Definition 2
Let
Ck[a, b] denote the set of continuous
functions defined on the interval
a≤x≤b which have their first
k-derivatives also continuous on
a≤x≤b.
The proof to follow requires the
integrand F(x, y, y') to be twice differentiable with respect to each
argument. What's more, the methods that we use in this module to solve
problems in the calculus of variations will only find those solutions
which are in C2[a, b]. More advanced
techniques (i.e. beyond MATH0043)
are designed to overcome this last restriction. This isn't just a
technicality: discontinuous
extremal functions are very important in optimal control problems, which
arise in engineering applications.
Theorem 1
If I(Y) is an extremum of the functional
I(
y) =
F(
x,
y,
y') d
x
defined on all functions
y∈C2[a, b] such that
y(a) = A, y(b) = B, then Y(x) satisfies the second order ordinary
differential equation
Definition 3
Equation () is the
Euler-Lagrange equation, or sometimes just Euler's
equation.
Warning 1
You might be wondering what
is suppose
to mean: how can we differentiate with respect to a derivative? Think
of it like this: F is given to you as a function of three variables,
say F(u, v, w), and when we evaluate the functional I we plug in
x, y(x), y'(x) for u, v, w and then integrate. The derivative
is just the partial derivative of F with respect to its second
variable v. In other words, to find
,
just pretend y' is a variable.
Equally, there's an important difference between
and
. The former is the derivative of F
with respect to x, taking into account the fact that y = y(x) and
y' = y'(x) are functions of x too. The latter is the partial
derivative of F with respect to its first variable, so it's found by
differentiating F with respect to x and pretending that y and y'
are just variables and do not depend on x. Hopefully the next example
makes this clear:
Warning 2
Y satisfying the Euler-Lagrange equation is a necessary, but not
sufficient, condition for I(Y) to be an extremum. In other
words, a function Y(x) may satisfy the Euler-Lagrange equation even
when I(Y) is not an extremum.
Proof.
Consider functions
Yε(x) of
the form
where
η(x)∈C2[a, b] satisfies
η(a) = η(b) = 0, so that
Yε(a) = A and
Yε(b) = B,
i.e.
Yε still
satisfies the boundary conditions. Informally,
Yε is a
function which satisfies our boundary conditions and which is `near to'
Y when
ε is small.
1
I(Yε) depends on the value of
ε, and we write
I[ε] for the value of
I(Yε):
I[
ε] =
F(
x,
Yε,
Yε') d
x.
When
ε = 0, the function
I[ε] has an extremum and so
= 0 when
ε = 0.
We can compute the derivative
by differentiating
under the integral sign:
=
F(
x,
Yε,
Yε') d
x =
(
x,
Yε,
Yε') d
x
We now use the multivariable chain rule to differentiate
F with
respect to
ε. For a general three-variable function
F(u(ε), v(ε), w(ε)) whose three arguments depend on
ε,
the chain rule tells us that
In our case, the first argument
x is independent of
ε, so
= 0, and since
Yε = Y + εη we
have
= η and
= η'. Therefore
(
x,
Yε,
Yε') =
η(
x) +
η'(
x).
Recall that
= 0 when
ε = 0. Since
Y0 = Y and
Y0' = Y',
0 = (x, Y, Y')η(x) + (x, Y, Y')η'(x) dx. |
(2) |
Integrating the second term in (
) by parts
η'(
x) d
x =
η(
x)
-
η(
x) d
x.
The first term on the right hand side vanishes because
η(a) = η(b) = 0.
Substituting the second term into (
),
The equation above holds for
any
η(x)∈C2[a, b] satisfying
η(a) = η(b) = 0, so the
fundamental
lemma of calculus of variations (explained on the next page) tells us
that
Y(x) satisfies
Definition 4
A solution of the Euler-Lagrange equation is called an
extremal of the functional.2
Exercise 1
Find an extremal y(x) of the functional
I(
y) =
(
y' -
y)
2 d
x,
y(0) = 0,
y(1) = 2,
Answer:
y(
x) = 2
.
Exercise 2
By considering y + g, where y is the solution from
exercise 1 and g(x) is a variation in
y(x) satisfying
g(0) = g(1) = 0, and then considering I(y + g), show explicitly
that y(x) minimizes I(y) in Exercise 1 above.
(Hint: use integration by parts, and the
Euler-Lagrange equation satisfied by y(x) to simplify the expression for
I(y + g)).
Exercise 3
Prove that the straight line y = x is the curve giving the
shortest distance between the points (0, 0) and (1, 1).
Exercise 4
Find an extremal function of
I[
y] =
x2(
y')
2 +
y d
x,
y(1) = 1,
y(2) = 1,
Answer:
y(
x) =
ln
x +
+1 - ln2
.