Math 5440 Problem Set 3 Solutions

Transcription

1 Mth 544 Mth 544 Problem Set 3 Solutions Aron Fogelson Fll, 25 1: Logn, 1.5 # 2) Repet the derivtion for the eqution of motion of vibrting string when, in ddition, the verticl motion is retrded by dmping force proportionl to the velocity of the string. Obtin the dmped wve eqution u tt c x) 2 u xx ku t where k is the constnt dmping coefficient. Let ux, t) be the verticl displcement of the string, nd θx, t) be the ngle between the tngent to the curve y ux, t) t x nd the positive x-xis. The steps from the book tht re the sme in this cse re: tn θx, t) u x x, t), nd ρx, t) q 1 + u x x, t) 2 ρ x), Tx, t) cosθx, t)) τt), Tb, t) sinθb, t)) T, t) sinθ, t)) τt)u x x, t). The derivtion chnges in tht the eqution for conservtion of momentum is d dt b ρx, t)u t x, t) q 1 + u x x, t) 2 dx Tb, t) sinθb, t)) T, t) sinθ, t)) b ku t x, t)ρx, t) where ku t x, t) is the dmping force per unit mss. This cn be rewritten s b ρ x)u tt x, t)dx τt)u x b, t) τt)u x, t) b τt) u xx x, t)dx b b q 1 + u x x, t) 2 dx, ku t x, t)ρ x)dx ku t x, t)ρ x)dx. Since the intervl [, b] is rbitrry, the integrnds must mtch so we obtin ρ x)u tt x, t) τt)u xx x, t) ku t x, t)ρ x). Assuming s before tht τt) c x)) 2 τ /ρ x), we obtin τ is constnt, dividing by ρ x), nd reclling tht u tt x, t) c x)) 2 u xx x, t) ku t x, t). 1

2 2: Logn, 1.5 # 5) Consider string whose motion is governed by the eqution ρ u tt τ u xx for x l with boundry conditions u, t) nd ul, t). Let the energy be defined by 1 2 ρ u 2 1 t + 2 τ u 2 x dx. l Et) Show tht the energy is constnt for ll t. l E t) ρ u t u tt + τ u x u xt ) dx. From ρ u tt τ u xx, we see tht ρ u t u tt τ u t u xx, so E t) l τ u t u xx + τ u x u xt ) dx l τ x u tu x )dx τ u t l, t)u x l, t) u t, t)u x, t)). The lst equlity follows becuse u, t) for ll t implies tht u t, t) for ll t, nd similrly u t l, t). 2

3 3: Logn, 1.5 # 8) At the end x ) of long tube x ), the density of ir chnges ccording to the formul 1 cos2t) t, ρ, t), t <. Find solution for the wve eqution ρ tt c 2 ρ xx in the domin x > nd < t < in the form of right-trveling wve tht stisfies the given boundry conditions. Tke c 1 nd plot the solution surfce. The generl solution to the 1-D wve eqution tells us tht ρx, t) Fx ct) + Gx + ct) for some functions F nd G. The term Gx + ct) represents wve trvelling to the left, nd we drop it. Hence ρx, t) Fx ct), nd our gol is to determine the fucntion F so tht the boundry condition t x is stisfied. Since ρ, t) F ct), we wnt 1 cos2t), t, F ct), t <. Letting z ct, we see tht we wnt 1 cos 2z/c), z, Hence, or Fz), z >. 1 cos 2x ct)/c), x ct, ρx, t) Fx ct), x ct >, 1 cos2x ct)/c), x ct, ρx, t), x ct >. 3

4 t x 15 2 Figure.1: Plot of solution for x 2 nd t 1. 4: Logn, 1.7 # 3) Suppose tht u ux, y, z) is solution of the Neumnn boundry vlue problem K4u f x, y, z) x, y, z) 2, Kru n gx, y, z) Show tht f nd g must stisfy the reltion f dv x, y, z) 2. gda. Suppose tht solution ux, y, z) to this problem exists. Integrte the PDE over the domin to get f x, y, z)dv K4u dv Kr ru dv Kn ru da gx, y, z) da. Hence we see tht R f x, y, z)dv R gx, y, z)da must be true for solution to this problem to exist. The physicl interprettion is tht for stedy-stte het flow with Neumnn boundry conditions, the net source of het in the domin equls the net flux of het cross the boundry of the domin. 4

5 5: Logn, 1.7 # 5) Prove tht if the Dirichlet problem 4u λu, x, y, z) 2, u, x, y, z) 2, hs nontrivil solution u ux, y, z), then λ must be negtive. λ u 2 dv u4u dv ur ru dv r uru) ru ru) dv n uru) da ru ru dv. ru ru dv Since ux, y, z) is nonzero for some points in, the integrl multiplying λ is positive, nd so R ru ru dv λ R. u2 dv The only wy tht the expression on the right-side could be zero is if the integrnd of the integrl in the numertor were lwys, tht is, rux, y, z) for ll x, y, z) 2. But then u would hve to be constnt for ll x, y, z) 2, nd since u is zero on nd u is continuous, this constnt would hve to be. Since we ssumed tht u is not identiclly zero, this cnnot hppen, nd so λ <. 5

6 6: Suppose tht ux, t) is solution of the diffusion eqution u t βu xx. Show tht ech of the following is lso solution of the diffusion eqution: ) vx, t) u x x, t) b) vx, t) u t x, t) c) vx, t) ux R s, t) for ny constnt s d) vx, t) gs)ux s, t)ds for ny function g. e) vx, p t) u x, t) for ny positive constnt. ) Let vx, t) u x x, t). Then v t x, t) u xt x, t) nd v xx x, t) u xxx x, t), so v t βv xx u xt βu xxx u t βu xx ) x ) x. b) Let vx, t) u t x, t). Then v t x, t) u tt x, t) nd v xx x, t) u txx x, t), so v t βv xx u tt βu xxt u t βu xx ) t ) t. c) Let vx, t) ux s, t) for constnt s. Then v t x, t) u t x s, t) nd v xx x, t) u xx x s, t), so v t βv xx u t x s, t) βu xx x s, t), since u stisfies the diffusion eqution for llrsptil points. d) Let vx, t) gs)ux s, t)ds. Then v R tx, t) gs)u tx s, t)ds nd v xx x, t) R gs)u R xx x s, t)ds, so v t βv xx gs)u R tx s, t) βu xx x s, t))ds ds. e) Let vx, p t) u x, t) for constnt >. Then, v t x, t) u p t x, t), v x x, t) u p x x, p t), nd v xx x, t) u p xx x, t). So v t βv xx u t βu xx ). 6