![Note: The matrix of coefficients is the fundamental matrix evaluated at t = 0
Using the solution method of your choice (row reduction, inverse, Cramer’s rule),
the solution is: c1 = 3, c2 = −1, c3 = 1. The solution of the initial-value problem is
x = 3e2t
1
−1
2
− et
3
−1
7
+ e−t
1
2
2
.
Two Difficulties
There are two difficulties that can arise:
1. A has complex eigenvalues.
If λ = a + bi is a complex eigenvalue with corresponding (complex) eigenvector
u + i x, then λ = a − bi (the complex conjugate of λ) is also an eigenvalue of A
and u − i x is a corresponding eigenvector. The corresponding linearly independent
complex solutions of x0
= Ax are:
w1 = e(a+bi)t
(u + i x) = eat
(cos bt + i sin bt)(u + i x)
= eat
[(cos bt u − sin bt x) + i(cos bt x + sin bt u)]
w2 = e(a−bi)t
(u − i x) = eat
(cos bt − i sin bt)(u − i x)
= eat
[(cos bt u − sin bt x) − i(cos bt x + sin bt u)]
Now
x1(t) = 1
2
[w1(t) + w2(t)] = eat
(cos bt u − sin bt x)
and
x2(t) = 1
2i
[w1(t) − w2(t)] = eat
(cos bt x + sin bt u)
are linearly independent solutions of the system, and they are real-valued vector
functions. It is worth noting that x1 and x2 are simply the real and imaginary
parts of w1 (or of w2).
Example 4. Determine a fundamental set of solution vectors of
x0
=
1 −4 −1
3 2 3
1 1 3
x.
102](https://image.slidesharecdn.com/systemsoflineardifferential-240919002245-2fcf38a1/85/Perdif-Systems-of-Linear-Differential-pdf-16-320.jpg)
![Example 5. Let A =
1 −3 3
3 −5 3
6 −6 4
.
det(A − λI) =
1 − λ −3 3
3 −5 − λ 3
6 −6 4 − λ
= −λ3
+ 12λ − 16 = −(λ − 4)(λ + 2)2
.
The eigenvalues are: λ1 = 4, λ2 = λ3 = −2.
As you can check, an eigenvector corresponding to λ1 = 4 is v1 =
1
1
2
.
We’ll carry out the details involved in finding an eigenvector corresponding to the
“double” eigenvalue −2.
[A − (−2)I]v =
3 −3 3
3 −3 3
6 −6 6
v1
v2
v3
=
0
0
0
.
The augmented matrix for this system of equations is
3 −3 3 0
3 −3 3 0
6 −6 6 0
which row reduces to
1 −1 1 0
0 0 0 0
0 0 0 0
The solutions of this system are: v1 = v2 −v3, v2, v3 arbitrary. We can assign values
to v2 and v3 independently and obtain two linearly independent eigenvectors. For
example, setting v2 = 1, v3 = 0, we get the eigenvector v2 =
1
1
0
. Reversing
the roles, we set v2 = 0, v3 = −1 to get the eigenvector v3 =
1
0
−1
. Clearly
v2 and v3 are linearly independent. You should understand that there is nothing
magic about our two choices for v2, v3; any choice which produces two independent
vectors will do.
The important thing to note here is that this eigenvalue of multiplicity 2 produced
two independent eigenvectors.
104](https://image.slidesharecdn.com/systemsoflineardifferential-240919002245-2fcf38a1/85/Perdif-Systems-of-Linear-Differential-pdf-18-320.jpg)
![Based on our work above, a fundamental set of solutions for the differential system
x0
=
1 −3 3
3 −5 3
6 −6 4
x
is
x1 = e4t
1
1
2
, x2 = e−2t
1
1
0
, x3 = e−2t
1
0
−1
.
Example 6. Let A =
0 1 0
0 0 1
12 8 −1
det(A − λI) =
−λ 1 0
0 −λ 1
12 8 −1 − λ
= −λ3
− λ2
+ 8λ − 12 = −(λ − 3)(λ + 2)2
.
The eigenvalues are: λ1 = 3, λ2 = λ3 = −2.
As you can check, an eigenvector corresponding to λ1 = 3 is v1 =
1
3
9
.
We’ll carry out the details involved in finding an eigenvector corresponding to the
“double” eigenvalue −2.
[A − (−2)I]v =
2 1 0
0 2 1
12 8 1
v1
v2
v3
=
0
0
0
.
The augmented matrix for this system of equations is
2 1 0 0
0 2 1 0
12 8 1 0
which row reduces to
2 1 0 0
0 2 1 0
0 0 0 0
The solutions of this system are v1 = 1
4
v3, v2 = −1
2
v3, v3 arbitrary. Here there is
only one parameter and so we’ll get only one eigenvector. Setting v3 = 4 we get the
eigenvector v2 =
1
−2
4
.
