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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 one-dimensional 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 separation-of-variables-type 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