1. Systems of Linear Differential Equations
CHAPTER 10
Systems of Linear Differential Equations

2. Contents 10.1 Preliminary Theory 10.2 Homogeneous Linear Systems
10.3 Solution by Diagonalization
10.4 Nonhomogeneous Linear Systems
10.5 Matrix Exponential

3. 10.1 Preliminary Theory
Introduction Recall that in Sec 3.11, we have the following system of linear DEs: (1)

4. We have also the normal form (2)

5. Linear Systems
When (2) is linear, we have the normal form as (3)

6. Matrix Form of a Linear Systems
If we let then (3) becomes (4) If it is homogeneous, (5)

7. Example 1
(a) If , the matrix form of is

8. (b) If , the matrix form of is

9. A solution vector on an interval I is any column
DEFINITION 10.1
A solution vector on an interval I is any column
matrix whose entries are differentiable functions satisfying (4)
on the interval.
Solution Vector

10. Example 2
Verify that on (−, ) are solutions of (6)

11. Example 2 (2)
Solution From we have

12. Initial Value Problem (IVP)
Let Then the problem Solve: Subject to: X(t0) = X0 (7) is an IVP.

13. Let the entries of A(t) and F(t) be functions continuous
on a common interval I that contains t0, Then there
exists a unique solution of (7) on I.
THEOREM 10.1
Existence of a Unique Solution
Let X1, X2,…, Xk be a set of solution of the
homogeneous system (5) on I, then X = c1X1 + c2X2 + … + ckXk is also a solution on I.
THEOREM 10.2
Superposition Principles

14. Example 3
Please verify that are solutions of (8)

15. Example 3 (2)
then is also a solution.

16. Let X1, X2, …, Xk be a set lf solution vectors of the
DEFINITION 10.2
Let X1, X2, …, Xk be a set lf solution vectors of the
homogeneous system (5) in an interval I. We say the
set is linear dependence in the interval of there exist
constants c1, c2, …, ck, not all zero, so that
c1X1 + c2X2 + … + ckXk = 0
for every t in the interval. If the set of vectors is not
linearly dependence on the interval, it is said to be
linearly independence.
Linear Dependence/Independence

17. Let be n solution vectors of the homogeneous system (5) on
THEOREM 10.3
Let be n solution vectors of the homogeneous system (5) on
an interval I. Then the set of solution vectors is linearly
Independent on I is and only if the Wronskian
Criterion for Linearly Independence
Solution

18. (continued) (9) for every t in the interval.
THEOREM 10.3
(9) for every t in the interval.
Criterion for Linearly Independence
Solution

19. Example 4
We saw that are solutions of (6). Since they are linearly independent for all real t.

20. Any set X1, X2, …, Xn of n linearly independent
DEFINITION 10.3
Any set X1, X2, …, Xn of n linearly independent
solution vectors of the homogeneous system (5) on an
Interval I is said to be a fundamental set of solutions
on the interval.
Fundamental Set of Solution

21. There exists a fundamental set of solutions for the
homogeneous system (5) on an interval I.
THEOREM 10.4
Existence of a Fundamental Set
Let X1, X2, …, Xn be a fundamental set of solutions of
the homogeneous system (5) on an interval I. Then the
general solution of the system in the interval is X = c1X1 + c2X2 + … + cnXn where the ci, i = 1, 2,…, n are arbitrary constants.
THEOREM 10.5
General Solution—Homogeneous
Systems

22. Example 5
We saw that are linearly independent solutions of (6) on (−, ). Hence they form a fundamental set of solutions. Then general solutions is then (10)

23. Example 6
Consider the vectors

24. Example 6 (2)
Their Wronskian is

25. Example 6 (2)
Then the general solution is

26. homogeneous system (5). Then the general solution of the
THEOREM 10.6
Let Xp be a given solution of the nonhomogeneous system (4) on the interval I, and let Xc = c1X1 + c2X2 + … + cnXn denote the general solution on the same interval of the associated
homogeneous system (5). Then the general solution of the
nonhomogeneous system on the interval is
X = Xc + Xp.
The general solution Xc of the homogeneous system (5) is called
the complementary function of the nonhomogeneous system (4).
General Solution—Nonhomogeneous
Systems

27. Example 7
The vector is a particular solution of (11) on (−, ). In Example 5, we saw the solution of is Thus the general solution of (11) on (−, ) is

28. 10.2 Homogeneous Linear Systems
A Question We are asked whether we can always find a solution of the form (1) for the homogeneous linear first-order system (2)

29. Eigenvalues and Eigenvectors
If (1) is a solution of (2), then X = Ket then (2) becomes Ket = AKet . Thus we have AK = K or AK – K = 0. Since K = IK, we have (A – I)K = 0 (3)
The equation (3) is equivalent to

30. If we want to find a nontrivial solution X, we must have
If we want to find a nontrivial solution X, we must have det(A – I) = 0 The above discussions are similar to eigenvalues and eigenvectors of matrices.

