Math 401 Class Notes: Test A Solutions and WKB Approximation - Prof. Stephen Fulling, Exams of Mathematics

Download Math 401 Class Notes: Test A Solutions and WKB Approximation - Prof. Stephen Fulling and more Exams Mathematics in PDF only on Docsity! Math. 401 (Fulling) 24 February 2006 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. (32 pts.) Consider x3 + x− 1 = 0 ( > 0, → 0). (a) Find the leading behavior of all roots. First let’s see what regular perturbation theory produces. x0 − 1 = 0 ⇒ x0 = 1. There should be two more roots, so we try the balancing-and-scaling trick. If the  term is to balance the highest-degree other term, then x3 ≈ −x ⇒ x ∼ −1/2. So we set x = −1/2x, getting x ≡ +1/2x and therefore x3 + x−√ = 0. The zeroth-order equation is then 0 = x0 3 + x0 = x0(x0 2 + 1) ⇒ x0 = ±i (since the root x0 = 0 would merely reproduce the regular root). Thus x0 = ±i√  . (b) For all real roots, find a perturbation expansion through order  . The only real root is the regular one. Make the ansatz x ∼ 1 + x1 (because we already know that x0 = 1). Then x 3 ∼ 1 + 3x1 + O(2) and hence the equation is 0 ∼  + 32x1 + · · ·+ 1 + x1 − 1 = (1 + x1) + O( 2). Thus x1 = −1, and x ∼ 1− . 401A-S06 Page 2 2. (24 pts.) Do ONE of these [(A) or (B)]. (A) Write down the (first-order) WKB approximation to the solution of d2y dt2 + [ω2 − t2]y = 0, ω → +∞, with initial data y(0) = 1, y′(0) = 0. (You are not expected to evaluate the integral that arises.) Then discuss the uniformity (or lack thereof) of the approximation. (Can the solution be extended to the whole domain −∞ < t <∞ ? If not, why not?) The ODE has two linearly independent solutions with WKB approximations y± ∼ [ω2 − t2]−1/4e±i ∫ t 0 √ ω2−t̃2 dt̃ (see p. 44 of class notes). Their derivatives are y′± ∼ ±i[ω2 − t2]+1/4e±i ∫ t 0 √ ω2−t̃2 dt̃ − 1 4 [ω2 − t2]−5/4(−2t)e±i ∫ t 0 √ ω2−t̃2 dt̃ . The last term of this expression is actually of higher order than the first one, so it can be dropped in constructing the first-order approximation; but since this point is not obvious, I will carry the term along for awhile. Consider y = c+y+ + c−y− and impose the initial conditions: 1 = c+ω −1/2 + c−ω−1/2, 0 = c+iω1/2 − c−iω1/2 + O ( ω−5/2 ) . (In fact, the O ( ω−5/2 ) term is zero in this problem because of the factor −2t, but for a potential with V ′(0) 6= 0 it would be nonzero but negligible.) Rewrite the system as c+ + c− = ω+1/2, c+ − c− = 0. The solution is c± = 1 2 ω1/2. Thus y(t) ∼ 1 2 ω1/2[ω2 − t2]−1/4 [ e i ∫ t 0 √ ω2−t̃2 dt̃ + e −i ∫ t 0 √ ω2−t̃2 dt̃ ] . (It could also be written as a hyperbolic cosine.) This approximation is valid in a neighborhood of t = 0, but it must go bad when t becomes close to ±ω, and it is completely wrong when ω2 − t2 is zero or negative. How large the “good” interval is depends on ω ; it can be made arbitrarily large by considering only very large ω, but it is small for small ω. In other words, for fixed ω the approximation is not uniform in t, and for fixed t the size of the interval where the approximation is useful is not uniform in ω. (B) Pronounce each of the following assertions true or false, and write something to explain your judgment. (a) et − 1 = O() as → 0 (for fixed t). True. This is just the statement that the exponential function has a Taylor series around  = 0 with leading term 1.