Consider the functional S[y] = ay(1)² + √² dx By, y(0) = 0, == with a natural boundary condition at x = 1 and subject to the constraint C[y] = √y(1)² + [* dx w(x) y² = 1, where a, ẞ and y are nonzero constants. Show that the stationary paths of this system satisfy the Euler-Lagrange equation d²y B +Aw(x) y = 0, y(0) = 0, (a-yλ) y(1) + ßy' (1) = 0, dx² where A is a Lagrange multiplier.
Consider the functional S[y] = ay(1)² + √² dx By, y(0) = 0, == with a natural boundary condition at x = 1 and subject to the constraint C[y] = √y(1)² + [* dx w(x) y² = 1, where a, ẞ and y are nonzero constants. Show that the stationary paths of this system satisfy the Euler-Lagrange equation d²y B +Aw(x) y = 0, y(0) = 0, (a-yλ) y(1) + ßy' (1) = 0, dx² where A is a Lagrange multiplier.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![Consider the functional
S[y] = ay(1)² + √²
dx By, y(0) = 0,
==
with a natural boundary condition at x = 1 and subject to the constraint
C[y] = √y(1)² + [* dx w(x) y² = 1,
where a, ẞ and y are nonzero constants.
Show that the stationary paths of this system satisfy the
Euler-Lagrange equation
d²y
B +Aw(x) y = 0, y(0) = 0, (a-yλ) y(1) + ßy' (1) = 0,
dx²
where A is a Lagrange multiplier.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c8ed7d-75cc-4e27-869e-3ad6a1efc0b4%2Fc36b2908-9a4f-43a0-bb29-18058f3c4a8f%2Fxc9fiwh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider the functional
S[y] = ay(1)² + √²
dx By, y(0) = 0,
==
with a natural boundary condition at x = 1 and subject to the constraint
C[y] = √y(1)² + [* dx w(x) y² = 1,
where a, ẞ and y are nonzero constants.
Show that the stationary paths of this system satisfy the
Euler-Lagrange equation
d²y
B +Aw(x) y = 0, y(0) = 0, (a-yλ) y(1) + ßy' (1) = 0,
dx²
where A is a Lagrange multiplier.
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