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Unformatted text preview: u ( x, t ) =4 t 2 + xx 2 + 8 tx from Example 4, because P 2 and Q 1 are in the region I . The point Q 1 is the intersection point of the characteristic lines x2 t = x2 t and x + 2 t = 1. Adding the equations and then solving for x , we get x = x + 12 t 2 . The second coordinate of Q 1 is then t = 1x + 2 t 4 . The point Q 2 is the intersection point of the characteristic line x + 2 t = x + 2 t and x = 1. Thus t = x + 2 t1 2 . The point P 2 is the intersection point of the characteristic lines x +2 t = 1 and x2 t = 1( x + 2 t1). Solving for x and t , we nd the coordinates of P 2 to be x = 3x2 t 2 and t =1 + x + 2 t 4 ....
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.
 Fall '11
 StuartChalk

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