In contrast to the preceding example, the “double” eigenvalue here has only one
(independent) eigenvector.
105](https://image.slidesharecdn.com/systemsoflineardifferential-240919002245-2fcf38a1/85/Perdif-Systems-of-Linear-Differential-pdf-19-320.jpg)
![Let’s look at [A − (−2)I]w = [A + 2I]w:
[A + 2I]w =
2 1 0
0 2 1
12 8 1
0
1
−4
=
1
−2
4
= v2;
A − (−2)I “maps” w onto the eigenvector v2. The corresponding solution of the
system has the form
x3 = e−2t
w + te−2t
v2
where v2 is the eigenvector corresponding to −2 and w satisfies
[A − (−2)I]w = v2.
General Result
Given the linear differential system x0
= Ax. Suppose that A has an eigenvalue λ
of multiplicity 2. Then exactly one of the following holds:
1. λ has two linearly independent eigenvectors, v1 and v2. Corresponding
linearly independent solution vectors of the differential system are x1(t) = eλt
v1
and x2(t) = eλt
v2.
2. λ has only one (independent) eigenvector v. Then a linearly independent pair
of solution vectors corresponding to λ are:
x1(t) = eλt
v and x2(t) = eλt
w + teλt
v
where w is a vector that satisfies (A − λI)w = v. The vector w is called a
generalized eigenvector corresponding to the eigenvalue λ.
Example 7. Find a fundamental set of solution vectors of x0
=
1 −1
1 3
!
x.
SOLUTION
det(A − λI) =
1 − λ −1
1 3 − λ
= λ2
− 4λ + 4 = (λ − 2)2
.
Characteristic values: λ1 = λ2 = 2.
Characteristic vectors:
(A − 2I)v =
−1 −1
1 1
!
v1
v2
!
=
0
0
!
;
107](https://image.slidesharecdn.com/systemsoflineardifferential-240919002245-2fcf38a1/85/Perdif-Systems-of-Linear-Differential-pdf-21-320.jpg)
Perdif Systems of Linear Differential.pdf
- 1. Chapter 4 Systems of Linear Differential Equations Introduction to Systems Up to this point the entries in a vector or matrix have been real numbers. In this section, and in the following sections, we will be dealing with vectors and matri- ces whose entries are functions. A vector whose components are functions is called a vector-valued function or vector function. Similarly, a matrix whose entries are functions is called a matrix function. The operations of vector and matrix addition, multiplication by a number and matrix multiplication for vector and matrix functions are exactly as defined in Chap- ter 5 so there is nothing new in terms of arithmetic. However, there are operations on functions other than arithmetic operations, e.g., limits, differentiation, and integra- tion, that we have to define for vector and matrix functions. These operation from calculus are defined in a natural way. Let v(t) = (f1(t), f2(t), . . . , fn(t)) be a vector function whose compo- nents are defined on an interval I. Limit: Let c ∈ I. If lim x→c fi(t) = αi exists for i = 1, 2, . . . n, then lim t→c v(t) = lim t→c f1(t), lim t→c f2(t), . . . , lim t→c fn(t) = (α1, α2, . . . , αn) . Limits of vector functions are calculated “component-wise.” Derivative: If f1, f2, . . ., fn are differentiable on I, then v is 87
- 2. differentiable on I, and v0 = ((f0 1(t), f0 2(t), . . . , f0 n(t)). That is, v0 is the vector function whose components are the derivatives of the components of v. Integration: Since differentiation of vector functions is done component- wise, integration must also be component-wise. That is Z v(t) dt = Z f1(t) dt, Z f2(t) dt, . . . , Z fn(t) dt . Limits, differentiation and integration of matrix functions is done in ex- actly the same way, component-wise. 4.1. Systems of Linear Differential Equations Consider the third-order linear differential equation y000 + p(t)y00 + q(t)y0 + r(t)y = f(t) where p, q, r, f are continuous functions on some interval I. Solving the equation for y000 , we get y000 = −r(t)y − q(t)y0 − p(t)y00 + f(t). Introduce new dependent variables x1, x2, x3, as follows: x1 = y x2 = x0 1 (= y0 ) x3 = x0 2 (= y00 ) Then y000 = x0 3 = −r(t)x1 − q(t)x2 − p(t)x3 + f(t) and the third-order equation can be written equivalently as a system of three first- order equations: x0 1 = x2 x0 2 = x3 x0 3 = −r(t)x1 − q(t)x2 − p(t)x3 + f(t) 88
