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Unformatted text preview: 1. Consider the principal part of the equation, u xx 2 u xy 3 u yy . Based on this, we construct the polynomial 2 2  3 and we note that it has roots r 1 = 3 and r 2 = 1 (the equation is hyperbolic). Consequently, we will suggest a linear coordinate transformation = r 1 x + y = 3 x + y, = r 2 x + y = x + y. If we define u ( , ) = u ( x,y ), then u x = 3 u  u u y = u + u u x + u y = 4 u and u xx = 9 u  6 u + u  2 u xy = 6 u  4 u + 2 u  3 u yy = 3 u  6 u  3 u 0 = 16 u We can conclude that u xx 2 u xy 3 u yy + u x + u y = 0 is equivalent to u  1 4 u = 0 . If we set v = u the PDE above becomes, v = 1 4 v , and so v = f ( ) e / 4 = u . Integrating with respect to we get u = e / 4 F ( ) + G ( ) or u ( x,y ) = e ( y x ) / 4 F (3 x + y ) + G ( y x ) X = e y/ 4 x/ 4 3 x/ 4+3 x/ 4 F (3 x + y ) + G ( y x ) = e x H (3 x + y ) + G ( y x ) Page 1 3. Exercise 4, page 89. We look for a solution of the form u ( x,t ) = X ( x ) T ( t ). If the solution has this form, then XT 00 + rXT = c 2 X 00 T. Dividing by c 2 XT we get 1 c 2 T 00 T r c 2 T T = X 00 X = . The homogeneous Dirichlet BCs tell us that the eigenfunctions in x are X n ( x ) = sin nx n = 1 , 2 , 3 ,.... Turning to T we see that for a given n , T 00 n + r T n + c 2 n T n = 0 where n = ( n/ ) 2 . Since 0 < r < 2 c/ , we can say that r 2 < 4( c/ ) 2 and so the discriminant r 2 4( nc/ ) 2 < ....
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This note was uploaded on 01/17/2012 for the course AM 3815 taught by Professor Llll during the Spring '11 term at UWO.
 Spring '11
 llll

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