Linear systems of ordinary differential equations

Transcription

1 Linear systems of ordinary differential equations (This is a draft and preliminary version of the lectures given by Prof. Colin Atkinson FRS on 2st, 22nd and 25th April 2008 at Tecnun Introduction. This chapter studies the solution of a system of ordinary differential equations. These kind of problems appear in many physical or chemical models where several variables depend upon the same independent one. For instance, Newton s second law applied to a particle. m d2 x dt 2 = F (x, x 2, x 3, ẋ, ẋ 2, ẋ 3, t m d2 x 2 dt 2 = F 2(x, x 2, x 3, ẋ, ẋ 2, ẋ 3, t m d2 x 3 dt 2 = F 3(x, x 2, x 3, ẋ, ẋ 2, ẋ 3, t Let us consider this system of ordinary differential equations: dx = 3x + 3y dt dy = xz + rx y dt dx dt = xy z where x, y, z are time-dependent variables, and r is a parameter. We find a system of three non-linear ODE s. The interesting thing in this example is the strong dependence of the solution on the paramater r and the initial value conditions (t 0, x 0, y 0, z 0. For the above reasons, the system presents as the chaos theory; a very known theory in Mathematics knows as the butterfly effect. These equations were posed by Lorenz in the study of metheorology. The phrase refers to the idea that a butterfly s wings might create tiny changes in the atmosphere that may ultimately alter the path of a tornado or delay, accelerate or even prevent the occurrence of a tornado in a certain location. The flapping wing represents a small change in the initial condition of the system, which causes a chain of events leading to large-scale alterations of events. Had the butterfly not flapped its wings, the trajectory of the system might have been vastly different. While the butterfly does not cause the tornado, the flap of its wings is an essential part of the initial conditions resulting in a tornado. Recurrence, the approximate return of a system towards its initial conditions, together with sensitive dependence on initial conditions are the two main ingredients for chaotic motion. They have the practical consequence of making complex systems, such as the weather, difficult to predict past a certain time range (approximately a week in the case of weather. ( (2 c 2008 Tecnun (University of Navarra

2 There is an important relationship between the system of ODE and the ODE s of any order superior to the first. It is a matter of fact, that an equation of order nth y (n = F (t, y, y, y,..., y (n (3 where y (n = d n y/dt n. This can be converted in a system of n equations of first order. With this change of variables or in general: x = y x 2 = y x 2 = x... x n = y (n x n = x n x n = F (t, x, x 2,..., x n x = F (t, x, x 2,..., x n x 2 = F 2 (t, x, x 2,..., x n... x n = F n (t, x, x 2,..., x n we reach n non-linear ODE s of first order. There are three questions to be answered: ( What about the existence of solutions? (2 What about uniqueness? (3 What is the sensitivity to the initial conditions? We are going to see in which cases we can assert that a system of ODE has a solution and this is unique. We must consider the following theorem Theorem. Let us assume that in a region A of the space (t, x, x 2,..., x n, the functions F, F 2,..., F n and F, F,..., F x x 2 x n F 2, F 2,..., F 2 x x 2 x n... F n, F n,..., F n x x 2 x n are continuous and such that the point (t 0, x 0, x 0 2,..., x 0 n is an interior point of A. Thus there exists an interval (4 (5 (6 t t 0 < ɛ (7 (local argument where there is a unique solution of the system given by eq. 5, x = Φ (t x 2 = Φ 2 (t... x n = Φ n (t (8 that fulfils the initial condition x 0 = Φ (t 0 x 0 2 = Φ 2 (t 0... x 0 n = Φ n (t 0 (9 c 2008 Tecnun (University of Navarra 2

