Solutions Chapter 9. u. (c) u(t) = 1 e t + c 2 e 3 t! c 1 e t 3c 2 e 3 t. (v) (a) u(t) = c 1 e t cos 3t + c 2 e t sin 3t. (b) du

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1 Solutions hapter 9 dode 9 asic Solution Techniques 9 hoose one or more of the following differential equations, and then: (a) Solve the equation directly (b) Write down its phase plane equivalent, and the general solution to the phase plane system (c) Plot at least four representative trajectories to illustrate the phase portrait (d) hoose two trajectories in your phase portrait and graph the corresponding solution curves u(t) Explain in your own words how the orbit and the solution graph are related (i) u + 4u =, (ii) u 4u =, (iii) u + u + u =, (iv) u + 4u + u =, (v) u u + u = (i) (a) u(t) = c cos t+c sin t (b) du = 4 (ii) (a) u(t) = c e t + c e t (b) du = 4 (iii) (a) u(t) = c e t +c te t (b) du = u (c) u(t) = c cos t + c sin t c sin t + c cos t c u (c) u(t) = e t + c e t c e t + c e t c u (c) u(t) = e t + c te t (c c )e t c te t (iv) (a) u(t) = c e t +c e t (b) du = c u (c) u(t) = e t + c e t 4 c e t c e t (v) (a) u(t) = c e t cos t + c e t sin t (b) du = u c (c) u(t) = e t cos t + c e t sin t (c + c )e t cos t + (c c )e t sin t 9 (a) onvert the third order equation d u + d u + 4 du + u = into a first order system in three variables by setting u = u, u = u, u = u (b) Solve the equation directly, and then use this to write down the general solution to your first order system (c) What is the dimension of the solution space? (a) du = u 4 (b) u(t) = c e t + c cos t + c sin t, u(t) = (c) dimension = c e t + c cos t + c sin t c e t c sin t + c cos t 9c e t 4c cos t 4c sin t 9 onvert the second order coupled system of ordinary differential equations u = a u + b v + cu + dv, v = p u + q v + r u + sv, into a first order system involving four variables, ds 9/9/4 469 c 4 Peter J Olver

2 Set u = u, u = u, u = v, u 4 = v Then du = c a d b u r p s q 94 True or false: The phase plane trajectories (99) for ( c, c ) T are hyperbolas False there is no quadratic equation in u, u that they satisfy because u, u u, u, u, u and are linearly independent functions 95 (a) Show that if u(t) solves u = u, then its time reversal, defined as v(t) = u( t), solves v = v, where = (b) Explain why the two systems have the same phase portraits, but the direction of motion along the trajectories is reversed (c) pply time reversal to the system(s) you derived in Exercise 9 (d) What is the effect of time reversal on the original second order equation? (a) Use the chain rule to compute dv = du ( t) = u( t) = v (b) Since v(t) = u( t) parametrizes the same curve as u(t), but in the reverse direction (c) (i) dv = 4 (ii) dv = 4 (iii) dv = (iv) dv = 4 (v) dv = u; solution: u(t) = u; solution: u(t) = u; solution: u(t) = u; solution: u(t) = u; solution: u(t) = c cos t c sin t c sin t + c cos t c e t + c e t c e t + c e t c e t c te t (c c )e t + c te t c e t + c e t c e t c e t c e t cos t c e t sin t (c + c )e t cos t (c c )e t sin t (vi) Time reversal changes u (t) = u(t) into v (t) = u ( t) = u( t) and u (t) = u(t) into v (t) = u ( t) = u( t) = v(t) The net effect is to change the sign of the coefficient of the first derivative term, so d u + a du + bu = becomes d v a dv + bv = 96 (a) Show that if u(t) solves u = u, then v(t) = u(t) solves v = v, where = (b) How are the solution trajectories of the two systems related? (a) Use the chain rule to compute dv = du (t) = u(t) = v(t), and so the coefficient matrix is multiplied by (b) The solution trajectories are the same, but the solution moves twice as fast (in the same direction) along them 97 True or false: Each solution to a phase plane system moves at a constant speed along its trajectory False If u = u then the speed at the point u(t) on the trajectory is u(t)) So the speed is constant only if u(t)) is constant (Later this will be shown to correspond to being a skew-symmetric matrix) 98 Use a three-dimensional graphics package to plot solution curves ( t, u (t), u (t) ) T of the phase plane systems in Exercise 9 Discuss their shape and explain how they are related the phase plane trajectories ds 9/9/4 47 c 4 Peter J Olver

3 The solution curves project to the phase plane trajectories 99 first order linear system u = au + bv, v = cu + dv, can be converted into a single second order differential equation by the following device ssuming b, solve the system for v and v in terms of u and u Then differentiate your equation for v with respect to t, and eliminate v from the resulting pair of equations The result is a second order ordinary differential equation for u(t) (a) Write out the second order equation in terms of the coefficients a, b, c, d of the first order system (b) Show that there is a one-to-one correspondence between solutions of the system and solutions of the scalar differential equation (c) Use this method to solve the following linear systems, and sketch the resulting phase portraits (i) u = v, v = u, (ii) u = u + 5v, v = u, (iii) u = 4u v, v = 6u v, (iv) u = u + v, v = u v, (v) u = v, v = (d) Show how to obtain a second order equation satisfied by v(t) by an analogous device re the second order equations for u and for v the same? (e) Discuss how you might proceed if b = (a) For b, we have v = u a b b u, Differentiating the first equation yields dv v = bc ad b u + d b u ( ) = u a u b b Equating this to the right hand side of the second equation yields leading to the second order differential equation u (a + d) u + (ad bc)u = (b) If u(t) solves ( ), then defining v(t) by the first equation in ( ) yields a solution to the first order system Vice versa, the first component of any solution u(t), v(t) to the system gives a solution u(t) to the second order equation (c) (i) u + u =, hence u(t) = c cos t + c sin t, v(t) = c sin t + c cos t (ii) u u + 5u =, hence u(t) = c e t cos t + c e t sin t, v(t) = (c + c )e t cos t + ( c + c )e t sin t (iii) u u 6u =, hence u(t) = c e t + c e t, v(t) = c e t + 6c e t (iv) u u =, hence u(t) = c t+c e e t, v(t) = ( )c e t ( +)c e t (v) u =, hence u(t) = c t + c, v(t) = c (d) For c we can solve for u = v c d d v, ad bc u = u + b v, leading to the same d d second order equation for v, namely, v (a + d) v + (ad bc)v = (e) If b = then u solves a first order linear equation; once we solve the equation, we can then recover v by solving an inhomogeneous first order equation Note that u continues to solve the same second order equation, but is no longer the most general solution and so the one-to-one correspondence between solutions breaks down ( ) 9 Find the solution to the system of differential equations du dv = u + 4v, with initial conditions u() = and v() = u(t) = 7 5 e 5 t e5 t, v(t) = 4 5 e 5 t e5 t 9 Find the general real solution to the following systems of differential equations: = 4u v, ds 9/9/4 47 c 4 Peter J Olver

