Chem Differential Eq HW Solutions Fall 2011 31

# Chem Differential Eq HW Solutions Fall 2011 31 - = 8 π 2...

This preview shows page 1. Sign up to view the full content.

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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ) = 8 π 2 ∞ ± k =0 (-1) k (2 k + 1) 2 sin((2 k + 1) πx ) cos(4(2 k + 1) πt ) . (b) Here is the initial shape of the string. Note the new Mathematica com-mand that we used to de±ne 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) πt ) is 1 / 2, the motion is periodic in t with period 1 / 2. This is illustrated by the following graphs. We use two diﬀerent 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....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online