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Show that $t\mapsto\frac{\sinh t}{\sinh T}$ extremises the functional $S[x] = \int_0^T \left( \dot{x}(t)^2 + x(t)^2 \right) \ \mathrm dt$ with boundary conditions $x(0) = 0$ and $ x(T) = 1$.

I can show that $\frac{\sinh t}{\sinh T}$ is one solution to the resultant ODE, $\ddot{x} = x$ after simplifying the functional in the title, but I can't think of a way to prove that it is an extremal solution (specifically, it is supposed to be a minimum, according to the solutions manual [1]).

How may I proceed?


[1]This problem is 3-5 of James Hartle's Gravity: A Gentle Introduction to Einstein's General Relativity.

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    $\begingroup$ If it is the only critical point (which it is by the Euler-Lagrange equation) and the domain is unbounded then it is necessarily extremal of one kind or the other. It is then easy to see that this functional is unbounded above so it must be a minimum. $\endgroup$
    – Ian
    Aug 17, 2021 at 2:38

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Let $ X $ be the real vector space that is the domain of $ S $ (it's usual to consider the space of twice continuously differentiable functions in variational problems, but you may take a space other than $ C ^ 2 ( [ 0 , T ] ) $ for your own intended application). If we define $ \langle . , . \rangle : X \times X \to \mathbb R $ with $$ \langle x , y \rangle = \int _ 0 ^ T \big( \dot x ( t ) \dot y ( t ) + x ( t ) y ( t ) \big) \ \mathrm d t $$ for all $ x , y \in X $, it's rather straightforward to check that it gives an inner product on $ X $, which induces the norm $ \| x \| = \sqrt { S [ x ] } $ for all $ x \in X $. This observation is not necessary, but makes everything simpler, and gives more compact notations.

Now, note that $$ S [ x + h ] = \| x + h \| ^ 2 = \| x \| ^ 2 + 2 \langle x , h \rangle + \| h \| ^ 2 \tag 0 \label 0 $$ for all $ x , h \in X $. Considering $ x _ * \in X $ given with $ x _ * ( t ) = \frac { \sinh t } { \sinh T } $ for all $ t \in [ 0 , T ] $, and taking any $ h \in X $ with $ h ( 0 ) = h ( T ) = 0 $, we can see that $$ \langle x _ * , h \rangle = \int _ 0 ^ T \left( \frac { \cosh t } { \sinh T } \dot h ( t ) + \frac { \sinh t } { \sinh T } h ( t ) \right) \ \mathrm d t = \\ \frac 1 { \sinh T } \int _ 0 ^ T \frac { \mathrm d } { \mathrm d t } \big( ( \cosh t ) h ( t ) \big) \ \mathrm d t = \frac { ( \cosh T ) h ( T ) - ( \cosh 0 ) h ( 0 ) } { \sinh T } = 0 \text . \tag 1 \label 1 $$ Together, \eqref{0} and \eqref{1} show that for all $ h \in X $ with $ h ( 0 ) = h ( T ) = 0 $, $$ S [ x _ * + h ] = S [ x _ * ] + \| h \| ^ 2 \ge S [ x _ * ] \text . \tag 2 \label 2 $$ As $ x _ * ( 0 ) = 0 $ and $ x _ * ( T ) = 1 $, \eqref{2} can be rephrased: for any $ x \in X $ with $ x ( 0 ) = 0 $ and $ x ( T ) = 1 $, $ S [ x _ * ] \le S [ x ] $. This means that $ x _ * $ is the global minimizer of $ S $.


The above argument was very specific to this problem, and not generalizable to a wide class of variational problems. If you're looking for more general methods, here are some useful tools.

Remember that:

A function $ f : \mathbb R \to \mathbb R $ has a local minimum at a point $ x _ * \in \mathbb R $ when $ \left. \frac { \mathrm d } { \mathrm d \epsilon } f ( x _ * + \epsilon ) \right| _ { \epsilon = 0 } = 0 $ and $ \left. \frac { \mathrm d ^ 2 } { \mathrm d \epsilon ^ 2 } f ( x _ * + \epsilon ) \right| _ { \epsilon = 0 } > 0 $.

