I need answers for 1c, 2d, 2e, 3b, 3c, 4f, 4g, 6e, 6f. I have provided answers for the other parts of the assignment as well for your convenience.
THE UNIVERSITY OF SYDNEY SCHOOL OF MATHEMATICS AND STATISTICS Assignment 2 ——————————————————————————————————————— MATH3078/3978: PDES AND WAVES Semester 2, 2017 ——————————————————————————————————————— Due 5pm, Wednesday, 18 October 2017. You must use LaTeX to type your solutions. Upload a PDF of your assignment with the Turnitin feature on Blackboard found at: https://elearning.sydney.edu.au. 1D Linear Waves We learned early in the course that ∂2t u = ∂ 2 xu =⇒ u(t, x) = f(x− t) + g(x+ t) (1) Define the initial conditions u(t = 0, x) = u0(x), ∂tu(t = 0, x) = v0(x). (2) The initial conditions determine the functions f(x) and g(x). Question 1(a): Write formulae for f(x) and g(x) in terms of u0(x) and v0(x). Hint: You answer will involve integrals. Solution 1(a): The initial conditions are: f(x) + g(x) = u0(x), and − f ′(x) + g′(x) = v0(x) (3) Therefore f(x) = u0(x) 2 − 1 2 ∫ x 0 v0(y) dy g(x) = u0(x) 2 + 1 2 ∫ x 0 v0(y) dy (4) Question 1(b): Solve the wave equation (1) on the real line for the the following initial condition, u(t = 0, x) ≡ u0(x) = 0 if x ≤ −1 1 + x if − 1 ≤ x ≤ 0 1− x if 0 ≤ x ≤ 1 0 if 1 ≤ x (5) and initial velocity ∂tu(t = 0, x) = 0. For each finite time t < ∞,="" assume="" the="" solution="" vanishes="" as="" |x|="" →="" ∞.="" copyright="" c©="" 2017="" the="" university="" of="" sydney="" 1="" solution="" 1(b):="" in="" the="" case="" where="" v0(x)="0," u(t,="" x)="u0(x+" t)="" +="" u0(x−="" t)="" 2="" (6)="" or="" u(t,="" x)="1" 2="" ="" 1="" +="" x+="" t="" if="" −="" 1−="" t="" ≤="" x="" ≤="" −t="" 1−="" x−="" t="" if="" −="" t="" ≤="" x="" ≤="" 1−="" t="" 0="" otherwise="" +="" 1="" 2="" ="" 1="" +="" x−="" t="" if="" −="" 1="" +="" t="" ≤="" x="" ≤="" t="" 1−="" x+="" t="" if="" t="" ≤="" x="" ≤="" 1="" +="" t="" 0="" otherwise="" (7)="" four="" regions="" exist="" where="" the="" two="" functions="" might="" overlap="" non-trivially.="" therefore="" define="" the="" four="" intervals="" a="" ≡="" [−t−="" 1,="" −t="" ],="" b="" ≡="" [−t,="" 1−="" t="" ],="" c="" ≡="" [="" t−="" 1,="" t="" ],="" d="" ≡="" [="" t,="" 1="" +="" t="" ]="" (8)="" assuming="" t=""> 0, we need to find the intersection of the different intervals. A ∩B = {−t}, C ∩D = {t}, A ∩D = ∅ (9) A ∩ C = { [ t− 1, −t ] if t ≤ 1/2 ∅ if t > 1/2 , B ∩D = { [ t, 1− t ] if t ≤ 1/2 ∅ if t > 1/2 (10) B ∩ C = [−t, t ] if t ≤ 1/2 [ t− 1, 1− t ] if 1/2 < t="" ≤="" 1="" ∅="" if="" t=""> 1 (11) Therefore we see that there are three important phases. Phase I: 0 ≤ t ≤ 1/2, u(t, x) = 0 if x ≤ −t− 1 1 2 (1 + x+ t) if − t− 1 ≤ x ≤ t− 1 1 + x if t− 1 ≤ x ≤ −t 1− t if − t ≤ x ≤ t 1− x if t ≤ x ≤ 1− t 1 2 (1− x+ t) if 1− t ≤ x ≤ 1 + t 0 if 1 + t ≤ x (12) Phase II: 1/2 ≤ t ≤ 1, u(t, x) = 0 if x ≤ −t− 1 1 2 (1 + x+ t) if − t− 1 ≤ x ≤ −t 1 2 (1− x− t) if − t ≤ x ≤ t− 1 1− t if t− 1 ≤ x ≤ 1− t 1 2 (1 + x− t) if 1− t ≤ x ≤ t 1 2 (1− x+ t) if t ≤ x ≤ 1 + t 0 