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Use of eigenvalues and eigenvectors for solving coupled differential equations Example: Consider a system of two first-order coupled linear ODEs for two functions of time, v(t) and w(t): ( dv dt = 4v(t) − 5w(t) dw dt = 2v(t) − 3w(t) with initial conditions ( v(0) = 8 w(0) = 5 . We can re-write this system in a matrix form: A d v(t) 4 −5 v(t) = 2 −3 w(t) dt w(t) | {z } | {z } ~ (t) U and the initial conditions ~0 = 8 U 5 Then we have d~ ~ (t), U (t) = AU dt a linear coupled system with constant coefficients (A6= A(t)). v(t) ~ A general solution for this type of system U (t) = represents a linear combination of its w(t) linearly independent particular solutions. In this case, i.e. for two unknown variables, we need two of them: ~ (t) = c1 U ~ 1 (t) + c2 U ~ 2 (t) U . The particular solutions for the linear coupled equations are sought in the same form as a solution for a scalar form of the first-order linear ODE: du = au dt u(0) = u0 u(t) = u0 eat Indeed, this exponent is a solution: d d u0 eat = a u0 eat = au(t). u(t) = dt dt Now let us assume particular solutions in the exponential form: ( v(t) = x1 eλt w(t) = x2 eλt 10 Then the system will look like this ( λx1 eλt = 4x1 eλt − 5x2 eλt λx2 eλt = 2x1 eλt − 3x2 eλt , where eλt is a common non-zero term. We can cancel it and obtain a system of two linear equations: ( λx1 = 4x1 − 5x2 . λx2 = 2x1 − 3x2 In matrix from we have A~x = λ~x, i.e. the definition of eigenvectors and eigenvalues! We have already solved such problems. First, find eigenvalues: det (A − λI) = 0 (4 − λ) −5 = (4 − λ)(−3 − λ) − 2(−5) = 0 2 (−3 − λ) λ1,2 λ2 − λ − 2 = 0 p √ 1 ± 1 + 4(1)(2) −b ± b2 − 4ac 1±3 = = = 2a 2(1) 2 λ1 = −1 λ2 = 2. Now let us find eigenvectors: (A − λI) ~x = 0. For λ1 = −1, the linearly dependent equations are: ( (4 + 1)x1 − 5x2 = 0 2x1 + (−3 + 1)x2 = 0 ( 5x1 − 5x2 = 0 2x1 − 2x2 = 0 The respective eigenvector ~xλ1 For λ2 = 2, the first equation is 1 = 1 (4 − 2)x1 − 5x2 = 2x1 − 5x2 = 0 and ~xλ2 Now we found two particular solutions: 1 = 0.4 λ1 t −t 1 ~ U1 = e ~xλ1 = e 1 11 ~ 2 = eλ2 t ~xλ2 = e2t U 1 0.4 Let us write the general solution as a linear combination of the particular ones 1 −t 1 2t ~ ~ ~ U = c1 U1 + c2 U2 = c1 e + c2 e 1 0.4 8 and use the initial conditions U (0) = to find the unknown scaling coefficients c1,2 . All expo5 nential terms will equate to 1, so we get a linear system again: 1 1 8 c1 + c2 = 1 0.4 5 or ( c1 + c2 = 8 c1 + 0.4c2 = 5 Subtract the second equation from the first one 0.6c2 = 3 → c2 = 5 Then c1 + 5 = 8 → c1 = 3. Finally, we have got 1 −t 1 2t ~ U = 3e + 5e 1 0.4 so that the solutions for v(t) and w(t) are ( v(t) = 3e−t + 5e2t w(t) = 3e−t + 2e2t . Note: for a system first-order coupled linear equations, the eigenvalues represent growth (positive power of the exponent) or decay (negative power) rates. This analysis can be extended to the second-order linear systems. 12 Decoupling coupled equations using matrix diagonalisation There is another way to apply eigenvalues and eigenvectors for solving systems of coupled linear ODEs. Let us consider a following example: ( dx dt = 4x + 2y dy dt = −x + y with initial conditions x(0) = 1, y(0) = 0. Let use the same notation as before ~ = x U y and write the matrix-vector form of the system d~ ~, U = AU dt where A= 4 2 . −1 1 Its eigenvalues and eigenvectors can be easily determined: 1 λ1 = 3 → ~xλ1 = −0.5 1 λ1 2 = 3 → ~xλ2 = −1 We can diagonalise matrix A: A = SΛS −1 where 3 0 1 1 . , Λ= S= 0 2 −0.5 −1 Now let us introduce another vector for which a(t) ~ V =S= , b(t) ~ = SV ~. U Back to the matrix equation: d~ d ~ d~ U= SU = S V dt dt dt because S is time-independent. On the