Chapter 6: Systems of Linear Differential. be continuous functions on the interval


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1 Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n firstorder differential equations
2 x = a (t)x + a 2 (t)x a n (t)x n + b (t) x 2 = a 2(t)x + a 22 (t)x a 2n (t)x n + b 2 (t). x n = a n(t)x + a n2 (t)x a nn (t)x n + b n (t). is called a firstorder linear differential system. (For us, a linear ddifferential system.) The system is homogeneous if b (t) b 2 (t) b n (t) 0 on I. 2
3 It is nonhomogeneous if the functions b i (t) are not all identically zero on I. 3
4 Set A(t) = a (t) a 2 (t) a 2 (t) a 22 (t)... a n (t) a n2 (t) a n (t) a 2n (t) a nn (t) and x = x x 2. x n, b(t) = b (t) b 2 (t). b n (t). The system can be written in the vectormatrix form x = A(t)x + b(t). (S) c.f., Section 2. 4
5 The matrix A(t) is called the matrix of coefficients or the coefficient matrix. The vector b(t) is called the nonhomogeneous term, or forcing function. 5
6 A solution of the linear differential system (S) is a differentiable vector function x(t) = x (t) x 2 (t). x n (t) that satisfies (S) on the interval I. 6
7 THEOREM: The initialvalue problem x = A(t)x + b(t), x(t 0 ) = c has a unique solution x = x(t). 7
8 Example : x = x + 2x 2 + 2e 4t x 2 = 2x + x 2 + e 4t Vector/matrix form x x 2 = 2 2 x x 2 + 2e4t e 4t or x = 2 2 x + 2e4t e 4t 8
9 x(t) = 8 5 e 4t 7 5 e 4t is a solution 9
10 x(t) = C e3t e 3t +C 2 e t e t e 4t 7 5 e4t is a solution for any numbers C, C 2. 0
11 Example 2: x = 3x x 2 x 3 x 2 = 2x + 3x 2 + 2x 3 x 3 = 4x x 2 2x 3 Vector/matrix form x x 2 x 3 = x x 2 x 3 or x = (a homogeneous system) x
12 x(t) = e 3t e 3t e 3t = e3t is a solution. 2
13 x(t) = C e 3t +C 2 e 2t 0 +C 3 e t 3 7 is a solution for any numbers C, C 2, C 3 3
14 Converting a linear equation to a linear system Consider the second order equation y + p(t)y + q(t)y = 0 Solve for y y = q(t)y p(t)y 4
15 Introduce new dependent variables x, x 2, as follows: x = y x 2 = x (= y ) 5
16 Vectormatrix form: x x 2 = 0 q p x x f Note that this system is just a very special case of the general system of two, firstorder differential equations: x = a (t)x + a 2 (t)x 2 + b (t) x 2 = a 2(t)x + a 22 (t)x 2 + b 2 (t) 6
17 Vectormatrix form: x x 2 = a (t) a 2 (t) a 2 (t) a 22 (t) x x 2 + b (t) b 2 (t) or x = A(t)x + b
18 Example : y 5y + 6y = 4e 4t which can be written y = 6y + 5y + 4e 4t Set 7
19
20 y 5y + 6y = 4e 4t Characteristic equation: Fundamental set: Particular solution General solution: 8
21 System: x x 2 = x x e 4t y = e 2t +2e 4t is a soln. of equation Corresponding solution of system???? 9
22
23 Consider the thirdorder equation y + p(t)y + q(t)y + r(t)y = f(t) or y = r(t)y q(t)y p(t)y + f(t). 20
24 Introduce new dependent variables x, x 2, x 3, as follows: x = y x 2 = x (= y ) x 3 = x 2 (= y ) Then y = x 3 = r(t)x q(t)x 2 p(t)x 3 +f(t) The thirdorder equation can be written equivalently as the system of three firstorder equations: 2
