Differential Equations Lecture Work Solutions 163

Differential Equations Lecture Work Solutions 163 - b. 2 u...

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b. 2 u = S ( x, y ) u (0 ,y )=0 u ( L, y )=1 u ( x, 0) = u ( x, H )=0 sin H y Use a Fourier sine series in y 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 )=0 u ( L, y )=1 X n =1 u n (0) sin H y =0 u n (0) = 0 X n =1 u n ( L )s in H y =1 u n ( L )= 1 H Z H 0 1 · sin H ydy u n ( L )= 4 for n odd and 0 for
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