527solns11

527solns11 - 642:527 SOLUTIONS: ASSIGNMENT 11 FALL 2008...

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Unformatted text preview: 642:527 SOLUTIONS: ASSIGNMENT 11 FALL 2008 Section 17.7: 7. Let us first consider > 0 and then write the solution of the ODE as y = A cos x + B sin x . Then y = A sin x + B cos x the boundary conditions y (0) y (1) = 0 and y (0) + y (1) = 0 become A ( A cos + B sin ) = 0 , B + ( A sin + B cos ) = 0 or bracketleftbigg 1 cos sin sin 1 + cos bracketrightbiggbracketleftbigg A B bracketrightbigg = . The determinant vanishes for all so any negationslash = 0 is an eigenvalue; a solution is A = sin , B = 1 cos , so the eigenfunction is y = sin cos x + (1 cos ) sin x . In fact this analysis applies for all nonzero , real or complex; has two com- plex square roots which we denote , and sin x and cos x are defined and have all the properties needed for the manipulations above: ( d/dx ) sin x = cos x , ( d/dx ) cos x = sin x , and cos 2 x + sin 2 x = 1. When = 0 we have y = A + Bx , y = B and so the boundary conditions become A ( A + B ) = 0, B + B = 0, so that B = 0 but A is free, i.e., y = 1 is an eigenfunction. This is not a regular Sturm-Liouville system because the boundary conditions are not separated (each equation gives conditions involving values at both ends of the interval). It is a bit like the case of periodic boundary conditions studied in Section 17.8; the moral is that we cannot write down nonseparated boundary conditions arbitrarily and expect to get a well-behaved solution. Section 17.8: 2 (a) See solution in text. (b) The equation is the same as in 17.7:1 (Assignment 10) above, so we again have the possibilities (I)(III), i.e., > 0, = 0, or < 0. Now, however, we have periodic boundary conditions on the interval 1 < x < 5. (I) If y ( x ) = A cos x + B sin x then y ( x ) = A sin x + B cos x and A and B must satisfy A cos( ) + B sin( ) = A cos(5 ) + B sin(5 ) , A sin( ) + B cos( ) = A sin(5 ) + B cos(5 ) , or bracketleftbigg cos( ) cos(5 ) sin( ) sin(5 ) , sin( ) + sin(5 ) cos( ) cos(5 ) bracketrightbiggbracketleftbigg A B bracketrightbigg = . The determinant of the matrix is, after simplification, 2[1 cos( ) cos(5 ) sin( ) sin(5 )] = 2[1 cos(6 )] , so that the eigenvalue equation is 6 n = 2 n or k n = n/ 3, n = 1 , 2 ,... . The eigenvalues are n = ( n/ 3) 2 and since for = n the coefficient matrix is identically zero, both A and B are arbitrary, i.e., we have two eigenfunctions for each n : cos( nx/ 3) and sin( nx/ 3)....
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This note was uploaded on 12/15/2009 for the course MATH 527 taught by Professor Staff during the Fall '08 term at Rutgers.

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527solns11 - 642:527 SOLUTIONS: ASSIGNMENT 11 FALL 2008...

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