25 x = x 2 x 2 = x 3 x 3 = r(t)x q(t)x 2 p(t)x 3 + f(t). That is x = 0x + x 2 + 0x x 2 = 0x + 0x 2 + x x 3 = r(t)x q(t)x 2 p(t)x 3 + f(t). 22
26 Vectormatrix form: x x 2 x 3 = r q p x x 2 x f 23
27 Note that this system is just a very special case of the general system of three, firstorder differential equations: x = a (t)x + a 2 (t)x 2 + a 3 (t)x 3 (t) + b (t) x 2 = a 2(t)x + a 22 (t)x 2 + a 23 (t)x 3 (t) + b 2 (t) x 3 = a 3(t)x + a 32 (t)x 2 + a 33 (t)x 3 (t) + b 3 (t). or in vectormatrix form: x = A(t)x + b 24
28 Example 2: y 3y 4y + 2y = 6e t. which can be written y = 2y + 4y + 3y + 6e t. Set x = y x 2 = x (= y ) x 3 = x 2 (= y ) Then x 3 = 2x +4x 2 +3x 3 +6e t 25
29 Equivalent system: x = x 2 x 2 = x 3 x 3 = 2x + 4x 2 + 3x 3 + 6e t. Which is x = 0x + x 2 + 0x x 2 = 0x + 0x 2 + x x 3 = 2x + 4x 2 + 3x 3 + 6e t. 26
30 Vectormatrix form: x x 2 x 3 = x x 2 x e t or x = Ax + b
31 y 3y 4y + 2y = 6e t Characteristic equation: Fundamental set: Particular solution General solution 27
32 y = e 3t + 2 e t is a solution of the equation. System: x x 2 x 3 Recall: x = = x x 2 x = e 3t 3e 3t 9e 3t y y y + x x 2 x 3 2 e t 2 e t 2 e t is a corresponding solution of the system. 28
33 II. Homogeneous Systems: General Theory: x = a (t)x + a 2 (t)x a n (t)x n (t) x 2 = a 2(t)x + a 22 (t)x a 2n (t)x n (t). x n = a n(t)x + a n2 (t)x a nn (t)x n (t). x = A(t)x. (H) Note: The zero vector z(t) 0 = 0 is a solution of (H). This solution is called the trivial 0. 0 solution. 29
34 THEOREM: If v and v 2 are solutions of (H), then u = v + v 2 is also a solutions of (H); the sum of any two solutions of (H) is a solution of (H). THEOREM: If v is a solution of (H) and α is any real number, then u = αv is also a solution of (H); any constant multiple of a solution of (H) is a solution of (H). 30
35 In general, THEOREM: If v, v 2,..., v k are solutions of (H), and if c, c 2,..., c k are real numbers, then c v + c 2 v c k v k is a solution of (H); any linear combination of solutions of (H) is also a solution of (H). 3
36 Linear Dependence/Independence in general Let v (t) = v v 2. v n, v 2 (t) = v 2 v 22. v n2,..., v k (t) = v k v 2ḳ. v nk be vector functions defined on some interval I. 32
37 The vectors are linearly dependent on I if there exist k real numbers c, c 2,..., c k, not all zero, such that c v (t)+c 2 v 2 (t)+ +c k v k (t) 0 on I. Otherwise the vectors are linearly independent on I. 33
38 THEOREM Let v (t), v 2 (t),..., v k (t) be k, kcomponent vector functions defined on an interval I. If the vectors are linearly dependent, then v v 2 v k v 2 v v 2k. v k v k2 v kk 0 on I. 34
39 The determinant v v 2 v k v 2 v v 2k. v k v k2 v kk is called the Wronskian of the vector functions v, v 2,..., v k. 35
40 Special case: n solutions of (H) THEOREM Let v, v 2,..., v n be n solutions of (H). Exactly one of the following holds:. W(v, v 2,..., v n )(t) 0 on I and the solutions are linearly dependent. 2. W(v, v 2,..., v n )(t) 0 for all t I and the solutions are linearly independent. 36