3 A way to prove this theorem is to use the Taylor s expansion of the functions. Note.- We must point out that this is a sufficient condition theorem. So, weakening the conditions we can define a stronger expression of this theorem to get a unique solution. The systems are classified in the same manner as the ODE s. They are linear and non-linear. If the functions F, F 2,..., F n can be written as x i = P i (t x + P i2 (t x P in (t x n + q i (t (0 with i =, 2,..., n, the system is called linear. If q i (t are equal to zero for all i, this system is called linear and homogeneous; if not, non-homogeneous. For this kind of systems the theorem of existence and uniqueness is simpler and to some extent, more satisfactory. This theorem has global character. Notice that in the general case this theorem it is defined in the neighbourhood of the initial conditions, consequently giving a local character to the existence and uniqueness of the solution. Recall: If an equation is linear, it means we can add solutions together and still satisfy the differential equation, e.g. x = H(x and H is linear, then If x = H(x y x 2 = H(x 2, then H(c x + c 2 x 2 = c H(x + c 2 H(x 2 ( (c x +c 2 x 2 = H(c x +c 2 x 2 = c H(x +c 2 H(x 2 = c x +c 2 x 2 (2 where c, c 2 are constants. This is a nice property fulfilled when an equation is linear. Basic theory of linear systems of ODE s Let us consider a system of n linear differential equations of first order x = P (t x + P 2 (t x + + P n (t x n + q (t x = P 2 (t x + P 22 (t x + + P 2n (t x n + q 2 (t... x = P n (tx + P n (t x + + P nn (t x n + q n (t (3 We can write this as a matrix where x = P(t x + q(t (4 x = (x x 2... x n T q(t = (q (t q 2 (t... q n (t T P (t P 2 (t... P n (t P 2 (t P 22 (t... P 2n (t P(t = P n (t P n2 (t... P nn (t (5 c 2008 Tecnun (University of Navarra 3

4 If q(t = 0, we have an homogeneous system and eq. 4 becomes x = P(t x (6 This notation emphasises the relationship between the linear systems of ODE s and the first order linear differential equations Theorem. If x and x 2 are solutions of eq. 6, so (c x + c 2 x 2 is solution as well. x = P(t x x (c x + c 2 x 2 = P(t (c x + c 2 x 2 (7 2 = P(t x 2 The question to be answered is: how many independent solutions of eq. 6 are there? Let us assume at the moment, if x, x 2,..., x n are solutions of the system, consider the matrix Ψ(t called fundamental matrix given by Ψ(t = (x x 2... x n (8 Its determinant will be x (t x 2 (t... x n (t x 2 (t x 22 (t... x 2n (t Ψ(t =..... = W (t (9. x n (t x n2 (t... x nn (t where W (t is called the wronskian of the system. These solutions will be linearly independent at each point t in an interval (α, β if Example. W (t 0, t (α, β (20 We will find it for a two by two system ( x P (t P = 2 (t x (2 P 2 (t P 22 (t whose solutions are x = (x (t x 2 (t T y x 2 = (x 2 (t x 22 (t T. They verify the system equations (see eq. 2 { x x = P(t x = P x + P 2 x 2 x 2 = P 2 x + P 22 x 2 { (22 x x 2 = P(t x 2 2 = P x 2 + P 2 x 22 x 22 = P 2 x 2 + P 22 x 22 The fundamental matrix, Ψ(t, is ( x (t x Ψ(t = 2 (t x 2 (t x 22 (t (23 c 2008 Tecnun (University of Navarra 4

