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Unformatted text preview: Solution to Homework Set 1, Math 716 1. Calculate and plot the caracteristic curves for u x 1 + 2 u x 2 + (2 x 1 − x 2 ) u = x 1 x 2 , ( x 1 , x 2 ) ∈ R 2 Afterwards, derive the general solution u ( x 1 , x 2 ) and in particular find solution satisfying u (0 , x 2 ) = e x 2 . Solution: Characteristic curves are determined from dx 1 dt = 1, dx 2 dt = 2 and are set of all straight lines with slope 2 in the ( x 1 , x 2 ) plane. (I am avoiding drawing the characteristics in this case since the set of straight lines with slope 2 is obvious.) Parametrically, they can be represented without loss of generality as x 1 = t + g ( s ), x 2 = 2 t + s , where g ( s ) is the value of x 1 when t = 0. The initial curve Γ here is parametrized by ( g ( s ) , s ). By taking g ( s ) arbitrary we can describe essentially all initial curves needed to characterize a general solution (In the nongeneric case when the initial curve is horizontal in the x 1 − x 2 plane, we can switch around the role of x 1 and x 2 and then the initial curve can be parametrized by ( s, g ( s )) ). On the characteric curves, we have from the equation du dt = − (2 x 1 − x 2 ) u − x 1 x 2 = − (2 g ( s ) − s ) u − ( t + g ( s ))(2 t + s ) or du dt + (2 g ( s ) − s ) u = 2 t 2 + [ s + 2 g ( s )] t + sg ( s ) We use initial condition u (0) = f ( s ), where f ( s ) is the initial value of u on the initial curve Γ, characterized by s . This ordinary differential equation can be solved for fixed s by first seeking a particular solution in the form of a quadratic in t and then using the fact that the associated homogeneous solution has a solution in the form exp[ − (2 g ( s ) − s ) t ]. One obtains through in a straight forward (but laborious) manner: u = U ( t, s ) = A ( s ) exp [ − (2 g − s ) t ] + 1 ( − 2 g + s ) 3 × ( − 2( s − 2 g ) 2 t 2 +(2 g − s )( s 2 − 4 g 2 + 4) t + 4 s 2 g 2 − s 3 g + 4 g 2 − 4 sg 3 − 4 − s 2 ) where f ( s ) = A ( s ) +...
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 Spring '08
 Tanveer,S
 Math, Boundary value problem, Lipschitz continuity, initial curve

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