Homework 3 Solutions Math 309, Fall 2015

Transcription

1 Homework 3 Solutions Math 39, Fall One easily checks that the only eigenvalue of the coefficient matrix is λ To find the associated eigenvector, we have 4 2 v v 8 4 (up to scalar multiplication) 2 v 2 Since the corresponding eigenspace is clearly one-dimensional, the eigenvalue λ is deficient, and so there exists an associated generalized eigenvector w To find it, we solve ( ) w w 2 2w 2 w 2 2 Therefore, our generalized eigenvector will have the form w + k /2 2 Here k is an arbitrary constant, and so we set it equal to and plug it into our formula for the solution: [ ] x(t) c + c 2 2 t + 2 /2 785 We compute the characteristic polynomial of the coefficient matrix to be λ 3 + 3λ 2 4, and looking for small integer roots we observe that λ is such a root Polynomial division shows that λ 2 is the only other root We first find the eigenvector associated to the (simple) eigenvalue λ : v v 2 v 3

2 Solving this system however you like, we find that, up to scalar multiplication, 3 v 4 2 We next compute the eigenspace associated to λ 2: v 2 v 2 v 3 Again solving this system any way you like, we conclude that the eigenspace is one-dimensional and generated by v 2 Since the eigenvalue λ 2 thus has geometric multiplicity and algebraic multiplicity 2, it is deficient and so there exists an associated generalized eigenvector: 2 w w 2 w 3 The last equation here implies w 2 + w 3, and plugging that into the second equation gives us w We thus conclude that our generalized eigenvector has the form w + k Plugging this all into our solution formula when n 3, we find that 3 x(t) c e t 4 + c 2 e 2t + c 3 te 2t + e 2t 2 2

3 Note: The answer in your textbook looks a bit different than this, and this is because they chose a generalized eigenvector which corresponds to k in my computation, that is The characteristic polynomial is given by λ det λ λ 3 + 3λ + 2 λ Checking for small integer roots and using polynomial division, we see that this factors as (λ 2)(λ + ) 2 and hence our eigenvalues are λ 2 and λ Computing the eigenvector for λ 2 amounts to solving v v 2 v 3 Since the third row is the sum of the first two, this is equivalent to solving the system { 2v + v 2 + v 3 v 2v 2 + v 3 It s easy to see (for example, by subtracting the first equation from the second to get 3v 3v 2 and then plugging that result into either equation) that the solution to this system is v v 2 v 3, so we set v 2 Now to find the eigenspace associated to λ we solve v v 2, v 3 3

4 which is easily seen to be equivalent to the equation v v 2 v 3 Our eigenspace thus consists of vectors of the form v v 2 v 3 v v 2 v 2 v 2 + v 3 v 3 v 3 This, of course, means that the eigenspace E is two-dimensional and spanned by the eigenvectors v () and v (2) Since the algebraic and geometric multiplicity of λ are both 2, this eigenvalue is not deficient and there is thus no need to look for generalized eigenvectors By our usual formula, the general solution to this problem thus has the form x(t) c e 2t + c 2 e t + c 3 e t 785 Let P be the coefficient matrix Observe that the characteristic polynomial for P is given by a λ b det λ 2 (a + d)λ + ad bc () c d λ Observe that in each of the formulas we have learned for the general solution of the homogeneous equation x P x, the only way we will always have x(t) as t is if all eigenvalues have Re(λ) < (ie, negative real part) Now if r and r 2 are the two (possibly identical) eigenvalues of P, the characteristic polynomial factors as (λ r )(λ r 2 ) λ 2 (r + r 2 )λ + r r 2 (2) There are two cases to consider: either r, r 2 are both real or they are complex conjugates In the former case, the roots are both negative if and only if their 4

5 sum is negative and their product is positive Comparing () and (2), this is equivalent to r + r 2 a + d < and r r 2 ad bc > as claimed Now if r and r 2 are complex conjugates, we automatically have r r 2 with equality holding if and only if r r 2 (recall that (a + bi)(a bi) a 2 + b 2 ) Since complex conjugates have the same real part, they will both have negative real part if and only if their sum is negative and their product, being nonzero, is positive, so the conclusion again follows by comparing () and (2) 794 We will do this and the following problems from Section 79 using diagonalization/jordan normal forms One easily computes the characteristic polynomial of the coefficient matrix P to be λ 2 +λ 6, and so the eigenvalues of P are λ 2 and λ 3 We find that the corresponding eigenspaces are spanned by v 2 and v 3 4 respectively We thus compute the diagonalizing matrix T and its inverse to be T T 4 4/5 /5 4 5 /5 /5 Setting x T y as usual, the given equation is equivalent to 2 y y + h 3 where (4/5)e h T g 2t (2/5)e t (/5)e 2t + (2/5)e t This, of course, implies that y satisfies y 2y e 2t 2 5 et d dt (e 2t y ) 4 5 e 4t 2 5 e t and so integrating yields e 2t y 5 e 4t e t + c y (t) 5 e 2t et + c e 2t 5