- 3. Example 1. (a) Consider the third-order nonhomgeneous equation y000 − y00 − 8y0 + 12y = 2et . Solving the equation for y000 , we have y000 = −12y + 8y0 + y00 + 2et . Let x1 = y, x0 1 = x2 (= y0 ), x0 2 = x3 (= y00 ). Then y000 = x0 3 = −12x1 + 8x2 + x3 + 2et and the equation converts to the equivalent system: x0 1 = x2 x0 2 = x3 x0 3 = −12x1 + 8x2 + x3 + 2et Note: This system is just a very special case of the “general” system of three, first-order differential equations: x0 1 = a11(t)x1 + a12(t)x2 + a13(t)x3(t) + b1(t) x0 2 = a21(t)x1 + a22(t)x2 + a23(t)x3(t) + b2(t) x0 3 = a31(t)x1 + a32(t)x2 + a33(t)x3(t) + b3(t) (b) Consider the second-order homogeneous equation t2 y00 − ty0 − 3y = 0. Solving this equation for y00 , we get y00 = 3 t2 y + 1 t y0 . To convert this equation to an equivalent system, we let x1 = y, x0 1 = x2 (= y0 ). Then we have x0 1 = x2 x0 2 = 3 t2 x1 + 1 t x2 which is just a special case of the general system of two first-order differential equa- tions: x0 1 = a11(t)x1 + a12(t)x2 + b1(t) x0 2 = a21(t)x1 + a22(t)x2 + b2(t) 89
- 4. General Theory Let a11(t), a12(t), . . ., a1n(t), a21(t), . . ., ann(t), b1(t), b2(t), . . . , bn(t) be contin- uous functions on some interval I. The system of n first-order differential equations x0 1 = a11(t)x1 + a12(t)x2 + · · · + a1n(t)xn(t) + b1(t) x0 2 = a21(t)x1 + a22(t)x2 + · · · + a2n(t)xn(t) + b2(t) . . . . . . x0 n = an1(t)x1 + an2(t)x2 + · · · + ann(t)xn(t) + bn(t) (S) is called a first-order linear differential system. The system (S) is homogeneous if b1(t) ≡ b2(t) ≡ · · · ≡ bn(t) ≡ 0 on I. (S) is nonhomogeneous if the functions bi(t) are not all identically zero on I; that is, if there is at least one point a ∈ I and at least one function bi(t) such that bi(a) 6= 0. Let A(t) be the n × n matrix A(t) = a11(t) a12(t) · · · a1n(t) a21(t) a22(t) · · · a2n(t) . . . . . . . . . an1(t) an2(t) · · · ann(t) and let x and b be the vectors x = x1 x2 . . . xn , b = b1 b2 . . . bn . Then (S) can be written in the vector-matrix form x0 = A(t) x + b. (S) The matrix A(t) is called the matrix of coefficients or the coefficient matrix. Example 2. The vector-matrix form of the system in Example 1(a) is: x0 = 0 1 0 0 0 1 −12 8 1 x + 0 0 2et , 90
- 5. a nonhomogeneous system. The vector-matrix form of the system in Example 1(b) is: x0 = 0 1 3/t2 1/t ! x + 0 0 ! = 0 1 3/t2 1/t ! x, where x = x1 x2 ! , a homogeneous system. A solution of the linear differential system (S) is a differentiable vector function x(t) = x1(t) x2(t) . . . xn(t) that satisfies (S) on the interval I. Example 3. Verify that x(t) = e2t 2e2t 4e2t + 1 2 et 1 2 et 1 2 et is a solution of the nonho- mogeneous system x0 = 0 1 0 0 0 1 −12 8 1 x + 0 0 2et of Example 2. 91
- 6. SOLUTION x0 = e2t 2e2t 4e2t + 1 2 et 1 2 et 1 2 et 0 = 2e2t 4e2t 8e2t + 1 2 et 1 2 et 1 2 et ? = 0 1 0 0 0 1 −12 8 1 e2t 2e2t 4e2t + 1 2 et 1 2 et 1 2 et + 0 0 2et ? = 0 1 0 0 0 1 −12 8 1 e2t 2e2t 4e2t + 0 1 0 0 0 1 −12 8 1 1 2 et 1 2 et 1 2 et + 0 0 2et = 2e2t 4e2t 8e2t + 1 2 et 1 2 et −3 2 et + 0 0 2et = 2e2t 4e2t 8e2t + 1 2 et 1 2 et 1 2 et . x is a solution. THEOREM 1. (Existence and Uniqueness Theorem) Let a be any point on the interval I, and let α1, α2, . . . , αn be any n real numbers. Then the initial-value problem x0 = A(t) x + b(t), x(a) = α1 α2 . . . αn has a unique solution. Exercises 4.1 Convert the differential equation into a system of first-order equations. 1. y00 − ty0 + 3y = sin 2t. 92
