M43(22) Solutions Problem Set (c) 22, Philip D. Loewen. The number λ C is an eigenvalue for A exactly when λ = det(a λi) = det = ( λ) 2 k, k λ i.e., when λ = ± k. (i) When k = 2, the eigenvalues are ± 2, or approximately.77 and.2929. Corresponding eigenvectors u and v are given by λ = + 2 = 2 u u 2 = = u = ( 2, ), λ = 2 = The general solution of ( ) in this case is x(t) = C e ( +/ 2)t 2 2 v v 2 = = v = ( 2, ), ] + C 2 e ( / 2)t ] 2, C,C 2 R. () (ii) When k = 2, the eigenvalues are ± 2, or approximately.442 and 2.442. Corresponding eigenvectors u and v are given by λ = + 2 = 2u u 2 = = u = (, 2), λ = 2 = 2v v 2 = = v = (, 2), The general solution of ( ) in this case is x(t) = C e ( + 2)t + C 2 2 e ( 2)t 2, C,C 2 R. (2) (iii) When k = 2, both eigenvalues of A are negative, so each trajectory in () converges to as t. By contrast, when k = 2, the first eigenvalue of A is positive. Hence each trajectory in (2) with C obeys x(t) as t. The general formula for the eigenvalues, λ = ± k, shows that transition between these situations happens when k = : for k < both eigenvalues have negative real part (they are complex when k < ), whereas whenever k >, one eigenvalue will be a positive real number. 2. (i) Here A = λi + M, where The ODE system ẋ = Mx, in components, says M =. ẋ =, ẋ 2 = x, ẋ 3 =. Therefore the general solution has x (t) = c and x 3 (t) = c 3 for some constants c and c 3, while x 2 (t) = c t + c 2 for some constant c 2. With this notation, x() = c, and it follows that x(t; ξ) = ξ + tξ ê 2. File hw, version of 7 September 22, page. Typeset at 9: September 7, 22.
2 UBC M43 Solutions for Problem Set # Thus we obtain e Mt = x(t;ê ) x(t;ê 2 ) x(t;ê 3 )] = t. Finally, since λi commutes with M, we can write e At = e (λi+m)t = e λit e Mt = e λt t = eλt te λt e λt. e λt (ii) Expanding the determinant along the top row, we find det(a λi) = det 3 λ 2 λ = (3 λ)(2 λ)( λ) + (2 λ) λ = (2 λ)(3 λ)( λ) + ] = (2 λ) 3. Thus A has a triple eigenvalue at 2. Since the matrix 2I commutes with any other matrix, the laws of matrix exponents give e At = e (A 2I)t+2It = e 2It e (A 2I)t. Writing M = A 2I, we have M =. Writing ẋ = Mx in components, we have ẋ = x + x 3, ẋ 2 =, ẋ 3 = x + x 2 x 3. Therefore x 2 (t) = c 2 (some constant) in general, while ẍ = ẋ + ẋ 3 = (x + x 3 ) + ( x + x 2 x 3 ) = x 2 (t) = c 2. Thus x (t) = 2 c 2t 2 + kt + c for some constants c and k, and then the first equation gives x 3 = ẋ x = (c 2 t + k) ( 2 c 2t 2 + kt + c ) = 2 c 2t 2 + (c 2 k)t + (k c ). Let s define c 3 = k c to eliminate k = c + c 3. This gives x() = c, and hence x(t; ξ) = ξ + (ξ + ξ 3 )t + 2 ξ 2t 2 ξ 2 = e Mt = x(t;ê ) x(t;ê 2 ) x(t;ê 3 )] = + t t2 /2 t. ξ 3 + (ξ 2 ξ ξ 3 )t 2 ξ 2t 2 t t 2 /2 t Consequently e At = e 2It e Mt = e 2t + t t2 /2 t. t t 2 /2 t Remark. In both parts, a power-series approach also works well: in part (i), one has M 2 = ; in part (ii), M 3 =. File hw, version of 7 September 22, page 2. Typeset at 9: September 7, 22.