41 THEOREM Let v, v 2,..., v n be n linearly independent solutions of (H). Let u be any solution of (H). Then there exists a unique set of constants c, c 2,..., c n such that u = c v + c 2 v c n v n. That is, every solution of (H) can be written as a unique linear combination of v, v 2,..., v n. 37
42 A set of n linearly independent solutions of (H) is called a fundamental set of solutions. A fundamental set is also called a solution basis for (H). 38
43 Let v, v 2,..., v n be a fundamental set of solutions of (H). Then x = c v + c 2 v c n v n, c, c 2,..., c n arbitrary constants, is the general solution of (H). 39
44 Example: y 3y 4y + 2y = 0 Fundamental set: { y = e 3t, y 2 = e 2t, y 3 = e 2t} Vectormatrix form x x 2 x 3 = x x 2 x 3 Solutions: 40
45 x = e 3t 3e 3t 9e 3t, x 2 = e 2t 2e 2t 4e 2t x 3 = e 2t 2e 2t 4e 2t is a fundamental set of solutions of the corresponding system x x 2 x 3 = x x 2 x 3 4
46 III. Homogeneous Systems with Constant Coefficients x = a x + a 2 x a n x n x 2 = a 2x + a 22 x a 2n x n.. x n = a nx + a n2 x a nn x n where a, a 2,..., a nn are constants. 42
47 The system in vectormatrix form is x x 2 x n = a a 2 a n a 2 a 22 a 2n a n a n2 a nn x x 2 x n or x = Ax. 43
48 Solutions of x = Ax: Example: x = e 3t 3e 3t 9e 3t = e3t 3 9 is a solution of x x 2 x 3 = x x 2 x 3 What is? 3 9 relative to
49 = 45
50 THAT IS: 3 is an eigenvalue of A and v = 3 9 is a corresponding eigenvector. 46
51 NOTE: y 3y 4y + 2y = 0 Characteristic equation r 3 3r 2 4r+2 = (r 3)(r 2)(r+2) = 0 Fundamental set: { y = e 3t, y 2 = e 2t, y 3 = e 2t} 47
52 Vectormatrix form x x 2 x 3 = x x 2 x 3 Characteristic equation: det(a λi) = λ 0 0 λ λ = λ 3 + 3λ 2 + 4λ 2 = 0 or λ 3 3λ 2 4λ+2 = (λ 3)(λ 2)(λ+2) = 0 48
53 Eigenvalues: λ = 3, λ 2 = 2, λ 3 = 2 Eigenvectors: 49
54 Given the homogeneous system with constant coefficients x = Ax. CONJECTURE: If λ is an eigenvalue of A and v is a corresponding eigenvector, then x = e λt v is a solution. 50
55 Proof: Let λ be an eigenvalue of A with corresponding eigenvecvtor v. Set x = e λt v 5
56 If λ, λ 2,, λ k are distinct eigenvalues of A with corresponding eigenvectors v, v 2,, v k, then x = e λ t v, x 2 = e λ 2t v 2,,x k = e λ kt v k are linearly independent solutions of x = Ax. 52
57 In particular: If λ, λ 2,, λ n are distinct eigenvalues of A with corresponding eigenvectors v, v 2,, v n, then x = e λ t v, x 2 = e λ 2t v 2,,x n = e λ kt v n form a fundamental set of solutions of x = Ax. and x = C x +C 2 x 2 + +C n x n is the general solution. 53
58 . Find the general solution of x = 4 2 x. Step. Find the eigenvalues of A: det(a λi) = 4 λ 2 λ = λ 2 5λ + 6. Characteristic equation: λ 2 5λ + 6 = (λ 2)(λ 3) = 0. Eigenvalues: λ = 2, λ 2 = 3. 54
59 Step 2. Find the eigenvectors: A λi = 4 λ 2 λ 55
60 λ = 2, v = 2 ; λ 2 = 3, v 2 =. Solutions: Fundamental set of solution vectors: x = e 2t 2, x 2 = e 3t General solution of the system: x = C e 2t 2 + C 2 e 3t. 56
