text:
\begin{document}

\author[1]{Ricardo Almeida}

\author[1]{Delfim F. M. Torres}

\runningauthor{R. Almeida and D. F. M. Torres}

\affil[1]{Center for Research and Development in Mathematics and Applications (CIDMA),
Department of Mathematics, University of Aveiro,
3810-193 Aveiro, Portugal,
e-mail: \{ricardo.almeida, delfim\}@ua.pt}

\title{A survey on fractional variational calculus}

\runningtitle{A survey on fractional variational calculus}

\abstract{Main results and techniques of the fractional
calculus of variations are surveyed.
We consider variational problems
containing Caputo derivatives and study them using both indirect
and direct methods. In particular, we provide necessary optimality conditions
of Euler--Lagrange type for the fundamental, higher-order, and isoperimetric problems,
and compute approximated solutions based on truncated Gr\"{u}nwald--Letnikov
approximations of Caputo derivatives.}

\keywords{calculus of variations;
fractional calculus;
direct and indirect methods.}

\classification[MSC 2010]{26A33; 34A08; 49K05; 49M05; 49M25.}

\maketitle

\section{Introduction}

Calculus of Variations is a prolific field of mathematics,
whose founders were Leonhard Euler (1707--1783) and Joseph Louis Lagrange
(1736--1813). The name of the field was given by Euler in his
\emph{Elementa Calculi Variationum} and is related with the main
technique used by Lagrange to derive first order necessary optimality
conditions. Its development is in intimate connection with various problems
occurring in biology, chemistry, control theory, dynamics,
economics, engineering, physics, etc. The current development
of the calculus of variations is based
on modern analysis techniques \cite{MR2191744,MR2191745}.

The Fractional Variational Calculus (FVC) is devoted to generalizations
of the classic calculus of variations and its modern face,
the theory of optimal control, to the case in which derivatives and integrals
are understood as fractional operators of arbitrary order.
The subject has been developed in the last two decades
and provides nontrivial generalizations of the calculus of variations and optimal control,
which open doors to new and interesting modern scientific problems.
Indeed, the combination of the calculus of variations and FC is not artificial,
coming naturally from more accurate descriptions of physical phenomena, in particular
to better describe non-conservative systems in mechanics, where the inclusion of non-conservatism
in the theory is extremely important \cite{MR1401316,MR1438729}.

This chapter is organized as follows.
In Section~\ref{sec:main}, we formulate
the fundamental problem of the FVC.
Then our survey proceed in two parts:
indirect methods, in Section~\ref{sec:03},
and a direct method, in Section~\ref{sec:04}.
In an indirect method, the Euler--Lagrange equations,
and possible other optimality conditions as well,
are used. Typically, the indirect approach
leads to a boundary-value problem that should be solved to determine
candidate optimal trajectories, called extremals.
Direct methods, in contrast, begin by approximating the variational problem,
reformulating it as a standard nonlinear optimization problem \cite{MyID:407}.

Here we restrict ourselves to scalar problems of Caputo type.
However, the techniques presented are easily adapted to the vectorial
case and other types of FDs and FIs. For a survey on the FVC with generalized
fractional operators see \cite{MR3221831}. For more on this fruitful area,
we refer the readers to the four books on the subject
and references therein: for a solid introduction to the FVC, see \cite{MR2984893};
for computational aspects, begin with \cite{MR3443073};
for advanced mathematical methods, including the question of existence of solutions,
see \cite{MR3331286}; for the variable order FVC and the state of the art see \cite{book4}.

\section{The fundamental problem}
\label{sec:main}

The calculus of variations deals with functional optimization problems
involving an unknown function $x$ and its derivative $x'$.
The general way of formulating the problem is as follows:
$$
\min \mathcal{J}(x)=\int_a^b L(t,x(t),x'(t))\,dt
$$
subject to the boundary conditions
\begin{equation}
\label{classicalBound}
x(a)=x_a \quad \mbox{and} \quad x(b)=x_b,
\quad \mbox{with}\, x_a,x_b\in\mathbb{R}.
\end{equation}
One way to solve this problem consists to determine the solutions
of the second order differential equation
\begin{equation}
\label{classicalEL}
\partial_2L (t,x(t),x'(t))=\frac{d}{dt}\partial_3L (t,x(t),x'(t)), \quad t\in[a,b],
\end{equation}
together with the boundary conditions \eqref{classicalBound}.
Equation \eqref{classicalEL} was studied by Euler, and later by Lagrange,
and is known as the Euler--Lagrange equation.