- 7. 2. y00 + y = 2e−2t . 3. y000 − y00 + y = et . 4. my00 + cy0 + ky = cos λt, m, c, k, λ are constants. Write the system in vector-matrix form. 5. x0 1 = −2x1 + x2 + sin t x0 2 = x1 − 3x2 − 2 cos t 6. x0 1 = et x1 − e2t x2 x0 2 = e−t x1 − 3et x2 7. x0 1 = 2x1 + x2 + 3x3 + 3e2t x0 2 = x1 − 3x2 − 2 cos t x0 3 = 2x1 − x2 + 4x3 + t 8. x0 1 = t2 x1 + x2 − tx3 + 3 x0 2 = −3et x2 + 2x3 − 2e−2t x0 3 = 2x1 + t2 x2 + 4x3 9. Verify that u(t) = t−1 −t−2 ! is a solution of the system in Example 1 (b). 10. Verify that u(t) = e−3t −3e−3t 9e−3t + 1 2 et 1 2 et 1 2 et is a solution of the system in Example 1 (a). 11. Verify that w(t) = te2t e2t + 2te2t 4e2t + 4te2t is a solution of the homogeneous system associated with the system in Example 1 (a). 93
- 8. 12. Verify that x(t) = − sin t − cos t − 2 sin t ! is a solution of the system x0 = −2 1 −3 2 ! x + 0 2 sin t ! . 13. Verify that x(t) = −2e−2t 0 3e−2t is a solution of the system x0 = 1 −3 2 0 −1 0 0 −1 −2 x. 4.2. Homogeneous Systems In this section we give the basic theory for linear homogeneous systems. This “theory” is simply a repetition results given in Sections 3.2 and 3.6, phrased this time in terms of the system x0 1 = a11(t)x1 + a12(t)x2 + · · · + a1n(t)xn(t) x0 2 = a21(t)x1 + a22(t)x2 + · · · + a2n(t)xn(t) . . . . . . x0 n = an1(t)x1 + an2(t)x2 + · · · + ann(t)xn(t) (H) or x0 = A(t)x. (H) Note first that the zero vector z(t) ≡ 0 = 0 0 . . . 0 is a solution of (H). As before, this solution is called the trivial solution. Of course, we are interested in finding nontrivial solutions. THEOREM 1. If x1, x2, . . ., xk are solutions of (H), and if c1, c2, . . . , ck are real numbers, then c1x1 + c2x2 + · · · + ckxk is a solution of (H); any linear combination of solutions of (H) is also a solution of (H). 94
- 9. DEFINITION 1. Let x1 = x11(t) x21(t) . . . xn1(t) , x2 = x12(t) x22(t) . . . xn2(t) , . . ., xk = x1k(t) x2k(t) . . . xnk(t) be n-component vector functions defined on some interval I. The vectors are linearly dependent on I if there exist k real numbers c1, c2, . . . , ck, not all zero, such that c1x1(t) + c2x2(t) + · · · + ckxk(t) ≡ 0 on I. Otherwise the vectors are linearly independent on I. THEOREM 2. Let x1, x2, . . . , xn be n, n-component vector functions defined on an interval I. If the vectors are linearly dependent, then W(t) = x11 x12 · · · x1n x21 x22 · · · x2n . . . . . . . . . . . . xn1 xn2 · · · xnn ≡ 0 on I. The determinant in Theorem 4 is called the Wronskian of the vector functions x1, x2, . . ., xn. COROLLARY Let x1, x2, . . . , xn be n, n-component vector functions defined on an interval I. If the Wronskian W(t) 6= 0 for at least one t ∈ I, then the vectors are linearly independent on I. Example 1. The vector functions u = t3 3t2 ! and x = t−1 −t−2 ! are solutions of the homogeneous system in Example 1(b), Section 4.1. Their Wron- skian is: W(t) = t3 t−1 3t2 −t−2 = −4t. The solutions are linearly independent. The vector functions x1 = e2t 2e2t 4e2t , x2 = e−3t −3e−3t 9e−3t , x3 = te2t e2t + 2te2t 4e2t + 4te2t 95
