E x 2y k2 dx 2 2 2 d 6 3 3 therefore the

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Unformatted text preview: = 36 dy 1 =1 =) y = 1+ C, 1i.e. x 2y = K2 . dx 2⌘ ⇠ )(2) =2 ⇠ ( ⌘ D= 6 3 3 Therefore the transformation we need is 4 4 1 (⌘ ⇠ )(8) = ⇠ ⌘ E= 6 ⇠ = 3x 2y,3 3 = x 2y ⌘ 4 1 ˆ f = sin(y (⇠ , ⌘ )) = sin( ⇠ ⌘) 3 (or the reverse). Next, we need several 3 applications of the chain rule to rewrite the PDE: the PDE becomes 36ˆ + 1 (⇠ ⌘ )ˆ + 4 (⇠ ⌘ )ˆ + sin( 4 ⇠ 1 ⌘ )ˆ = 0. u3 u Thus u 3 3 3u 1 ⇠x 1u 1 4 1 ux = u⇠(⇠ + ⌘ )ux = 3ˆ⇠ + u⌘)ˆ ˆ u⌘ ⌘ ˆ ˆ .u Canonical form: u ˆ ⌘ u 108 27 (⇠ 36 sin( 3 ⇠ 3 ⌘ )ˆ = 0. u y = u ⇠ ⇠ y + u ⌘ ⌘y = 2 u ⇠ 2 u ⌘ . ˆ ˆ ˆ ˆ uxx = uxy = @ @ ( ux ) = (3ˆ⇠ + u⌘ ) = 3 (ˆ⇠⇠ ⇠x + u⇠⌘ ⌘x ) + (ˆ⌘⇠ ⇠x + u⌘⌘ ⌘x ) u ˆ u ˆ u ˆ @x @x = 3 (3ˆ⇠⇠ + u⇠⌘ ) + (3ˆ⌘⇠ + u⌘⌘ ) = 9ˆ⇠ + 6ˆ⇠⌘ + u⌘⌘ . u ˆ u ˆ u u ˆ @ @ ( ux ) = (3ˆ⇠ + u⌘ ) = 3 (ˆ⇠⇠ ⇠y + u⇠⌘ ⌘y ) + (ˆ⌘⇠ ⇠y + u⌘⌘ ⌘y ) u ˆ u ˆ u ˆ @y @y = 3 ( 2u⇠⇠ ˆ uyy = 2u⇠⌘ ) + ( 2u⌘⇠ ˆ ˆ @ @ ( uy )...
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This document was uploaded on 02/09/2014.

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