Sol-HW12 - Solutions for Assignment 12 Math Methods Partial...

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Unformatted text preview: Solutions for Assignment 12 Math Methods Partial Differential Equations Dec. 1, 2010 Fall 2010 Calculus of Variations Due Dec. 8, 2010 1 (6). Find the solution of the equation subject to the boundary conditions, using method of characteristics. 8 x xx- 6 x xy + yy + 4 x = 0 ,x > ,y > 0 ; | y =0 = x 2 , y | y =0 = 0. The equation is hyperbolic: A ( x,y ) = 8 x, B ( x,y ) =- 3 x, C ( x,y ) = 1. D = 4 B 2- 4 AC = 36 x- 4 8 x = 4 x > The equation of characteristics: dx dy 2 + 6 x dx dy + 8 x = 0 dx dy =- 3 x x New variables: = x + y, = x + 2 y Derivatives of and x = 1 2 x , x = 1 2 x , y = 1 , y = 2 Derivatives of x = 1 2 x ( + ) , y = + 2 xx =- 1 4 x 3 / 2 ( + ) + 1 4 x ( + 2 + ) xy = 1 2 x ( + 3 + 2 ) , yy = + 4 + 4 The equation in canonical form is = 0 The general solution is ( x,y ) = f ( x + y ) + g ( x + 2 y ) where functions can be found from the initial conditions: ( x, 0) = f ( x ) + g ( x ) = x 2 and y ( x, 0) = f ( x ) + 2 g ( x ) = 0 Let t = x , then f ( t ) + g ( t ) = t 2 2...
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Sol-HW12 - Solutions for Assignment 12 Math Methods Partial...

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