- 10. are solutions of the homogeneous system x0 = 0 1 0 0 0 1 −12 8 1 x. Their Wronskian is: W(t) = e2t e−3t te2t 2e2t −3e−3t e2t + 2te2t 4e2t 9e−3t 4e2t + 4te2t = −25et . These solutions are linearly independent. THEOREM 3. Let x1, x2, . . ., xn be n solutions of (H). Exactly one of the following holds: 1. W(x1, x2, . . ., xn)(t) ≡ 0 on I and the solutions are linearly dependent. 2. W(x1, x2, . . ., xn)(t) 6= 0 for all t ∈ I and the solutions are linearly independent. It is easy to construct sets of n linearly independent solutions of (H). Simply pick any point a ∈ I and any nonsingular n × n matrix A. Let α1 be the first column of A, α2 the second column of A, and so on. Then let x1 be the solution of (H) such that x1(a) = α1, let x2 be the solution of (H) such that x2(a) = α2, . . ., and let xn be the solution of (H) such that xn = αn. The existence and uniqueness theorem guarantees the existence of these solutions. Now W(x1, x2, . . ., xn)(a) = det A 6= 0. Therefore, W(t) 6= 0 for all t ∈ I and the solutions are linearly independent. A particularly nice set of n linearly independent solutions is obtained by choosing A = In, the identity matrix. THEOREM 4. Let x1, x2, . . . , xn be n linearly independent solutions of (H). Let u be any solution of (H). Then there exists a unique set of constants c1, c2, . . . , cn such that u = c1x1 + c2x2 + · · · + cnxn. That is, every solution of (H) can be written as a unique linear combination of x1, x2, . . ., xn. 96
- 11. DEFINITION 2. A set {x1, x2, . . ., xn} of n linearly independent solutions of (H) is called a fundamental set of solutions. A fundamental set of solutions is also called a solutions basis for (H). If x1, x2, . . . , xn is a fundamental set of solutions of (H), then the n × n matrix X(t) = x11(t) x12(t) · · · x1n(t) x21(t) x22(t) · · · x2n(t) . . . . . . . . . xn1(t) xn2(t) · · · xnn(t) (the vectors x1, x2, . . . , xn are the columns of X) is called a fundamental matrix for (H). DEFINITION 3. Let x1, x2, . . . , xn be a fundamental set of solutions of (H). Then x = c1x1 + c2x2 + · · · + cnxn, where c1, c2, . . . , cn are arbitrary constants, is the general solution of (H). Exercises 4.2 Determine whether or not the vector functions are linearly dependent. 1. x1 = 2t − 1 −t ! , x2 = −t + 1 2t ! 2. x1 = cos t sin t ! , x2 = sin t cos t ! 3. x1 = 2 − t t −2 , x2 = t −1 2 , x3 = 2 + t t − 2 2 . 4. x1 = et −et et , x2 = −et 2et −et , x3 = 0 et 0 . 5. x1 = et 0 ! , x2 = 0 0 ! , x3 = 0 et ! 97
- 12. 6. Given the linear differential system x0 = 5 −3 2 0 ! x. Let x1 = e2t e2t ! and x2 = 3e3t 2e3t ! . (a) Show that x1, x2 are a fundamental set of solutions of the system. (b) Let X be the corresponding fundamental matrix. Show that X0 = AX. (c) Give the general solution of the system. (d) Find the solution of the system that satisfies x(0) = 1 0 ! . 7. Let X be the matrix function X(t) = 0 4te−t e−t 1 e−t 0 1 0 0 (a) Verify that X is a fundamental matrix for the system x0 = −1 4 −4 0 −1 1 0 0 0 x. (b) Find the solution of the system that satisfies x(0) = 0 1 2 . 4.3. Homogeneous Systems with Constant Coefficients A homogeneous system with constant coefficients is a linear differential system having the form x0 1 = a11x1 + a12x2 + · · · + a1nxn x0 2 = a21x1 + a22x2 + · · · + a2nxn . . . . . . x0 n = an1x1 + an2x2 + · · · + annxn 98
- 13. where a11, a12, . . . , ann are constants. The system in vector-matrix form is x0 1 x0 2 − x0 n = a11 a12 · · · a1n a21 a22 · · · a2n − − − − an1 an2 · · · ann x1 x2 − xn or − → x0 = A− → x . Example 1. Consider the 3rd order linear homogeneous differential equation y000 − y00 − 8y0 + 12y = 0. The characteristic equation is: r3 − r2 − 8r + 12 = (r − 2)2 (r + 3) = 0 and {e2t , te2t , e−3t } is a solution basis for the equation. The corresponding linear homogeneous system is x0 = 0 1 0 0 0 1 −12 8 1 x and x1(t) = e2t 2e2t 4e2t = e2t 1 2 4 is a solution vector. Similarly, x2(t) = e−3t 1 3 9 is a solution vector. The example suggests that homogeneous systems with constant coefficients might have solution vectors of the form x(t) = eλt v, for some number λ and some constant vector v. If x(t) = eλt v is a solution vector of (H), then x0 = Ax which implies λeλt v = Aeλt v and so Av = λ v. The latter equation is an eigenvalue-eigenvector equation for A. Thus, we look for solutions of the form x(t) = eλt x where λ is an eigenvalue of A and c is a corresponding eigenvector. 99
