

n!

which converges for all t. Let us now replace the scalar a by then × n constant matrix A and consider the corresponding series





I +∞∑n=1

Antn

n!= I + At +

A2t2

2!+ ...+

Ant2

n!+ ...

Each term in the series is an n × n matrix. It is possible to showthat each element of this matrix sum converges for all t asn→∞. Thus, we have a well defined n × n matrix, which will bedenote by eAt

eAt = I +∞∑n=1

Antn

n!

By differentiating the above series term by term, we obtain

d

dt

[eAt]

=∞∑n=1

Antn−1

(n − 1)!= A

[I +

∞∑n=1

Antn

n!

]= AeAt





Therefore, eAt satisfies the differential equation

d

dt

[eAt]

= AeAt

Further, by setting t = 0 in the definition of eAt we find that eAt

satisfies the initial condition

eAt∣∣∣t=0

= I

In this way, we have that the fundamental matrix Φ satisfies thesame initial value problem as eAt , namely,

Φ′ = AΦ, Φ(0) = I





Then, by uniqueness of an IVP (extended to matrix differentialequations), we conclude that eAt and the fundamental matrix Φ(t)are the same. Thus we can write the solution of the initial valueproblem

x = Ax, x(0) = x0

in the form

x = eAtx0





Diagonalizable Matrices.

The basic reason why a system of linear (algebraic or differential)equations presents some difficulty is that the equations are usuallycoupled.

Hence the equations in the system must be solved simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved independently of all the others.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.





Let’s assume that the matrix A has n eigenvectors x(1), x(2), ...,x(n) linearly indepedent, then

Ax(1) = λ1x(1); Ax(2) = λ2x(2); ...Ax(n) = λnx(n)

and considering the matrix

T =

x(1)1 · · · x(n)

......

x(1)n · · · x

(n)n

we have

AT =

Ax(1) · · · Ax

(n)1

......

... · · ·...

=

λ1x

(1)1 · · · λnx

(n)1

λ1x(1)2

...

λ1x(1)n λnx

(n)n

= TD





where D is the diagonal matrix

D =

λ1

λ2. . .

λn

whose diagonal elements are the eigenvalues of A. From the lastequations we have that it follows that

T−1AT = D

Thus, if the eigenvalues and eigenvectors of A are known, A canbe transformed into a diagonal matrix by the process shown in theabove equation.




eAt Matrix and DiagonalizationThis process is known as a similarity transformation.Alternatively, we may say that A is diagonalizable.

If A is Hermitian, then the determination of T−1 is very simple.The eigenvectors v(1), ..., v(n) of A are known to be mutuallyorthogonal, so let us choose them so that they are also normalizedby < v(i), v(i) >= 1 for each i . It is easy verify that T−1 = T∗. Inother words, the inverse of T is the same as its adjoint (thetranspose of its complex conjugate).

Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix T such that T−1AT = D. Inthis case, A is not diagonalizable.


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