527solns11

527solns11 - 642:527 SOLUTIONS ASSIGNMENT 11 FALL 2007...

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Unformatted text preview: 642:527 SOLUTIONS: ASSIGNMENT 11 FALL 2007 Section 17.7:1. In all sections of this problem the equation is y ′′ + λy = 0; comparing with the standard Sturm-Liouville form (1.a) we see that p ( x ) = 1, q ( x ) = 0, and w ( x ) = 1. The solutions of the differential equation are (I) y ( x ) = A cos κx + B sin κx, if λ = κ 2 > , (II) y ( x ) = A + Bx, if λ = 0 , (III) y ( x ) = A cosh κx + B sinh κx, if λ = − κ 2 < , (b) This is just the derivation of the QRC series as a Sturm-Liouville problem. (c) Similarly, this is the derivation of the HRC series as a Sturm-Liouville problem. (d) Here the boundary conditions are y ′ (0) = 0, y ( L )+ y ′ ( L ) = 0, i.e., α = 0, β = γ = δ = 1. We analyze the possibilities (I)–(III): (I) If y ( x ) = A cos κx + B sin κx then y ′ (0) = 0 implies that B = 0; then y ( L ) + y ′ ( L ) = 0 implies that A (cos κL − κ sin κL ) = 0. Since we do not want A = 0 (which leads to a trivial solution) we must have cos κL − κ sin κL = 0, that is, κ = cot κL . To solve this equation graphically we let u = κL , which means that we should solve u/L = cot u . A graphical solution is indicated below (the graph is drawn with L = 2, but would look much the same with any value of L ); we see that there is an infinite sequence 0 < u < u 1 < u 2 ··· of solutions. Correspondingly we have solutions κ n = u n /L of κ = tan κL and eigenvalues λ n = κ 2 n . It is clear from the figure that u n ≈ nπ for large n , so that then λ n ≈ ( nπ/L ) 2 . u u 0 u 1 u 2 u 3 u 4 u 5 u 6 1 2 p 3 2 p 5 2 p 7 2 p 9 2 p 1 1 2 p y K 2 2 4 6 8 1 0 (II) If y ( x ) = A + Bx then y ′ (0) = 0 implies that B = 0, and then y ( L )+ y ′ ( L ) = 0 implies that A = 0. Only the zero solution exists so λ = 0 is not an eigenvalue. (III) If y ( x ) = A cosh κx + B sinh κx then y ′ (0) = 0 implies that B = 0; then y ( L )+ y ′ ( L ) = 0 implies that A (cosh κL + κ sinh κL ) = 0. Since for u > 0 we have cosh u > 0 and sinh u > we must have A = 0. We reject this identically zero solution and conclude that no λ < can be an eigenvector. In fact, this also follows from Theorem 17.7.2 of the text. Now we have found all the eigenvalues (they are all positive); the eigenfunction cor- responding to λ n = κ 2 n is φ n ( x ) = cos κ n ( x ). Any function f ( x ) defined for 0 < x < L has a corresponding expansion f ( x ) ∼ ∞ summationdisplay n =0 c n φ n ( x ) , c n = integraltext L f ( x ) φ n ( x ) dx integraltext L φ n ( x ) 2 dx . If f ( x ) and f ′ ( x ) are piecewise continuous then this expansion converges at each value of x to f ( x +) + f ( x − ) / 2....
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  • Fall '07
  • SPEER
  • Boundary value problem, Eigenvalue, eigenvector and eigenspace, Eigenfunction, Boundary conditions, Sturm–Liouville theory, nπx nπx

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527solns11 - 642:527 SOLUTIONS ASSIGNMENT 11 FALL 2007...

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