- 14. Example 2. Find a fundamental set of solution vectors of x0 = 1 5 3 3 ! x and give the general solution of the system. SOLUTION First we find the eigenvalues: det(A − λI) = 1 − λ 5 3 3 − λ = (λ − 6)(λ + 2). The eigenvalues are λ1 = 6 and λ2 = −2. Next, we find corresponding eigenvectors. For λ1 = 6 we have: (A−6I)x = −5 5 3 −3 ! x1 x2 ! = 0 0 ! which implies x1 = x2, x2 arbitrary. Setting x2 = 1, we get the eigenvector 1 1 ! . Repeating the process for λ2 = −2, we get the eigenvector 5 −3 ! . Thus x1 = e6t 1 1 ! and x2 = e−2t 5 −3 ! are solution vectors of the system. The Wronskian of x1 and x2 is: W(t) = e6t 5e−2t e6t −3e−2t = −8e4t 6= 0. Thus x1 and x2 are linearly independent; they form a fundamental set of solutions. The general solution of the system is x(t) = c1x1 + c2x2 = c1e6t 1 1 ! + c2e−2t 5 −3 ! . Example 3. Find a fundamental set of solution vectors of x0 = 3 −1 −1 −12 0 5 4 −2 −1 x 100
- 15. and find the solution that satisfies the initial condition x(0) = 1 0 1 . SOLUTION det(A − λI) = 3 − λ −1 −1 −12 −λ 5 4 −2 −1 − λ = −λ3 + 2λ2 + λ − 2. Now det(A − λI) = 0 implies λ3 − 2λ2 − λ + 2 = (λ − 2)(λ − 1)(λ + 1) = 0. The eigenvalues are λ1 = 2, λ2 = 1, λ3 = −1. As you can check, corresponding eigenvectors are: v1 = 1 −1 2 , v2 = 3 −1 7 , v3 = 1 2 2 . A fundamental set of solution vectors is: x1 = e2t 1 −1 2 , x2 = et 3 −1 7 , x3 = e−t 1 2 2 . since distinct exponential vector-functions are linearly independent (calculate the Wronskian to verify.) To find the solution vector satisfying the initial condition, solve c1x1(0) + c2x2(0) + c3x3(0) = 1 0 1 which is: c1 1 −1 2 + c2 3 −1 7 + c3 1 2 2 = 1 0 1 or 1 3 1 −1 −1 2 2 7 2 c1 c2 c3 = 1 0 1 . 101
- 16. Note: The matrix of coefficients is the fundamental matrix evaluated at t = 0 Using the solution method of your choice (row reduction, inverse, Cramer’s rule), the solution is: c1 = 3, c2 = −1, c3 = 1. The solution of the initial-value problem is x = 3e2t 1 −1 2 − et 3 −1 7 + e−t 1 2 2 . Two Difficulties There are two difficulties that can arise: 1. A has complex eigenvalues. If λ = a + bi is a complex eigenvalue with corresponding (complex) eigenvector u + i x, then λ = a − bi (the complex conjugate of λ) is also an eigenvalue of A and u − i x is a corresponding eigenvector. The corresponding linearly independent complex solutions of x0 = Ax are: w1 = e(a+bi)t (u + i x) = eat (cos bt + i sin bt)(u + i x) = eat [(cos bt u − sin bt x) + i(cos bt x + sin bt u)] w2 = e(a−bi)t (u − i x) = eat (cos bt − i sin bt)(u − i x) = eat [(cos bt u − sin bt x) − i(cos bt x + sin bt u)] Now x1(t) = 1 2 [w1(t) + w2(t)] = eat (cos bt u − sin bt x) and x2(t) = 1 2i [w1(t) − w2(t)] = eat (cos bt x + sin bt u) are linearly independent solutions of the system, and they are real-valued vector functions. It is worth noting that x1 and x2 are simply the real and imaginary parts of w1 (or of w2). Example 4. Determine a fundamental set of solution vectors of x0 = 1 −4 −1 3 2 3 1 1 3 x. 102
- 17. SOLUTION det(A−λI) = 1 − λ −4 −1 3 2 − λ 3 1 1 3 − λ = −λ3 +6λ2 −21λ+26 = −(λ−2)(λ2 −4λ+13). The eigenvalues are: λ1 = 2, λ2 = 2+3i, λ3 = 2−3i. The corresponding eigenvectors are: v1 = 1 0 −1 , v2 = −5 + 3i 3 + 3i 2 = −5 3 2 + i 3 3 0 v3 = −5 − 3i 3 − 3i 2 = −5 3 2 − i 3 3 0 . Now e(2+3i)t −5 3 2 + i 3 3 0 = e2t (cos 3t + i sin 3t) −5 3 2 + i 3 3 0 = e2t cos 3t −5 3 2 − sin 3t 3 3 0 + i e2t cos 3t 3 3 0 + sin 3t −5 3 2 . A fundamental set of solution vectors for the system is: x1 = e2t 1 0 −1 , x2 = e2t cos 3t −5 3 2 − sin 3t 3 3 0 , x3 = e2t cos 3t 3 3 0 + sin 3t −5 3 2 . 2. A has an eigenvalue of multiplicity greater than 1 We’ll look first at the case where A has an eigenvalue of multiplicity 2. 103
