(a) For the functional บ S[y] = [[ dr F(x, y, y ) the types of the boundary conditions can be expressed in terms of a generalised transversality condition [(F - y' Fy)dx + Fydy]”, = 0 (1) where at each endpoint the terms involving the independent infinitesimals dx and dy must vanish identically. Suppose that, at the left-hand endpoint x = u, there is a fixed endpoint u = a, a constant, with y free to move, while at the right-hand endpoint the stationary path is constrained to lie on the curve 7(x, y) = 0, with τy ‡ 0. Show that the stationary path satisfies a natural boundary condition at x = a together with the standard transversality condition at the right-hand endpoint. Hint: If √(x, y) = 0 then τdx + Tydy = 0. (b) Let S[y] = dx 3y/3 4x2, y(1) = 2, with the right-hand endpoint (v, y(v)), v > 1, constrained to lie on the curve 7(x, y) = 0, where t(x, y) = y² — 4xy - 2x² - 2y +8x+8. - Find the stationary path, and the value of S on the stationary path. You may use the fact that the value of at the right-hand endpoint is 2. Sketch the stationary path and the curve √(x, y) = 0 on the same diagram. (a) For the functional S[y] = √ dx F(x, y, y'), the generalised transversality condition is [(F - y' Fy) dx + Fydy]” = 0, where at each endpoint the terms involving the independent infinitesimals vanish. Consider the left-hand endpoint x = u, where there is a fixed endpoint condition u = a with y free to move in the y-direction. Then at x = u = a, dx = 0 (since x is fixed) and so Fydy = 0 at x = u = a. Since dy is arbitrary it is required that Fy = 0 at x = a, which is the natural boundary condition at x = a. 9y/2 So = K where K > is constant, and so y' = K × where K ≥ 0 is a 4.x2 constant. Thus the stationary path is y(x) = C x² + D where C, D are constants K and C = ≥ 0. 2 Using the left-hand boundary condition y(1) = 2 gives D+C=2. Now the right-hand endpoint lies on the curve 7(x, y) = 0. Since y(v) = C v² + D, (3) (Cv² + D)² −4v (C'v² + D) − 2 v² − 2 (C v² + D)+8v+8=0. (4) - Now consider the right-hand endpoint at x = v. The point is free to move along the curve 7(x, y) = 0. Consider an infinitesimal perturbation x + dx, y + dy at x = v. Then, to first order, we have 7(x + dx, y + dy) = T(x, y) + Txdx + Tydy = 0 with √(x, y) = 0 so that page 7 of 11 Txdx + Tydy = 0. (1) This is the relationship between dx and dy at x = v. Now the generalised transversality condition at x = v is Finally there is the transversality condition (F-yFy) dx + Fydy = 0 TxFy + Ty(y'Fy – F') = 0. - (2) A necessary and sufficient condition for the relations (1) and (2) to be consistent is that the determinant Now y(x) = C x² + D, t(x, y) = y² − 4xy −2x² - 2y+8x+8 and 35/3 F(x,y,y') = It follows that: y′ = 2Cx; Tx = -4y-4x+8; 4x2 Ty = 2y-4x-2; and Fy = = 0 9y² 4x2 Thus y' Fy - F = 3 yr³ 2x2 = = 120³ x. (5) Τη Ty F-yFy Fy so that TxFy · Ty(F' — y′ Fy) = 0, which is the standard transversality condition at the right-hand endpoint x = v. An alternative approach is to use Ty 0 to obtain from (1) the relation dy = −(Tx/Ty)dx. Substituting for dy in (2) gives the result. (b) The Euler-Lagrange equation is So d 92 ¿ (or) = dx 4x2 9y/2 4x2 constant. = =0 = K where K > is constant, and so y' = K x where à ≥ 0 is a Thus the stationary path is y(x) = C x² + D where C, D are constants K and C = ≥ 0. 2 02:57 AM 74% (6) Evaluating at the point x = v expanding and collecting terms, and removing the common factor 120², the transversality condition becomes 20² v³-7C v² + (2 CD - 2C-3) v − 3D + 6 = 0. We are told that v is 2. Substituting v = 2 and D = 2 - C into equation (4) we get, after cancellation, 402-70-2=0. (7) ११ how come ! The two possible solutions are C = -1/4 and C = 2. We know C≥ 0, so C = 2 and hence D = 0, and so the stationary path is y(x) = 2x². The value of S on the stationary path is 3(4x)3 Sy] = 3y3 4x2 = √² dx 3² = f² dz ³(42)³ = [48] = 2 = 72. 1 The stationary path and constraint curve are illustrated in Figure 1. 10 63.64% 02:57 AM 174% (8) (9) [15] 63.64%