# Pde

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: r equivalently, Ax − λx = 0. We also obtain the condition ∂H = −g (x) = 0, ∂λ which is just our constraint. Thus the point on the sphere at which f assumes its maximum is a unit-length eigenvector b1 , the eigenvalue being the value λ1 of the variable λ. Let W be the “linear subspace” of R n deﬁned by the homogeneous linear equation b1 · x = 0. The intersection S n−1 ∩ W is a sphere of dimension n − 2. 31 We next use the method of Lagrange multipliers to ﬁnd a point on S n−1 ∩ W at which f assumes its maximum. To do this, we let g1 (x) = xT x − 1, g2 (x) = b1 · x. The maximum value of f on S n−1 ∩ W will be assumed at a critical point for the function H (x, λ, µ) = f (x) − λg1 (x) − µg2 (x). This time, diﬀerentiation shows that ∂H = 2ai1 x1 + 2ai2 x2 + · · · + 2ain xn − 2λxi − µbi = 0, ∂xi or equivalently, Ax − λx − µb1 = 0. (2.2) It follows from the constraint equation b1 · x = 0 that b1 · (Ax) = bT (Ax) = (bT A)x = (bT AT )x 1 1 1 = (Ab1 )T x = (λ1 b1 )T x = λ1 b1 · x = 0. Hence...
View Full Document

## This document was uploaded on 01/12/2014.

Ask a homework question - tutors are online