The fractional calculus of variations is a generalization of the
ordinary variational calculus, where the integer-order derivative
is replaced by a fractional derivative $D^\a x$:
$$
\min \mathcal{J}(x)=\int_a^b L(t,x(t),D^\a x(t))\,dt.
$$
As Riewe noted in \cite{MR1401316},
``traditional Lagrangian and Hamiltonian mechanics cannot
be used with nonconservative forces such as friction''.
FVC allows it in an elegant way.

Since there are several definitions of fractional derivatives,
one finds several works on the calculus of variations for these
different operators. Here we consider the Caputo fractional derivative \cite{Caputo:1967},
but analogous results can be formulated for other types of fractional derivatives.
Our choice of derivative is based on the following important fact:
the Laplace transform of the Caputo fractional derivative depends on integer-order
derivatives evaluated at the initial time $t=a$. Thus, we have an obvious physical
interpretation for them. For other types of differentiation, this may not be so clear.
For example, when dealing with the Riemann--Liouville fractional derivative,
the initial conditions are fractional.

The main problem of the fractional calculus of variations, in the context
of the Caputo fractional derivative, is stated as follows. Given a functional
\begin{equation}
\label{funct1}
\mathcal{J}(x)=\int_a^b L\left(t,x(t),\LD x(t)\right)\,dt, \quad x\in\Omega,
\end{equation}
where $\Omega$ is  a given class of functions, find a curve $x$ for
which $\mathcal{J}$ attains a minimum value. We suppose that $L:[a,b]\times \mathbb{R}^2\to\mathbb{R}$
is continuously differentiable with respect to the second and third variables
and, for every $x\in\Omega$, the function $\LD x$ exists and is continuous
on the interval $[a,b]$. On the set of admissible functions of the problem,
the boundary conditions
\begin{equation}
\label{bound}
x(a)=x_a \quad \mbox{and} \quad x(b)=x_b,
\quad \mbox{with}\, x_a,x_b\in\mathbb{R},
\end{equation}
may be imposed. We observe that a curve $x^\star\in\Omega$ is a (local)
minimizer of functional \eqref{funct1} if there exists some positive real
$\delta$ such that, whenever $x\in\Omega$ and $\|x^\star-x\|<\delta$,
one has $\mathcal{J}(x^\star)-\mathcal{J}(x) \leq 0$.