4 (a) u = u + 9u, u = u + u (d) y = y, y = y + y, y = y (b) x = 4x + x, x = x 4x (e) (c) x = x 8x + x, x = x + x + x, x = x 4x + x y = y y, y = y + y (a) u (t) = ( )c e (+ ) t ( +)c e ( ) t, u (t) = c e (+ ) t +c e ( ) t (b) x (t) = c e 5 t + c e 5 t, x (t) = c e 5 t + c e 5 t (c) y (t) = e th c cos t (c + c ) sin t i, y (t) = e th c cos t + (c + c ) sin t i (d) y (t) = c e t c e t c, y (t) = c e t c e t, y (t) = c e t + c e t + c ; (e) x (t) = c e t + c e t + c e 4 t, x (t) = c e t + c e t, x (t) = c e t + c e t + c e 4 t 9 Solve the following initial value problems: (a) du = (b) du = u, u() =, (c) du 4 = (d) du = (f ) du = 7 u, u() = 4 u, u() =, (e) du = du, (g) = (a) u(t) = e t + e + t, e t + e + t T ; (b) u(t) = e t e t, e t + e t T ; (c) u(t) = (d) u(t) = u, u() =,, u, u() = 6 4 u, u(π) =, u, u() = e t cos t, e t sin T t«; e t cos 6 t, e t cos 6 t + q sin 6 t, e t cos 6 t «T ; (e) u(t) = ( 4 6 cos t 9 sin t, + cos t + 6 sin t, sin t ) T ; (f ) u(t) = e t + e +t, e4 t + e 4+ t, e t + e +t, e4 t + e 4+ t T ; (g) u(t) = e t + cos t sin t, e t 5 cos t + sin t, cos t, cos t + sin t T 9 (a) Find the solution to the system dx dy = x + y, = x y, that has initial conditions x() =, y() = (b) Sketch a phase portrait of the system that shows several typical solution trajectories, including the solution you found in part (a) learly indicate the direction of increasing t on your curves x(t) = e t cos t, y(t) = e t sin t; the orbits spiral in a clockwise dirfection, approaching the origin exponentially fast as t 94 planar steady state fluid flow has velocity vector v = ( x y, x y ) T The motion of the fluid is described by the differential equation dx = v floating object starts out at the point (, ) T Find its position after time unit ds 9/9/4 47 c 4 Peter J Olver

5 x(t) = e t/ cos t sin t, y(t) = e t/ cos t sin t, so ( x(), y() ) T = ( 79, 48 ) T 95 steady state fluid flow has velocity vector v = ( y, x, z ) T Describe the motion of the fluid particles as governed by the differential equation dx = v x(t) = c cos t c sin t, c sin t + c cos t, c e t T The fluid particles circle counterclockwise around the z axis at angular velocity, while spiraling closer and closer, at an exponential rate, to the xy plane In the limit, they mover around the unit cirle in the xy plane at constant speed 96 Solve the initial value problem du = 6 u, u() = Explain how orthogonality can 6 help The coefficient matrix has eigenvalues λ = 5, λ = 7, and, since the coefficient matrix is symmetric, orthogonal eigenvectors v =, v = The general solution is u(t) = c e 5 t + c e 7 t For the initial conditions we can use orthogonality to find Therefore, the solution is u() = c + c =, c = u(), v v =, c = u(), v v = u(t) = e 5 t + e 7 t 97 (a) Find the eigenvalues and eigenvectors of K = (b) Verify that the eigenvectors are mutually orthogonal (c) ased on part (a), is K positive definite, positive semi-definite or indefinite? (d) Solve the initial value problem du = K u, u() =, using orthogonality to simplify the computations (a) eigenvalues: λ =, λ =, λ =, eigenvectors: v =, v =, v = (b) The matrix is positive semi-definite since it has one zero eigenvalue and the rest are positive (c) The general solution is For the initial conditions u(t) = c u() = c + c e t + c + c + c e t = = u, ds 9/9/4 47 c 4 Peter J Olver

6 we can use orthogonality to find c = u, v v =, c = u, v v =, c = u, v v = Therefore, the solution is u(t) = + e t et, + 4 et, et et T 98 Demonstrate that one can also solve the initial value problem in Example 98 by writing the solution as a complex linear combination of the complex eigensolutions, and then using the initial conditions to specify the coefficients The general complex solution to the system is u(t) = c e t + c e (+ i ) t i + c e ( i) t Substituting into the initial conditions, u() = c + c + c c + i c i c c + +c + c Thus, we obtain the same solution: u(t) = e t i e(+ i ) t = i + i e( i) t and so i = i c =, c = i, c = i e t + e t sin t e t + e t cos t e t + e t sin t 99 Determine whether the following vector-valued functions are linearly dependent or linearly independent: t + t (a),, (b), t t t t t t t, (c),,, (d) e t t t e t, e t (e) (h) e t cos t e t sin t e t e t, e t, e t e t, e t e t sin t e t cos t e t e t e t, (f ), (i) e t te t t e t Only (d), (g) are linearly dependent cos t sin t,, t e t e t te t sin t cos t,, (g) te t t e t e t, t t e t e t e t,, e t 9 Let be a constant matrix Suppose u(t) solves the initial value problem u = u, u() = b Prove that the solution to the initial value problem u = u, u(t ) = b, is equal to eu(t) = u(t t ) How are the solution trajectories related? d Using the chain rule, u(t) e = du (t t ) = u(t t ) = eu, and hence eu(t) solves the differential equation Moreover, eu(t ) = u() = b has the correct initial conditions The trajectories are the same curves, but eu(t) is always ahead of u(t) by an amount t 9 Prove that the general solution to a linear system u = Λu with diagonal coefficient matrix Λ = diag (λ,, λ n ) is given by u(t) = c e λ t,, c n e λ n t T du = λ c e λ t,, λn c n e λ n t T = Λ u 9 Show that if u(t) is a solution to u = u, and S is a constant, nonsingular matrix of, + t, t ds 9/9/4 474 c 4 Peter J Olver

