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**1 The second variation**

** **be a nonlinear functional, with x(a) = A and x(b) = B ﬁxed. As usual, we will assume that F is as smooth as necessary.

The ﬁrst variation of J is

where h(t) is assumed as smooth as necessary and in addition satisﬁes h(a) = h(b) = 0. We will call such h admissible.

The idea behind ﬁnding the ﬁrst variation is to capture the linear part of the J [x]. Speciﬁcally, we have

J[x + εh] = J [x] + εδJ_{x}[h] + o(ε),

where o(ε) is a quantity that satisﬁes

The second variation comes out of the quadratic approximation in ε,

It follows that

To calculate it, we note that

Applying the chain rule to the integrand, we see that

where the various derivatives of F are evaluated at Setting ε = 0 and inserting the result in our earlier expression for the second variation, we obtain

Note that the middle term can be written as Using this in the equation above, integrating by parts, and employing h(a) = h(b) = 0, we arrive at

**2 Legendre’s trick**

Ultimately, we are interested in whether a given extremal for J is a weak (relative) minimum or maximum. In the sequel we will always assume that the function x(t) that we are working with is an extremal, so that x(t) satisﬁes the Euler-Lagrange equation, makes the ﬁrst variation δJ_{x}[h] = 0 for all h, and ﬁxes the functions

To be deﬁnite, we will always assume we are looking for conditions for the extremum to be a weak minimum. The case of a maximum is similar.

Let’s look at the integrand It is generally true that a function can be bounded, but have a derivative that varies wildly. Our intuition then says that is the dominant term, and this turns out to be true. In looking for a minimum, we recall that it is necessary that δ^{2}J_{x} [h] ≥ 0 for all h. One can use this to show that, for a minimum, it is also necessary, but not suﬃcient, that P ≥ 0 on [a, b]. We will make the stronger assumption that P > 0 on [a, b]. We also assume that P and Q are smooth.

Legendre had the idea to add a term to δ2J to make it nonnegative. Speciﬁcally, he added to the integrand in (1). Note that Hence, we have this chain of equations,

where we completed the square to get the last equation. If we can ﬁnd w(t) such that

then the second variation becomes

Equation (4) is called a Riccati equation. It can be turned into the second order linear ODE below via the substitution

which is called the Jacobi equation for J . Two points t = α and are said to be conjugate points for Jacobi’s equation if there is a solution u to (6) such that between α and and such that

When there are no points conjugate to t = a in the interval [a, b], we can construct a solution to (6) that is strictly positive on [a, b]. Start with the two linearly indepemdent solutions u_{0} and u_{1} to (6) that satsify the initial conditions

Since there is no point in [a, b] conjugate a, u_{0} (t) ≠ 0 for any a < t ≤ b. In particular, since u˙ 0 (a) = 1 > 0, u(t) will be strictly positive on (a, b]. Next, because u_{1}(a) = 1, there exists t = c, a < c ≤ b, such that u_{1}(t) ≥ 1/2 on [a, c]. Moreover, the continuity of u_{0} and u_{1} on [c, b] implies that min_{c≤t≤b} u_{0}(t) = m_{0} > 0 and min_{c≤t≤b} u_{1}(t) = m_{1} ∈ R. It is easy to check that on [a, b],

and, of course, u solves (6).

This means that the substitutuion yields a solution to the Riccati equation (4), and so the second variation has the form given in (5).

It follows that δ^{2}Jx[h] ≥ 0 for any admissible h. Can the second variation vanish for some h that is nonzero? That is, can we ﬁnd an admissible h ≡ 0 such that δ^{2}J_{x}[h] = 0? If it did vanish, we would have to have

and, since P > 0, this implies that This ﬁrst order linear equation has the unique solution,

However, since h is admissible, h(a) = h(b) = 0, and so h(t) ≡ 0. We have proved the following result.

Proposition 2.1. If there are no points in [a, b] conjugate to t = a, the the second variation is a positive deﬁnite quadratic functional. That is, δ^{2}J_{x} [h] > 0 for any admissible h not identical ly 0.

**3 Conjugate points**

There is direct connection between conjugate points and extremals. Let x(t, ε) be a family of extremals for the functional J depending smoothly on a parameter ε. We will assume that x(a, ε) = A, which will be independent of ε. These extremals all satisfy the Euler-Lagrange equation

If we diﬀerentiate this equation with respect to ε, being careful to correctly apply the chain rule, we obtain

Cancelling and rearranging terms, we obtain

Set ε = 0 and let u(t) Observe that the functions in the equation above, which is called the variational equation, are just and Q =

Consequently, (7) is simply the Jacobi equation (6). The diﬀerence here is that we always have the initial conditions,

We remark that if u˙ (a) = 0, then u(t) ≡ 0.

What do conjugate points mean in this context? Suppose that is conjugate to t = a. Then we have

which holds independently of how our smooth family of extremals was constructed. It follows that at , we have . Thus, the family either crosses again at or comes close to it, accumulating to order higher than ε there.

**4 Suﬃcient conditions **

A suﬃcient condition for an extremal to be a relative minimum is that the second variation be strongly positive deﬁnite. This means that there is a c > 0, which is independent of h, such that for all admissible h one has

where H^{1} = H^{1}[a, b] denotes the usual Sobolev space of functions with distributional derivatives in L^{2} [a, b].

Let us return to equation (2), where we added in terms depending on an arbitrary function w. In the integrand there, we will add and subtract where σ is an arbitary constant. The only requirement for now is that 0 < σ < min_{t∈[a,b]} P (t). The result is

For the ﬁrst integral in the term on the right above, we repeat the argument that was used to arrive at (5). Everything is the same, except that P is replaced by P − σ. We arrive at this:

We continue as we did in section 2. In the end, we arrive at the new Jacobi equation,

The point is that if for the Jacobi equation (6) there are no points in [a, b] conjugate to a, then, because the solutions are continuous functions of the parameter σ, we may choose σ small enough so that for (9) there will be no points conjugate to a in [a, b]. Once we have fouund σ small enough for this to be true, we ﬁx it. We then solve the corresponding Riccati equation and employ it in (8) to obtain

Now, for an admissble h, it is easy to show that that we have

Consequently, we obtain this inequality:

which is what we needed for a relative minimum. We summarize what we found below.

**Theorem 4.1.** A suﬃcient condition for an extremal x(t) to be a relative minimum for the functional where x(a) = A and x(b) = B , is that P (t) and that the interval [a, b] contain no points conjugate to t = a.

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