61 2. Solve x = x. Step. Find the eigenvalues of A: det(a λi) = 3 λ 2 λ λ = λ 3 + 2λ 2 + λ 2. Characteristic equation: λ 3 2λ 2 λ+2 = (λ 2)(λ )(λ+) = 0. Eigenvalues: λ = 2, λ 2 =, λ 3 =. 57
62 Step 2. Find the eigenvectors: A λi = 3 λ 2 λ λ λ = 2: 58
63 λ = 2 : v = 2, λ 2 = : v 2 = 3 7, λ 3 = : v 3 = 2 2. Fundamental set of solutions: x = e 2t 2, x 3 = e t x 2 = e t , 59
64 The general solution of the system: x = C e 2t 2 +C 2 e t 3 7 +C 3 e t
65 3. Solve the initialvalue problem x = x, x(0) = 0. To find the solution vector satisfying the initial condition, solve C v (0)+C 2 v 2 (0)+C 3 v 3 (0) = 0 which is: C 2 +C C = 0 6
66 or C C 2 C 3 = 0. Augmented matrix:
67 Solution: C = 3, C 2 =, C 3 =. The solution of the initialvalue problem is: x = 3e 2t 2 e t 3 7 +e t
68 TWO DIFFICULTIES: I. A has complex eigenvalues and complex eigenvectors. II. A has an eigenvalue of multiplicity greater than. 64
69 I. Complex eigenvalues/eigenvectors Examples:. Find the general solution of x = x. det(a λi) = 3 λ 2 4 λ = λ2 +2λ+5. The eigenvalues are: λ = + 2i, λ 2 = 2i. 65
70 A λi = 3 λ 2 4 λ For λ = + 2i: Solve 2 2i i 0 66
71 The solution set is: x 2 = ( + i)x, x arbitrary Set x =. Then, for λ = +2i: v = i = +i 0. and, for λ 2 = 2i: v 2 = i 0. 67
72
73 Solutions u = e λ t v = e ( +2i)t [( ) + i ( 0 )] = e t (cos 2t + i sin 2t) [( ) + i ( 0 )] = e t [cos 2t ( ) sin 2t ( 0 )] + i e t [cos 2t ( 0 ) + sin 2t ( )]. 68
74 u 2 = e λ 2t v 2 = e ( 2i)t [( ) i ( 0 )] = e t (cos 2t + i sin 2t) [( ) + i ( 0 )] = e t [cos 2t ( ) sin 2t ( 0 )] i e t [cos 2t ( 0 ) + sin 2t ( )]. 69
75 Fundamental set: x = u + u 2 2 = e t cos 2t sin 2t 0 x 2 = u + u 2 2i = e t cos 2t 0 + sin 2t General solution: x = C e t cos 2t sin 2t 0 + C 2 e t cos 2t 0 + sin 2t 70
76 Summary: x = Ax, A n n const. a+ib, a ib complex eigenvalues. α +i β, α i β corresponding eigenvectors. Independent (complexvalued) solutions: u = e (a+ib)t ( α +i β ) u 2 = e (a ib)t ( α i β ) 7
77
78 Corresponding realvalued solutions: x = e at [cos bt α sin bt β ] x 2 = e [cos at bt β +sin bt ] α General solution: x = C e at [cos bt α sin bt β ] + C 2 e [cos at bt β +sin bt ] α 72
79 2. Determine a fundamental set of solution vectors of x = x. det(a λi) = λ λ 3 3 λ = λ 3 +6λ 2 2λ+26 = (λ 2)(λ 2 4λ+3). The eigenvalues are: λ = 2, λ 2 = 2 + 3i, λ 3 = 2 3i. 73
80 A λi = λ λ 3 3 λ λ = 2: Solve v = 0. 74
81 A λi = λ λ 3 3 λ For λ 2 = 2 + 3i: Solve 3i i 3 0 3i 0 75
82
83 The solution set is: x = ( i ) x 3, x 2 = ( ) 2 i x 3,, x 3 arbitrary. v 2 = 5 + 3i 3 + 3i 2 = i and v 3 = 5 3i 3 3i 2 = i
84 Now u = e (2+3i)t i and u 2 = e (2 3i)t i convert to: x = e 2t cos 3t sin 3t and x 2 = e 2t cos 3t sin 3t
85 Fundamental set of solution vectors: x 2 = e 2t x 3 = e 2t x = e 2t cos 3t cos 3t , sin 3t + sin 3t ,. General solution: x = C x + C 2 x 2 + C 3 x 3 78
86 II. Repeated eigenvalues Examples:. Find a fundamental set of solutions of x = x. det(a λi) = λ λ λ = 6+2λ λ 3 = (λ 4)(λ+2) 2. Eigenvalues: λ = 4, λ 2 = λ 3 = 2 79