7 the same size as, then v(t) = S u(t) solves the linear system v = v, where = S S is similar to dv = S du = S u = S S v = v 9 (i) ombine Exercises 9 to show that if = S ΛS is diagonalizable, then the solution to u = u can be written as u(t) = S c e λ t,, c n e λ n t T, where λ,, λ n are its eigenvalues and S = ( v v v n ) is the corresponding matrix of eigenvectors (ii) Write the general solution to the systems in Exercise 9 in this form (i) This is an immediate consequence of the preceding two exercises (ii) (a) u(t) = c e t c e t ; (b) u(t) = c e t c e t (c) u(t) = (d) u(t) = (e) u(t) = (g) u(t) = i i c e(+ i ) t + i q i q 4 + i i i + i c e ( i ; )t c e t c e ( + i 6) t c e ( i 6)t c c e i t c e i t i i i + i + i i ; ; (f ) u(t) = c e t c e t c e i t c 4 e i t ; c e t c e t c e t c 4 e t ; 94 Find the general solution to the linear system du = u for the following incomplete 4 coefficient matrices: (a), (b), (c), (d), (e), (f ) (a) u(t) = (c) u(t) = c e t + c t e t c e t c e t + c + t e t c e t + c te t, (g), (b) u(t) =, (d) u(t) =, (h) c e t + c + t e t, c e t + c te t c e t + c e t + c t e t c e t c e t c e t + c e t + c t e t, ds 9/9/4 475 c 4 Peter J Olver

8 (e) u(t) = (g) c e t + c t e t + c + t e t c e t + c t e t c e t + c te t + c t e t c cos t + c sin t + c t cos t + c 4 t cos t c sin t + c cos t c t sin t + c 4 t cos t c cos t + c 4 sin t c sin t + c 4 cos t (a) du =, (f ) = 9 u, u c e t + c te t 4 c e t 4 c t 4 (t + )e c e t + c 4 te t c e t 4 c 4 e t c 4 e t 95 Find a first order system of ordinary differential equations that has the indicated vectorvalued function as a solution: (a) e t + e t e t, (b) e t cos t e t, (c), sin t t sin t cos t e t sin t t e t sin t (d), (e) sin t + cos t e t, (f ) e t e t cos t, (g) t, (h) e t cos t + t e t sin t u, (b) du u, (c) du u, (d) du = 5 (g) du = u, (e) du u, = (h) du = (f ) du = = u, 96 Which sets of functions in Exercise 99 can be solutions to a common first order, homogeneous, constant coefficient linear system of ordinary differential equations? If so, find a system they satisfy; if not, explain why not (a) No, since neither du i is a linear combination of u, u Or note that the trajectories described by the solutions cross, violating uniqueness (b) No, since polynomial solutions a two-dimensional system can be at most first order in t (c) No, since a two-dimensional system has at most linearly independent solutions (d) Yes: u = u (e) Yes: u = u (f ) No, since neither du i is a linear combination of u, u Or note that both solutions have the unit circle as their trajectory, but traverse it in opposite directions, violating uniqueness (g) Yes: u = u (h) Yes: u = u (i) No, since a three- dimensional system has at most linearly independent solutions 97 Solve the third order equation d u + d u + 4 du + u = by converting it into a first order system ompare your answer with what you found in Exercise 9 Setting u(t) = u(t) u(t), the first order system is du u(t) eigenvalues of the coefficient matrix are, ± i with eigenvectors = u The 4, ± i and the 9 4, ds 9/9/4 476 c 4 Peter J Olver

9 resulting solution is u(t) = found in Exercise 9 c e t + c cos t + c sin t c e t c sin t + c cos t 9c e t 4c cos t 4c sin t, which is the same as that 98 Solve the second order coupled system of ordinary differential equations u = u + u v, v = v u + v, by converting it into a first order system involving four variables Setting u(t) = u(t) u(t) v(t) v(t) du, the first order system is = The co-,, Thus efficient matrix has eigenvalues,,, and eigenvectors u(t) = c e t + c + c e t + c 4 e t c e t + c e t + c 4 e t c e t + c + c e t c 4 e t,, whose first and third components give the general solu- c e t + c e t c 4 e t tion u(t) = c e t + c + c e t + c 4 e t, v(t) = c e t + c + c e t c 4 e t to the second order system 99 Suppose that u(t) R n is a polynomial solution to the constant coefficient linear system u = u What is the maximal possible degree of u(t)? What can you say about when u(t) has maximal degree? The degree is at most n, and this occurs if and only if has only one Jordan chain in its Jordan basis 9 (a) Under the assumption that u,, u k form a Jordan chain for the coefficient matrix, prove that the functions (95) are solutions to the differential equation u = u (b) Prove that they are linearly independent (a) y direct computation, which equals u j = e λ t du j j X i = = λe λ t j X t j i i = (j i) w i = eλ t t j i (j i) w i + eλ t j X i = 4 tj (j ) w + j t j i (j i ) w i, X i = t j i (j i) (λw i + w i ) (b) t t =, we have u j () = w j, and the Jordan chain vectors are linearly independent 9 (a) Explain how to solve the inhomogeneous ordinary differential equation du = u+b when b is a constant vector belonging to rng Hint: Look at v(t) = u(t) u where u is an equilibrium solution (b) Use your method to solve (i) du dv du dv = u v +, = u v, (ii) = 4v +, = u (a) The equilibrium solution satisfies u = b, and so v(t) = u(t) u satisfies v = u = u + b = (u u ) = v, which is the homogeneous system 5 ds 9/9/4 477 c 4 Peter J Olver

10 (b) (i) u(t) = c e t + c e t 4, v(t) = c e t + c e t + 4 (ii) u(t) = c cos t + c sin t, v(t) = c sin t + c cos t stability 9 Stability of Linear Systems 9 lassify the following systems according to whether the origin is (i) asymptotically stable, (ii) stable, or (iii) unstable: (a) du dv du = u v, = u v (b) = u 5v, dv = u v (c) du dv = u v, = u 5v (d) du dv = v, = 8u (e) du dv dw du = u v + w, = u v + w, = u v + w (f ) = u v, dv dw = 6u + v 4w, = 4u w (g) du dv = u v + w, = u v + w, dw du = 4u + v 5w (h) = u + v w, dv dw = u v + w, = v + w (a) asymptotically stable eigenvalues ± i, (b) unstable eigenvalues ± i, (c) asymptotically stable eigenvalue (d) stable eigenvalues ±4 i (e) stable eigenvalues,, with complete (f ) unstable eigenvalues, ± i, (g) asymptotically stable eigenvalues,, (h) unstable eigenvalues,, with incomplete 9 Write out the formula for the general real solution to the system in Example 96 and verify its stability r u(t) = e th c + c r cos 6 t + c + c i sin 6 t + c e t, v(t) = e th c cos 6 t + c sin 6 t i + c e t, w(t) = e th c cos 6 t + c sin 6 t i + c e t 9 Write out and solve the gradient flow system corresponding to the following quadratic forms: (a) u + v, (b) uv, (c) 4u uv + v, (d) u uv uw + v v w + w (a) u = u, v = v, with solution u(t) = c e t, v(t) = c e t (b) u = v, v = u, with solution u(t) = c e t + c e t, v(t) = c e t + c e t (c) u = 8u+v, v = u v, with solution u(t) = c + e (5+)t +c e (5 )t, v(t) = c e (5+)t + c e (5 )t (d) u = 4u + v + v, v = u 4v + w, w = u + v 4w, with solution u(t) = c e 6 t + c e (+ ) t + c e ( ) t, v(t) = ( + )c e (+ ) t + ( )c e ( ) t, w(t) = c e 6 t + c e (+ ) t + c e ( ) t ds 9/9/4 478 c 4 Peter J Olver

