math54 - quiz 12 - solns

math54 - quiz 12 - solns - constant say k Then we get the...

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Math 54 Quiz 12 Solutions 1. (5 pts) Find the Fourier series of the function f ( x ) = ± - 1 , - 2 x 0 1 , 0 < x 2 Solution : The interval we will compute this Fourier Series on is [-2,2]. Notice that the function given is odd, so the constant term and all the cosine terms will go to zero, since the product of an odd function and an even function is odd, and integrating an odd function on [-2,2] will yield zero. Now, b n = 1 2 Z 2 - 2 f ( x ) sin ² nπx 2 ³ dx = Z 2 0 sin ² nπx 2 ³ dx = ´ - 2 cos ² nπx 2 ³ µ 2 0 = - 2(( - 1) n - 1) . In the second step we used that f ( x ) is odd, so the integrand is even, so the integral on [ - 2 , 2] is just twice the integral on [0 , 2]. Thus, the Fourier Series is f ( x ) 4 π sin π 2 + 4 3 π sin 3 π 2 + 4 5 π sin 5 π 2 + ... = X n =1 4 (2 n - 1) π sin (2 n - 1) π 2 . 2. (5 pts) Use separation of variables to convert the PDE u tt = - u xx into two ordinary differential equations. Use this to find a solution that satisfies u ( x, 0) = sin x . Solution : We substitute a solution of the form u ( x, t ) = X ( x ) T ( t ) into the equation, and obtain X 00 T = - XT 00 ⇒ - X 00 X = T 00 T . Since the two sides depend on different variables, they both must be some
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Unformatted text preview: constant, say k . Then we get the two ODEs X 00 + kX = 0 , T 00-kT = 0 The characteristic equation of the X ODE is r 2 + k = 0 ⇒ r = ± √-k. Since we are looking for an initial condition of sin x , we probably want sines in the solution to this ODE. This will occur when the roots are complex, which happens when k > 0. In this case the general solution is X general ( x ) = c 1 sin( √ kx ) + c 2 cos( √ kx ) . In order to get the initial condition we want, let’s take c 2 = 0 (no cosines needed), c 1 = 1, and k = 1 (to get the right coefficient inside the sine). When k = 1, the T differential equation is T 00-T = 0 . This has the general solution T g ( t ) = b 1 e t + b 2 e-t . Let’s take the solution T ( t ) = e t , so that u ( x, t ) = X ( x ) T ( t ) = e t sin x. Note that this has the initial condition required....
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This note was uploaded on 09/17/2011 for the course MATH 54 taught by Professor Chorin during the Spring '08 term at Berkeley.

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math54 - quiz 12 - solns - constant say k Then we get the...

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