There is the following generalization of this theorem to the case of functionals (see here for references):

A functional $ S : X \to \mathbb R $ has a local minimum at $ x _ * \in X $ if its first variation vanishes at $ x _ * $, and its second variation is strongly positive at $ x _ * $.

For any nonnegative integer $ k $, the $ k $-th variation $ \delta ^ k S [ x , h ] $ is defined to be the quadratic part of the change of value of $ S $ at $ x $ in the direction of $ h $; i.e. $$ \delta ^ k S [ x , h ] = \left. \frac { \mathrm d ^ k } { \mathrm d \epsilon ^ k } S [ x + \epsilon h ] \right| _ { \epsilon = 0 } \text . $$ Strong positivity of the second variation at $ x $ is the property of existence of a positive constant $ a \in \mathbb R $ such that $ \delta ^ 2 S [ x , h ] \ge a \| h \| ^ 2 $ for all $ h $.

The part related to the first variation being zero is the part you already know of; it results in the Euler-Lagrange equations, and you find candidates for $ x _ * $. The part related to the second variation was fairly trivial in the case of your problem; we had $$ S [ x + \epsilon h ] - S [ x ] = 2 \epsilon \langle x , h \rangle + \epsilon ^ 2 \| h \| ^ 2 \text , $$ and thus $ \delta S [ x , h ] = 2 \langle x , h \rangle $ and $ \delta ^ 2 S [ x , h ] = 2 \| h \| ^ 2 $, which shows that the second variation is strongly positive at any $ x \in X $. This is in fact the result of the clever choice of the norm on $ X $. We could choose other norms on $ X $, for example $$ \| x \| _ 2 = \int _ 0 ^ T x ( t ) ^ 2 \ \mathrm d t \text . $$ With some more effort, this could also work, since we know that $ \| \dot x \| _ 2 \ge \frac 2 L \| x \| _ 2 $ (see here). What I'm trying to point out is in case the norm is not given, choosing it wisely will help a lot in solving the problem.

For showing that a given local minimum is in fact a global minimum, let's again remember that:

If a function $ f : \mathbb R \to \mathbb R $ is convex and $ x _ * \in \mathbb R $ is a local minimizer of $ f $, then it is also a global minimizer of $ f $.

This same statement is also true for a functional $ S : X \to \mathbb R $:

If a functional $ S : X \to \mathbb R $ is convex and $ x _ * \in X $ is a local minimizer of $ S $, then it is also a global minimizer of $ S $.

For example, in the case of your problem, you could note that for any $ x , y \in X $ and any $ \alpha \in [ 0 , 1 ] $, the properties of norm imply that $$ \| ( 1 - \alpha ) x + \alpha y \| \le \| ( 1 - \alpha ) x \| + \| \alpha y \| = ( 1 - \alpha ) \| x \| + \alpha \| y \| \text , $$ and thus the norm itself is a convex functional. As $ \| x \| $ and $ S [ x ] = \| x \| ^ 2 $ have the same local/global minima, $ x _ * $ is a local minimizer of the norm (on the set of functions $ x : [ 0 , T ] \to \mathbb R $ with $ x ( 0 ) = 0 $ and $ x ( T ) = 1 $). By convexity, $ x _ * $ is the global minimizer of $ \| . \| $, and thus of $ S $ as well.

The property of convexity can be weakened, so that the theorem works for a larger class of functionals. Some examples of such properties are being semilocally convex or star-shaped. You might like to take a look at the following article by Ewing for such examples:

Ewing, George M., Sufficient conditions for global minima of suitably convex functionals from variational and control theory, SIAM Rev. 19, 202-220 (1977). ZBL0361.49011.

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