87 λ = 4 : (A 4I) =
88 λ 2 = λ 3 = 2: A ( 2)I =
89 which row reduces to Solution set: x = a b, x 2 = a, x 3 = b a, b any real numbers. Set a =, b = 0 : v 2 = 0 ; Set a = 0, b = : v 3 = 0. 82
90 Fundamental set: e 4t 2, e 2t 0, e 2t 0. 83
91 2. Find a fundamental set of solutions of x = x. det(a λi) = 5 λ λ λ = 36+5λ+2λ 2 λ 3 = (λ+4)(λ 3) 2. Eigenvalues: λ = 4, λ 2 = λ 3 = 3. 84
92 λ = 4 : v = A ( 4)I =
93 λ 2 = λ 3 = 3: Solve A 3I =
94 which row reduces to x = 5x 3, x 2 = 2x 3, x 3 arbitrary Set x 3 = : v 2 = 5 2 Problem: Only one eigenvector! 87
95 Solutions: x = e 4t 6 8 3, x 2 = e 3t 5 2. We need a third solution x 3 which is independent of x, x 2. 88
96 3. y + y 8y 2y = 0 Char.eqn. r 3 +r 2 8r 2 = (r 3)(r+2) 2 = 0. Fundamental set: { e 3t, e 2t, te 2t} 89
97 Equivalent system: x = det(a λi) = λ 0 0 λ 2 8 λ = λ 3 λ 2 + 8λ + 2λ x char. eqn.: λ 3 + λ 2 8λ 2 = (λ 3)(λ + 2) 2 Eigenvalues: λ = 3, λ 2 = λ 3 = 2 90
98 Fundamental set: x = e 3t 3 9, x 2 = e 2t 2 4, x 3 = e 2t te 2t 2 4 Question: What is the vector? 0 4 9
99 [A ( 2)I] 0 4 = = 92
100 A ( 2I) maps the eigenvector onto 0 4 is called a generalized eigenvector. 93
101
102 2. continued. Solve (A 3I)w =
103 / Solution set: x = +5x 3, x 2 = 2 2x 3, x 3 arbitrary Set x 3 = 0: w = /2 0 95
104 Fundamental set: x = e 4t 6 8 3, x 2 = e 3t 5 2, x 3 = e 3t /2 0 + te 3t
105 Eigenvalues of multiplicity 2: Given x = Ax. Suppose that A has an eigenvalue λ of multiplicity 2. Then exactly one of the following holds: 97
106 . λ has two linearly independent eigenvectors, v and v 2. Corresponding linearly independent solution vectors of the differential system are x (t) = e λt v and x 2 (t) = e λt v 2. (See Example.) 98
107 2. λ has only one (independent) eigenvector v. (See Examples 2 and 3.) Then a linearly independent pair of solution vectors is: x (t) = e λt v and x 2 (t) = e λt w+te λt 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. 99
108 Examples:. Find a fundamental set of solutions and the general solution of x = x. det (A λ I) = 2 λ 5 4 λ = λ 2 6λ + 3 Eigenvalues: 3 + 2i, 3 2i 00
109
110 (A λ I) = 2 λ 5 4 λ λ = 3 + 2i, v = + i 2 0 0
111
112 Fundamental set: x = e 3t cos 2t sin 2t 2 0 x 2 = e 3t cos 2t 2 0 sin 2t General solution: x(t) = C x + C 2 x 2 02
113 2. Find a fundamental set of solutions and the general solution of x = x. HINT: 3 is an eigenvalue and 2 is an eigenvalue of multiplicity 2 Characteristic eqn: (λ+3)(λ+2) 2 =0 03
114 (A λi) = 4 λ λ 2 2 λ λ = 3 : 04
115 (A λi) = 4 λ λ 2 2 λ λ 2 = λ 3 = 2 : 2 0,
116 Fundamental set: e 3t, e 2t 2 0, e 2t 0 2 General solution: x = C e 3t +C 2 e 2t 2 0 +C 3 e 2t
117 3. Find a fundamental set of solutions and the general solution of x = x. HINT: 4 is an eigenvalue and 2 is an eigenvalue of multiplicity 2 Characteristic eqn: (λ 4)(λ+2) 2 =0 07
118 (A λi) = 3 λ 7 5 λ λ λ = 4 : v = 0 08
119 (A λi) = 3 λ 7 5 λ λ λ 2 = λ 3 = 2 : v 2 = 0 09
