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Unformatted text preview: Lecture 27 Boundary Value/Eigenvalue Problems for Linear Second Order Dif ferential Equations (cont’d) In the last lecture, we examined the generalized boundary value/eigenvalue problem y ′′ + λq ( x ) y = 0 , y ( a ) = y ( b ) = 0 , q ( x ) > 0 for x ∈ [ a, b ] , (1) and showed that there exists an infinite set of solutions – the eigenfunctions u n ( x ) – with associated eigenvalues λ n , n = 1 , 2 , ··· such that < λ 1 < λ 2 < ··· , with λ n → ∞ as n → ∞ . (2) Now consider two distinct eigenfunctions u m and u n , with m negationslash = n so that λ m negationslash = λ n , satisfying the equations u ′′ m + λ m q ( x ) u m = 0 ( a ) (3) u ′′ n + λ n q ( x ) u n = 0 . ( b ) If we multiply (a) by u n and (b) by u m , subtract the latter from the former and integrate over [ a, b ], we arrive at the result (Exercise): ( λ m − λ n ) integraldisplay b a u m ( x ) u n ( x ) q ( x ) dx = 0 , (4) which is similar to the result obtained in a previous lecture for the special case q ( x ) = 1. The functions u m and u n are said to be orthogonal with respect to the weight function q ( x ). This allows us to define an associated scalar product for f, g : [ a, b ] → R as follows: < f, g > = integraldisplay b a f ( x ) g ( x ) q ( x ) dx. (5) In the same way as for the Fourier series, if we define N n = < u n , u n >, n = 1 , 2 , ··· , (6) 1 then the set of functions φ n = 1 √ N n u n , n = 1 , 2 , ··· , (7) forms an orthonormal basis on [ a, b ] with respect to the weight function q ( x ), i.e. < φ m , φ n > = δ mn . (8) Just to give an idea of where such problems can occur in practice, let us go back to the onedimensional vibrating string problem. However, we now relax the assumption that both the tension T and the mass density ρ of the string are constant. Following the same type of analysis that was done at the beginning of this course, i.e. considering the string as a collection of beads of mass Δ m , etc., it is not too difficult to show that the partial differential equation describing the displacement of the string u ( x, t ) from equilibrium is given by ∂ ∂x parenleftBigg T ( x ) ∂u ∂x parenrightBigg = ρ ( x ) ∂ 2 u ∂t 2 . (9) Now let us make the assumption that T ( x ) = T constant. If this assumption is reasonable (and I emphasize the word “If”), then we can pull T out of the brackets. If, as before, we now assume a separationofvariablestype solution u ( x, t ) = v ( x ) cos ωt, (10) then, after some rearrangement, we see that v ( x ) must satisfy the ODE v ′′ + ω 2 T ρ ( x ) v = 0 , v (0) = v ( L ) = 0 , ρ ( x ) > , x ∈ [0 , L ] . (11) which has the form of the eigenvalue problem in Eq. (1) with a = 0 and b = L and λ = ω 2 T . (12) The mass density function ρ ( x ) will play the role of the weighting function q ( x ). Solutions to the vibrating string problem will then have the form u ( x, t ) = ∞ summationdisplay k =1 c n u n ( x ) cos ω n t, (13) 2 where the mode frequencies...
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This note was uploaded on 09/20/2010 for the course AMATH 351 taught by Professor Sivabalsivaloganathan during the Spring '08 term at Waterloo.
 Spring '08
 SivabalSivaloganathan

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