{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

527solns11

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

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

642:527 SOLUTIONS: ASSIGNMENT 11 FALL 2008 Section 17.7: 7. Let us first consider λ > 0 and then write the solution of the ODE as y = A cos λ x + B sin λ x . Then y = A λ sin λ x + B λ cos λ x the boundary conditions y (0) y (1) = 0 and y (0) + y (1) = 0 become A ( A cos λ + B sin λ ) = 0 , B λ + ( A λ sin λ + B λ cos λ ) = 0 or bracketleftbigg 1 cos λ sin λ sin λ 1 + cos λ bracketrightbigg bracketleftbigg A B bracketrightbigg = 0 . The determinant vanishes for all λ so any λ negationslash = 0 is an eigenvalue; a solution is A = sin λ , B = 1 cos λ , so the eigenfunction is y = sin λ cos λ x + (1 cos λ ) sin λ x . In fact this analysis applies for all nonzero λ , real or complex; λ has two com- plex square roots which we denote ± λ , and sin λ x and cos λ x are defined and have all the properties needed for the manipulations above: ( d/dx ) sin λ x = cos λ x , ( d/dx ) cos λ x = sin λ x , and cos 2 λ x + sin 2 λ x = 1. When λ = 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., λ > 0, λ = 0, or λ < 0. Now, however, we have periodic boundary conditions on the interval 1 < x < 5. (I) If y ( x ) = A cos κx + B sin κx then y ( x ) = sin κx + cos κx and A and B must satisfy A cos( κ ) + B sin( κ ) = A cos(5 κ ) + B sin(5 κ ) , sin( κ ) + cos( κ ) = sin(5 κ ) + cos(5 κ ) , or bracketleftbigg cos( κ ) cos(5 κ ) sin( κ ) sin(5 κ ) , sin( κ ) + sin(5 κ ) cos( κ ) cos(5 κ ) bracketrightbigg bracketleftbigg A B bracketrightbigg = 0 . The determinant of the matrix is, after simplification, 2[1 cos( κ ) cos(5 κ ) sin( κ ) sin(5 κ )] = 2[1 cos(6 κ )] , so that the eigenvalue equation is 6 κ n = 2 or k n = nπ/ 3, n = 1 , 2 , . . . . The eigenvalues are λ n = ( nπ/ 3) 2 and since for κ = κ n the coefficient matrix is identically zero, both A and B are arbitrary, i.e., we have two eigenfunctions for each κ n : cos( nπx/ 3) and sin( nπx/ 3). (II) If y ( x ) = A + Bx then it is easily seen that the boundary conditions imply that B = 0 and A is arbitrary: λ 0 = 0 is an eigenvalue with eigenfunction 1. (III) As in examples done in earlier homework we find that there are no negative eigen- vectors.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online