- 18. Example 5. Let A = 1 −3 3 3 −5 3 6 −6 4 . det(A − λI) = 1 − λ −3 3 3 −5 − λ 3 6 −6 4 − λ = −λ3 + 12λ − 16 = −(λ − 4)(λ + 2)2 . The eigenvalues are: λ1 = 4, λ2 = λ3 = −2. As you can check, an eigenvector corresponding to λ1 = 4 is v1 = 1 1 2 . We’ll carry out the details involved in finding an eigenvector corresponding to the “double” eigenvalue −2. [A − (−2)I]v = 3 −3 3 3 −3 3 6 −6 6 v1 v2 v3 = 0 0 0 . The augmented matrix for this system of equations is 3 −3 3 0 3 −3 3 0 6 −6 6 0 which row reduces to 1 −1 1 0 0 0 0 0 0 0 0 0 The solutions of this system are: v1 = v2 −v3, v2, v3 arbitrary. We can assign values to v2 and v3 independently and obtain two linearly independent eigenvectors. For example, setting v2 = 1, v3 = 0, we get the eigenvector v2 = 1 1 0 . Reversing the roles, we set v2 = 0, v3 = −1 to get the eigenvector v3 = 1 0 −1 . Clearly v2 and v3 are linearly independent. You should understand that there is nothing magic about our two choices for v2, v3; any choice which produces two independent vectors will do. The important thing to note here is that this eigenvalue of multiplicity 2 produced two independent eigenvectors. 104
- 19. Based on our work above, a fundamental set of solutions for the differential system x0 = 1 −3 3 3 −5 3 6 −6 4 x is x1 = e4t 1 1 2 , x2 = e−2t 1 1 0 , x3 = e−2t 1 0 −1 . Example 6. Let A = 0 1 0 0 0 1 12 8 −1 det(A − λI) = −λ 1 0 0 −λ 1 12 8 −1 − λ = −λ3 − λ2 + 8λ − 12 = −(λ − 3)(λ + 2)2 . The eigenvalues are: λ1 = 3, λ2 = λ3 = −2. As you can check, an eigenvector corresponding to λ1 = 3 is v1 = 1 3 9 . We’ll carry out the details involved in finding an eigenvector corresponding to the “double” eigenvalue −2. [A − (−2)I]v = 2 1 0 0 2 1 12 8 1 v1 v2 v3 = 0 0 0 . The augmented matrix for this system of equations is 2 1 0 0 0 2 1 0 12 8 1 0 which row reduces to 2 1 0 0 0 2 1 0 0 0 0 0 The solutions of this system are v1 = 1 4 v3, v2 = −1 2 v3, v3 arbitrary. Here there is only one parameter and so we’ll get only one eigenvector. Setting v3 = 4 we get the eigenvector v2 = 1 −2 4 . In contrast to the preceding example, the “double” eigenvalue here has only one (independent) eigenvector. 105
- 20. Suppose that we were asked to find a fundamental set of solutions of the linear differential system x0 = 0 1 0 0 0 1 12 8 −1 x. By our work above, we have two independent solutions x1 = e3t 1 3 9 and x2 = e−2t 1 −2 4 . We need a third solution which is independent of these two. Our system has a special form; it is equivalent to the third order equation y000 + y00 − 8y0 − 12y = 0. The characteristic equation is r3 + r2 − 8r − 12 = (r − 3)(r + 2)2 = 0 (compare with det(A−λI).) The roots are: r1 = 3, r2 = r3 = −2 and a fundamental set of solutions is {y1 = e3t , y2 = e−2t , y3 = te−2t }. The correspondence between these solutions and the solution vectors we found above should be clear: e3t −→ e3t 1 3 9 , e−2t −→ e−2t 1 −2 4 . The solution y3 = te−2t of the equation corresponds to the solution vector x3 = y3 y0 3 y00 3 = te−2t e−2t − 2te−2t −4e−2t − 4te−2t = e−2t 0 1 −4 + te−2t 1 −2 4 . The appearance of the te−2t v2 term should not be unexpected since we know that a characteristic root of multiplicity 2 produces a solution of the form tert . You can check that x3 is independent of x1 and x2. Therefore, the solution vectors x1, x2, x3 are a fundamental set of solutions of the system. The question is: What is the significance of the vector w = 0 1 −4 ? How is it related to the eigenvalue −2 which generated it, and to the corresponding eigenvector? 106
