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Unformatted text preview: 642:527 SOLUTIONS: ASSIGNMENT 12 FALL 2007 Section 17.7: 8. Solution not currently available; may be posted later. Section 18.3: 10. (a) u s ( x ) = u 1 + Q 2 x . (b) u s ( x ) = ( u 2 − Q 1 L ) + Q 1 x . (c) If Q 1 negationslash = Q 2 then we are imposing different heat flows on the two ends of the rod; the rod will heat up or cool down indefinitely, depending on whether the net heat flow Q 1 − Q 2 is positive or negative. If Q 1 = Q 2 = Q then u s ( x ) = C + Qx satisfies the boundary conditions for any choice of C . But E ( t ) = integraldisplay L u ( x, t ) dx (which, up to a constant factor, is the total heat energy of the rod at time t ) is constant, as one would expect on physical grounds: d dt integraldisplay L u ( x, t ) dx = integraldisplay L u t ( x, t ) dx = α 2 integraldisplay L u xx ( x, t ) dx = α 2 u x ( x, t ) vextendsingle vextendsingle vextendsingle vextendsingle x = L x =0 = α 2 ( Q − Q ) = 0 . So E ( ∞ ) = integraltext L ( C + Qx ) dx = CL + QL 2 / 2 = E (0) = integraltext t f ( x ) dx if we impose the initial condition u ( x, 0) = f ( x ); thus C = E (0) /L − QL/ 2. 14. This is essentially the same problem as 28, solved below. I will make a few comments there on the specific questions in 14. 19. Suppose that we are looking for a function u ( x, t ) satisfying Equation : α 2 u xx = u t , < x < L, t > 0; Boundary conditions : u x (0 , t ) = Q 1 , u x ( L, t ) = Q 2 Initial condition : u ( x, 0) = f ( x ) . (12.1) The book takes f ( x ) = 0, Q 1 = − 1, and Q 2 = 0, but that is not really necessary; however, we do assume that Q 1 negationslash = Q 2 . Since Q 1 negationslash = Q 2 these boundary conditons are precisely those for which our usual method does not work, because we cannot find a stationary solution u = A + Bx which satisfies the boundary conditions; this is covered in 10(c). The text suggests that we instead use a quadratic function z ( x ) which satisfies the boundary condtions: z ′ (0) = Q 1 , z ′ ( L ) = Q 2 . This means that we must take z ( x ) = A + Q 1 x +( Q 2 − Q 1 ) x 2 / 2 L , with A some arbitrary constant; suppose in the future that we have chosen one such z ( x ). For the boundary conditions given in the problem the text makes z ( x ) = ( x − L ) 2 / 2 L . We can think of z ( x ) as a particular solution of the given boundary conditions; this means the function v ( x, t ) = u ( x, t ) − z ( x ) will satisfy homogeneous boundary conditions....
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This note was uploaded on 09/29/2009 for the course 642 527 taught by Professor Speer during the Fall '07 term at Rutgers.
- Fall '07