11 94 Write out and solve the Hamiltonian systems corresponding to the first three quadratic forms in Exercise 9 Which of them are stable? (a) u = v, v = u, with solution u(t) = c cos t + c sin t, v(t) = c sin t + c cos t Stable (b) u = u, v = v, with solution u(t) = c e t, v(t) = c e t Unstable (c) u = u + v, v = 8u + v, with solution u(t) = 4 (c c ) cos t + 4 ( c + c ) sin t, v(t) = c cos t + c sin t Stable 95 Which of the following systems are gradient flows? Which are Hamiltonian systems? In each case, discuss the stability of the zero solution u = u + v, u = u v, u = v, u = v, u = u v, (a) (b) (c) (d) (e) v = u v, v = u + v, v = u, v = u, v = u v (a) Gradient flow symptotically stable (b) Neither Unstable (c) Hamiltonian flow Unstable (d) Hamiltonian flow Stable (e) Neither Unstable 96 True or false: nonzero linear gradient flow cannot be a Hamiltonian flow True If K = a b, then we must have H b c v H by equality of mixed partials, u v positive, a, c >, which is a contradiction = au+bv, H u = bu cv Therefore, = a = c ut if K >, both diagonal entries must be 97 (a) Show that the matrix = has λ = ± i as incomplete complex conjugate eigenvalues (b) Find the general real solution to u = u (c) Explain the behavior of a typical solution Why is the zero solution not stable? (a) The characteristic equation is λ 4 + λ + =, and so ± i are double eigenvalues However, each has only one linearly independent eigenvector, namely (, ± i,, ) T The general solution is u(t) = c cos t + c sin t + c t cos t + c 4 t cos t c sin t + c cos t c t sin t + c 4 t cos t c cos t + c 4 sin t c sin t + c 4 cos t (b) ll solutions with c + c 4 spiral off to as t ± Since these can start out arbitrarily close to, the zero solution is not stable 98 Let be a real matrix, and assume that the linear system u = u has a periodic solution of period P Prove that every periodic solution of the system has period P What other types of solutions can there be? Is the zero solution necessarily stable? Every solution to a real first order system of period P comes from complex conjugate eigenvalues ±π i /P real matrix has at least one real eigenvalue λ Therefore, if the system has a solution of period P, its eigenvalues are λ and ±π i /P If λ =, every solution has period P Otherwise, the solutions with no component in the direction of the real eigenvector all have period P, and are the only periodic solutions, proving the result The system is stable (but never asymptotically stable) if and only if the real eigenvalue λ 99 re the conclusions of Exercise 98 valid when is a 4 4 matrix? ds 9/9/4 479 c 4 Peter J Olver

12 No, since a 4 4 matrix could have two distinct complex conjugate pairs of purely imaginary eigenvalues, ±π i /P, ±π i /P, and would then have periodic solutions of periods P and P The general solution in such a case is quasi-periodic; see Section 95 for details 9 Let be a real 5 5 matrix, and assume that has eigenvalues i, i,, (and no others) Is the zero solution to the linear system x = x necessarily stable? Explain Does your answer change if is 6 6? The system is stable since ± i must be simple eigenvalues since a 5 5 matrix has 5 eigenvalues counting multiplicities, and the multiplicities of complex conjugate eigenvalues are the same 6 6 system can have a complex conjugate pair of incomplete double eigenvalues ± i in addition to the simple real eigenvalues,, and in that case the origin is unstable 9 True or false: The system u = H n u, where H n is the n n Hilbert matrix (69), is asymptotically stable True since H n > by Proposition 4 9 True or false: If K is positive semi-definite, then the zero solution to u = K u is stable True, because all eigenvalues, including, of a symmetric matrix are complete 9 Let u(t) solve u = u Let v(t) = u( t) be its time reversal (a) Write down the linear system v = v satisfied by v(t) Then classify the following statements as true or false Explain your answers (b) If u = u is asymptotically stable, then v = v is unstable (c) If u = u is unstable, then v = v is asymptotically stable (d) If u = u is stable, then v = v is stable (a) v = v = v (b) True, since the eigenvalues of = are minus the eigenvalues of, and so will all have positive real parts (c) False For example, a saddle point, with one positive and one negative eigenvalue is still unstable when going backwards in time (d) False, unless all the eigenvalues of and hence are complete and purely imaginary or zero 94 True or false: If is a symmetric matrix, then the system u = u has an asymptotically stable equilibrium solution The eigenvalues of are all of the form λ < where λ is an eigenvalue of Thus, if is nonsingular, the result is true, while if is singular, then the equilibrium solutions are stable, but not asymptotically stable 95 True or false: (a) If tr >, then the system u = u is unstable (b) If det >, then the system u = u is unstable (a) True, since the sum of the eigenvalues equals the trace, so at least one must be positive or have positive real part in order that the trace be positive (b) False = is an example of a asymptotically stable system with positive determinant 96 onsider the differential equation u = K u, where K is positive semi-definite (a) Find all equilibrium solutions (b) Prove that all non-constant solutions decay exponentially fast to some equilibrium What is the decay rate? (c) Is the origin (i) stable, (ii) asymptotically stable, or (iii) unstable? (d) Prove that, as t, the solution u(t) converges to the orthogonal projection of its initial vector a = u() onto ker K (a) Every v ker K gives an equilibrium solution u(t) v ds 9/9/4 48 c 4 Peter J Olver

