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AMath350.F13.A7.Sol

# Subject to the 0 ucondition uxx 0 cos x 2ux uy u

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Unformatted text preview: rtial integration (since it contains derivatives with respect to each variable). So, we’ll use the method of characteristics instead. To ﬁnd the “base characteristics”, we need to solve the DE dy 1 =. dx 2 That’s easy: y = x/2+C . Expressing this in implicit form, we have x 2y = C2 , (these are the characteristics) which tells us that we should let one of our new variables be x 2y . So, let’s let ⇠=x 2y, ⌘ = y. With this choice, we have u x = u ⇠ ⇠ x + u ⌘ ⌘x = u ⇠ ˆ ˆ ˆ u y = u ⇠ ⇠ y + u ⌘ ⌘y = ˆ ˆ 2 u⇠ + u⌘ ˆ ˆ and this converts the original PDE into the form 2ˆ⇠ + ( 2u⇠ + u⌘ ) + u = 0, u ˆ ˆ ˆ or simply u⌘ = ˆ u. ˆ This is an equation we can solve! We just need to observe how similar it is du ˆ to the ODE = u. Since we know that the general solution to this is (by ˆ d⌘ inspection) u = Ce ⌘ , we know immediately that the solution to our PDE is ˆ u = f (⇠ ) e ˆ ⌘ where f is an arbitrary function of one variable. Returning to the original variables, we 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 Page 9 Assignment #7 Solutions - Fall 2013 9. Solve the initial value problem 3 y 2 ux on the domain x 2xuy = x, u(0, y ) = y 2 0. Solution: We start by solving the equation dy = dx which is separable: 3 Z 2x , 3y 2 2 y dy = 2 y3 = =) Z xdx x2 + C. Solving for C gives the characteristic curves: C = x2 + y 3 . We introduce a change of coordinates which will map these curves to straight horizontal (or vertical1 ) lines: Let ⇠ = x2 + y 3 , Then @u @ u @⇠ ˆ @u ˆ = = 2x = 2x u⇠ ˆ @x @⇠ @ x @⇠ ux = and uy = ⌘ = y. @u @ u @⇠ @ u @⌘ ˆ ˆ @u @u ˆ ˆ = + = 3y 2 + = 3y 2 u⇠ + u⌘ . ˆ ˆ @y @ y @ y @⌘ @ y @⇠ @⌘ These allow us to rewrite the given PDE: 3 y 2 ux =) 3y 2 (2xu⇠ ) ˆ =) 2xuy = x 2 x 3 y 2 u⇠ + u⌘ = x ˆ ˆ u⌘ = ˆ...
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