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2.2 The Euler-Lagrange Equation
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d
(2y ! ) + 2ky = 0;
dx
i.e.,
y !! + ky = 0.
The general solution to the Euler-Lagrange equation is
√
√
y(x) = c1 cos( kx) + c2 sin( kx).
√
Now
√ y(0) = 0 implies that c1 = 0, and y(π) = 0 implies that c2 sin( kπ)√= 0.
If k is not an integer,
√ then c2 = 0, and the only extremal is y = 0. If k is
an integer, then sin( kπ) = 0 and c2 can be any number. In the √
latter case
we have an infinite number of extremals of the form y(x) = c2 sin( kx).
Exercises 2.2:
1. Alternative Proof of Condition (2.6): Let y ∈ S and η ∈ H be fixed
functions. Then the quantity J(y + #η) can be regarded as a function of
the single real variable #. Show that the equation dJ/d# = 0 at # = 0 leads
to condition (2.6) under the same hypotheses for f .
2. The First Variation: Let J : S → Ω and K : S → Ω, be functionals
defined by
! x1
! x1
J(y) =
f (x, y, y ! ) dx, K(y) =
g(x, y, y ! ) dx,
x0
x0
where f and g are smooth functions of the indicated arguments and Ω ⊆
R.
(a) Show that for any real numbers A and B,
δ(AJ + BK)(η, y) = AδJ(η, y) + BδK(η, y)
(2.11)
(i.e., δ is a linear operator), and
δ(JK)(η, y) = K(η, y)δJ(η, y) + J(η, y)δK(η, y)
(2.12)
(a product rule).
(b) Suppose that G : Ω × Ω → R is differentiable on Ω × Ω. Show that
δG(J, K)(η, y) =
∂G
∂G
δJ(η, y) +
δK(η, y)
∂J
∂K
(2.13)
(a “chain rule” for the δ operator).
3. Let n be any positive integer. Extend Lemma 2.2.1 by showing that there
exists a ν ∈ C n (R) such that ν(x) > 0 for all x ∈ (α, β) and ν = 0 for all
x ∈ R \ (α, β).
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2 The First Variation
4. Let J be the functional defined by
! 1
" !2
#
J(y) =
y + y 2 + 4yex dx,
0
with boundary conditions y(0) = 0 and y(1) = 1. Find the extremal(s) in
C 2 [0, 1] for J.
5. Consider the functional defined by
! 1
J(y) =
x4 y !2 dx.
−1
(a) Show that no extremals in C 2 [−1, 1] exist which satisfy the boundary
conditions y(−1) = −1, y(1) = 1.
(b) Without resorting to the Euler-Lagrange equation, prove that J cannot have a local minimum in the set
S = {y ∈ C 2 [−1, 1] : y(−1) = −1 and y(1) = 1}.
2.3 Some Special Cases
The Euler-Lagrange equation is a second-order nonlinear differential equation,
and such equations are usually difficult to simplify let alone solve. There
are, however, certain cases when this differential equation can be simplified.
We examine two such cases in this section. We suppose throughout that the
functional satisfies the conditions of Theorem 2.2.3.
2.3.1 Case I: No Explicit y Dependence
Suppose that the functional is of the form
! x1
J(y) =
f (x, y ! ) dx,
x0
where the variable y does not appear explicitly in the integrand. Evidently,
the Euler-Lagrange equation reduces to
∂f
= c1 ,
∂y !
(2.14)
where c1 is a constant of integration. Now ∂f /∂y ! is a known function of x
and y ! , so that equation (2.14) is a first-order differential equation for y. In
principle, equation (2.14) is solvable for y ! , provided ∂ 2 f /∂y !2 #= 0, 5 so that
equation (2.14) could be recast in the form
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One can invoke a variant of the implicit function theorem (Appendix A.2).