� 18.034 PROBLEM SET 8 Due date: Friday, April 23 in lecture. Late work will be accepted only with a medical note or for another Institute-approved reason. You are s trongly encouraged to work with others, but the final write-up should be entirely your own and based on your own understanding. This problem set is essentially a reading assignment. I had originally intended to present the material in the problem set in lecture. However, this material is less relevant to other topics in this course, and there is no time to present it in lecture. Each student will receive 10 points simply for reading through this problem set. At several places, you are asked to work-through and write-up details in the derivation. You will turn in this write-up to be graded. Problem 1(30 p oints) A collection of N identical particles of mass m0 = = mN−1 = m··· are allowed to oscillate about their equilibrium positions. Denote by x0(t), x1(t), . . . , xN−1(t) the displacement of the masses from equilibrium. The mass m0 is c onnected to a motionless base by a spring. It is also connected to mass m1 by a spring. For i = 1, . . . , N − 2, mass mi is connected to mass mi−1 and mass mi+1 by springs. Finally, mass mN−1 is connected to mass mN−2 and to a motionless base by a spring. Each spring is identical and has spring constant κ. Time is measured in units so that the frequency κ/m equals 1. The equations of motion for this system are, ⎤⎡ x�� = ANx, x = ⎢⎢⎢⎣ x0 x1 . . . xN−1 ⎥⎥⎥⎦ , where AN is the N × N matrix, ⎤⎡ −2 1 0 0 . . . 0 0 1 −2 1 0 . . . 0 0 0 1 −2 1 . . . 0 0 0 0 1 −2 . . . 0 0 . . ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ AN = . . . . . . . . . . .. . . . ... . 0 0 0 0 . . . −2 1 0 0 0 0 . . . 1 −2 In other words, (AN)i,j = ⎧ ⎪⎪⎨ ⎪⎪⎩ 1, j = i + 1, −2, j = i, 1, j = i − 1, otherwise 0, In this problem, you will determine the general solution of this system of linear differential equa-tions. Because of the s pecial nature of this problem (namely, every eigenvalue of A has multiplicity 1), it is not necessary to reduce to a system of first-order linear differential equations. (a), St ep 1(10 points) In this s tep, you will find, for each integer n, all the roots of a polynomial pn that is defined inductively be low. Please read the whole derivation. In your writeup, only fill in the missing steps in the third paragraph below. You do not need to fill in the miss ing steps in the other paragraphs (but you are encouraged to work them out for yourself). 1� � � � �Let u be an indeterminate. Let p0, p1, . . . , pn, . . . be a sequence of real polynomials in u that satisfy the linear difference equation, ⎧⎨ ⎩ pn+2 − 2upn+1 + pn = 0, p0 = 1, (1) p1 = 2u 3= 4u2 = 8uwhere n varies among all nonnegative integers. In particular, p2 − 1, and p3 n is a polynomial with integer coefficients.) In this step, you will prove that a closed − 4u. (In fact, each pformula for pn is, npn = 2n (u − cos(kπ/(n + 1))). k=1 As a reality check, observe this gives, p1 = 2(u − cos(π/2)) = 2u 2p2 = 4(u − cos(π/3))(u − cos(2π/3)) = 4(u − 1/2)(u + 1/2) = 4u − 1, 3p3 = 8(u − cos(π/4))(u − cos(2π/4))(u − cos(3π/4)) = 8(u − 1/√2)(u)(u + 1/√2) = 8u− 4u Write up a careful proof of the missing steps in the next paragraph. To prove the formula, observe that it suffices to check when |u| > 1; two polynomials in u that agree for infinitely many n n � values of are equal (you don’t have to write a proof of this). Assume that +C C=u p r r ,+ −n + −where are continuous functions in that do not depend For C , C on = =, r , r u n r r , r r. ,+ + +− − −prove that r satisfies the characteristic polynomial, 2 r − 2ur + 1 = 0. Conclude that, 2r = a ± b, a = u, b = u− 1.± Plugging this in to the equations for p0 and p1, conclude that, C+ + C = 1,−(C+ + C−)a + (C+ − C−)b = 2a, Solve this system of linear equations to get, C+ = (a + b)/2b, C = −(a − b)/2b− Therefore one solution of Equation 1 for |u| > 1 is, (a + b)n+1 − (a − b)n+12pn = [ ]/2b, a = u, b = u − 1. But, of course, there is a unique solution: for each n ≥ 2, pn can be determined recursively in terms of pn−1 and pn−2. Therefore, this is the solution of Equation 1. For the rest of this part, do not write up the missing details. The equation above also makes sense and is correct if u < 1, where b = √u2 − 1 is interpreted as a complex number. Let |u| < 1. Then pn(u) = 0 iff (a|+|b)n+1 = (a − b)n+1. B oth a + b and a − b are nonzero (because the imaginary part of each complex number is nonzero). Therefore (a + b)n+1 = (a − b)n+1 iff one of the following equations holds, i2πk/(n+1)(a + b) = ζk(a − b), ζ = e , as k varies among the integers 0, . . . , n. Of course for k = 0, the equation is a + b = a − b, i.e. b = 0. Since b = 0, this case is ruled out. Let k = 1, . . . , n. If (a + b) = ζk(a − b), then (ζk − 1)a = (ζk + 1)b. 2� � Squaring both sides, 2(ζk − 1)2 a = (ζk + 1)2b2 , i.e., 2 2(ζk − 1)2 u = (ζk + 1)2(u − 1). Solving for u2 gives, 22(2ζk)u = ζ2 + 2ζk + 1,k i.e., 2 u = (ζk + ζ−1 + 2)/4.k Simplifying, i2πk/(n+1) + e−i2πk/(n+1) ζk + ζk−1 = e = 2 cos(2πk/(n + 1)). Also, 2 + 2 cos(θ) = 4 cos2(θ/2). Therefore, u2 = cos2(πk /(n + 1)). So u = ± cos(πk/(n + 1)). But of course − cos(πk/(n + 1)) = cos(π(n + 1 − k )/(n + 1)). Therefore, every root of pn(u) = 0 with |u| < 1 is of the form cos (πk/(n + 1)) for some integer k = 1, . . . , n. Reversing the steps above, cos(πk/(n+1)) is a root of pn for every k = 1, . . . , n. Also, for …
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