PHYSICS 116C Homework 5 Solutions. y 2 = 0. x T

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1 PHYSICS 116C Homework 5 Solutions 1. a The temperature T, y satisfies aplace s equation T + T y. The boundary conditions are T along,, and y, with T T at y. We look for a separable solution, T,y XYy. Following the steps described in class we get d X d k X, which have to be solved with the boundary conditions d Y dy k Y, X X, Y. This requires k nπ/, with n 1,,3,, and X sinnπ/, Y sinhnπy/. Forming a linear combination of such solutions gives T,y a n sinnπ/sinhnπy/. The coefficients a n are determined by the boundary condition that T T at y, i.e. T a n sinnπ/sinhnπ. Using standard Fourier series techniques we get Hence T,y 4T π 1 a n T sinnπ/ d sinh nπ 1 T sinh nπ nπ [ cosnπ/] T nsinhnπ 1 1n 4T n odd, nπsinhnπ n even. 1 n 1π n 1πy n 1sinhn 1π sin sinh. 1

2 b At the center of the square T/,/ T 1 n 1 sinhn 1/π n 1/π sinhn 1π T sinhπ/ π/sinhπ sinh3π/ 3π/sinh3π + sinh5π/ 5π/sinh5π, where we noted that sinn 1/π 1 n 1. Evaluating the first few terms numerically I get Hence I find that no. of terms T/T T/,/.5T. Note that the series converges rapidly, and it looks suspiciously as though the eact answer is T /4. This is indeed the case, as we see in the net part. c et the eact value for the temperature at the center be ct so the goal is to determine c. Consider the 4 distinct solutions T α,y,α 1,,3,4, each of which is for the case where one of the sides has temperature T and the other three have temperature. Using superposition, T 1 +T +T 3 +T 4 is a solution of aplace s equation for boundary conditions where T T on all boundaries. Clearly the solution is trivial: T 1,y+T,y+T 3,y+T 4,y T a constant. 1 By symmetry, all the T α /,/ must be equal to ct. Substituting into Eq. 1 gives 4cT T, so c 1/4, i.e. T/,/ T 4.. We denote the temperature by T,t. We have to solve the diffusion equation for t > and < < with boundary conditions T,t T,t, T, T. ooking for a separable solution, T,t XΘt, and dividing by T, we get X +k X Θ +Dk Θ, where k is the separation constant. The boundary conditions on X are X X. Consider first the k solution, Θ 1, X c 1 +c. For t the k solutions decay to zero and so we must choose c 1 and c such that X X, i.e. we just have the trivial solution of zero. Net consider the k solutions: X Acosk+Bsink Θ ep Dk t

3 The boundary conditions X X give A and k nπ/ n 1,,3,, and so the solution is [ nπ ] T,t a n sin ep D t. Imposing the boundary condition at t gives The Fourier coefficient a n is given by Integrating by parts twice gives T a n T a n sin. sin d. { [ ] a n T cos + cos nπ T nπ cos d nπ T { [ ] sin sin nπ nπ 4T nπ nπ sin d 4T [ ] nπ cos 4T nπ nπ nπ [1 1n ] n even, 8 nπ 3 T n odd. Hence the solution is writing n m 1 with m 1,,3, T,t 8 π 3 T m1 1 m 1 3 sin m 1π [ ep D } d } d m 1π t]. Note that the first term, m 1 is by far the largest. The net largest, which is for m, is 7 times smaller. This is because the first sine wave, sinπ/, matches the parabola,, pretty well. To see this, I plot below the solution at t with parameters T 1, both the full solution, 1, and the first term in the series, 8/π 3 sinπ. The two are fairly close. 3

4 .3.5. T, full solution: 1- first term: 8/π 3 sinπ et us denote the temperature by u,t. For t < it is given by the steady state solution u,t T 1 +T T 1 t <. For t > the temperatures at the ends are reversed, and so, for t we have a new steady state solution u,t T +T 1 T t. 3 The goal of the question is to see how the temperature changes from Eq. to Eq. 3 at intermediate times, see the figure. We need to solve the diffusion equation u t u D. 4

