section 17.7

section 17.7 - 640:527: Homework 16: Solutions 17.7,...

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640:527: Homework 16: Solutions 17.7, Problem 1 (c). The Sturm Liouville problem is y 00 + λy = 0, y 0 (0) = 0, y 0 ( L ) = 0. In the standard notation, p = 1, q = 0 and w = 1. For any function φ satisfying the boundary conditions φ 0 (0) = φ 0 ( L ) = 0, p (1) φ (1) φ 0 (1) - p (0) φ (0) φ 0 (0) = 0, and, since q = 0, Theorem 17.7.2 implies that there are no negative eigenvalues. It is easy to check that the constant function ˜ φ = 1 satisfies the equation and its boundary condition for λ = 0, and hence λ = 0 is an eigenvalues. The positive eigenvalues are easily found to be λ n = ( nπ/L ) 2 for integers n 1, with corresponding, unnormalized eigenfunctions ˜ φ n ( x ) = cos(( nπ/L ) x ). With normalization using the inner product h h,g i = R L 0 h ( x ) g ( x ) dx , the eigenfunctions are φ 0 ( x ) = 1 L , φ n ( x ) = 2 L cos(( nπ/L ) x ) , n 1 . Any function f with || f || < has the expansion h f,φ 0 i φ 0 ( x ) + X 1 h f,φ n i φ n ( x ) = a 0 + X 1 a n cos(( nπ/L ) x ) . where a 0 = 1 L Z L 0 f ( x ) dx, a n = 2 L Z L 0 f ( x )cos(( nπ/L ) x ) dx. We obtain therefore the half range Fourier cosine series in this problem. For the function f ( x ), where f ( x ) = 1, for 0 x L/ 2 and f ( x ) = 0, for L/ 2 < x L , calculation of the coe±cients leads to the eigenfunction expansion 1 2 + 2 π X 1 sin( nπ/ 2) n cos(( nπ/L ) x ) = 1 2 + 2 π X k =0 ( - 1) k 2 k + 1 cos(((2 k + 1) π/L ) x ) . Problem 1 (d). The problem is y 00 + λy = 0, y 0 (0) = 0, y ( L ) + y 0 ( L ) = 0. This is a Sturm-Liouville problem with p = 1, q = 0 and w = 1. For any function φ satisfying the boundary conditions φ 0 (0) = 0 and φ ( L ) + φ 0 ( L ) = 0, p ( L ) φ ( L ) φ 0 ( L ) - p (0) φ (0) φ 0 (0) = - φ ( L ) 2 0, Since q = 0 also, Theorem 17.7.2 implies that there are no negative eigenvalues. It is easy to check that λ = 0 is not an eigenvalue. The positive eigenvalues are found by substituting the general solution A cos( x λ )+ B sin( x λ ) into the boundary conditions and looking for a condition which ensures there is a non-trivial solution for A and B . In this case, substitution in the boundary conditions gives λB = 0 ( A + λB )cos( L λ ) + ( B - λA )sin( L λ ) = 0 .
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section 17.7 - 640:527: Homework 16: Solutions 17.7,...

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