Ux 0 x h x gx x2 h x x gx apply second

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Unformatted text preview: ntradiction. y 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. y So the BVP has no solution. So the BVP has no solution. So the BVP has no solution. So the BVP has no solution. 2 x 2 (ii) u(x, 0) = x22,, uy (x, y )||y=x = exx.. As above, H (x) = x22 G(x). (ii) u(x, 0) = x22 uy (x, y ) y=x = ex As above, H (x) = x22 G(x). x x (ii) u(x, 0) = x2,, uy (x, y )||y=x = ex .. As above, H (x) = x2 G(x). (ii) u(x, 0) = x uy (x, y ) y=x = ex As above, H (x) = x x G(x). y 3 y y=x =x 2 x3 Apply second BC: uy (x, y )||yy=x = e x ) x22e xx33G(x) = e x.. Thus Apply second BC: uy (x, y ) y=x = ex ) x22e x 33G(x) = ex Thus y 2x x x x y =x = e x ) Apply second3 BC: uy (x, y )||y=xx+=3 e ) x e x G(x) = e x.. Thus Apply second BC: uy (x, y ) y=xx+x3 x e G(x) = e Thus y 3 3 y =x x3 e x +x 3 e e x +x 3 2 ex+x 3, H (x) = x2 + e x+x 3 , and the BVP has the unique solution: x+x , H (x) = x2 + e x+x 3 , and the BVP has the unique solution: x+x 3 +x x+x +x G(x) = e x 22 G(x) = ex 2 ex 2 x G(x) = ex 22 ,, H (x) = x2 + e xx222 ,, and the BVP has the unique solution: H (x) = x2 + exx22 and the BVP has the unique solution: G(x) = x2 x2 x x x 3 x+x3 exx+xx33 ex+ x 3 2 2 + 2 2 x2 y x+x (1 2 + ex+x 3(1 u(x, y ) = x 2 + e 2 e xx22yy).. u(x, y ) = x 2 e x 2yy) xy 2 x ). 2 (1 2 (1 u(x, y ) = x + x 2 e u(x, y ) = x + x 2 e ). x2 x 2 2y 2 2y y 2 (iii) u(x, 0) = x 22,, uy (1,, y ) = e 22y.. As above H (x) = x22 G(x). (iii) u(x, 0) = x2 uy (1 y ) = e2 y As above H (x) = x22 G(x). (iii) u(x, 0) = x 2,, uy (1,, y ) = ey y.. As above H (x) =y x2 G(x). (iii) u(x, 0) = x uy (1 y ) = 2 y As above H (x) = x e2 G(x). y y 2 3y Apply second BC: uy (1,, y ) = e 22y ) e yyG(1) = e22yy ) G(1) = e33yy.. Apply second BC: uy y y ) = e2y ) e yyG(1) = e22yy ) G(1) = e33yy y (1 2y y 2y y Apply second BC: uy (1,, y ) = e 2 ) e G(1) = e ) G(1) = e3y .. Apply second BC: uy (1 y ) = e ) e G(1) = e ) G(1) = e y This is impossible, so the BVP has no solutions. This is impossible, so the BVP has no solutions. This is impossib...
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This note was uploaded on 02/08/2014 for the course AMATH 350 taught by Professor Davidhamsworth during the Fall '12 term at University of Waterloo, Waterloo.

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