31. Let 1, 2,…, n be n distinct eigenvalues of the matrix
THEOREM 10.7
Let 1, 2,…, n be n distinct eigenvalues of the matrix
A of (2), and let K1, K2,…, Kn be the corresponding
eigenvectors. Then the general solution of (2) is
General Solution—Homogeneous
Systems

32. Example 1
Solve (4)
Solution we have 1 = −1, 2 = 4.

33. Example 1 (2)
For 1 = −1, we have 3k1 + 3k2 = 0 2k1 + 2k2 = 0 Thus k1 = – k2. When k2 = –1, then
For 1 = 4, we have −2k1 + 3k2 = 0 2k1 − 2k2 = 0 Thus k1 = 3k2/2. When k2 = 2, then

34. Example 1 (3)
We have and the solution is (5)

35. Example 2
Solve (6)
Solution

36. Example 2 (2)
For 1 = −3, we have Thus k1 = k3, k2 = 0. When k3 = 1, then (7)

37. Example 2 (3)
For 2 = −4, we have Thus k1 = 10k3, k2 = − k3. When k3 = 1, then (8)

38. Example 2 (4)
For 3 = 5, we have Then (9)

39. Example 2 (5)
Thus

40. Example 3: Repeated Eigenvalues
Solve
Solution

41. Example 3 (2)
We have – ( + 1)2(– 5) = 0, then 1 = 2 = – 1, 3 = 5. For 1 = – 1, k1 – k2 + k3 = 0 or k1 = k2 – k3. Choosing k2 = 1, k3 = 0 and k2 = 1, k3 = 1, in turn we have k1 = 1 and k1 = 0.

42. Example 3 (3)
Thus the two eigenvectors are For 3 = 5,

43. Example 3 (4)
Implies k1 = k3 and k2 = – k3. Choosing k3 = 1, then k1 = 1, k2 = –1, thus Then the general solution is

44. Second Solution
Suppose 1 is of multiplicity 2 and there is only one eigenvector. A second solution will be of the form (12) Thus X = AX becomes we have (13) (14)

45. Example 4
Solve
Solution First solve det (A – I) = 0 = ( + 3)2,  = －3, －3, and then we get the first eigenvector Let

46. Example 4 (2)
From (14), we have (A + 3 I) P = K. Then Choose p1 = 1, then p2 = 1/6. However, we choose p1 = ½, then p2 = 0. Thus

47. Example 4 (3)
From (12), we have The general solution is

48. Eigenvalue of Multiplicity 3
When there is only one eigenvector associated with an eigenvalue of multiplicity 3, we can find a third solution as and

49. Example 5
Solve
Solution (1 – 2)3 = 0, 1 = 2 is of multiplicity 3. By solving (A – 2I)K = 0, we have a single eigenvector

50. Example 5 (2)
Next, solving (A – 2I) P = K and (A – 2I) Q = P, then Thus

51. Let A be the coefficient matrix having real entires of
THEOREM 10.8
Let A be the coefficient matrix having real entires of
the homogeneous system (2), and let K1 be an
eigenvector corresponding to the complex eigenvalue
1 =  + i ,  and  are real. Then and are solutions of (2).
Solution Corresponding to a
Complex Eigenvalue

52. Let 1 =  + i be a complex eigenvalue of the
THEOREM 10.9
Let 1 =  + i be a complex eigenvalue of the
coefficient matrix A in the homogeneous system (2),
and let B1 and B2 denote the column vectors defined in
(22). Then (23) are linearly independent solutions of (2) on (－, ).
Real Solutions Corresponding to a
Complex Eigenvalue