13 (b) The general solution has the form u(t) = c e λ t v + + c r e λ r t vr + c r+ v r+ + + c n v n, where λ,, λ r > are the positive eigenvalues of K with (orthogonal) eigenvectors v,, v r, while v r+,, v n form a basis for the null eigenspace, ie, ker K Thus, as t, u(t) c r+ v r+ + + c n v n ker K, which is an equilibrium solution (c) The origin is asymptotically stable if K is positive definite, and stable if K is positive semi-definite (d) Note that a = u() = c v + + c r v r + c r+ v r+ + + c n v n Since the eigenvectors are orthogonal, c r+ v r+ + +c n v n is the orthogonal projection of a onto ker K 97 (a) Let H(u, v) = au + b uv + cv be a quadratic function Prove that the nonequilibrium trajectories of the associated Hamiltonian system and those of the gradient flow are mutually orthogonal, ie, they always intersect at right angles Verify this result for (i) u + v, (ii) uv by drawing representative trajectories of both systems on the same graph (a) The tangent to the Hamiltonian trajectory at a point ( u, v ) T is v = ( H/ v, H/ u ) T while the tangent to the gradient flow trajectory is w = ( H/ u, H/ v ) T Since v w =, the tangents are orthogonal (b) 98 True or false: If the Hamiltonian system for H(u, v) is stable, then the corresponding gradient flow u = H is stable False Only the positive definite Hamiltonians lead to stable gradient flows 99 Suppose that u(t) satisfies the gradient flow system (9) (a) Prove that d q(u) = K u (b) Explain why if u(t) is any nonconstant solution to the gradient flow, then q(u(t)) is a strictly decreasing function of t, thus quantifying how fast a gradient flow decreases energy (a) When q(u) = ut K u then d q(u) = u T K u + ut K u = u T K u = (K u) T K u = K u d (b) Since K u for any u, assuming u is not the equilibrium solution, q(u) = K u < and hence q(u) is a decreasing function of t Indeed, for K >, the solution u, and q(u) exponentially fast 9 The law of conservation of energy states that the energy in a Hamiltonian system is constant on solutions (a) Prove that if u(t) satisfies the Hamiltonian system (9), then H(u(t)) = c is a constant, and hence solutions u(t) move along the level sets of the Hamiltonian or energy function Explain how the value of c is determined by the initial conditions (b) Plot the level sets of H in the particular case H(u, v) = u uv + v and verify that they coincide with the solution trajectories ds 9/9/4 48 c 4 Peter J Olver

14 (a) y the multivariable calculus chain rule d H du H(u(t), v(t)) = u + H dv v = H H u v + H H v u Since H(u(t), v(t)) c, its value is c = H(u, v ) where u(t ) = u, v(t ) = v are the initial conditions (b) 9 Prove Lemma 94 Let us do the case f(t) = t k e µ t cos ν t; replacing the cosine by a sine is a trivial modification of the proof If µ > then f(t) since e µ t, t k for k > and is equal to for k =, while cosine term is bounded cos ν t If µ =, then f(t) when k =, while f(t) if k > If µ <, then f(t) t k e µ t = e µ t+k log t as t, since µt + k log t when µ < QED 9 Prove Proposition 97 n eigensolution u(t) = e λ t v with λ = µ+ i ν is bounded by u(t) v e µ t Moreover, since exponentials grow faster than polynomials, any solution of the form u(t) = e λ t p(t) where p(t) is a vector of polynomials can be bounded by e a t for any a > µ = Re λ and some > Since every solution can be written as a linear combination of such solutions, every term is bounded by a multiple of e a t provided a > a = max Re λ and so, by the triangle inequality, is their sum If the maximal eigenvalues are complete, then there are no polynomial terms, and we can use the eigensolution bound, so we can set a = a QED dode 9 Two-Dimensional Systems 9 For each the following: (a) Write the system as u = u (b) Find the eigenvalues and eigenvectors of (c) Find the general real solution of the system (d) Draw the phase portrait, indicating its type and stability properties: (i) u = u, u = 9u, (ii) u = u u, u = u u, (iii) u = u u, u = u u (i) = ; λ 9 = i, v = i, λ = i, v = i, u (t) = c cos t + c sin t, u (t) = c sin t c cos t; center (ii) = ; λ = + i, v = + i, λ = i, v = i, u (t) = e t/» c «c cos t + c sin t + t c + «c sin t u (t) = e t/» c cos ; stable focus (iii) =, λ =, λ =, v =, and v =, u (t) = c e t + c e t, u (t) = c e t + c e t ; saddle point 9 For each of the following systems (i) u = u, (ii) u = 5 u, (iii) u =, 5/ u 5/ ds 9/9/4 48 c 4 Peter J Olver

15 (a) Find the general real solution (b) Using the solution formulas obtained in part (a), plot several trajectories of each system On your graphs, identify the eigenlines (if relevant), the direction of increasing t on the trajectories (c) Write down the type and stability properties of the system (i) u(t) = c e t (ii) u(t) = c e t (iii) u(t) = c e t/ + c e t cos t sin t 5 cos t ; saddle point + c e t sin t + cos t 5 sin t + c e t/ t t + 5 ; stable focus ; stable improper node 9 lassify the following systems, and sketch their phase portraits du du du = u + 4v, = u + v, = 5u + 4v, (a) (b) (c) dv = u v dv = u 4v dv = u + v (a) For the matrix = 4 an unstable saddle point (b) For the matrix = 4 stable node (c) For the matrix = 5 4 unstable node (d) For the matrix = line (d) du = u v, dv = u + v, tr = <, det = <, = 7 >, so this is, tr = 6 <, det = 7 >, = 8 >, so this is a, tr = 7 >, det = 6 >, = 5 >, so this is an, tr = <, det =, = >, so this is a stable 94 Sketch the phase portrait for the following systems: (a) (b) u = u 4u, u = u u (c) u = u + u, u = 4u u (d) u = u + u, u = u u = u u, u = u + u (e) u = u + 5 u, u = 5 u u 95 (a) Solve the initial value problem du = u, u() = (b) Sketch a picture of your solution curve u(t), indicating the direction of motion (c) Is the origin (i) stable? (ii) asymptotically stable? (iii) unstable? (iv) none of these? Justify your answer (a) u(t) = e t cos t + 7e t sin t, e t cos t 4e t sin t T ; (b) (c) asymptotically stable since the coefficient matrix has tr = 4 <, det = 5 >, = 4 <, and hence it is a stable focus, or, equivalently, the eigenvalues ± i have negative real part 96 Justify the solution formulas (9) and (9) For (9), the complex solution e λ t v = e (µ+ i ν) t (w + i z) = e µ th cos(ν t) w sin(ν t) z i + e µ th sin(ν t) w + cos(ν t) z i ds 9/9/4 48 c 4 Peter J Olver

16 leads to the general real solution u(t) = c e µ th cos(ν t) w sin(ν t) z i + c e µ th sin(ν t) w + cos(ν t) z i = e µ th c cos(ν t) + c sin(ν t) i w + e µ th c sin(ν t) + c cos(ν t) i z = r e µ th cos(ν t σ) w + sin(ν t σ) z i, where r = q c + c and tan σ = c /c To justify (9), we differentiate du = d h (c + c t)e λ t v + c e λ t w i = λ h (c + c t)e λ t v + c e λ t w i + c e λ t v, while u = (c + c t)e λ t v + c e λ t w = (c + c t)e λ t λ v + c e λ t (λ w + v) by the Jordan chain condition Therefore u = u 97 Which of the 4 possible two-dimensional phase portraits can occur for the phase plane equivalent (97) of a second order scalar ordinary differential equation? ll except for IV(a c), the stars and the trivial case ME 94 Matrix Exponentials 94 Find the exponentials e t of the following matrices: (a), (b) 4, (c), (d), (e), (f ) (a) e t e t et + e t ; (b) 4 et 4 e t et + 4 e t e t + e t et e t cosh t sinh t e t e t et + e t = sinh t cosh t cos t sin t t (c) ; (d) ; (e) e t cos t e t sin t e t sin t sin t cos t 5e t sin t e t cos t + e t ; sin t (f ) e t + te t te t te t e t te t ; 94 Determine the matrix exponential e t for the following matrices: (a), (b), (c), (d) (a) (b) sin t cos t sin t cos t sin t cos t ; 6 et + e t + e4 t et e4 t 6 et e t + e4 t et e4 t et + e4 t et e4 t 6 et e t + e4 t et e4 t 6 et + e t + e4 t ; ds 9/9/4 484 c 4 Peter J Olver

