MAE294B/SIOC203B: Methods in Applied Mechanics Winter Quarter sgls/mae294b Solution IV

Transcription

1 MAE9B/SIOC3B: Methods in Applied Mechanics Winter Quarter 8 sgls/mae9b 8 Solution IV (i The equation becomes in T Applying standard WKB gives ɛ y TT ɛte T y T + y = φ T Te T φ T + =, φ TT + φ T φ T Te T φ T = The first equation can be solved to give T τe φ = τ ± i τ e T dτ Once again there are two roots The equation for Φ! can be solved using an integrating factor, which gives a double integral, and is not nearly as simple as in the standard WKB case Liouville normal form (LNF for the general equation y + ay + by = comes from writing y = e I z and removing the term in z One obtains I = a/ and Here one obtains z + z TT + ( T e T (b a a z = + e T ( T ɛ z = The WKB solution will be the same (taking into account the prefactor e I One could try keeping the O(ɛ term in φ and get the usual WKB expansion, then expand that in ɛ That would avoid double integrals (ii The MMS solution requires deciding how to write the y term in terms of t or T, and also whether to use LNF To use standard MMS, one can try ɛte ɛt y or ɛte T y, otherwise the term comes in at O( Neither of these forms gives a uniformly valid MMS solution One can also try ɛte T y T ; this too fails Finally Te T y t leads to Te T ± i ] T exp e T This has some potential, but is suspicious since one normally epxects integrals rather than straight exponentials It s likely this fails at the next order (we saw something similar in class Try the LNF equation with µ(t = T e T + e T ( T ɛ = µ (T + ɛµ (T

2 Now the procedure from class can be applied at leading order, with t = T S e S ds ɛ The solution e it recovers the WKB phase The equation at the next order becomes ( µt A T + + a A =, µ which gives A = A µ / e I, recovering the WKB solution Presumably this can be done from the original form of the equation without LNF The quartic (x (x has a maximum of at x =, a minimum of 9/ and zeros at ±, ± There are hence three possibilities Note that these are all asymptotic results and there may be exact eigenvalues that do not fit this classification, in particular when σ is not very big Write Q(x = E/σ (x (x (i E/σ < 9/, in which case there are no oscillatory solutions and hence no eigenvalues (ii E/σ >, in which case there is one oscillatory region between a + and a +, where a + is the positive version of 5 ± ] / 9 + E/σ a ± = The usual connection result give an equation for E n : a+ a + Q(u] / du = (n + π It has to be checked a posteriori that numerical solutions E n to this equation satisfy the constraint E n /σ > (iii 9/ < E/σ <, in which case there are two oscillatory regions in ( a +, a and (a, a + Several connection problems are required Write the solution as ay + by in x < a +, cy + + dy = CY + + DY in a + < x, a, AY + BY = EC + FS in a < x <, where a + y = e x Q(u] / du, y + = e i x a + Q(u]/ du, Y+ = e i a x Q(u] / du, a Y = e x Q(u] / a ] du, C = cosh Q(u] / du and so on There are two equation relating (c, d to (C, D, since they represent the same expansion, and similarly two equations relating (A, B to (E, F The connection formulae x

3 are the usual ones for a d and the same for A D, ie equations in all We have a = since the solution is bounded For even solutions we have F = and for odd solutions E = Hence we have unknowns and linear equations: requiring a non-trivial solution leads to an equation in E If E is large enough, the solution grows so rapidly in ( a, that it is a good approximation to set the growing solution to zero This is now the usual connection problem at a, so in that case the eigenvalue equation is simply using the fact that Q(x is even a+ a Q(u] / du = (n + π 3 There are two regions: local when ɛ θ and global where θ = O( Divide the range at δ with ɛ δ Then δ/ɛ ɛ I L = ɛ + sin ɛu du = ɛ = π ɛ ( ɛ δ + O δ 3, ɛ δ δ/ɛ + u + O(ɛ u ] du = ɛ tan u] δ/ɛ + O(ɛ δ and I G = π/ δ dθ sin θ + O ( ɛ ] θ dθ = cot θ] π/ δ ( ɛ + O δ 3 = δ (δ, + O ɛ δ 3 Putting this together gives the answer Divide and conquer The division point is at to δ with ɛ δ, since x log x becomes small when x approaches The global contribution runs from a to δ: δ I G = ( ɛ a x ln x x ln x = = ln ln x ] δ a + O(ɛδ In the local contribution, write x = + ɛy: I L = δ/ɛ dx = δ ɛ dy ( + ɛy log ( + ɛy ɛ = = log ( + y] δ/ɛ + O(log δɛ a δ/ɛ ( ɛ ] + O x ln x x ln x ( ] dy ɛy + O y y Since the terms containing δ must vanish in the final result, I ln ln a for small ɛ This decays when ln ln a =, ie a = e ± Given the constraint on a, we take a = e 3

4 5 The function in the exponential is h(t = t t It has a maximum at t =, with h( = 3, h ( =, h ( = 3, h ( = 3/ So the integral becomes ( I = exp x 3 3 (t + ] (t 3 + dt = e 3x e 3v x/ + ] xv3 + dv We can extend the lower integration limit to while remaining asymptotic The change of variable u = (v/3x / gives I e 3x e v ( ] v 3/ x v + / dv 3x 3x ] e3x Γ( 6x 3/ 3 3/ x / Γ( + π e 3x 6x ] 8x 6 This is stationary phase if we take the real part of J = dt exp ix( f (t + a + f (t a] + t We need to find the critical points of f (t + a + f (t a There are no endpoint contributions We have f (t + a + f (t + a (t a = (t + a + ] (t a (t a + ] = t t + t ( + a (3a ( + a (t + a + ] (t a + ] This vanishes at t = and t = u = a ± a( + a / The u are real, their sum is negative and their product is positive if 3a < and negative otherwise Hence if 3a >, there is a negative u and a positive u; the square root of the positive root gives two critical points ±t c in addition to the one at the origin However, for this integral, infinity also contributes Split the integral at ±A, where A is larger than t c if t c is real and arbitrary otherwise The answer will not depend on these endpoints so we do not need to consider their contributions Then the contribution from the interval ( A, A is found as described above However, the contribution from (, A and (A, can be transformed using s = t into A J = exp ix( f (s + a + f (s ds a] A + s

5 This has a critical point at s = We find J ( π / exp ix(s + ]( + ds e iπ/ x For t =, we have h( = + a, h ( = (3a ( + a 3 For the non-zero critical point t c ( + a h(±t c = / a( a + a( + a /, h (±t c < and similarly for t c Add the contributions: I ( π( + a 3 / ( x (3a cos x + a + π ( π / ( + h (t c x + t cos xh(t c π + ( π / c x 5