5 and the wronskian, W (t, The derivative of W (t with respect to t is W (t = x x 22 x 2 x 2 (24 W (t = x x 22 + x x 22 x 2 x 2 x 2 x 2 (25 and substituting x, x 2, x 2, x 22 as a function of x, x 2, x 2, x 22, we obtain W (t = (P + P 22 (x x 22 x 2 x 2 = (trace P(t W (t (26 This is the Abel s formula. Solving the differential equation W (t = W (t 0 e t t 0 (trace P(s ds (27 As e x is never zero, W (t 0 for all finite value of t if trace P(t is integrable and W (t 0 0. In n-dimensions the same happens. This generalises to n-dimensions to give dw = (P + + P nn W (t W (t = W (t 0 e dt Example 2. A Sturm-Liouville equation has the form t t 0 (trace P(s ds d ( dy a(t + b(t y = 0 (28 dt dt This is a second order differential equation of the form a 2 (t d2 y dt 2 + a (t dy dt + a 0(t y = 0 This kind of equations were studied in 8th/9th century and early 20th century, where a 2 (t, a (t and a 0 (t are linear in t. For instance, the vibration of a plate is given by these equations. We now write eq. (28 as a system Then it gives x = y (29 x 2 = dy dt (30 x = x 2 (3 x 2 = a (t a(t x 2 b(t a(t x (32 We could use our theory of systems to get the connection wronskian between x and x 2 independent solutions of the system. However, we consider eq. (28 directly assuming that y and y 2 are two possible solutions. So d ( dy a(t + b(t y = 0 (33 dt dt d ( dy 2 a(t + b(t y2 = 0 (34 dt dt c 2008 Tecnun (University of Navarra 5

6 We now multiply y 2 by eq. (33 and y by eq. (34 and substracting both expressions we get a(t (y y 2 + a (t (y y 2 y 2 y = 0 (35 but and Then eq. (35 becomes then d ( y2 y y y dt 2 = y2 y y y 2 (36 W (t = y 2 y y y 2 a(t dw dt + a (t W = 0 d dt( a(t W (t = 0 dw W = (t dt a a(t [ t a (s ] W (t = W (t 0 exp ds a(s Homogeneous linear system with constant coefficients. Let us consider the system t 0 x = A x (37 where A is real, n n constant matrix. As solution, we will try where a is a constant vector. Then Then we have a solution provided that x = e r t a (38 x = r e r t a (39 r e r t a = A e r t a (A r I a = 0 (40 and for non-trivial solution (i.e. a 0, r must satisfy Procedure. A r I = 0 (4 Finding eigenvalues, r, r 2,..., r n, solution of A r I = 0 and corresponding eigenvectors, a, a 2,..., a n. Then if the n eigenvectors are linearly independent, we have a general solution x = c e r t a + c 2 e r2 t a c n e rn t a n (42 c 2008 Tecnun (University of Navarra 6

7 where c, c 2,..., c n are arbitrary constants. Recall a i = (a i a 2i... a ni and x i = a i e ri t, with i =, 2,..., n. Then the wronskian will be a a 2... a n a 2 a a 2n W (t =..... e (r+r2+ +rn t 0 (43. a n a n2... a nn since a, a 2,..., a n are linearly independent. In a large number of problems, getting the eigenvalues can be very difficult problem. Example 3. ( x = x (44 4 Consider x = a e r t. Then we require ( ( a 0 (A r I = a 2 0 A r I = r 4 r = 0 ( r 2 4 = 0 ( r 2 = 4 r = ±2 r = 3,. So the eigenvalues are 3 and. With r = 3 ( ( ( 2 a 0 = 4 2 a 2 0 So 2 a + a 2 = 0 implies that a = and a 2 = 2. Hence, the eigenvector will be ( With r = ( ( a ( 0 = a 2 0 So 2 a + a 2 = 0 implies that a = and a 2 = 2. Therefore, the eigenvector will be ( The general solution is x = C 2 ( ( e 2 3 t + C 2 e 2 t (45 The equation (45 is a family of solutions since C and C 2 are arbitrary (i.e., if x and x 2 are known at t = 0, we can solve eq. (45 to get C and C 2 for c 2008 Tecnun (University of Navarra 7

8 Figure : Phase plane of example 3 specific solution. Note: eq. (44 does not involve time explicitly. It could be written as dx = x + x 2 (46 dx 2 4 x + x 2 So we can study the problem in 2 d space (x, x 2. This is often called the phase plane. Note that eq. (46 defines dx /dx 2 uniquely except for points where top and bottom terms of the quotient are zero simultaneously. dx dx dt ( ( = a x + a 2 x 2 dx = a a 2 x dt (47 a 2 a 22 x 2 2 dx 2 = a 2 x + a 22 x 2 dt dt dx 2 = a 2 x + a 22 x 2 (48 dx a x + a 2 x 2 { a x + a 2 x 2 = 0 (49 a 2 x + a 22 x 2 = 0 In general, what about A?. If A is hermitian (i.e., A H = A T = A, where A is the complex conjugate of the matrix, then the eigenvalues are real and we can find n linearly-independent eigenvectors. 2. A is non-hermitian. We have the following possibilities (2.a. n real and disttinct eigenvalues and n independent eigenvectors. (2.b. Complex eigenvalues. (2.c. Repeated eigenvalues. c 2008 Tecnun (University of Navarra 8