17 (c) (d) e t + te t te t te t + e t e t + e t e t te t te t te t et + e t/ cos t e t e t/ cos t + e t/ sin t et e t/ cos t e t/ sin t ; e t e t/ cos t + e t/ sin et e t/ cos t e t/ sin et + e t/ cos t et e t/ cos et e t/ cos t t et + e t/ cos t e t/ sin t t + e t/ sin 94 Verify the determinant formula of Lemma 98 for the matrices in Exercises 94 and (a) det e t = e t = e t tr, (b) det e t = = e t tr, (c) det e t = = e t tr, (d) det e t = = e t tr, (e) det e t = e 4 t = e t tr, (f ) det e t = e t = e t tr 94 (a) det e t = = e t tr, (b) det e t = e 8 t = e t tr, (c) det e t = e 4 t = e t tr, (d) det e t = = e t tr t t 944 Find e when = 5 (a), (b) 5 (a) (d) (e + e 7 ) (e e 7 ) e e, (e) e 5, (c), (d) 4, (e) 5 (e e 7 ) e cos e sin, (b) (e + e 7 ) e sin e cos, (c), cos cos + sin 9 9 cos sin cos sin cos 9 9 cos + sin 9 9 cos + sin 9 9 cos sin cos 945 Solve the indicated initial value problems by first exponentiating the coefficient matrix and then applying formula (94): (a) du = u, u() =, (b) du = 6 u, u() =, (c) du = 8 5 6u, u() = cos t sin t cos t + sin t (a) u(t) = = ; sin t cos t sin t cos t (b) u(t) = e t e t e t + e t e t e t e t + e t = 6e t + 5e t 4e t + 5e t ; ds 9/9/4 485 c 4 Peter J Olver

18 u(t) = (c) e t cos t sin t e t cos t sin t sin t e t + cos t + sin t e t + cos t + sin t sin t e t cos t e t cos t + sin t cos t + sin t = e t cos t sin t e t + cos t + sin t e t cos t + sin t 946 What is e t O when O is the n n zero matrix? e t O = I for all t 947 Find all matrices such that e t = O There are none, since e t is always invertible π 948 Let = Show that e π = I e t = cos πt sin πt, and hence, when t =, e sin πt cos πt = cos π sin π = sin π cos π 949 (a) Let be a matrix such that tr = and δ = det > Prove that e = (cos δ) I + sin δ Hint: Use Exercise 85 (b) Establish a similar formula when δ det < (c) What if det =? (a) ccording to Exercise 85, = δ I since tr =, det = δ Thus, by induction, m = ( ) m δ m I, m+ = ( ) m δ m e t = X n = t n n n = X m = ( ) m (δ t) m (m) I + X m = ( ) m t m+ δ m (m + ) = cos δ t + sin δ t δ Setting t = proves the formula (b) e = (cosh δ) I + sinh δ where det = δ δ (c) e = I + since = O by Exercise Explain in detail why the columns of e t form a basis for the solution space to the system u = u ssuming is an n n matrix, since e t is a matrix solution, its n individual columns must be solutions Moreover, the columns are linearly independent since e = I is nonsingular Therefore, they form a basis for the n-dimensional solution space 94 True or false: (a) e = e, (b) e + = e e (a) False, unless = (b) True, since and commute 94 Prove formula (94) Hint: Fix s and prove that, as functions of t, both sides of the equation define matrix solutions with the same initial conditions Then use uniqueness Fix s and let U(t) = e (t+s), V (t) = e t e s Then, by the chain rule, U = e (t+s) = U, while, by the matrix Leibniz rule (99), V = e t e s = V Moreover, U() = e s = V () Thus U(t) and V (t) solve the same initial value problem, hence, by uniqueness, U(t) = V (t) for all t QED ds 9/9/4 486 c 4 Peter J Olver

19 94 Prove that commutes with its exponential: e t = e t Hint: Prove that both are matrix solutions to U = U with the same initial conditions Set U(t) = e t, V (t) = e t Then, by the matrix Leibniz formula (99), U = e t = U, V = e t = V, while U() = = V () Thus U(t) and V (t) solve the same initial value problem, hence, by uniqueness, U(t) = V (t) for all t QED 944 Prove that e t ( λ I ) = e t λ e t by showing that both sides are matrix solutions to the same initial value problem Set U(t) = e t λ e t Then, U = λe t λ e t + e t λ e t = ( λ I )U Moreover, U() = I Therefore, by the the definition of matrix exponential, U(t) = e t ( λ I ) 945 (a) Prove that the exponential of the transpose of a matrix is the transpose of its exponential: e t T = (e t ) T (b) What does this imply about the solutions to the linear systems u = u and v = T v? (a) Let V (t) = (e t ) T Then du = d et T = (e t ) T = T (e t ) T = T V, and V () = I Therefore, by the the definition of matrix exponential, V (t) = e t T (b) The columns of e t form a basis for the solutions to u = u, while its rows are a basis for the solutions to v = T v 946 Prove that if = S S are similar matrices, then so are their exponentials: e t = S e t S First note that n = S n S Therefore, e t X t n X t n = n n = n S n S = S n = n = X n = t n n n S = S e t S n alternative proof uses the fact that e t and S e t S both satisfy the initial value problem U = U = S S U, U() = I, and hence, by uniqueness, they are equal 947 Diagonalization provides an alternative method for computing the exponential of a complete matrix (a) First show that if D = diag (d,, d n ) is a diagonal matrix, so is e t D = diag (e t d,, e t d n ) (b) Second, prove that if = S D S is diagonalizable, so is e t = S e t D S (c) When possible, use diagonalization to compute the exponentials of the matrices in Exercises 94 d (a) diag (et d, e t d n ) = diag (d e t d,, d n e t d n ) = D diag (e t d,, e t d n ) Moreover, at t =, we have diag (e d,, e d n ) = I Therefore, diag (e t d,, e t d n ) satisfies the defining properties of e t D (b) See Exercise 946 (c) 94: (a) (b) e t 4 e t e t e t = 4 4 = et e t et + e t ; 4 et 4 e t et + 4 e t e t + e t et e t ; e t e t et + e t (c) i i e i t e i t i i = cos t sin t ; sin t cos t ds 9/9/4 487 c 4 Peter J Olver

