Differential Equations Lecture Work Solutions 161

Differential Equations Lecture Work Solutions 161 - 1 a 2 u...

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1. a. 2 u = s ( x, y ) u (0 ,y )= u ( L, y )=0 u ( x, 0) = u ( x, H )=0 sin H y Use a Fourier sine series in y (wecanalsouseaFouriersineseriesin x or a double Fourier sine series, because of the boundary conditions) u ( x, y )= X n =1 u n ( x )s in H y S ( x, y )= X n =1 s n ( x )s in H y s n ( x )= R H 0 s ( x, y )s in H ydy R H 0 sin 2 H ydy X n =1 H ± 2 u n sin H y | {z } u yy u n sin H y | {z } u xx = X n =1 s n ( x )s in H y ¨ u n ( x ) H 2 u n ( x )= s n ( x ) Boundary conditions are coming from u (0 ,y )= u ( L, y )=0 X n =1 u n (0) sin H y =0 u n (0) = 0 X n =1 u n ( L )s in H y =0 u n ( L )=0 (*) u n ( x )= sinh H ( L x )
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