- 21. Let’s look at [A − (−2)I]w = [A + 2I]w: [A + 2I]w = 2 1 0 0 2 1 12 8 1 0 1 −4 = 1 −2 4 = v2; A − (−2)I “maps” w onto the eigenvector v2. The corresponding solution of the system has the form x3 = e−2t w + te−2t v2 where v2 is the eigenvector corresponding to −2 and w satisfies [A − (−2)I]w = v2. General Result Given the linear differential system x0 = Ax. Suppose that A has an eigenvalue λ of multiplicity 2. Then exactly one of the following holds: 1. λ has two linearly independent eigenvectors, v1 and v2. Corresponding linearly independent solution vectors of the differential system are x1(t) = eλt v1 and x2(t) = eλt v2. 2. λ has only one (independent) eigenvector v. Then a linearly independent pair of solution vectors corresponding to λ are: x1(t) = eλt v and x2(t) = eλt w + teλt v where w is a vector that satisfies (A − λI)w = v. The vector w is called a generalized eigenvector corresponding to the eigenvalue λ. Example 7. Find a fundamental set of solution vectors of x0 = 1 −1 1 3 ! x. SOLUTION det(A − λI) = 1 − λ −1 1 3 − λ = λ2 − 4λ + 4 = (λ − 2)2 . Characteristic values: λ1 = λ2 = 2. Characteristic vectors: (A − 2I)v = −1 −1 1 1 ! v1 v2 ! = 0 0 ! ; 107
- 22. −1 −1 0 1 1 0 ! −→ 1 1 0 0 0 0 ! . The solutions are: v1 = −v2, v2 arbitrary; there is only one eigenvector. Setting v2 = −1, we get v = 1 −1 ! . The vector x1 = e2t 1 −1 ! is a solution of the system. A second solution, independent of x1 is x2 = e2t w+te2t v where w is a solution of (A − 2I)z = v: (A − 2I)z = −1 −1 1 1 ! z1 z2 ! = 1 −1 ! ; −1 −1 1 1 1 1 ! −→ 1 1 −1 0 0 0 ! . The solutions of this system are z1 = −1 − z2, z2 arbitrary. If we choose z2 = 0 (any choice for z2 will do), we get z1 = −1 and w = −1 0 ! . Thus x2(t) = e2t −1 0 ! + te2t 1 −1 ! is a solution of the system independent of x1. The solutions x1(t) = e2t 1 −1 ! , x2(t) = e2t −1 0 ! + te2t 1 −1 ! are a fundamental set of solutions of the system. Eigenvalues of Multiplicity 3. Given the differential system x0 = Ax. Suppose that λ is an eigenvalue of A of multiplicity 3. Then exactly one of the following holds: 1. λ has three linearly independent eigenvectors v1, v2, v3. Then three linearly independent solution vectors of the system corresponding to λ are: x1(t) = eλt v1, x2(t) = eλt v2, x3(t) = eλt v3. 108
- 23. 2. λ has two linearly independent eigenvectors v1, v2. Then two linearly inde- pendent solutions of the system corresponding to λ are: x1(t) = eλt v1, x2(t) = eλt v2 A third solution, independent of x1 and x2 has the form x3(t) = eλt w + teλt x where x is an eigenvector corresponding to λ and (A − λI)w = x. 3. λ has only one (independent) eigenvector v. Then three linearly independent solutions of the system have the form: x1 = eλt v, x2 = eλt w + teλt v, x3(t) = eλt z + teλt w + t2 eλt v where (A − λI)w = v and (A − λI)z = w. Exercises 4.3 Find the general solution of the system x0 = Ax where A is the given matrix. If an initial condition is given, also find the solution that satisfies the condition. 1. −2 4 1 1 ! . 2. −1 1 4 2 ! , x(0) = −1 1 ! . 3. −2 2 1 0 −1 0 2 −2 −1 . Hint: −3 is an eigenvalue. 4. 3 0 −1 −2 2 1 8 0 −3 , x(0) = −1 2 −8 . Hint: 2 is an eigenvalue. 5. 1 −2 2 1 ! . 109
- 24. 6. −1 2 −1 −3 ! . 7. 3 2 −8 −5 ! . 8. −3 0 −3 1 −2 3 1 0 1 . Hint: −2 is an eigenvalue. 9. 2 −1 −1 −1 2 −1 1 1 4 . Hint: 3 is an eigenvalue 10. −2 1 −1 3 −3 4 3 −1 2 . Hint: 1 is an eigenvalue. 11. −3 1 −1 −7 5 −1 −6 6 −2 . 110