*This preview shows
pages
1–2. Sign up
to
view the full content.*

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **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...

View
Full
Document