9 Example 4. Put x = e r t a to get x = ( 4 r 4 r we obtain r = y r 2 = 3. We need the eigenvectors x (50 = 0 (5 r = a = ( 2 T r 2 = 3 a 2 = ( 2 T (52 We have two solutions ( ( x = e 2 t x 2 = e 2 3t (53 To find the general solution, we construct it via a linear combination by adding these two linearly independent solutions together ( ( x = c e t + c 2 2 e 3 t (54 2 c and c 2 are arbitrary constants to be determined by intial conditions or other conditions on x. We can plot (see Figure 2 the family of the solutions in the (x, x 2 plane with an arrow to signify the direction of time (means time increasing. Our solution in components looks like x =c e 3 t + c 2 e t x 2 =2 c e 3 t 2 c 2 e t we can study the motion of a pendulum, we can use systems in order to study position and velocity. Suppose c 2 0 and we are interested in the point x = x 2 = 0 and note x /x 2 = /2 with c 0. We only reach (0, 0 if t since we need e 3 t 0. If c 0 and c 2 0, then x /x 2 = /2. To get (0, 0 we need t +. So (0, 0 is a special point. I can represent the time by an arrow { x c t e t x 2 2 c e t x 2 2 x { x c t + 2 e 3 t (55 x 2 2 c 2 e 3 t x 2 2 x Point (0, 0 is a saddle point. deta = 3 < 0. Thinking of the future, we can observe that c 2008 Tecnun (University of Navarra 9

10 Example 5. Figure 2: Phase plane of Example 4 Let us consider this problem. ( x 3 2 = x ( In order to solve this problem, we obtain the eigenvalues 3 r 2 = 0 ( r and r = 4 y r 2 =. The eigenvectors are r = 4 a = ( 2 T r 2 = a 2 = ( 2 T (58 they are linearly independent. Hence the solutions are ( ( 2 x = e 4 t x 2 = 2 e t (59 And the general solution is ( ( x = c e 4 t 2 + c 2 e t 2 (60 Plotting the paths on the phase plane (x, x 2 { x c t 2 e 4 t x 2 2 c 2 e 4t x 2 2 x { x 0 t + x 2 0 (x, x 2 (0, 0 (6 Point (0, 0 is a stable node. Thinking of the future again, deta = 2 > 0, tracea = 5 < 0 and deta < (tracea 2 /4. c 2008 Tecnun (University of Navarra 0

11 Example 6. Figure 3: Phase plane of Example 5 Let us consider another system x = 0 0 x (62 0 where A is a hermitian matrix r r r = 0 (63 we obtain r = (double and r 2 = 2. The eigenvectors are r = a = ( 0 T, a 2 = ( 2 T r 2 = 2 a 3 = ( T (64 They are linearly independent. Hence, the solutions are x = 0 e t x 2 = 2 e t x 3 = e 2 t (65 And the general solution is x = c e t + c 2 e t c 3 e 2 t (66 Complex eigenvalues. If A is non hermitian (but real and has complex eigenvalues then the determinant A r I = 0 takes complex conjugates eigenvalues A r I = 0 (67 c 2008 Tecnun (University of Navarra