if 1 + t ≤ x (13) 2 Phase III: t ≥ 1, u(t, x) = 0 if x ≤ −t− 1 1 2 (1 + x+ t) if − t− 1 ≤ x ≤ −t 1 2 (1− x− t) if − t ≤ x ≤ 1− t 0 if 1− t ≤ x ≤ t− 1 1 2 (1 + x− t) if t− 1 ≤ x ≤ t 1 2 (1− x+ t) if t ≤ x ≤ 1 + t 0 if 1 + t ≤ x (14) Circular Drum The two-dimensional wave equation 1 c2 ∂2t u = 1 r ∂r (r∂ru) + 1 r2 ∂2θu (15) governs small displacements, u = u(r, θ, t) of a circular drum. The coordinate r represents the radial direction, and θ represents the angular direction. We want clamped boundary conditions on the outer radius, u(r = a, θ, t) = 0. (16) Question 2(a): Solve for the time-periodic characteristic modes of equation (15) using separation of variables. That is, find u(r, θ, t) = R(r)Θ(θ)eiωt, (17) where ω represents a characteristic frequency. Solution 2(a): Full form on the solution is: u(r, θ, t) = A0Jm(kr)e imθeiωt where k = ω c , m = 0, 1, 2, . . . , (18) and A0 is a constant. The function, Jm(x) is the mth-order Bessel function. It satisfies the differential equations J ′′m(x) + 1 x J ′m(x)− m2 x2 Jm(x) = −Jm(x) (19) with a series solution Jm(x) = ∑ n≥0 (−1)n n!(m+ n)! (x 2 )2n+m . (20) Knowing that the solution is a Bessel function is enough. Question 2(b): Apply boundary conditions at the outer rim to determine a formula for the fre- quencies ω. 3 Solution 2(b): The boundary condition requires Jm(ka) = 0 =⇒ ω = c a zn,m where Jm(zm,n) = 0 with n = 1, 2, . . . . (21) Question 2(c): What are the frequencies (in Hz) of the 4 slowest modes for a drum with a 60 cm diameter and wave speed c =200m/s. Solution 2(c): A table of the first few zeros of the Bessel functions: zm,n n = 1 n = 2 n = 3 n = 4 m = 0 2.40483 5.52008 8.65373 11.7915 m = 1 3.83171 7.01559 10.1735 13.3237 m = 2 5.13562 8.41724 11.6198 14.796 m = 3 6.38016 9.76102 13.0152 16.2235 (22) For a = 0.6 m and c = 200 m/s, the for slowest frequencies are approximately ω0,1 = 801.6 Hz, for m = 0, n = 1 (23) ω0,2 = 1277.2 Hz, for m = 1, n = 1 (24) ω0,3 = 1711.9 Hz, for m = 2, n = 1 (25) ω2,1 = 1840.0 Hz, for m = 0, n = 2 (26) The zeros of the Bessel function provide an isomorphism between N and N2; albeit not a very convenient one. Orthogonal polynomials Consider the generating function 1 1− 2tx+ t2 = ∞∑ n=0 yn(x)t n. (27) Question 3(a): Using equation (27), find a second-order differential equation of the form, p(x)y′′n(x) + q(x)y ′ n(x) + λn yn(x) = 0. (28) Where p(x) and q(x) are polynomials in x (at most 2nd degree), and λn is an eigenvalue that depends on n. What are p(x), q(x), and λn? Solution 3(a): The generating function satisfies the following partial differential equation[ (1− x2)∂2x − 3x∂x + ϑ2 + 2ϑ ] G = 0, where ϑ = t ∂t (29) The ϑ operator is multiplication by n on the sequence. This means that 4 p(x) = (1− x2), q(x) = −3x λn = n2 + 2n. (30) Question 3(b): Prove that∫ 1 −1 ym(x)yn(x)w(x) dx = Γn δn,m it m 6= n. (31) for some weight function w(x), and normalisation Γn. What are w(x) and Γn? Solution 3(b): Using an integrating factor, we can re-write the differential equation in the