2 i ux 0 x 2 uy 0 y y i ux 0 x u 0

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Unformatted text preview: = x2 ) H (x) = x2 G(x). x u(x, 0) = x2 ) H (x) + G(x) = x2 ) H (x) = x2 G(x). Apply second BC: uy (0, y ) = y ) 0 = y , which is a contradiction. Apply second BC: uy (0, y ) = y ) 0 = y , which is a contradiction. So the BVP has no solution. So the BVP has no solution. (ii) u(x, 0) = x2, uy (x, y )|y=x = ex . As above, H (x) = x2 G(x). x (ii) u(x, 0) = x2, uy (x, y )|y=x = ex . As above, H (x) = x2 G(x). 3 Apply second BC: uy (x, y )|y=x = e x ) x2e x 3G(x) = ex . Thus 2x Apply second BC: uy (x, y )|y=xx+x3 e ) x e G(x) = ex . Thus = x +x 3 G(x) = e x+x3, H (x) = x2 + e x+x3, and the BVP has the unique solution: e x2 e x2 2 G(x) = x2 , H (x) = x + x2 , and the BVP has the unique solution: 3 ex+x 3 2 x+x 2 u(x, y ) = x 2 + e 2 (1 e x 2y ). xy u(x, y ) = x + x 2 (1 e ). x (iii) u(x, 0) = x2 , uy (1, y ) = e2y . As above H (x) = x2 G(x). 2 (iii) u(x, 0) = x2 , uy (1, y ) = ey y . As above H (x) = x2 G(x). 2 Apply second BC: uy (1, y ) = e 2y ) e y G(1) = e2y ) G(1) = e3y . Apply second BC: uy (1, y ) = e...
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This document was uploaded on 02/09/2014.

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