AMath350.F12.A7.Sol

# U f x applying the initial condition we discover that

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Unformatted text preview: e have 2y ) e y . u = f (x Applying the initial condition, we discover that f (x) = cos x, so the solution we seek is u (x, y ) = e y cos (x 2y ) . AMATH 350 9. 3y 2 ux Assignment #7 Solutions - Fall 2012 xuy = x, u(x, 0) = e x , y Page 9 0. We start by solving the equation dy = dx which is separable: 3 Z x , 3y 2 Z 2 y dy = xdx x2 + C. 2 Solving for C gives the characteristic curves: C = x2 + 2y 3 . We introduce a change of coordinates which will map these curves to straight horizontal (or vertical1 ) lines: Let ⇠ = x2 + 2y 3 , ⌘ = y. y3 = =) Then ux = and uy = @u @ u @⇠ ˆ @u ˆ = = 2x = 2x u⇠ ˆ @x @⇠ @ x @⇠ @u @ u @⇠ @ u @⌘ ˆ ˆ @u @u ˆ ˆ = + = 6y 2 + = 6 y 2 u ⇠ + u⌘ . ˆ ˆ @y @⇠ @ y @⌘ @ y @⇠ @⌘ These allow us to rewrite the given PDE: 3 y 2 ux x uy = x 3y 2 (2xu⇠ ) ˆ =) =) x 6 y 2 u⇠ + u⌘ = x ˆ ˆ u⌘ = ˆ 1. Now we can use partial integration: u= ˆ =) ⌘ + g (⇠ ) u= y + g x2 + 2 y 3 (You can easily verify that this solves the DE, for any diﬀerentiable function g ). All that remains is to apply the initial condition: u(x, 0) = e x2 =) e x2 = g x2 Therefore the solution to the given BVP is u(x, y ) = 1 =) g (x) = e x . 2 3 y + e (x +2y ) . The transformation ⇠ = x, ⌘ = x2 + 2y 3 works just as well....
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