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Unformatted text preview: , which satisÂ¯es the boundary [12] conditions w (0 ; t ) = w ( L; t ) = 0, has the form w ( x; t ) = 1 X n =1 b n sin Â³ nÂ¼x L Â´ e Â¡ Â® 2 n 2 Â¼ 2 L 2 t : 2 Find the solution u ( x; t ) of u xx = 1 9 u t ; < x < 3 ; which satisÂ¯es the boundary conditions u (0 ; t ) = 1 ; u (3 ; t ) = 4, and the initial condition u ( x; 0) = 3 x + 1. Write down the complete solution u ( x; t ). Solution: u ( x; t ) = v ( x ) + w ( x; t ), where v ( x ) = x + 1 and w ( x; t ) = 1 X n =1 b n sin Â³ nÂ¼x 3 Â´ e Â¡ n 2 Â¼ 2 t , with w ( x; 0) = u ( x; 0) Â¡ v ( x ) = 2 x ) b n = 2 3 Z 3 2 x sin Â³ nÂ¼x 3 Â´ dx = Â¡ 4 nÂ¼ x cos Â³ nÂ¼x 3 Â´ Â¯ Â¯ Â¯ Â¯ 3 + Z 3 4 nÂ¼ cos Â³ nÂ¼x 3 Â´ dx = Â¡ 12 nÂ¼ cos( nÂ¼ ). Thus, u ( x; t ) = x + 1 + 1 X n =1 12( Â¡ 1) n Â¡ 1 nÂ¼ sin Â³ nÂ¼x 3 Â´ e Â¡ n 2 Â¼ 2 t ....
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 Winter '09
 Dr.E.Devdariani
 Math, Fourier Series, Periodic function, Partial differential equation, Joseph Fourier, Fourier cosine series

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