UBC M43 Solutions for Problem Set # 3 3. In components, the ODE system ẋ = Ax becomes ẋ = x 2, ẋ 2 = ω2 R x 3, ẋ 3 = R x 2. The third equation implies ẍ 3 = Rẋ2, which the second equation reduces to ẍ 3 = R ( ) ω2 R x 3, i.e., ẍ 3 + ω2 R 2x 3 =. The general solution is x 3 (t) = αcos(ωt/r)+β sin(ωt/r) for arbitrary α,β R. Now the third equation gives x 2 (t) = Rẋ 3 (t) = αω sin(ωt/r) + βω cos(ωt/r), and the first equation gives some constant γ such that ẋ = x 2 = αω sin(ωt/r) + βω cos(ωt/r), so x (t) = αr cos(ωt/r) + βr sinωt/r) + γ. The general solution for x involves the three parameters α,β,γ: αr cos(ωt/r) + βr sin(ωt/r) + γ x = αω sin(ωt/r) + βω cos(ωt/r) α cos(ωt/r) + β sin(ωt/r) Note that x() = (αr + γ, ωβ, α). Therefore x() = e α =, β =, γ = x(t) = (,, ), x() = e 2 α =, β = /ω, γ = x(t) = ((R/ω)sin(ωt/R), cos(ωt/r), (/ω)sin(ωt/r)), x() = e 3 α =, β =, γ = R x(t) = (Rcos(ωt/R) ], ω sin(ωt/r), cos(ωt/r)), These provide the columns of the matrix e At : (R/ω)sin(ωt/R) R cos(ωt/r) R e At = cos(ωt/r) ω sin(ωt/r) (/ω) sin(ωt/r) cos(ωt/r) This reduces correctly to I when t =, and the comparison below confirms that the calculation is correct: (R/ω)sin(ωt/R) R cos(ωt/r) R cos(ωt/r) ω sin(ωt/r) d cos(ωt/r) ω sin(ωt/r) = (ω/r)sin(ωt/r) (ω 2 /R)cos(ωt/R), dt (/ω) sin(ωt/r) cos(ωt/r) (/R) cos(ωt/r) (ω/r) sin(ωt/r) (R/ω)sin(ωt/R) R cos(ωt/r) R cos(ωt/r) ω sin(ωt/r) ω 2 /R cos(ωt/r) ω sin(ωt/r) = (ω/r)sin(ωt/r) (ω 2 /R)cos(ωt/R). /R (/ω) sin(ωt/r) cos(ωt/r) (/R) cos(ωt/r) (ω/r) sin(ωt/r) 4. (a) The vector-matrix system ẋ = Ax expands to ẋ = x 2, ẋ 2 = x 2, giving ẍ = ẋ 2 = x 2 = ẋ, i.e., ( ) ẍ + ẋ =. Equation ( ) is solved by x = e st when = s 2 + s, i.e., when s = or s =, giving x (t) = α + βe, α,β R. File hw, version of 7 September 22, page 3. Typeset at 9: September 7, 22.
4 UBC M43 Solutions for Problem Set # Recalling x 2 = ẋ leads to the general solution of ẋ = Ax: e x(t) = α + β, α,β R. e This has x() = (α + β, β). Thus the initial condition x() = ê leads to α =, β =, and x(t;ê ) = (,). The initial condition x() = ê 2 leads to α =, β =, so x(t;ê 2 ) = ( e,e ). In summary, ] e At = x(t;ê ) e x(t;ê 2) = e. (b) Expanding the standard formula produces the component equations x(t) = e At ξ + e A(t s) Bu(s)ds e ξ e (t s) = e + ξ 2 e (t s) ξ + ξ = 2 ξ 2 e 2 e (t s) ξ 2 e + x (t) = ξ + ξ 2 ξ 2 e + x 2 (t) = ξ 2 e + e (t s) e (t s) u(s)ds. ] u(s) ds ] u(s) ds (2 e (t s) ) (c) When (ξ,ξ 2 ) = (,3), we get x() = exactly when u satisfies () = x () = 2 3e + (2) = x 2 () = 3e + (2 e ( s) ) e ( s) u(s)ds. An equivalent system is produced by adding the equations, and by multiplying e into (2): ( ) = 2 + 2 (2 ) = 3 + e s u(s)ds. To choose a function u that solves this system of two equations, one might try to introduce just two unknowns by choosing u from some two-parameter family of functions. Endless variety exists. For example, one might try u(t) = mt + b for some constants m,b. Integration by parts leads to = 3 = m (ms + b)ds = m + b, i.e., m = 2(b + ), 2 se s ds + b e s ds = m (s )e s + b(e ) = m + b(e ). s= File hw, version of 7 September 22, page 4. Typeset at 9: September 7, 22.
UBC M43 Solutions for Problem Set # 5 Substituting the first equation into the second gives 3 = 2b 2 + (e )b, i.e., = b(e 3), i.e., b = 3 e ; ( ) 4 e Hence m = 2(b + ) = 2, and one possible control function is 3 e ( ) 4 e u(t) = 2 t + 3 e 3 e. Alternatively, one could use a piecewise-constant assumption like u(t) = { c, if t 2, c 2, if 2 < t. ( ) (Transition times other than 2 can also work.) Easy integrals lead to the system c + c 2 = 4, (e /2 )c + (e e /2 )c 2 = 3, so the constants that work with ( ) are c = 3 + 4 e 4e ( e ) 2, c 2 = 4 e 7 ( e ) 2. File hw, version of 7 September 22, page 5. Typeset at 9: September 7, 22.