AMath350.F12.A7.Sol

# 2ux uy u 0 ux 0 cos x this is a rst order linear

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Unformatted text preview: t by partial 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, w...
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