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oldex1sol - Math 425 Midterm 1 Dr DeTurck February 8 2007 1...

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Math 425 Dr. DeTurck Midterm 1 February 8, 2007 1. Solve u x + yu y + u = 0, u (0 , y ) = y . In what domain in the plane is your solution valid? As usual, we construct the graph of the solution by propagating the initial data off the line in the xy -plane where the data are given, namely x = 0. To get the part of the solution emanating from the point (0 , b ), we need to solve the system of ODEs: dx ds = 1 , dy ds = y, du ds + u = 0 together with initial data x (0) = 0 , y (0) = b, u (0) = b. There is no “coupling” in this system, and each of the three ODEs can be solved by separation of variables to yield: x ( s ; b ) = s, y ( x ; b ) = be s , u ( s ; b ) = be - s . So we have u as a function of b and s . We use the solutions for x and y to calculate b and s as functions of x and y : s = x, b = y e s = y e x . Now substitute this into the equation for u ( s ; b ) to get u ( x, y ) = y e x e - x = ye - 2 x . Since any point ( x, y ) can be expressed in terms of ( b, s ) via x = s , y = be s , our solution is valid for all x and y . 2. Let u ( x, t ) be the temperature in a rod of length L that satisfies the partial differential equation: u t = ku xx for ( x, t ) (0 , L ) × (0 , ), where k is a positive constant, together with the initial condition u ( x, 0) = φ ( x ) for x [0
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