5 Following the method of separation of variables discussed in class, we look for a separable solution u,t XTt. Substituting and dividing by u and for convenience by D we get 1 D T T X X. Both sides must be a constant so X and T satisfy the following equations: where k is the separation constant. X +k X, T +Dk T, Consider first the k solution, T 1, X c 1 + c. For t the k solutions decay to zero and so we must choose c 1 and c such that u XTt gives Eq. 3, i.e. c 1 T and c T 1 T /. Net consider the k solutions: X Acosk+Bsink T ep Dk t Since the these solutions decay with time, whereas the boundary conditions at and for t > are time independent and satisfied by the k solution, the k solutions must vanish at and, i.e. A and k nπ/. This gives u sinnπ/ep[ Dnπ/ t]. Forming a linear combination of such solutions gives u [ nπ ] t a n sin ep D The solution for t > is the sum of Eq. 3 and Eq. 4, i.e. k pieces. 4 u T +T 1 T + [ nπ ] t a n sin ep D. 5 The coefficents a n are chosen so that, at t, Eq. 5 reduces to Eq.. This gives T 1 T 1 Evaluating the Fourier coefficients in the usual way, a n T 1 T 4T 1 T nπ a n sin. 1 n even n odd. sin d 6 5

6 To conclude, the solution is writing n m with m 1,,3, u,t T +T 1 T + 4T 1 T π m1 1 m sin mπ The solution is shown in the figure above for t.,d/ 1. [ ep D mπ t]. 4. We look for a solution of the form y,t XTt. Substituting into the wave equation and dividing by y gives X X 1 Y v Y. As usual, each term must be a constant which we write as k the negative sign because we are looking for oscillatory solutions. The solutions are X Asink+Bcosk, T Csinkvt+Dcoskvt. 7 Because X we have B and because t y,t t we have C. Forming a linear combination of such solutions gives nπvt y,t c n sin cos. 8 The initial condition on y is y, y sin3π/. Comparing with the above solution we see by inspection that c 3 y and all the other c n vanish. Hence the solution which satisfies the initial conditions is 3π 3πvt y,t y sin cos. 5. a We have sin3 Ime 3i Imcos+isin 3 Imcos 3 +3icos sin 3cossin isin 3 3cos sin sin 3 31 sin sin sin 3 3sin 4sin 3. b Proceeding as in Qu. 4, the general solution with zero time derivative at the initial time is given by Eq. 8. Here the initial condition is y, y sin 3 π y 4 3sin sin3. Comparing with Eq. 8 we see by inspection that c 1 3y /4,c 3 y /4 and all the other c n vanish. Hence the solution which satisfies the initial conditions is y,t y 4 [ 3sin π cos πvt sin ] 3π 3πvt cos. 6

7 6. The boundary condition is now y rather than y at t. Hence we need the sinkvt solution in Eq. 7. Following the solution given in Qu. 4 we have y,t nπvt a n sin sin. n The a n are determined from the boundary condition on the velocity at t, i.e. v nπv a nsin, n where v is the initial velocity, v v / a tent -like curve which is given mathematically by v v / 1 v / 9 The Fourier coefficients are given by We write this as where nπv a n v / sin d+ v / a n nπv v I 1 +I 4v nπv I 1 +I 1 I 1 I / / sin d 1 sin 1 sin d. d Evaluating I 1 by integrating by parts gives / I 1 d nπ d cos d 1 [ ] / cos + 1 / cos. nπ nπ nπ nπ cos + [ ] / nπ sin nπ nπ cos + nπ nπ sin 7

8 In the same way we find so where we note that Hence the final result is y,t 8v π 3 v I nπ cos nπ nπ sin + nπ sin nπ I 1 +I nπ sin nπ, 1, n 1,5,9, 1, n 3,7,11,, n,4,6, sin π πvt sin 1 3π 3πvt sin sin π 5πvt sin sin a The initial displacement of the string is φ and the initial velocity if ψ, so φ f+g, 1 ψ vf +vg, 11 where the primes indicate differentiation with respect to the argument of the function. Integrating Eq. 11 from to gives f g 1 v ψzdz +c, 1 where the constant of integration c is equal to f g. Adding and subtracting Eqs. 1 and 1 gives f 1φ 1 v g 1φ+ 1 v ψzdz + c ψzdz c Substituting these epressions for f and g into the general solution given in the question gives u,t 1 [φ vt+φ+vt]+ 1 +vt ψzdz. v b According to the boundary conditions, φz sinπz/, ψz. Hence, from part a, the solution is u,t 1 [ ] π vt π+vt sin +sin Using the trigonometric relation sina±b sinacosb ±sinbcosa this can be written as π πvt u,t sin cos, which is a standing wave as required. vt. 8