120 (A λi) = 3 λ 7 5 λ λ [A ( 2)I]w = 0 0
121 Fund. Set: e 4t 0, e 2t 0 e 2t + te 2t 0 General solution: x = C e 4t 0 + C 2 e 2t 0 + C 3 e 2t + te 2t 0
122 Eigenvalues of Multiplicity 3. Given the differential system x = Ax. Suppose that λ is an eigenvalue of A of multiplicity 3. Then exactly one of the following holds: 2
123 . λ has three linearly independent eigenvectors v, v 2, v 3. Then three linearly independent solution vectors of the system corresponding to λ are: x (t) = e λt v, x 2 (t) = e λt v 2, x 3 (t) = e λt v 3. 3
124 2. λ has two linearly independent eigenvectors v, v 2. Then three linearly independent solutions of the system corresponding to λ are: x (t) = e λt v, x 2 (t) = e λt v 2 and x 3 (t) = e λt w + te λt v where v is an eigenvector corresponding to λ and (A λi)w = v. That is: (A λi) 2 w = 0. 4
125 3. λ has only one (independent) eigenvector v. Then three linearly independent solutions of the system have the form: x = e λt v, x 2 = e λt w + te λt v, v 3 (t) = e λt z + te λt w + t 2 e λt v where (A λi)z = w & (A λi)w = v, i.e. (A λi) 3 z = 0 & (A λi) 2 w = 0 5
126 Example: y 6y + 2y 8y = 0 Char. eqn.: (r 2) 3 = 0 Char. roots: r = r 2 = r 3 = 2 Fundamental set: { e 2t, te 2t, t 2 e 2t} 6
127 Corresponding system: x = x Fundamental set: e 2t 2 4, e 2t te 2t 2 4, e 2t te 2t t 2 e 2t 2 4 7
128 Nonhomogeneous Linear Differential Systems Let A(t) be the n n matrix A(t) = a (t) a 2 (t) a 2 (t) a 22 (t)... a n (t) a n2 (t) a n (t) a 2n (t) a nn (t) and let x and b(t) be the vectors x = x x 2. x n, b(t) = b (t) b 2 (t). b n (t). Nonhomogeneous system: x = A(t)x + b(t) (N) Reduced system: x = A(t)x (H) 8
129 Theorem. If z (t) and z 2 (t) are solutions of (N), then x(t) = z (t) z 2 (t) is a solution of (H). (C.f. Theorem, Section 3.4, and Theorem 7, Section 6..) Theorem 2. Let x (t), x 2 (t),...,x n (t) be a fundamental set of solutions the reduced system (H) and let z = z(t) be a particular solution of (N). If u = u(t) is any solution of (N), then there exist constants c, c 2,..., c n such that u(t) = c x (t)+c 2 x 2 (t)+ +c n x n (t)+z(t) (C.f. Theorem 2, Section 3.4, and Theorem 8, Section 6..) 9
130 General Solution of (N): x = C x (t) + C 2 x 2 (t) + + C n x n (t) }{{} general solution of (H) + z(t). }{{} particular solution of (N) 20
131 Variation of Parameters x (t), x 2 (t),...,x n (t) a fundamental set of solutions of (H). V (t) the corresponding fundamental matrix. x(t) = V (t)c where C = C C 2.. C n is the general solution of (H). V satisfies the matrix differential system X = A(t)X. That is, V (t) = A(t)V (t). (Exercises 6.3, Problem 9.) 2
132 Replace the constant vector C by a vector function u(t) which is to be determined so that z(t) = V (t)u(t) is a solution of (N). u (t) = V (t)b(t) u(t) = V (t)b(t) dt. and z(t) = V (t) V (t)b(t) dt is a solution of (N). General solution of (N): x(t) = V (t)c + V (t) V (t)b(t) dt. 22
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