This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 5.3.8 Multiplying the equation by φ yields: 0 = φ d 2 φ dx 2 + ( λ x 2 ) φ 2 = d dx φ dφ dx dφ dx 2 + λφ 2 x 2 φ 2 . Integrating this identity over the interval 0 ≤ x ≤ 1 allows us to conclude: φ dφ dx 1 Z 1 " dφ dx 2 + x 2 φ 2 # dx = λ Z 1 φ 2 dx. If φ is an eigenfunction, and hence not identically zero, we can now divide by R 1 φ 2 dx to obtain the standard Rayleigh quotient λ = φ dφ dx 1 + R 1 h ( dφ/dx ) 2 + x 2 φ 2 i dx R 1 φ 2 dx The first derivatives of φ vanish at the boundary which means λ = R 1 h ( dφ/dx ) 2 + x 2 φ 2 i dx R 1 φ 2 dx ≥ . If λ = 0, we would then have 0 = Z 1 dφ dx 2 dx + Z 1 x 2 φ 2 dx ≥ since each term in the integrand is nonnegative. This implies that each integral vanishes, and that each integrand vanishes. In particular, x 2 φ 2 ( x ) = 0 for all 0 ≤ x ≤ 1, which implies that φ ( x ) = 0 for all < x ≤ 1 and hence φ (0) = 0 by continuity. Thus φ is the trivial function and hence λ = 0 cannot be an eigenvalue.an eigenvalue....
View Full
Document
 Spring '10
 sim

Click to edit the document details