other hand, d~ ~ U = AU dt so that S d~ ~ ~) → dV ~ = S −1 AS V V = A(S V | {z } . dt dt Λ 13 Now we have a new system with a diagonal system matrix: d~ ~, V = ΛV dt which means that the variables a(t) and b(t) are actually decoupled d a(t) 3 0 a(t) = 0 2 b(t) dt b(t) ( d dt a(t) = 3a(t) d dt b(t) = 2b(t) We can write separate solutions a(t) = k1 e3t , b(t) = k2 e2t . Now let us find x and y from a and b. By definition, ~ = SV ~, U so that ( x(t) a(t) 1 1 a(t) =S = , y(t) b(t) −0.5 −1 b(t) x(t) = a + b = k1 e3t + k2 e2t y(t) = −0.5a − b = −0.5k1 e3t − k2 e2t . In order to find the unknown coefficients k1,2 , we need to use the initial conditions at t = 0. The exponents will be equal to 1 again, so that we get: ( x(0) = 1 = k1 + k2 y(0) = 0 = −0.5k1 − k2 From the second equation, k1 = −2k2 so that k2 = −1 and k1 = 2. The general solution is x(t) = 2e3t − e2t y(t) = −e3t + e2t . Again, the eigenvalues here are growth/decay rates. 14 Solving second-order coupled systems of linear ODE It is possible to perform almost exactly the same procedure to solve a second-order coupled system of linear ordinary differential equations. However, in this case the eigenvalues will represent characteristic frequencies of the system rather than its growth/decay rates, in contrast to first-order ODEs. Consider a mechanical system consisting of two masses, m1 and m2 , on a horizontal surface. The masses are connected by a spring with a spring constant k. Let us find time dependence of their displacements ∆x1 (t) and ∆x2 (t) from the respective equilibrium positions x1,0 and x2,0 (see Fig.1) assuming that the friction forces are negligible. Figure 1: Spring and two masses. Recalling the Hooke’s Law F = −k∆x, we can write the balance of forces equations for both masses accounting for only for the force exerted by the spring: ( 2 1 = −k∆x1 + k∆x2 m1 d dt∆x 2 . 2 d ∆x2 m2 dt2 = k∆x1 − k∆x2 Re-write the system as a matrix-vector product: # " k − mk1 d2 ∆x1 ∆x1 m1 = 2 , k − mk2 ∆x2 dt ∆x2 m2 d2 ∆~x. dt2 Now, by analogy with the scalar second-order linear DE (wave equation), write particular solutions for both displacements as periodic functions (complex exponents): A∆~x = ∆x1 (t) = c1 eiωt ∆x2 (t) = c2 eiωt . The equations will look like " − mk1 k m2 k m1 − mk2 # ✟ iωt ✟ e c1 ∆x1 ✟ 2 iωt = −ω ✟ e , c2 ∆x2 so that −ω 2 will represent an eigenvalue, A∆~x = −ω 2 ∆~x. 15 The characteristic frequencies an be found as square roots of the eigenvalues from det A + ω 2 I = 0, which in this case would represent a fourth-order polynomial with respect to ω but only a secondorder one with respect to λ = −ω 2 : k k k2 2 2 ω − = 0, ω − − m1 m2 m1 m2 ω4 − k 1 k2 1 k2 + − = 0, ω2 + m1 m1 m1 m2 m1 m2 1 1 2 2 + ω ω −k = 0. m1 m2 The eigenvalues are ω12 = 0, 1 1 2 + ω2 = k m1 m2 and the respective characteristic frequencies ω1 = 0, s 1 1 + . ω2 = k m1 m2 The eigenvectors are calculated from the system equation: A + ω 2 I ∆~xω = 0, by substituting ω12 and ω22 . For ω12 = 0 we have " − mk1 so that − k m2 k m1 − mk2 ∆~xω1 1 m1 + 1 m2 ∆x1 =0 ∆x2 k k ∆x1 + ∆x2 = 0 m1 m2 so that ∆x1 = ∆x2 and For ω22 = k # 1 = . 1 , the respective eigenvector will be (check it yourself!) ∆~xω2 = 1 . −1 Then the general solution for the system will be a linear combination of the particular solutions: ~ = b1 e−iω1 t ∆~xω1 + b2 e−iω2 t ∆~xω2 U 16 r −i k m1 + m1 t 1 1 1 2 , + b2 e = b1 −1 1 where unknown coefficients b1,2 can be found from specifying the initial conditions at t = 0, i.e. ∆x1 (0) and ∆x2 (0). The first term is time-independent (ω = 0) and corresponds to a constant displacement of the masses from their initial position. In our case, b1 = ∆x1 (0) + ∆x2 (0) , 2 b2 = ∆x1 (0) − ∆x2 (0) . 2 17