12 As A and I are real, if we work out the conjugate of eq. 67, we obtain A r I = 0 (68 this means that if r is an eigenvalue, its complex conjugate is eigenvalue as well. Therefore the eigenvectors will be complex conjugates, too: The solution associated to r is (A r I x = 0 (A r I x = 0 (69 x = e r t a = e (λ+i µ t (u + i v = = e λ t (cos(µ t + i sin(µ t (u + i v = = e λ t (u cos(µ t v sin(µ t + i e λ t (u sin(µ t + v cos(µ t (70 And the other one x = e λ t (u cos(µ t v sin(µ t i e λ t (u sin(µ t+v cos(µ t (7 We are looking for real solution to this system. We know that a linear combination of these solutions will be a solution as well. Hence, we can take the real and the imaginary parts of these ones x = c e λ t (u cos(µ t v sin(µ t+c 2 e λ t (u sin(µ t+v cos(µ t (72 Example 7.. ( x /2 = x (73 /2 where A is not symmetric. Solving /2 r /2 r = 0 (74 we get r = /2 ± i. The eigenvector is ( a = i Hence the solution is ( ( x = e ( /2+i t = i i (75 e t/2 (cos(t + i sin(t (76 Then x = ( cos(t + i sin(t sin(t + i cos(t ( e t/2 cos(t = sin(t ( e t/2 sin(t +i cos(t e t/2 (77 c 2008 Tecnun (University of Navarra 2

13 The general solution is x = c e t/2 ( cos(t sin(t ( + c 2 e t/2 sin(t cos(t (78 Plotting this family of curves on the phase plane (x, x 2 Figure 4: Phase plane of Example 7 { x 0 t + x 2 0 (x, x 2 (0, 0 (79 Let us study the points (x 0 T y (0 y T. Substituting into eq. 73 x = (x 0 T x = ( x/2 x T dy dx = 2 x = (0 y T x = (y y/2 T dy dx = 2 (80 Point (0, 0 is a stable spiral point. Thinking of the future once again, det A = 5/4 > 0, tracea = < 0 and deta > (tracea 2 /4. Repeated eigenvalues. Let us consider the system given by eq. 37, with A a real non-symmetric matrix. Let us assume that one of its eigenvalues, r, is repeated (m = m >, and there are not m linearly independent eigenvectors, k < m. We should obtain (m k additional linearly independent solutions. How do we amend the procedure to deal with cases when there are no m linearly independent eigenvectors? For example, the exponential of a square matrix: e At = I + At + A2 t An t n ! n! (8 c 2008 Tecnun (University of Navarra 3

14 We recall the Cayley Hamilton theorem. If f(λ is the characteristic poynomial of a matrix A, f(λ = det (A λ I (82 This theorem says that f(a = 0. For instance, f(λ = r 2 i (trace Ar i + det A = r 2 i p r i + q = 0 (83 Hence, by Cayley-Hamilton theorem, f(a = A 2 p A + q I = 0 (84 using this theorem, any expansion function like e A can be reduced to polynomials on A. This is very useful theorem in problems of materials and continuum mechanics. A 2 = p A q I A 3 = p A 2 q A = ( p 2 q A p q I A 4 = p A 3 q A 2 = p (p 2 2q A (p 2 q q I... So, we can use this theorem to get a finite expansion of eq. 8. Differentiating with respect to t that expression Hence (85 de At = A + A 2 t + A3 t An t n dt (2! (n! + = = A (I + At + + An t n ( = Ae At (n! x = e At v = (I + At + A2 t An t n +... v (87 2! n! where v is a constant vector, then So it looks all right. Hence, dx dt = AeAt v = Ax (88 Av = A a i = r i a i A 2 v = r i A a i = r 2 i a i... A n v = r n i a i (89 Hence, substituing into eq. 87 x = ( + r i t + r2 i t2 2! + + rn i tn n! +... a i = e ri t a i (90 c 2008 Tecnun (University of Navarra 4

