Differential Equations Exercise Set

Exercise Set 4 (HW for 7.6, 7.8) 1. Solve x0 = Ax, where A is given below: Tr(A) = 2 λ = 1 + 2i det(A) = 5 Spiral Source (a) A = ∆ = 4 − 20 = −16 v = [1 + i, 2]T The solution will be C1 Re(eλt v) + C2 Im(eλt v)  3 −2 4 −1       cos(2t) − sin(2t) cos(2t) + sin(2t) x(t) = e C1 + C2 2 cos(2t) 2 sin(2t) t Tr(A) = 2 λ = 1, 1 det(A) = 1 Degen Source (b) A = ∆=4−4=0 v = [2, 1]T We need to compute the generalized eigenvector w that satisfies the equation:   (3 − 1)w1 − 4w2 = 2 1 ⇒ w= w1 + (−1 − 1)w2 = 1 0  3 −4 1 −1  The solution is therefore x(t) = eλt (C1 v + C2 (tv + w)), which is         2 2 1 t x(t) = e C1 + C2 t + 1 1 0 Tr(A) = 0 λ=i det(A) = 1 Center ∆ = −4 v = [5, 2 − i]T The solution will be C1 Re(eλt v) + C2 Im(eλt v)  (c) A = 2 −5 1 −2    x(t) = C1 5 cos(t) 2 cos(t) + sin(t)   + C2 5 sin(t) 2 sin(t) − cos(t)  Tr(A) = −2 λ = −1, −1 det(A) = 1 Degen Sink ∆=4−4=0 v = [2, 1]T We need to compute the generalized eigenvector w that satisfies the equation:   (−3/2 + 1)w1 + w2 = 2 0 ⇒ w= (−1/4)w1 + (−1/2 + 1)w2 = 1 2  (d) A = −3/2 1 −1/4 −1/2  The solution is therefore x(t) = eλt (C1 v + C2 (tv + w)), which is         2 2 0 −t x(t) = e C1 + C2 t + 1 1 2 1 Tr(A) = −2 λ = −1 + i det(A) = 2 Spiral Sink (e) A = ∆ = 4 − 8 = −4 v = [2 + i, 5]T The solution will be C1 Re(eλt v) + C2 Im(eλt v)  1 −1 5 −3  x(t) = e −t   C1 2 cos(t) − sin(t) 5 cos(t)   + C2 2 sin(t) + cos(t) 5 sin(t)  2. For 1(a) and 1(b), convert to a second order differential equation and solve by methods from Chapter 3. SOLUTION: Looking at the solution in 1(a), I should probably solve for the differential equation in x2 . However, in this case, we’ll solve for x1 and then just point to how we would compute x2 - It will not be in exactly the same form as our solution to 1(a), but it will be equivalent. For 1(a), solving the first equation for x2 , we get 1 x2 = − (x01 − 3x1 ) 2 Substitution into the second equation gives us: 1 1 − (x001 − 3x01 ) = 4x1 + (x01 − 3x1 ) 2 2 Multiply both sides by −2 and simplify to get: x001 − 2x01 + 5x1 == 0 ⇒ r2 − 2r + 5 = 0 ⇒ r = 1 ± 2i Therefore, x1 (t) = et (C1 cos(2t)+C2 sin(2t)). To find x2 , substitute x1 into the equation above- This is a bit of a pain, which is why we are computing these solutions using eigenvalues and eigenvectors. For 1(b), I will try solving for the other function first. Then: x1 = x02 + x2 , and substitution into the first equation: x002 + x02 = 3(x02 + x2 ) − 4x2 ⇒ x002 − 2x02 + x2 = 0 The characteristic polynomial is r2 − 2r + 1 = 0, so we get r = 1, 1 and x2 (t) = et (C1 + C2 t) To find x1 , we compute x02 + x2 to get: x1 = x02 + x2 = et (2C1 + C2 (2t + 1)) and you can verify that this is what we had before in 1(b). 2 3. Given the eigenvalues and eigenvectors for some matrix A, write the general solution to x0 = Ax. Furthermore, classify the origin as a sink, source, spiral sink, spiral source, saddle, or none of the above.   1−i (a) λ = −1 + 2i v= 2      cos(2t) + sin(2t) sin(2t) − cos(2t) −t C1 + C2 x(t) = e 2 cos(2t) 2 sin(2t) (Spiral Sink)  (b) λ = −2, 3 v1 =  1 , v2 = −1     1 1 3t −2t + C2 e x(t) = C1 e −1 2 1 2   (Saddle)  (c) λ = −2, −2 v=  1 1 x(t) = e  , −2t w=   C1 1 1 3 1        1 3 + C2 t + 1 1 (Degenerate Sink)  (d) λ = 2, −3 v1 = −1 2   2 −1   −1 2  v2 = x(t) = C1 e 2t + C2 e −3t  2 −1  (Saddle)   1 (e) λ = 1 + 3i v= 1−i      cos(3t) sin(3t) t x(t) = e C1 + C2 cos(3t) + sin(3t) sin(3t) − cos(3t) (Spiral Source)  (f) λ = 2i  1+i v= 1      cos(2t) − sin(2t) sin(2t) + cos(2t) x(t) = C1 + C2 cos(2t) sin(2t) (Center) 3 4. (Also see 13-20, p. 410) For each system x0 = Ax, the matrix A will depend upon the parameter α: (i) Determine the eigenvalues in terms of α, (ii) Find the critical values of α where the behavior of the solution to the system changes significantly. We’ll go through one or two in class. Tr(A) = 2α det(A) = α2 + 1 (a) ∆ = (2α)2 − 4(α2 + 1) = −4 In the Poincaré diagram, the determinant is positive and the discriminant is negative puts us “inside” the parabola. Therefore:  α 1 −1 α  • If α < 0, • If α = 0, • If α > 0,   0 −5 (b) 1 α the origin is a spiral sink. the origin is a center. the origin is a spiral source. Tr(A) = α det(A) = 5 ∆ = α2 − 20 In the Poincaré diagram, you can think of us √ being on the horizontal line at det(A) = 5. We see that ∆ = 0 when α = ±2 5, which tells us when we are inside or outside the parabola: √ • If α < −2 5, the origin is a sink. √ • If α = −2 5, the origin is a degenerate sink. √ • If −2 5 < α < 0, the origin is a spiral sink. • If α = 0, the origin is a center. √ • If 0 < α < 2 5, the origin is a spiral source. √ • If α = 2 5, the origin is a degenerate source. √ • If α > 2 5, the origin is a source.   Tr(A) = 0 2 −5 det(A) = −4 + 5α (c) α −2 ∆ = −4(−4 + 5α) On the diagram, we are on the vertical axis, so the classification depends only on whether the determinant is positive or negative: If α > 4/5, then we have a center. If α = 4/5, we have “uniform motion” (we have a double zero eigenvalue), and if α < 4/5, we have a saddle.    Tr(A) = −1 3 α 2 −5 det(A) = 6(α − 2) (d) −6 −4 α −2 ∆ = 49 − 24α We are on a vertical line where the trace is −1, and we have two critical points for α (α = 2 and α = 49/24 ≈ 2.04). • For α < 2, the determinant is negative and the origin is a saddle 4 • • • • For For For For α = 2, we have a “line of stable fixed points” 2 < α < 2.04, we have a sink. α = 2.04, we have a degenerate sink. α > 2.04, we have a spiral sink. 5