following form, d dx [ (1− x2)w(x)y′n(x) ] = −λnw(x)yn(x) (32) where w(x) = √ 1− x2 (33) which vanishes at the end points, x = ±1. Multiplying equation (32) by ym(x) and subtracting from the same for m and n reversed, d dx [ (1− x2)w(x) ( ym(x)y ′ n(x)− yn(x)y′m(x) )] = (λm − λn)w(x)yn(x)yn(x) (34) Integrating over x < |1|, (m− n)(2 +m+ n) ∫ 1 −1 w(x)yn(x)yn(x) dx = 0 (35) this implies that ∫ 1 −1 w(x)yn(x)yn(x) dx = 0 if m 6= n. (36) we need to use the generating function to get the normalisation.∫ 1 −1 g(x, t)2w(x) dx = ∑ n≥0 ∑ m≥0 tn+m ∫ 1 −1 yn(x)ym(x)w(x) dx (37) but the right-hand side double sum collapses into a single sum because the yn(x) are orthogonal. therefore ∫ 1 −1 g(x, t)2w(x) dx = ∑ n≥0 γnt 2n (38) you can do the integral using the residue theorem, or mathematica, π 2 1 1− t2 = ∑ n≥0 γnt 2n. (39) the left-hand side function is a geometric series in terms of t2. therefore, γn = π 2 , and w(x) = √ 1− x2. (40) 5 question 3(c): use the generating function to find a three-term recurrence relation of the form, anyn+1(x) + bnyn(x) + cnyn−1(x) = x yn(x) (41) find an, bn, and cn. solution 3(c): the generating functions satisfies the differential equation (1− 2xt+ t2)∂tg+ 2(t− x)g = 0 (42) using ϑ = t∂t, this is the same as t2(ϑ+ 2)g− 2xt(ϑ+ 1)g+ ϑg = 0. (43) in terms of yn(x), nyn−2(x)− 2xnyn−1(x) + nyn(x) = 0 (44) or yn+1(x) + yn−1(x) = 2x yn(x), =⇒ an = cn = 1 2 , bn = 0. (45) question 3(d): prove that yn(x) = n∑ `=0 p`(x)pn−`(x) (46) where p`(x) are the legendre polynomials. hint: this question is easy if you use a specific property of generating functions. solution 3(d): suppose we have two generating functions g(t) = ∑ n≥0 ynt n, and h(t) = ∑ n≥0 pnt n (47) it’s one of the important properties of generating functions that g(t) = h(t)2 ⇐⇒ yn = n∑ `=0 p`pn−` (48) the generating functions for legendre polynomials is h(t, x) = 1√ 1− 2xt+ t2 = ∑ `≥0 p`(x)t ` (49) therefore h(t, x)2 = ∑ n≥0 yn(x)t n (50) 6 burgers shock the goal in this question is to solve the nonlinear burgers equation ∂tu+ u∂xu = 0 (51) for the same initial condition as in question 1, i.e., equation (5). x u figure 1: hint: the (symmetric) dotted triangle shows the shape of the initial condition; the (asym- metric) dashed triangle shows the shape of the solution for some time after t = 0 and before a shock forms; the solid (right) triangle shows the shape of the solution for some time after a shock forms. question 4(a): find the characteristic curves, x = x(t, a), of the solution and invert for the initial position of a curve a = a(x, t). solution 4(a): the characteristic curves are x = a+ u0(a)t (52) for u0(x) |1|,="" (m−="" n)(2="" +m+="" n)="" ∫="" 1="" −1="" w(x)yn(x)yn(x)="" dx="0" (35)="" this="" implies="" that="" ∫="" 1="" −1="" w(x)yn(x)yn(x)="" dx="0" if="" m="" 6="n." (36)="" we="" need="" to="" use="" the="" generating="" function="" to="" get="" the="" normalisation.