53. Example 6
Solve
Solution First, For 1 = 2i, (2 – 2i)k1 + 8k2 = 0, – k1 + (–2 – 2i)k2 = 0, we get k1 = –(2 + 2i)k2.

54. Example 6 (2)
Choosing k2 = –1, then
Since  = 0, then

55. Fig 10.4

56. 10.3 Solution by Diagonalization
Formula If A is diagonalizable, then there exists P, such that D = P-1AP is diagonal. Letting X = PY then X = AX becomes PY = APY, Y = P-1APY, that is, Y = DY, the solution will be

57. Example 1
Solve
Solution From det (A – I) = – ( + 2)(– 1)(– 5), we get 1 = – 2, 2 = 1 and 3 = 5. Since they are distinct, the eigenvectors are linearly independent. For i = 1, 2, 3, solve (A –iI)K = 0, we have

58. Example 1 (2)
then and Since Y = DY, then Thus

59. 10.4 Nonhomogeneous Linear Systems
Example 1 Solve
Solution First solve X = AX,  = i, −i,

60. Example 1 (2)
Now let we have Thus Finally, X = Xc + Xp

61. Example 2
Solve
Solution First we solve X = AX. By similar procedures, we have 1 = 2, 2 = 7, and Then

62. Example 2 (2)
Now let After substitution and simplification, or

63. Example 2 (3)
Solving the first equations , and substitute these values into the least two equations, we obtain The general solution of the system on (－, ) is X = Xc + Xp or

64. Example 3
Determine the form pf Xp for dx/dt =5x + 3y – 2e-t dy/dt =−x + y + e-t – 5t + 7
Solution Since then

65. Fundamental Matrix
If X1, X2,…, Xn is a fundamental set of solutions of X = AX on I, its general solution is the linear combination X = c1X1 + c2X2 +…+ cnXn, or (1)

66. (1) can be also written as. X = Φ(t)C
(1) can be also written as X = Φ(t)C (2) where C is the n  1 vector of arbitrary constants c1, c2,…, cn, and is called a fundamental matrix.
Two Properties of (t): (i) nonsingular; (ii) (t) = A(t) (3)

67. Variation of Parameters
In addition, we to find a function such that Xp = Φ(t)U(t) (4) is a particular solution of (5) Since (6) then (7)

68. Using (t) = A(t), then or (8) Thus and so
Since Xp = Φ(t)U(t), then (9) Finally, X = Xc + Xp (10)

69. Example 4 Find the general solution of on (−, ).
Solution First we solve the homogeneous system The characteristic equation of the coefficient matrix is

70. Example 4 (2) We can get  = −2, −5, and the eigenvectors are
Thus the solutions are
Thus

71. Example 4 (3)
Now

72. Example 4 (4)
Hence

73. Example 5
Solve
Solution From the same method, we have Then

74. Example 5 (2)
Let X = PY, We have two DEs:

75. Example 5 (3)
The solutions are y1 = 1/5 et + c1 and y2 = –7/20 et + c2 e5t. Thus

76. 10.5 Matrix Exponential Matrix Exponential For any n  n matrix A, (3)
DEFINITION10.4
For any n  n matrix A, (3)
Matrix Exponential
Also A0 = I, A2 = AA, ….

77. Derivative of eAt
(4) Since

78. Besides, (5) then eAt is also the solution of X = AX.

79. Computation of eAt

80. Using the Laplace Transform
X = AX, X(0) = I (7) If x(s) = L{X(t)} = L{eAt}, then sx(s) – X(0) = Ax(s) or (sI – A)x(s) = I Now x(s) = (sI – A)–1I = (sI – A)–1. Thus L{eAt} = (sI – A)– (8)

81. Example 1
Use the Laplace Transform to compute eAt, where
Solution

82. Example 1 (2)

83. Using Powers Am
(10) where the coefficient cj are the same in each and the last expression is valid for the eigenvalues 1, 2, …, n of A. Assume that the eigenvalues of A are distinct. By setting  = 1, 2, …n in second expression in (10), we found the cj in first expression by solving n equations in n unknown. From (3) and (2), we have

84. Replacing Ak and k as finite sums followed by an interchange of the order of summations

85. Example 2: Using Powers Am
Compute eAt, where
Solution We already know 1= −1 and 2 = 2, then eAt = b0I + b1A and (14) Using the values of , then we have b0 = (1/3)[e2t + 2e– t], b1 = (1/3)[e2t – e–t].

86. Example 2 (2)
Thus

87. Thank You !