Chem Differential Eq HW Solutions Fall 2011 31

Chem Differential Eq HW Solutions Fall 2011 31 - ) = 8 2 k...

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Section 3.3 Wave Equation, the Method of Separation of Variables 31 Solutions to Exercises 3.3 1. The solution is u ( x, t )= ± n =1 sin nπx L ² b n cos c nπt L + b * n sin c nπt L ³ , where b n are the Fourier sine coefficients of f and b * n are L cnπ times the Fourier coefficients of g . In this exercise, b * n = 0, since g =0 , b 1 =0 . 05; and b n = 0 for all n> 1, because f is already given by its Fourier sine series (period 2). So u ( x, t )=0 . 05 sin πx cos t. 5. (a) The solution is u ( x, t )= ± n =1 sin( nπx )( b n cos(4 nπt )+ b * n sin(4 nπt )) , where b n is the n th sine Fourier coefficient of f and b * n is L/ ( cn ) times the Fourier coefficient of g , where L = 1 and c = 4. Since g = 0, we have b * n =0 for all n . As for the Fourier coefficients of f , we can get them by using Exercise 17, Section 2.4, with p =1 , h = 1, and a =1 / 2. We get b n = 8 π 2 sin 2 n 2 . Thus u ( x, t )= 8 π 2 ± n =1 sin 2 n 2 sin( nπx
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Unformatted text preview: ) = 8 2 k =0 (-1) k (2 k + 1) 2 sin((2 k + 1) x ) cos(4(2 k + 1) t ) . (b) Here is the initial shape of the string. Note the new Mathematica com-mand that we used to dene piecewise a function. (Previously, we used the If command.) Clear f f x_ : 2 x ; 0 x 1 2 f x_ : 2 1 x ; 1 2 x 1 Plot f x , x, 0, 1 Because the period of cos(4(2 k + 1) t ) is 1 / 2, the motion is periodic in t with period 1 / 2. This is illustrated by the following graphs. We use two dierent ways to plot the graphs: The rst uses simple Mathematica commands; the second one is more involved and is intended to display the graphs in a convenient array....
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