20 (d) not diagonalizable; (e) 5 5 i i e (+ i )t e ( i )t 5 5 i i = e t cos t e t sin t e t sin t 5e t sin t e t cos t + e t sin t ; 94: (f ) not diagonalizable (a) (b) i i (c) not diagonalizable; (d) e i t e i t e t e t e 4 t i + i + i i which is the same as before = i i = sin t cos t sin t cos t sin t cos t 6 et + e t + e4 t et e4 t 6 et e t + e4 t et e4 t et + e4 t et e4 t 6 et e t + e4 t et e4 t 6 et + e t + e4 t e ( i )t e ( + i )t ; ; i + i + i i 948 Justify the matrix Leibniz rule (99) using the formula for matrix multiplication Let M have size p q and N have size q r The derivative of the (i, j) entry of M(t) N(t) is, d q X k = m ik (t) n kj (t) = q X k = dm ik n kj (t) + q X k = m ik (t) dn kj The first sum is the (i, j) entry of dm N while the second is the (i, j) entry of M dn QED 949 Let be a real matrix (a) Explain why e is a real matrix (b) Prove that det e > (a) The exponential series is a sum of real terms lternatively, one can choose a real basis for the solution space to construct the real matrix solution U(t) before substituting into formula (94) (b) ccording to Lemma 98, det e = e tr > since a real scalar exponential is always positive 94 Show that tr = if and only if det e t = for all t Lemma 98 implies det e t = e t tr = if and only if tr = 94 Prove that if λ is an eigenvalue of, then e t λ is an eigenvalue of e t What is the eigenvector? ds 9/9/4 488 c 4 Peter J Olver

21 Let u(t) = e t λ v where v is the eigenvector of Then, du = λ et λ v = λ u = u, and hence, by (94), u(t) = e t u() = e t v Therefore, equating the two formulas for u(t), we conclude that e t v = e t λ v, which proves that v is an eigenvector of e t with eigenvalue e t λ 94 Show that the origin is an asymptotically stable equilibrium solution to u = u if and only if lim t et = O The origin is an asymptotically stable if and only if all solutions tend to zero as t Thus, all columns of e t tend to as t, and hence lim t et = O onversely, if lim t et = O, then any solution has the form u(t) = e t c, and hence u(t) as t, proving asymptotic stability 94 Let be a real square matrix and e its exponential Under what conditions does the linear system u = e u have an asymptotically stable equilibrium solution? ccording to Exercise 94, the eigenvalues of e are e λ = e µ cos ν + i e µ sin ν, where λ = µ + i ν are the eigenvalues of For e λ to have negative real part, we must have cos ν <, and so ν = Im λ must lie between k + π < ν < k + π for some k Z 944 Prove that if U(t) is any matrix solution to du = U, so is U(t) e = U(t) where is any constant matrix (of compatible size) Indeed, the columns of U(t) e are linear combinations of the columns of U(t), and hence automatically solutions to the linear system lternatively, we can prove this directly using the Leibniz rule (99) Therefore, de U = d h i du U(t) = = U = e U, since is constant 945 (a) Show that U(t) satsifies the matrix differential equation U = U if and only if U(t) = e t where = U() (b) Show that if U() is nonsingular, then U(t) also satisfies a matrix differential equation of the form U = U Is =? Hint: Use Exercise 946 (a) If U(t) = e t, then du = et = U, and so U satisfies the differential equation Moreover, = U() Thus, U(t) is the unique solution to the initial value problem U = U, U() = (b) y Exercise 946, U(t) = e t QED = e t where = Thus, U = U as claimed Note that = if and only if commutes with U() 946 (a) Suppose u (t),, u n (t) are vector-valued functions whose values at each point t are linearly independent vectors in R n Show that they form a basis for the solution space of a homogeneous constant coefficient linear system u = u if and only if each du j / is a linear combination of u (t),, u n (t) (b) Show that if u(t) solves u = u, so do its derivatives d j u/ j, j =,, (c) Show that a function u(t) belongs to the solution space of a homogeneous constant coefficient linear system u = u if and only if dn u n linear combination of u, du,, dn u n (a) Let U(t) = ( u (t) u n (t) ) be the corresponding matrix-valued function Then du j is a = ds 9/9/4 489 c 4 Peter J Olver

22 nx i = b ij u i for all j =,, n, if and only if U = U where b ij are the entries of Therefore, by Exercise 945, U = U, where = QED (b) y induction, if dj+ u j+ = d dj u j = dj u, then, differentiating the equation with j (c) respect to t, we find d/ d j+ u/ j+ = d/ d j u/ j = d j+ u/ j+, which proves the induction step 947 Prove that if = O O is a block diagonal matrix, then so is e t = et O O e t Write the matrix solution to the initial value problem du = U, U() = I, in block form U(t) = e t V (t) W (t) = Then the differential equation decouples into dv = V, Y (t) Z(t) dw dy = O, = O, dz = Z, with initial conditions V () = I, W () = O, Y () = O, Z() = I Thus, W, Y are constant, while V, Z satisfy the initial value problem for the matrix exponential, and so V (t) = e t, W (t) = O, Y (t) = O, Z(t) = e t QED 948 (a) Prove that if J,n is an n n Jordan block matrix with diagonal entries, cf (847), then e t J,n = t t t t 6 t t t n n t n (n ) t n (n ) t (b) Determine the exponential of a general Jordan block matrix J λ,n Hint: Use Exercise 944 (c) Explain how you can use the Jordan canonical form to compute the exponential of a matrix Hint: Use Exercise 947 ds 9/9/4 49 c 4 Peter J Olver

23 (a) d t t t t 6 t t t n n t n (n ) t n (n ) t = = t t t t n (n ) t n (n ) t n (n ) t t t t 6 t t t n n t n (n ) t n (n ) t Thus, U(t) satisfies the initial value problem U = J,n U, U() = I, that characterizes the matrix exponential, so U(t) = e t J,n (b) Since J λ,n = λ I + J,n, by Exercise 944, e t J λ,n QED = e t λ e t J,n, ie, multiply all entries in the previous formula by e t λ (c) ccording to Exercises 947, 7, if = S J S where J is the Jordan canonical form, then e t = S e t J S, and e t J is a block diagonal matrix given by the exponentials of its individual Jordan blocks, computed in part (b) 949 Prove that if λ is an eigenvalue of with multiplicity k, then e t λ is an eigenvalue of e t with the same multiplicity Hint: ombine the Jordan canonical form (849) with Exercises 947, 8 If J is a Jordan matrix, then, by the arguments in Exercise 948, e t J is upper triangular with diagonal entries given by e t λ where λ is the eigenvalue appearing on the diagonal of the corresponding Jordan block of In particular, the multiplicity of λ, which is the number of times it appears on the diagonal of J, is the same as the multiplicity of e t λ for e t J Moreover, since e t is similar to e t J, its eigenvalues are the same, and of the same multiplicities QED 94 y a (natural) logarithm of a matrix we mean a matrix such that e = (a) Explain why only nonsingular matrices can have a logarithm (b) omparing Exercises 946 8, explain why the matrix logarithm is not unique (c) Find all real logarithms of the identity matrix I = Hint: Use Exercise 94 (a) ll matrix exponentials are nonsingular by the remark after (94) (b) oth, ds 9/9/4 49 c 4 Peter J Olver

