Singular Perturbation Theory: Solutions to Test A Problems - Prof. Stephen Fulling, Exams of Mathematics

Download Singular Perturbation Theory: Solutions to Test A Problems - Prof. Stephen Fulling and more Exams Mathematics in PDF only on Docsity! Math. 401 (Fulling) 25 February 2005 Test A – Solutions “Up to [a certain order]” means “stop right before calculating the term of that order.” “Up through [a certain order]” means “do calculate a term of that order.” Calculators may be used for simple arithmetic operations only! 1. (30 pts.) Here are three problems, each of which requires a different flavor of singular perturbation theory. For each problem, tell which method you would expect to work. For full credit, state the ansatz in as much detail as you can, but don’t try to solve for the unknown elements (because you won’t have time). (a) d2y dt2 + ω2(100 + t2)y = 0 (ω → +∞). The WKB approximation applies. The general solution is a linear combination of the two solutions approximated by (100 + t2)−1/4e±iω ∫ t√ 100+t̃2 dt̃ . Because 100 + t2 is always positive, this approximation is valid for all t. (b) d2y dt2 + 36y + y5 = 0 (→ 0). The distorted-time (Poincaré) method is the best approach to a nonlinear oscillator. Change variable to τ = t+ω1t+ 2ω2t+ · · ·, where the ωn are constants in this case because there are no t-dependent coefficients in the differential equation. Then let y = y0(τ)+y1(τ)+ · · ·. When solving for yn, choose ωn to eliminate any resonant forcing terms (proportional to cos(6τ) or sin(6τ) in this case). (c) d2y dt2 +  ( dy dt )3 + 9y = 0 (→ 0+). The damping suggests that a two-time ansatz is necessary. Let u = t and y = y0(t, u)+y1(t, u)+· · ·. The solution for y0 will involve “constants” of integration that depend on u and are not completely determined by the zeroth-order problem. To get first-order differential equations for these functions of u one needs to look at the first-order problem and require that no resonant forcing terms (cos(3τ) or sin(3τ)) appear. 2. (35 pts.) Consider x4 − x2 + 3x− 2 +  = 0 (→ 0). (a) Find the leading behavior of all roots. First let’s try regular perturbation theory, just setting  = 0 to get an unperturbed problem: 0 = −x2 + 3x− 2 = −(x− 2)(x− 1). Therefore, we expect two roots with the behavior x = 1 + · · · and x = 2 + · · · . Since this is a fourth-order equation, there should be two more roots. Balancing the x4 term with the largest -independent term, x2, we expect that x2 ∼ 1, or x ∼ −1/2. Therefore, define x by x = −1/2x and substitute into the equation, getting −1x4 − −1x2 + 3−1/2x− 2 +  = 0. 401A-S05 Page 2 Multiply by : x4 − x2 + 3+1/2x− 2 + 2 = 0. In the lowest order we need to solve 0 = x4 − x2 = x2(x2 − 1) = x2(x− 1)(x + 1). The two zero roots will simply reproduce the two solutions we found earlier. The others are x = ±1, or x = −1/2 + · · · and x = −−1/2 + · · · . (b) For the smallest root, find a perturbative solution through order  . (More precisely, “smallest root” means “the real root that is smallest in the limit → 0+.”) The root in question is the one with x0 = 1. So we set x ∼ 1 + x1 and calculate x4 = 1 + O() , x2 = 1 + 2x1 + O( 2) , so 0 =  + (−1− 2x1) + (3 + 3x1)− 2 +  + O(2). The zeroth-order terms, −1+3−2, cancel of course, since we already solved the zeroth-order equation to get x0 = 1. The equation of order  is 0 = 1− 2x1 + 3x1 + 1 = 2 + x1 , whence x1 = −2. Finally, therefore, x = 1− 2 + O(2) . 3. (35 pts.) Let’s treat by regular perturbation theory (as far as it will take us) the problem d2y dt2 +  ( dy dt )2 + y = 0, y(0) = 0, dy dt (0) = 1. (a) Find the first-order solution (i.e., through order  ). Since we’ll need it in part (b), let’s install the full second-order ansatz, y ∼ y0 + y1 + 2y2. y′′ = y′′0 + y ′′ 1 +  2y′′2 + O( 3), (y′)2 = (y′0 + y ′ 1 + O( 2))2 = (y′0) 2 + 2y′0y ′ 1 + O( 2). Thus 0 = y′′0 + y ′′ 1 +  2y′′2 + (y ′ 0) 2 + 22y′0y ′ 1 + y0 + y1 +  2y2 + O( 3). The expansion of the initial conditions is trivial: y′0(0) = 1 and all the other initial values are 0. The zeroth-order problem is y′′0 + y0 = 0, y0(0) = 0, y ′ 0(0) = 1. The solution is y0(t) = sin t.