∫="" 1="" −1="" g(x,="" t)2w(x)="" dx="∑" n≥0="" ∑="" m≥0="" tn+m="" ∫="" 1="" −1="" yn(x)ym(x)w(x)="" dx="" (37)="" but="" the="" right-hand="" side="" double="" sum="" collapses="" into="" a="" single="" sum="" because="" the="" yn(x)="" are="" orthogonal.="" therefore="" ∫="" 1="" −1="" g(x,="" t)2w(x)="" dx="∑" n≥0="" γnt="" 2n="" (38)="" you="" can="" do="" the="" integral="" using="" the="" residue="" theorem,="" or="" mathematica,="" π="" 2="" 1="" 1−="" t2="∑" n≥0="" γnt="" 2n.="" (39)="" the="" left-hand="" side="" function="" is="" a="" geometric="" series="" in="" terms="" of="" t2.="" therefore,="" γn="π" 2="" ,="" and="" w(x)="√" 1−="" x2.="" (40)="" 5="" question="" 3(c):="" use="" the="" generating="" function="" to="" find="" a="" three-term="" recurrence="" relation="" of="" the="" form,="" anyn+1(x)="" +="" bnyn(x)="" +="" cnyn−1(x)="x" yn(x)="" (41)="" find="" an,="" bn,="" and="" cn.="" solution="" 3(c):="" the="" generating="" functions="" satisfies="" the="" differential="" equation="" (1−="" 2xt+="" t2)∂tg+="" 2(t−="" x)g="0" (42)="" using="" ϑ="t∂t," this="" is="" the="" same="" as="" t2(ϑ+="" 2)g−="" 2xt(ϑ+="" 1)g+="" ϑg="0." (43)="" in="" terms="" of="" yn(x),="" nyn−2(x)−="" 2xnyn−1(x)="" +="" nyn(x)="0" (44)="" or="" yn+1(x)="" +="" yn−1(x)="2x" yn(x),="⇒" an="cn" =="" 1="" 2="" ,="" bn="0." (45)="" question="" 3(d):="" prove="" that="" yn(x)="n∑" `="0" p`(x)pn−`(x)="" (46)="" where="" p`(x)="" are="" the="" legendre="" polynomials.="" hint:="" this="" question="" is="" easy="" if="" you="" use="" a="" specific="" property="" of="" generating="" functions.="" solution="" 3(d):="" suppose="" we="" have="" two="" generating="" functions="" g(t)="∑" n≥0="" ynt="" n,="" and="" h(t)="∑" n≥0="" pnt="" n="" (47)="" it’s="" one="" of="" the="" important="" properties="" of="" generating="" functions="" that="" g(t)="H(t)2" ⇐⇒="" yn="n∑" `="0" p`pn−`="" (48)="" the="" generating="" functions="" for="" legendre="" polynomials="" is="" h(t,="" x)="1√" 1−="" 2xt+="" t2="∑" `≥0="" p`(x)t="" `="" (49)="" therefore="" h(t,="" x)2="∑" n≥0="" yn(x)t="" n="" (50)="" 6="" burgers="" shock="" the="" goal="" in="" this="" question="" is="" to="" solve="" the="" nonlinear="" burgers="" equation="" ∂tu+="" u∂xu="0" (51)="" for="" the="" same="" initial="" condition="" as="" in="" question="" 1,="" i.e.,="" equation="" (5).="" x="" u="" figure="" 1:="" hint:="" the="" (symmetric)="" dotted="" triangle="" shows="" the="" shape="" of="" the="" initial="" condition;="" the="" (asym-="" metric)="" dashed="" triangle="" shows="" the="" shape="" of="" the="" solution="" for="" some="" time="" after="" t="0" and="" before="" a="" shock="" forms;="" the="" solid="" (right)="" triangle="" shows="" the="" shape="" of="" the="" solution="" for="" some="" time="" after="" a="" shock="" forms.="" question="" 4(a):="" find="" the="" characteristic="" curves,="" x="x(t," a),="" of="" the="" solution="" and="" invert="" for="" the="" initial="" position="" of="" a="" curve="" a="a(x," t).="" solution="" 4(a):="" the="" characteristic="" curves="" are="" x="a+" u0(a)t="" (52)="" for="">