24 π = O and = have the identity matrix as their exponential e π = I (c) If e = I and λ is an eigenvalue of, then e λ =, since is the only eigenvalue of I Therefore, the eigenvalues of must be integer multiples of π i Since is real, the eigenvalues must be complex conjugate, and hence either both, or ±nπ i for some positive integer n In the latter case, since the characteristic equation of is λ + 4n π, must have zero trace and determinant 4n π, hence = a b with a c a + bc = 4n π If has both eigenvalues zero, it must be complete, and hence = O 94 True or false: The solution to the non-autonomous initial value problem u = (t)u, u() = b, is u(t) = er t (s) ds b Even though this formula is correct in the scalar case, it is false in general Would that life were so simple 94 Solve the following initial value problems: (a) (b) (d) u = u + u + e t, u () =, u = u u + e t, u () = u = u + v +, u() =, v = 4u + t, v() = (e) (c) u = u u, u () =, u = 4u u + e t, u () = u = u, u () =, u = 4u + cos t, u () = p = p + q + t, p() =, q = p q + t, q() = (a) u (t) = et e t 4 e t, u (t) = et e t ; (b) u (t) = e t e t + t e t, u (t) = e t e t + t e t ; (c) u (t) = cos t sin t cos t, u (t) = cos t + sin t sin t; (d) u(t) = 6 e4 t t, v(t) = 6 e4 t t; (e) p(t) = t + t, q(t) = t t 94 Solve the following initial value problems: u = u + u, u () =, (a) u = u + u u + t, u () =, (b) u = u + u u +, u () = u = u u, u () =, u = u + e t, u () =, u = 4u 4u u, u () = (a) u (t) = cos t + 4 sin t + t, u (t) = e t cos t 4 sin t + t, u (t) = e t 4 cos t 4 sin t t; (b) u (t) = et + e t + te t, u (t) = te t, u (t) = e t + e t + te t 944 (a) Write down an integral formula for the solution to the initial value problem du = u + b, u() =, where b is a constant vector (b) Suppose b rng Do you recover the solution you found in Exercise 9? Z t (a) u(t) = e(t s) b ds ds 9/9/4 49 c 4 Peter J Olver

25 (b) Yes, since if b = c, then the integral can be evaluated as u(t) = Z t e(t s) c ds = e (t s) c t s = = et c c = v(t) u, where v(t) = e t c solves the homogeneous system v = v, while u = c is the equilibrium solution 945 Suppose that λ is not an eigenvalue of Show that the inhomogeneous system u = u + e λ t v has a solution of the form u (t) = e λ t w, where w is a constant vector What is the general solution? Set w = ( λ I ) v Then u (t) = e λ t w is a solution The general solution is u(t) = e λ t w + z(t) = e λ t w + e t b, where b is any vector 946 Find the one-parameter groups generated by the following matrices and interpret geometrically: What are the trajectories? What are the fixed points? (a), (b), (c), (d), (e) 4 (a) e t scalings in the x direction, which expand when t > and contract when t < The trajectories are half-lines parallel to the x axis Points on the y axis are left fixed (b) shear transformations in the y direction The trajectories are lines parallel t to the y axis Points on the y axis are fixed cos t sin t (c) rotations around the origin, starting in a clockwise direction for sin t cos t t > The trajectories are the circles x + y = c The origin is fixed cos t sin t (d) elliptical rotations around the origin The trajectories are the sin t cos t ellipses x + 4 y = c The origin is fixed cosh t sinh t (e) hyperbolic rotations These are area-preserving scalings: for t >, sinh t cosh t expanding in the direction x = y and contracting by the reciprocal factor in the direction x = y; the reverse holds for t < The trajectories are the semi-hyperbolas x y = c and the four rays x = ±y The origin is fixed 947 Write down the one-parameter groups generated by the following matrices and interpret What are the trajectories? What are the fixed points? (a), (b), (c), (d), (e) e t (a) e t scalings by a factor λ = e t in the y direction and λ = e t in the x direction The trajectories are the semi-parabolas x = cy, z = d for c, d constant, and the half-lines x, y =, z = d and x =, y, z = d Points on the z axis are left ds 9/9/4 49 c 4 Peter J Olver

26 fixed t (b) shear transformations in the x direction, with magnitude proportional to the z coordinate The trajectories are lines parallel to the x axis Points on the xy plane are fixed cos t sin t (c) rotations around the y axis The trajectories are the circles sin t cos t x + z = c, y = d Points on the y axis are fixed cos t sin t (d) sin t cos t e t spiral motions around the z axis The trajectories are the positive and negative z axes, circles in the xy plane, and exponential cylindrical spirals (helices) winding around the z axis The only fixed point is the origin cosh t sinh t (e) hyperbolic rotations in the xz plane, cf Exercise 946(e) The sinh t cosh t trajectories are the semi-hyperbolas x z = c, y = d, and the rays x = ±z, y = d The points on the y axis are fixed 948 (a) Find the one-parameter group of rotations generated by the skew-symmetric matrix = (b) s noted above, e t represents a family of rotations around a fixed axis in R What is the axis? (a) + cos t cos t sin t cos t + sin t + cos t sin t + cos t + sin t (b) The axis is the null eigenvector: (,, ) T + cos t + sin t + cos t + cos t sin t + cos t 949 Let v R (a) Show that the cross product L v [x] = v x defines a linear transformation on R (b) Find the matrix representative v of L v and show that it is skew-symmetric (c) Show that every non-zero skew-symmetric matrix defines such a cross product map (d) Show that ker v is spanned by v (e) Justify the fact that the matrix exponentials e t v are rotations around the axis v Thus, the cross product with a vector serves as the infinitesimal generator of the one-parameter group of rotations around v (a) Given c, d R, x, y R, we have L v [cx + dy ] = v (cx + dy) = cv x + dv y = cl v [x] + dl v [y ], proving linearity c b (b) If v = ( a, b, c ) T, then v = c a = T v b a (c) Since v v =, also e t v v = v, and hence the rotations fix v r cos r t sin r t (d) If v = r e, then r e = r and hence e t r e = sin r t cos r t, which represent rotations around the z axis more generally, given v with r = v, let ds 9/9/4 494 c 4 Peter J Olver

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