15 Note that e (A+B t = e A t e B t A B = B A (9 Let us consider e At v = e (A λi t e λi t v, λ (92 I can do this for all λ since (A λ I(λ I = λ A λ 2 I = (λ I (A λ I (93 Let us come back to eq. 87 e λ I t v = (I + λ I t + λ2 I 2 t λn I n t n +... v = e λ t v (94 2! n! hence e At v = e λ t e (A λ I t v (95 But x = e A t v is a solution of the system. Moreover, let us assume that v = a i where a i is an eigenvector of r i, then ( x = e ri t I + t(a r i I + t2 2! (A r i I a i = ( = e ri t a i + t (A r i I a i + t2 2! (A r i I 2 a i +... = But, as a i is eigenvector, it verifies (96 (A r i I a i = 0 = (A r i I 2 a i = (A r i I 3 a i =... (97 Then eq. 96 becomes We can look for another vector v such that x = e ri t a i (98 (A r i I v 0 y (A r i I 2 v = 0 = (A r i I 3 v =... (99 Lemma. Let us assume that the characteristic polynomial of A, non-hermitian, of order n, have repeated roots (r, r 2,..., r k, k < n, of order (m, m 2,..., m k (m + m m k = n, respectively such that f(λ = (λ r m (λ r 2 m2... (λ r k m k (00 if A has only n j < m j eigenvectors of the eigenvalue r j (i.e., (A r j I v = 0 has n j independent solutions, then (A r j I 2 v = 0 has at least n j + independent solutions. In general, if (A r j I m v = 0 has got n j < m j independent solutions, (A r j I m+ v = 0 has at least n j + independent solutions c 2008 Tecnun (University of Navarra 5

16 Example 8. Let it be the system ( x = 0 x (0 In order to solve, we take r 0 r = 0 (02 the root is r =, double, with only one eigenvector obtained from (A r I a = 0 r = a = ( 0 T (03 We cannot solve the second solution we need. So, we have ( x = e t (04 0 We must determine another vector a 2 such that (A r I a 2 0 (05 (A r I 2 a 2 = 0 = (A r I 3 a 2 =... (06 As (A r I 2 = (A r I 3 = 0, any vector a 2 verifies eq. 06. However, according to the inequality given by eq. 05, a 2 cannot be a linearly dependent vector of a (eq. 03. So ( ( ( a 2 = (0 T 0 0 = 0 ( From eq. 96, we obtain x 2 = e t (a 2 + t (A r I a 2 + t2 2! (A r I 2 a = (( ( ( (08 = e t 0 + t + 0 = e t t 0 The general solution is x = c e t ( 0 ( + c 2 e t t (09 Thinking of next paragraph, deta = > 0, tracea = 2 > 0 and deta = (tracea 2 /4. c 2008 Tecnun (University of Navarra 6

17 Figure 5: Phase plane of Example 8 Résumé of the study of a system of two homogeneous equations with constant coefficients. Let us consider the system its characteristic polynomial is x = a x + b y y = c x + d y (0 f(r = r 2 p r + q ( where p = a + d (the trace y q = a d b c (the determinant. Its eigenvalues are r = p ± p 2 4 q (2 2 The study of eigenvalues and eigenvectors is very useful in order to classify the critical point (0, 0 and to know the trajectories on the phase plane.. q < 0. The eigenvalues are positive and negative, respectively. Saddle point. 2. q > 0. 2.a. p 2 > 4q. Both eigenvalues are either positive or negative. - Stable node, if p < 0. - Unstable node, if p > 0. 2.b. p 2 < 4 q. The eigenvalues are complex conjugates. - Stable spiral, if p < 0. - Unstable spiral, if p > 0. - Center, if p = 0. 2.c. p 2 = 4 q. The eigenvalue is a double root of the characteristic polynomial. - Stable node, if p < 0 and there is only one eigenvector. - Unstable node, if p > 0 and there is only one eigenvector. c 2008 Tecnun (University of Navarra 7