\section{Indirect methods}
\label{sec:03}

The literature about the fractional calculus
of variations is very vast \cite{MR3443073,book4,MR3331286,MR2984893}.
The usual procedure in the calculus of variations, to find a candidate
to minimizer, consists to take variations of the optimal curve $x^\star$.
Let $\epsilon$ be a real number close to zero and $\eta$ a function chosen
so that $x^\star + \epsilon \eta$ belongs to $\Omega$. For example,
if the functional is defined in a class of functions with the boundary
conditions \eqref{bound}, then the curve $\eta$ must satisfy the conditions
$\eta(a)=0$ and $\eta(b)=0$. The curve $x^\star + \epsilon \eta$
is called a variation of the curve $x^\star$. For simplicity of notation,
we will use the operator $[\, \cdot \, ]^\alpha$ defined by
$$
[x]^\alpha(t):=\left(t,x(t),\LD x(t)\right).
$$
Suppose that $x^\star$ minimizes $\mathcal{J}$, and consider
a variation $x^\star + \epsilon \eta$. Depending on the existence
of boundary conditions, some constraints over $\eta(a)$ and $\eta(b)$
may be imposed. Consider a function $j$, defined on a neighborhood of zero,
and given by
$$
j(\epsilon):=\mathcal{J}(x^\star+\epsilon \eta)
=\int_a^b L(t,x^\star(t)+\epsilon \eta(t),\LD x^\star(t)+\epsilon\LD \eta(t))\,dt.
$$
Since $\mathcal{J}$ attains a minimum value at $x^\star$, then we conclude that
$\epsilon=0$ is a minimizer of $j$, and thus $j'(0)=0$. Since
$$
j'(\epsilon)=\int_a^b \partial_2 L[x^\star+\epsilon \eta]^\alpha(t)\eta(t)
+ \partial_3 L[x^\star+\epsilon \eta]^\alpha(t)\LD \eta(t)\,dt,
$$
we conclude that
\begin{equation}
\label{auxi1}
\int_a^b \partial_2 L[x^\star]^\alpha(t)\eta(t)+ \partial_3 L[x^\star]^\alpha(t)\LD \eta(t)\,dt=0.
\end{equation}
Using now fractional integration by parts \cite{MR2218073}, we obtain that
\begin{multline}
\label{auxi2}
\int_a^b \partial_3 L[x^\star]^\alpha(t)\LD \eta(t)\,dt\\
=\int_a^b \RRD \left( \partial_3 L[x^\star]^\alpha(t)\right) \eta(t)\,dt
+\left[{I_{b-}^{1-\a}}\left(\partial_3 L[x^\star]^\alpha(t)\right)\eta(t)\right]_a^b.
\end{multline}
Combining equations \eqref{auxi1} and \eqref{auxi2},
we prove that
\begin{multline}
\label{auxi3}
\int_a^b \left[ \partial_2 L[x^\star]^\alpha(t)
+\RRD \left( \partial_3 L[x^\star]^\alpha(t)\right)\right]^\alpha \eta(t)\,dt\\
+\left[{I_{b-}^{1-\a}}\left(\partial_3 L[x^\star]^\alpha(t)\right)\eta(t)\right]_a^b=0.
\end{multline}
Assume that the set of admissible functions have to satisfy
the boundary constraints \eqref{bound}.
In this case, the variation function $\eta$ must satisfy the conditions
$\eta(a)=0$ and $\eta(b)=0$. Thus, equation \eqref{auxi3} becomes
$$
\int_a^b \left[ \partial_2 L[x^\star]^\alpha(t)
+\RRD \left( \partial_3 L[x^\star]^\alpha(t)\right)\right] \eta(t)\,dt=0.
$$
Assume now that the function
$t\mapsto \RRD \left( \partial_3 L[x^\star]^\alpha(t)\right)$
is continuous on $[a,b]$. Since $\eta$ is an arbitrary function
on the open interval $]a,b[$, we conclude, by the fundamental
lemma of the calculus of variations \cite{MR0160139}, that
\begin{equation}
\label{auxi4}
\partial_2 L[x^\star]^\alpha(t)+\RRD \left( \partial_3 L[x^\star]^\alpha(t)\right)=0
\end{equation}
for all $t\in[a,b]$. Assume now that $x(a)$ and $x(b)$ are free.
By the arbitrariness of $\eta$, we can first consider variations such that
$\eta(a)=0$ and $\eta(b)=0$, and in this case we prove equation \eqref{auxi4}.
Therefore, replacing  \eqref{auxi4} in \eqref{auxi3}, we deduce that
$$
\left[{I_{b-}^{1-\a}}\left(\partial_3 L[x^\star]^\alpha(t)\right)\eta(t)\right]_a^b\,dt=0.
$$
If $\eta(a)=0$ and $\eta(b)\not=0$, then
$$
{I_{b-}^{1-\a}}\left(\partial_3 L[x^\star]^\alpha(t)\right)=0 \quad \mbox{at} \quad t=b,
$$
and if  $\eta(a)\not=0$ and $\eta(b)=0$, then
$$
{I_{b-}^{1-\a}}\left(\partial_3 L[x^\star]^\alpha(t)\right)=0 \quad \mbox{at} \quad t=a.
$$
In conclusion, we proved the following result, which provides a necessary optimality
condition, known as the fractional Euler--Lagrange equation,
and two subsidiary conditions, known as transversality conditions.

\begin{theorem}
Let $x^\star$ be an admissible function and suppose that
$\mathcal{J}$ attains a minimum value at $x^\star$.
If there exists and is continuous the function
$t\mapsto \RRD \left(\partial_3 L[x^\star]^\alpha(t)\right)$, then
\begin{equation}
\label{ELeq}
\partial_2 L[x^\star]^\alpha(t)+\RRD \left(\partial_3 L[x^\star]^\alpha(t)\right)=0
\end{equation}
for every $t\in[a,b]$. Moreover, if $x(a)$ is free, then
$$
{I_{b-}^{1-\a}}\left(\partial_3 L[x^\star]^\alpha(t)\right)=0 \quad \mbox{at} \quad t=a
$$
and, if $x(b)$ is free, then
$$
{I_{b-}^{1-\a}}\left(\partial_3 L[x^\star]^\alpha(t)\right)=0 \quad \mbox{at} \quad t=b.
$$
\end{theorem}

