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Mathematics MATC46, Assignment 2 Solutions to Selected Problems First question: dx dy = x(y − 1) = − (2y + (x − 1)(x − 4)) dt dt Critical points: x = 0 or y = 1, and 2y + (x − 1)(x − 4) = 0 implies if x = 0 then y = −2). If y = 1 then (x − 1)(x − 4) + 2 = 0 or x2 − 5x + 6 = 0 or (x − 2)(x − 3) = 0, so x = 2 or x = 3. So there are three critical points (0, −2), (2, 1), (3, 1) The matrix is A= " ∂F ∂x ∂G ∂x ∂F ∂y ∂G ∂y # " = " −3 0 5 −2 y−1 x −2x + 5 −2 # At the critical point (0, −2), A= # The eigenvalues are −2 and −3, so this is a stable node. At the critical point (3, 1), A= " 0 6 −7 −2 # The eigenvalues λ satisfy λ2 + 2λ + 3 = 0 √ √ so λ = 12 (−2 ± 4 − 4 · 3) = −1 ± 2i This is a stable spiral. 1 At the critical point (2, 1), the matrix is A= " 0 2 1 −2 # The eigenvalues λ satisfy λ2 + 2λ − 2 = 0 so λ = −1 ± point. Sect. 9.2 #5(a) dx = x(1 − y), dy = y(1 + 2x) dt dt Critical points: x = 0 OR y = 1 √ 3. This is a saddle (1) AND y = 0 OR x = −1/2 (2) From( 1), if x = 0 then the possibility x = −1/2 in (2) is impossible, so for (2) to be satisfied, we require y = 0. Thus one critical point is (0, 0). On the other hand, if y = 1 is satisfied (in (1)) then y = 0 is impossible in (2), so we must have x = −1/2. Thus there is one more critical point (−1/2, 1). Sect. 9.2 #7(a) The critical points are: x(1 − x − y) = 0 → x = 0 OR x + y = 1 (3) 1 y 3x y( − − ) = 0 → y = 0 OR 3x + y = 2 (4) 2 4 4 First consider x = 0 in (3). This forces y = 0 or y = 2 in (4). So there are two critical points (0, 0) and (0, 2). If on the other hand x + y = 1 in (3), then y = 0 in (4) yields the critical point (1, 0). We can also solve (4) by 3x+y = 2 which gives x = 1/2, y = 1/2. So the critical points are (0, 0), (0, 2), (1, 0) and (1/2, 1/2). Sect. 9.2 #24(a) dx = y, dt dy − x + x3 /6 dt This leads to 2 dy −x + x3 /6 = dx y or ydy = (−x + x3 /6)dx or y 2 /2 = −x2 /2 + x4 /24 + C or x2 + y 2 − x4 /12 = C. Sect. 9.3 #3 dx = (1 + x) sin(y) (5) dt dy = 1 − x − cos(y) (6) dt (0, 0) is a critical point because the right hand sides of equations (5) and (6) vanish at this value. Using the Taylor expansion of sin(y), sin(y) = y − y 3 /3! + . . . so the linear part of the right hand side of (5) is y. The Taylor expansion of cos(y) is 1 − y 2 /2 + . . . so the linear part of the right hand side of (6) is −x. Note that |sin(y) − y| ≤ |y|3 (by the Taylor expansion) Also |1 − cos(y)| ≤ |y|2 √ Since |y| ≤ x2 + y 2 it follows that the system is almost linear. The corresponding linear system is d dt x y where A= ! " =A 0 1 −1 0 x y # ! . The eigenvalues are ±i, so (0, 0) is a center. Sect. 9.3 #26 3 (a)(0, 0) is a critical point because the right hand side of both equations (i) and (ii) is 0 when x = y = 0. The linear part is !′ ! x x =A y y where A= " 0 1 −1 0 # . So the eigenvalues are ±i and the critical point is a center. (b)The system is almost linear because the nonlinear part is ~v = x y ! r2 (equation (i)) and x y ~v = − ! r2 (equation (ii)) the norm of this is r3 so |~v |/| x y ! | = r2 → 0 as r → 0. (c) Equation (i): r so computing r dr dx dy =x +y dt dt dt dx dy dr =x +y = r4 dt dt dt so 1 =t+C 2r2 and since r = r0 at t = 0 we find C = − 2r12 so − 0 r=q 1 2(C − t) which becomes unbounded as t → C. Hence the critical point is unstable. Equation (ii): rdr = −r4 4 so dr = −dt r3 and 1 =t+C 2r2 or which tends to 0 as t → ∞. Sect. 9.3 #30 Substituting y = dx dt 1 r=q 2(t + C) dx d2 x + c(x) + g(x) = 0 2 dt dt this becomes the system dx =y dt dy = −c(x)y − g(x) dt (b) The right hand side is 0 when x = 0 and y = 0 since g(0) = 0. Subtract ! ! x x A = −c(0)y − g ′ (0)x y y we obtain the linear system dx =y dt dy = −c(0)y − g ′ (0)x dt The nonlinear part is (−c(x) + c(0))y − (g(x) − g ′ (0)x) The first term tends to 0 as x → 0 because c(x) is continuously differentiable and |y|/r is bounded. The second term tends to 0 as x → 0 because the Taylor expansion of g(x) is |g(x) − g ′ (0)x| ≤ |g ′′ (0)/2|x2 5 so |g(x) − g ′ (0)x|/r → 0 as r → 0. (c) The linear part is A= " 0 1 ′ −g (0) −c(0) # The eigenvalues are therefore λ± = −c(0) ± q c(0)2 − 4g ′ (0) 2 q If g ′ (0) > 0, and c(0) > 0, then c(0)2 − 4g ′ (0) < c(0) so both λ± are real < 0, and hence the critical point is asymptotically stable. If c(0) < 0 or g ′ (0) < 0 then either λ+ > 0 and λ2 < 0 or both λ+ and λ− are > 0 and hence the critical point is unstable. Sect. 9.6 #2 x′ = F (x, y), y ′ = G(x, y) where F (x, y) = −(1/2)x3 + 2xy 2 G(x, y) = −y 3 . The associated vector field is W = (F (x, y), G(x, y)) We study Liapunov functions of the form V (x, y) = ax2 + cy 2 . Then ▽V · W = (2ax, 2cy) · (−1/2x3 + 2xy 2 , −y 3 ) = a(−x4 + 4x2 y 2 ) − 2cy 4 Set u = x2 , v = y 2 and define α = −a, β = 4a, γ = −2c. Then this expression becomes αu2 + βuv + γv 2 . If we choose a > 0 and c > 0, then V is positive definite and α < 0.. According to Theorem 9.6.4, if 4αγ − β 2 > 0 then αu2 + βuv + γv 2 is negative definite. If we choose c > 2a, ▽V · W is negative definite. For example, we could choose V (x, y) = x2 + 3y 2 . Then (0, 0) is asymptotically stable, by Theorem 1. Sect. 9.6 #5 The vector field is (F (x, y), G(x, y)) where F (x, y) = y − xf (x, y), 6 G(x, y) = −x − yf (x, y). As above, choose V (x, y) with ▽V = (2ax, 2cy) so ▽V · W = (2ax, 2cy) · (y − xf (x, y), −x − yf (x, y)) (7) = 2(a − c)xy − 2(ax2 + cy 2 )f (x, y) . If we choose a = c > 0, then if f > 0 ⇒ the inner product (7) is negative definite (asymptotically stable) while if f < 0 then the inner product is positive definite (unstable). 7