Math 102. Rumbos Spring 2015 1 Solutions to Assignment #4 Do
Math 102. Rumbos Spring 2015 1 Solutions to Assignment #4 Do the following problems 1. Explain why the following linear system of differential equations cannot be put in diagonal form. dx dt = 2x + y; dy = −x + 4y. dt Solution: The linear system cannot be put in diagonal form because the matrix 2 1 A= −1 4 is not diagonalizable. Indeed, the characteristic polynomial of A is −A (λ) = λ2 − 6λ + 9 = (λ − 3)2 , and therefore λ = 3 is the only eigenvalue. To find the eigenspace corresponding to λ, solve the equation (A − λI)v = 0, 1 to get that solutions are generated by the single eigenvector v = . Hence, 1 there is no basis for R2 made up of eigenvectors of A. Hence, A is not diagonalizable. 2. Explain why the following linear system of differential equations cannot be put in diagonal form. dx dt = ay; dy = −bx, dt where a and b are positive constants 0 a Solution: In this case, the characteristic polynomial of the matrix A = −b 0 2 associated with √ the system is pA (λ) = λ + ab. Thus, the solutions of pA (λ) = 0 are λ = ±i ab, since ab > 0. Thus, the eigenvalues of A are not real, and therefore A cannot be put in diagonal form over the reals. Math 102. Rumbos Spring 2015 2 3. Explain why the following linear system of differential equations cannot be put in diagonal form. dx dt = −2x − 3y; dy = dt 3x − 2y. Solution: In this case, the characteristic polynomial of the matrix −2 −3 A= 3 −2 associated with the system is pA (λ) = λ2 + 4λ + 13 = (λ + 2)2 + 9; so that, pA (λ) = 0 has no real solutions. Therefore, A doesn’t have real eigenvalues. Hence, the system can not be put in diagonal form. 4. Determine whether or not the linear system dx dt = y; (1) dy = 8x − 2y dt can be put in diagonal form. If so, give the general solution to the system and sketch the phase portrait. Solution: Write the system in matrix form to get x˙ x =A , y˙ y where A= 0 1 8 −2 is the matrix associated with the system. Its characteristic polynomial is pA (λ) = λ2 + 2λ − 8, which factors into pA (λ) = (λ + 4)(λ − 2). Hence, the eigenvalues of A are λ1 = −4 and λ2 = 2. We find eigenvectors associated with λ1 and λ2 to get 1 1 v1 = and v2 = , −4 2 (2) Math 102. Rumbos Spring 2015 respectively. Solutions of the system in (1) are then given by x(t) = c1 e−4t v1 + c2 e2t v2 , y(t) 3 (3) where v1 and v2 are given in (2) In order to sketch the phase portrait, we use the information in (3) to sketch trajectories along lines spanned by the eigenvectors in (2). We first note that, if y x Figure 1: Sketch of Phase Portrait of System (1) c1 = c2 = 0, (3) yields the equilibrium solution, (0, 0). If c1 6= 0 and c2 = 0, the trajectories will lie on the line spanned by the vector v1 ; both trajectories will point towards the origin because the exponential e−4t decreases with increasing t. On the other hand, if c1 = 0 and c2 6= 0, the trajectories lie on the line spanned by the vector v2 and point away from the origin because the exponential e2t increases with increasing t. These five trajectories are shown in Figure 1. In order to sketch the solution curves off the lines spanned by the eigenvectors of A, we use the information contained in the eigenvalues of A; in particular, since λ1 and λ2 have opposite signs, the origin has the structure of a saddle point. Accordingly, we sketch a few of those trajectories in Figure 1. The directions Math 102. Rumbos Spring 2015 4 along those trajectories has to be consistent with the directions of trajectories along the lines spanned by v1 and v2 . 5. Determine whether or not the linear system dx dt = −3x + 2y; (4) dy = 4x − 5y dt can be put in diagonal form. If so, give the general solution to the system and sketch the phase portrait. Solution: Write the system in matrix form to get x˙ x =A , y˙ y where −3 2 A= 4 −5 is the matrix associated with the system. Its characteristic polynomial is pA (λ) = λ2 + 8λ + 7, which factors into pA (λ) = (λ + 7)(λ + 1). Hence, the eigenvalues of A are λ1 = −7 and λ2 = −1. We find eigenvectors associated with λ1 and λ2 to get 1 1 v1 = and v2 = , −2 1 respectively. Solutions of the system in (1) are then given by x(t) = c1 e−7t v1 + c2 e−t v2 , y(t) (5) (6) where v1 and v2 are given in (5) In order to sketch the phase portrait, it is helpful to sketch the phase portrait in the uv–plane of the system resulting from the transformation u −1 x =Q ; (7) v y Math 102. Rumbos Spring 2015 5 namely, du dt = −7u; (8) dv = −v. dt Solutions to the system in (8) are given by −7t u(t) ce = 1 −t , for t ∈ R. v(t) c2 e (9) It follows from (9) that the phase portrait of the system in (8) has trajectories along the u and v axes pointing towards the origin. These, as well as the equilibrium solution, (0, 0), are shown in green in Figure 2. To get the trajectories v u Figure 2: Sketch of Phase Portrait of System (8) off the axes in the uv–plane, eliminate the parameter t from the parametric equations u = c1 e−7t and v = c2 e−t , Math 102. Rumbos Spring 2015 6 to get u = cv 7 , (10) where we have written c for c1 /c72 . A few of the trajectories on the curves in (10) are shown in the phase portrait in Figure 2. In this case, all the trajectories point towards the origin. We use the information in (6) and the phase portrait in Figure 2 to sketch the phase portrait of the system in (4). We first note that, if c1 = c2 = 0, (6) y x Figure 3: Sketch of Phase Portrait of System (4) yields the equilibrium solution, (0, 0). If c1 6= 0 and c2 = 0, the trajectories will lie on the line spanned by the vector v1 ; both trajectories will point towards the origin because the exponential e−7t decreases with increasing t. Similarly, if c1 = 0 and c2 6= 0, the trajectories lie on the line spanned by the vector v2 and point towards the origin because the exponential e−t increases with increasing t. These five trajectories are shown in Figure 3. In order to sketch the solution curves off the lines spanned by the eigenvectors of A, we use the information contained in the phase portrait in Figure 2. A few of the transformed trajectories are shown in the phase portrait in Figure 3.
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