Pde

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

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