MATH 285 E1 F1 GRADED HOMEWORK SET 7 DUE MONDAY DECEMBER 8 IN LECTURE This time the homework has just one part Please staple your homework together and put your name and section on it Thank you 1 15 points Find the solution of following wave problem for vibrations in a string of length 10 corresponding to a struck string 2 2 u 16 xu2 t2 u 0 t 0 u 10 t 0 u x 0 0 u x 0 x 10 x t Although this problem was essentially done in lecture for general initial conditions we will recall the solution The separable solutions u x t X x T t of the wave equation and the boundary conditions satisfy d2 X X 0 dx2 X 0 0 X 10 0 d2 T c2 T 0 dt2 We recognize the conditions on X as an eigenvalue problem with eigenvalues n n 10 2 and eigenfunctions Xn x sin n x 10 where n 1 2 3 ranges over the positive integers The corresponding Tn solves d2 Tn c2 n 10 2 Tn 0 dt2 whence Tn t An cos cn t 10 Bn sin cn t 10 The separable solutions are un x t An cos cn t 10 Bn sin cn t 10 sin n x 10 and the general solution is u x t X An cos cn t 10 Bn sin cn t 10 sin n x 10 n 1 1 2 GRADED HOMEWORK 7 Now to apply the initial conditions Plugging t 0 into the general solution X An sin n x 10 u x 0 n 1 The initial condition says that this ought to be zero so we conclude that all An are zero Thus u x t X Bn sin cn t 10 sin n x 10 n 1 where have yet to determine Bn Differentiate with respect to t X u cn x t Bn cos cn t 10 sin n x 10 t 10 n 1 Plugging in t 0 makes the cosine factors become 1 and we get something that looks like a sine series X u cn x 0 Bn sin n x 10 t 10 n 1 The initial condition says that this ought to equal x 10 x So we need to find the sine series of this function on the interval 0 x 10 Thus we must compute Z 10 2 bn x 10 x sin n x 10 dx 10 0 This can be done directly by repeated integration by parts The result is 800 n is odd 400 n 3 bn 1 cos n 3 n 0 n is even We have the relation Bn cn bn 10 so Bn 8000 c n 4 n is odd 0 n is even Noting that actually c 4 which could have been put in earlier if desired X 2000 u x t sin 4n t 10 sin n x 10 n 4 n odd GRADED HOMEWORK 7 3 2 5 points The function u x t sin 20t cos 5x solves a wave equation Write u x t as a sum of a right moving wave and a left moving wave and determine the speed of these waves Hint Scour the article List of trigonometric identities on Wikipedia for a relevant identity The relevant identity is what is called the product to sum formula 1 sin A cos B sin A B sin A B 2 Thus u x t sin 20t cos 5x 1 1 sin 20t 5x sin 20t 5x 2 2 We would like to see expressions like x ct and x ct inside the sines so we factor out the 5 and use the fact that sin z sin z u x t 1 1 sin 5 x 4t sin 5 x 4t 2 2 Thus if we define F z 1 2 sin 5z we find u x t F x 4t F x 4t The term F x 4t 12 sin 5 x 4t is a left moving wave with speed c 4 and the term F x 4t 21 sin 5 x 4t is a right moving wave with speed 4 3 15 points Find the solution of the Laplace equation problem on the square 0 x 1 0 y 1 2 2 u yu2 0 x2 u x 0 1 u x 1 0 u 0 y 0 u 1 y 0 Taking into account Laplace s equation and the homogeneous boundary conditions we find that the separable solutions u x y X x Y y satisfy d2 X X 0 dx2 X 0 0 d2 Y Y 0 dy 2 X 1 0 Y 1 0 4 GRADED HOMEWORK 7 The set of conditions on X is an eigenvalue problem with solutions n n 2 Xn x sin n x where n 1 2 3 The corresponding function Yn y must satisfy d2 Yn n 2 Yn 0 dy 2 So Yn y Aen y Be n y The condition Yn 1 0 means 0 Aen Be n So we can write B Ae2n and thus Yn y Aen y Ae2n e n y A en y en 2 y This is actually good enough to work with but it can also be written in terms of sinh Let s factor out a factor of en Yn y Aen en y 1 en 1 y 2Aen sinh n 1 y Since 2Aen is just a constant we can take un x y sinh n 1 y sin n x as our basic separable solution The general solution to Laplace s equation and the three homogeneous boundary conditions is therefore X u x y cn sinh n 1 y sin n x n 1 Last is to satisfy the condition u x 0 1 Plugging in y 0 we get X u x 0 cn sinh n sin n x n 1 which has the form of a sine series with L 1 take the sine series of 1 on the interval 0 x are given by Z 4 2 1 2 bn 1 sin n x dx 1 cos n n 1 0 n 0 These coefficients are related to cn by cn sinh n bn Thus cn 4 n sinh n 0 n is odd n is even Thus we need to 1 the coefficients n is odd n is even GRADED HOMEWORK 7 We conclude u x y X 4 sinh n 1 y sin n x n sinh n n odd 5
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