hw 13 - 642:527 SOLUTIONS: ASSIGNMENT 13 FALL 2007 Section...

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Unformatted text preview: 642:527 SOLUTIONS: ASSIGNMENT 13 FALL 2007 Section 18.3:6 In all sections of this problem we will find the steady-state solution v ( x ) of the equation and boundary conditions; then solve a homogeneous boundary value problem for w ( x, t ) = u ( x, t )- v ( x ). Note that if u ( x, 0) = f ( x ) then the initial condition for w ( x, t ) is w ( x, 0) = f ( x )- v ( x ). The steady state solution is always of the form v ( x ) = A + Bx , and we must choose A and B to satisfy the given boundary conditions. I wont bother to do the integrals giving the Fourier coefficients. 6. (a) Here the steady-state solution must satisfy v (0) = 20, v ( ) = 3, so that v ( x ) = 20 + 3 x . Then u ( x, t ) = 10 + 3 x + w ( x, t ), where w satisfies Equation : 2 w xx = w t , < x < , t > 0; Boundary conditions : w (0 , t ) = 0 , w x ( , t ) = 0 Initial condition : w ( x, 0) =- 20- 3 x, < x < . (12.1) This problem has a solution as a quarter-range sine series: w ( x, t ) = summationdisplay n =1 , 3 , 5 ,... b n sin nx 2 e ( n/ 2) 2 t ; b n =- 2 integraldisplay (20 + 3 x ) sin nx 2 dx. 6. (i) Here the steady-state solution must satisfy v (0) = v (10) = 5, so that it has the form A + 5 x . But now we see that the initial condition is of this form, so we obtain the solution u ( x, t ) = 45 + 5 x by inspection! If we did not notice this simple observation then we could solve the problem more systematically as follows. The constant A in the steady state soluion could be determined by the method discussed in Problem 10(c); however, mathematically it is convenient to take v ( x ) = 5 x . Even though this is not the steady state, we can still write u ( x, t ) = v ( x ) + w ( x, t ), where w satisfies homogeneous boundary conditions: Equation : 2 w xx = w t , < x < 10 , t > 0; Boundary conditions : w x (0 , t ) = 0 , w x (10 , t ) = 0 Initial condition : w ( x, 0) = (45 + 5 x )- v ( x ) = 45 , < x < 10 ....
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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.

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hw 13 - 642:527 SOLUTIONS: ASSIGNMENT 13 FALL 2007 Section...

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