6 A very similar computation gives us that y 2 (t) 5 e 2t + et + c 2 e 3t, and so our general solution is given by (/5)e x(t) T y 2t + (2/5)e t + c e 2t 4 (/5)e 2t + (/)e t + c 2 e 3t (/2)e t + c e 2t + c 2 e 3t e 2t + c e 2t 4c 2 e 3t We express our answer in the more illuminating form (/2)e t x(t) e 2t + c e 2t + c 2 e 3t The characteristic polynomial of the coefficient matrix P is λ 2 2λ 3, and so the eigenvalues of P are λ 3 and λ We easily find associated eigenvectors: v 3 2 and v 2 We thus compute the diagonalizing matrix T and its inverse: T T 2 /2 / /2 /4 Letting x T y, we thus have where y 3 y + h (3/4)e h T t g (5/4)e t Solving for y and y 2 as in the previous problem, we find that (3/8)e y t + c e 3t (5/8)e t + c 2 e t 6

7 and hence (3/8)e x(t) T y y t + c e 3t (5/8)e t + c 2 e t (/4)e t + c e 3t + c 2 e t 2e t + 2c e 3t 2c 2 e t We again record our final answer in the form (/4)e t x(t) 2e t + c e 3t + c 2 2 e t 2 79 (BONUS!) This problem would be straightforward using the method of undetermined coefficients, but it is a bit tricky to do using a complex diagonalizing matrix One easily checks that the eigenvalues of the coefficient matrix are λ i and λ i To find the corresponding (complex) eigenvector for λ i, we must solve ( 2 i 5 2 i ) v v 2 5 v i 2 i (up to scalar mult) Now we learned that the eigenvector corresponding the the conjugate eigenvalue is simply the conjugate of this, and so a diagonalizing matrix for the coefficient matrix is 5 5 T 2 i 2 + i whose inverse is given by T i ( 2 + i i 5 ) (/5)i + / (/2)i (/5)i + / (/2)i Hence, defining h as in the previous problems, we have (/5)i + / (/2)i h T g (/5)i + / (/2)i cos t (/2)i cos t (/2)i cos t At this point, we recall the following two important identities, both of which follow directly from Euler s formula: cos t 2 (eit + e it ) (3) sin t 2 i(eit e it ) (4) 7

8 Now setting x T y as usual (observe that y could thus be complex), we have that i (/2)i cos t y y + i (/2)i cos t Let s solve the first component equation using (3): y iy + 4 ieit + 4 ie it d dt (e it y ) 4 i + 4 ie 2it Integrating yields e it y 4 it 8 e 2it + c For the purposes of this problem, we will assume all arbitrary constants are zero because we are only interested in finding the particular solution (we learned how to find the general solution in Section 76) We thus conclude that y (t) 4 iteit 8 e it Rather than going through a nearly identical computation to compute y 2, observe by conjugating the differential equation for y 2 that y 2 : y is a solution for the second component equation We thus conclude (with zero arbitrary constants) that We thus conclude that x(t) T y y 2 (t) 4 ite it 8 eit ( 5 5 ) 4 iteit 8 e it 2 i 2 + i 4 ite it 8 eit Multiplying this out, we see that ( ) 5 x(t) 4 ti(eit e it ) 5 8 (eit + e it ) 2 ti(eit e it ) + 4 t(eit + e it ) 4 (eit + e it ) 8 i(eit e it ) Though this looks messy, I ve suggestively grouped terms so that we can apply the identities (3) and (4): ( 5 x(t) t sin t 5 cos t ) 2 4 t sin t + t cos t cos t + sin t

9 Note: As has happened before, this particular solution is not the precise particular solution given in the back of your textbook However, recall that if one adds any solution to the homogeneous problem x P x to a particular solution of the nonhomogeneous equation, then we will obtain a new particular solution So what s happened here is that my particular solution and the book s differ by a solution to the homogeneous problem Seeing that the book expresses this homogeneous part of the solution as ( ) ( ) 5 cos t 5 sin t x hom (t) c + c 2 cos t + sin t 2, cos t + 2 sin t we observe that if we choose c /4 and c 2 and add the resulting function to the book s particular solution, you will obtain mine 796 Since φ is a general solution to the nonhomogeneous problem, it satisfies φ P φ + g Now v, being a particular solution to the same nonhomogeneous problem, satisfies v P v + g Subtracting the second equation from the first gives us φ v P (φ v), and so clearly u : φ v is a solution to the homogeneous equation x P x Then since φ φ v + v u + v, the assertion is proved 9