Consider the functional S[y] = ay(1)² + [[* dx ßy², y(0) = 0, with a natural boundary condition at x = = 1 and subject to the constraint C[y]=ry(1)² + [* dx w(x) y² = 1, where a, ẞ and y are nonzero constants. Euler-Lagrange equation β d²y dx2 +\w(x)y=0, y(0) = 0, (a− yλ) y(1) + ßy' (1) = 0, where is a Lagrange multiplier. Euler- Lagrange eft. is py" + 1 way=0 7 y" + dy , since A = 1, W(X) = 1 The general solution is y (x) = As in (√xx) + B cos (√xx) where A and B when y =0 B = 0 > are constants. heme, yox) = A sin (Dx) 110=0 When (α-rd) y (1) + By '10. (1-1) A sin √π + y'α> (1) = 0 since α = , x = 8 = p²l ⇒ (1-1) A sint + A√ cos t = 0 A Sin√√ C[y]=1 1 √ cos√λ = 0 1 - A² So Sin ² (√XX) = = ⇒ 1 = A² The nontrivial Az √2 stationary parth is g(x)= √2 sin (√xx)

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter9: Multivariable Calculus
Section9.3: Maxima And Minima
Problem 20E
Question

What is the value of λ? 

 

Consider the functional
S[y] = ay(1)² + [[* dx ßy², y(0) = 0,
with a natural boundary condition at x = = 1 and subject to the constraint
C[y]=ry(1)² + [* dx w(x) y² = 1,
where a, ẞ and y are nonzero constants.
Euler-Lagrange equation
β
d²y
dx2
+\w(x)y=0, y(0) = 0, (a− yλ) y(1) + ßy' (1) = 0,
where is a Lagrange multiplier.
Transcribed Image Text:Consider the functional S[y] = ay(1)² + [[* dx ßy², y(0) = 0, with a natural boundary condition at x = = 1 and subject to the constraint C[y]=ry(1)² + [* dx w(x) y² = 1, where a, ẞ and y are nonzero constants. Euler-Lagrange equation β d²y dx2 +\w(x)y=0, y(0) = 0, (a− yλ) y(1) + ßy' (1) = 0, where is a Lagrange multiplier.
Euler- Lagrange eft. is
py" + 1 way=0
7
y" + dy
,
since A = 1, W(X) = 1
The general solution is
y (x) = As in (√xx) + B cos (√xx) where A and B
when y =0
B = 0
>
are constants.
heme, yox) = A sin (Dx)
110=0
When (α-rd) y (1) + By '10.
(1-1) A sin √π +
y'α>
(1) = 0
since α =
,
x = 8 = p²l
⇒ (1-1) A sint + A√ cos t = 0
A
Sin√√
C[y]=1
1
√ cos√λ = 0
1 - A² So Sin ² (√XX) =
=
⇒ 1 = A²
The nontrivial
Az √2
stationary
parth is
g(x)=
√2 sin (√xx)
Transcribed Image Text:Euler- Lagrange eft. is py" + 1 way=0 7 y" + dy , since A = 1, W(X) = 1 The general solution is y (x) = As in (√xx) + B cos (√xx) where A and B when y =0 B = 0 > are constants. heme, yox) = A sin (Dx) 110=0 When (α-rd) y (1) + By '10. (1-1) A sin √π + y'α> (1) = 0 since α = , x = 8 = p²l ⇒ (1-1) A sint + A√ cos t = 0 A Sin√√ C[y]=1 1 √ cos√λ = 0 1 - A² So Sin ² (√XX) = = ⇒ 1 = A² The nontrivial Az √2 stationary parth is g(x)= √2 sin (√xx)
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