18 - Sink point, if p < 0 and there are two independent eigenvectors. - Source point, if p > 0 and there are two independent eigenvectors. 3. q = 0. It means that the matrix rank is and therefore, one row can be obtained multipliying the other one by a constant (c/a = d/b = k. Then, (0, 0 is not an isolated critical point. There is a line y = a x/b, b 0, of critical points. The trajectories on the phase plane are y = k x+e, E being a constant. - If p > 0, paths start at the critical points and go to the infinity. - If p < 0, conversely, the trajectories end at the critical points. It is very convenient and useful to know the plane trace-determinant (p, q. Fundamental matrix of a system Let us consider the homogeneos system given by eq. 6 and let x, x 2,..., x n be a set of its independent solutions. We know that we can build the general solution via a linear combination of them. We denote fundamental matrix, Ψ(t, a matrix whose columns are the solution vectors of eq. 8. The determinant of this matrix is not zero (eq. 9. This determinant is called wronskian (eq. 20. Let us assume that we are looking for a solution x such that x(t 0 = x 0. Then x 0 = c x + c 2 x c n x n = Ψ(t 0 c (3 where c = (c c 2... c n T. As Ψ(t = 0, t, c can be obtained using the inverse matrix of Ψ(t 0 : c = Ψ (t 0 x 0 (4 and the solution will be This matrix is very useful when x = Ψ(t Ψ (t 0 x 0 (5 Ψ(t 0 = I (6 This special set of solutions builds the matrix Φ(t. It verifies x = Φ(t x 0 (7 We obtain that Φ(t = Ψ(t Ψ (t 0. Moreover, with constant coefficients: (. e A t (eq. 8 is a fundamental matrix of fundamental solutions since it verifies eq. 86 and e A 0 = I. (2. If we know two fundamental matrices of the system, Ψ and Ψ 2, there is always a constant matrix C such that Ψ 2 = Ψ C, since each column of Ψ 2 can be obtained by a linear combination of the columns of Ψ. (3. It can be shown that e A t = Ψ(t Ψ (t 0 (see eq. 5. According to paragraphs and 2, there exists a matrix C such that This expression at t = 0 e At = Ψ(t C (8 I = Ψ(0 C C = Ψ (0 (9 c 2008 Tecnun (University of Navarra 8

19 Nonhomogeneous systems. We have x = P(t x + q(t (20 We assume that we have solved x = P(t x. We actually have a procedure for P(t = A, a constant matrix. Consider special cases ( If P(t = A and A has n independent eigenvectors, the procedure is to build the matrix T with the eigenvectors of A: T = (a a 2... a n (2 Then, we change variables x = T y with x = T y. Going back to the system x = T y = A x + q = A T y + q (22 As T is built with n eigenvectors, this is a regular matrix (det T 0, so we can work out T, the inverse of T. Hence, from eq. 4 y = T A T y + T q = D y + h (23 where D is the diagonal matrix of the eigenvalues of A. Therefore, Hence y i(t = r i y i (t + h i (t, i =, 2,..., n (24 y i (t = e rit e ri t h i (t dt + c i e ri t, i =, 2,..., n (25 After obtaining y i, we can get x = T y. This method is only possible if there are n linearly independent eigenvectors, i.e., A is a diagonalizable constant matrix. Because we can reduce the matrix to its diagonal form, the above procedure works. In cases where we do not have the n independent eigenvectors, the matrix A can only be reduced to its Jordan canonical form. (2 Variation of the parameters. Let us consider the system of eq. 4, and we know the solution of the associated homogeneous system (eq. 6. Then we can build the fundamental matrix of the system, Ψ(t (eq. 8, whose columns are the linearly independent of the homogeneous system solutions. We are looking for a solution like x = Ψ(t u(t (26 c 2008 Tecnun (University of Navarra 9

20 where u(t is a vector to be determined such that eq. 8 is a solution of eq. 4. Substituting Ψ (t u(t + Ψ(t u (t = P(t Ψ(t u(t + q(t (27 as we know that Ψ (t = P(t Ψ(t, Ψ(tu (t = q(t u(t = Ψ (tq(tdt + c (28 where c is a constant vector and there exists Ψ (t since the n columns of matrix Ψ(t are linearly independent (eq. 20. The general solution is x = Ψ(t Ψ (t q(t dt + Ψ(t c (29 c 2008 Tecnun (University of Navarra 20