Solved: Consider the SturmLiouville problem [p(x)y ] + q(x)y = r(x)y, 1y(0) + 2y (0) =

Chapter 11, Problem 21

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Consider the SturmLiouville problem [p(x)y ] + q(x)y = r(x)y, 1y(0) + 2y (0) = 0, 1y(1) + 2y (1) = 0, where p, q, and r satisfy the conditions stated in the text. (a) Show that if is an eigenvalue and a corresponding eigenfunction, then 1 0 r2 dx = 1 0 (p2 + q2 ) dx + 1 2 p(1)2 (1) 1 2 p(0)2 (0), provided that 2 = 0 and 2 = 0. How must this result be modified if 2 = 0 or 2 = 0? (b) Show that if q(x) 0 and if 1/2 and 1/2 are nonnegative, then the eigenvalue is nonnegative. (c) Under the conditions of part (b) show that the eigenvalue is strictly positive unless 1 = 1 = 0 and q(x) = 0 for each x in 0 x 1.