Now we can use partial integration u g 2 y g

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Unformatted text preview: 1/2. Now we can use partial integration: u= ˆ ⌘ + g (⇠ ) 2 y + g x2 + y 3 2 (You can easily verify that this solves the DE, for any differentiable function g ). =) u= All that remains is to apply the initial condition: y y u(0, y ) = , =) = 2 2 =) 1 y + g y 3 =) g y 3 = y 2 p g (y ) = 3 y Therefore the solution to the given BVP is y p2 u(x, y ) = + 3 x + y3. 2 The transformation ⇠ = x, ⌘ = x2 + y 3 also works, but is a bit more difficult. AMATH 350 Assignment #7 Solutions - Fall 2013 Page 10 BONUS: If we try to satisfy the condition u (x, 0) = x, we obtain the expression x = g (x2 ). If our domain were restricted to x 0, we could conclude that p p g (x) = x, while if we had x < 0 we could conclude that g (x) = x, but we cannot have both of these at once. The real reason becomes apparent if we plot the characteristics. The curves x2 + y 3 = C look like this: Recall that for this kind of problem, information from the initial condition is transported along the characteristics. Because the characteristics loop around in the 1st and 2nd quadrants, if the initial condition were given along the positive x-axis, that would actually determine the values of u along the negative x-axis as well. In other words, we will only be able to satisfy an initial condition given along the x-axis if the values of u for negative values of x are consistent with the values of u for positive values of x. If f (x) were even, we’d be ok....
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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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