set9 - Lecture 24 Boundary Value and Eigenvalue Problems...

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Unformatted text preview: Lecture 24 Boundary Value and Eigenvalue Problems for Linear Second Order Differential Equations We start with a model problem that you examined in Problem Set No. 1, the linear second order DE, y ′′ + λy = 0 , λ ∈ R . (1) The nature of the general solution for this DE depends on the parameter λ : 1. λ < 0: y ( x ) = c 1 e √ − λx + c 2 e − √ − λx (nonoscillatory) 2. λ = 0: y ( x ) = c 1 + c 2 x (nonoscillatory) 3. λ > 0: y ( x ) = c 1 cos √ λx + c 2 sin √ λx (oscillatory) As discussed in the early part of this course, a unique solution to the following initial value condition: y ( x ) = A, y ′ ( x ) = B, A, B ∈ R , (2) is guaranteed. When A = B = 0, then the unique solution is the trivial solution y ( x ) = 0. When we consider boundary value problems of the form y ( a ) = A, y ( b ) = B, (3) there are many possibilities. For example, consider the special cse y (0) = 0 , y ( π ) = 0 . (4) 1. If λ < 0, no nontrivial solution of (1) and (4) can exist. This follows from the fact, studied earlier, that when q ( x ) < 0, nontrivial solutions can have at most one zero. 2. If λ = 0, then y ( x ) = c 1 + c 2 x . But y (0) = 0 implies that c 1 = 0 and y ( π ) = c 2 π = 0 implies that c 2 = 0. Therefore the trivial solution y ( x ) = 0 is the only solution. 1 3. If λ > 0, then nontrivial solutions of (4) are possible. As mentioned earlier, the general solution of (1) is y ( x ) = c 1 cos √ λx + c 2 sin √ λx. (5) The condition y (0) = 0 implies that c 1 = 0 so solutions will have the form y ( x ) = c 2 sin √ λx. (6) But y ( π ) = 0 implies that sin √ λπ = 0 which, in turn, implies that √ λπ = nπ, n = 1 , 2 , ··· (7) or λ = n 2 , n = 1 , 2 , ··· . (8) Thus the solutions of the BVP (4) are (up to a constant) y n ( x ) = sin nx, n = 1 , 2 , ··· . (9) • The discrete λ-values λ n = n 2 , n = 1 , 2 , ··· , are called eigenvalues . • The corresponding functions y n ( x ) = sin nx , n = 1 , 2 , ··· , are called eigenfunctions . The function y n ( x ) has n − 1 nodes between 0 and π . The reason for the “eigenvalue/eigenfunction” terminology is the following: Following stan- dard practice, let us rewrite Eq. (1) as Ly + λy = 0 , (10) where L now represents the differential operator d 2 dx 2 . We may rearrange this equation to have the form − Ly = λy, (11) which may be viewed as an eigenvalue/eigenfunction equation: When operator − L acts on the function y , it produces a multiple of y . This is analogous to the matrix eigenvalue/eigenvector problems that you have studied in linear algebra: Av = λ v , v ∈ R n . (12) 2 Later, we shall generalize L to include a general family of linear second order differential operators....
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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.

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set9 - Lecture 24 Boundary Value and Eigenvalue Problems...

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