As observed in \cite{MR2453739}, using the well--know relation
$$
\RRD f(t) = \RD f(t) + \RRD f(b)
= \RD f(t) + \frac{f(b)}{\Gamma(1-\a) (b-t)^\a},
$$
between Riemann--Liouville and Caputo derivatives, one can easily
rewrite the the Euler--Lagrange equation \eqref{ELeq} in the equivalent form
\begin{equation*}
\partial_2 L[x^\star]^\alpha(t)+\RD \left(\partial_3 L[x^\star]^\alpha(t)\right)
+ \frac{\left.\partial_3 L[x^\star]^\alpha(t)\right|_{t = b}}{\Gamma(1-\a) (b-t)^\a}
=0.
\end{equation*}
Following similar ideas, we can study the case for several dependent variables
$(x_1(t),\ldots,x_m(t))$, with $m\in\mathbb{N}$. In this case, we obtain $m$
fractional differential equations, one for each function $x_i$, $i=1,\ldots,m$.

\begin{theorem}
Let $\mathcal{J}$ be the functional
$$
\mathcal{J}(x_1,\ldots,x_m):=\int_a^b L[x]^\alpha_m(t)\,dt
$$
with
$$
[x]^\alpha_m(t):=\left(t,x_1(t),
\ldots,x_m(t),{^CD_{a+}^{\a_1}} x_1(t),\ldots,{^CD_{a+}^{\a_m}} x_m(t)\right),
$$
where $\a_1,\ldots,\a_m\in \, ]0,1[$ and $m\in\mathbb N$, defined on the set of functions
$$
\left\{x\in C^1([a,b],\mathbb R^m)\,: x(a)=x_a \wedge x(b)=x_b\right\},
$$
where $x_a,x_b$ are two fixed vectors in $\mathbb{R}^m$.
Suppose also that $L:[a,b]\times \mathbb{R}^{2m}\to\mathbb{R}$
is continuously differentiable with respect to its $i$th variable, $i=2,\ldots,2m+1$.
Let $(x_1^\star,\ldots,x_m^\star)$ be a minimizer of $\mathcal{J}$, and suppose that
the functions $t\mapsto {D_{b-}^{\a_i}}\left(\partial_{i+1+m} L[x^\star]^\alpha_m(t)\right)$
exist and are continuous on $[a,b]$, $i=1,\ldots,m$. Then,
$$
\partial_{i+1} L[x^\star]^\alpha_m(t)+{D_{b-}^{\a_i}}\left(\partial_{i+1+m}
L[x^\star]^\alpha_m(t)\right)=0
$$
for all $i=1,\ldots,m$ and for all $t\in[a,b]$.
\end{theorem}

In the next result, the lower bound of the cost functional is a real $A$,
where $A>a$, that is, $A$ is greater than the lower bound
of the fractional derivative. For simplicity of presentation, we restrict
ourselves to the case $m = 1$.

\begin{theorem}
Define the functional $A$ as
$$
\mathcal{J}(x):=\int_A^b L[x]^\alpha(t)\,dt,
$$
where $A>a$. If $\mathcal{J}$ achieves a minimum value at $x^\star$,
and if the maps $t\mapsto {D_{A-}^{\a}}\left(\partial_3 L[x^\star]^\alpha(t)\right)$,
for $t\in[a,A]$, and $t\mapsto \RRD \left(\partial_3 L[x^\star]^\alpha(t)\right)$,
for $[a,b]$, exist and are continuous, then
$$
\RRD \left(\partial_3 L[x^\star]^\alpha(t)\right)-{D_{A-}^{\a}}\left(\partial_3
L[x^\star]^\alpha(t)\right)=0, \quad t\in[a,A],
$$
and
$$
\partial_2 L[x^\star]^\alpha(t)+\RRD \left(\partial_3 L[x^\star]^\alpha(t)\right)=0,
\quad t\in[A,b].
$$
Moreover, the following transversality conditions are fulfilled:
$$
\left\{
\begin{array}{l}
{I_{b-}^{1-\a}}\left(\partial_3 L[x^\star]^\alpha(t)\right)
-{I_{A-}^{1-\a}}\left(\partial_3 L[x^\star]^\alpha(t)\right)=0
\quad \mbox{at $t=a$, if $x(a)$ is free;}\\
{I_{A-}^{1-\a}}\left(\partial_3 L[x^\star]^\alpha(t)\right)=0
\quad \mbox{at $t=A$, if $x(A)$ is free;}\\
{I_{b-}^{1-\a}}\left(\partial_3 L[x^\star]^\alpha(t)\right)=0
\quad \mbox{at $t=b$, if $x(b)$ is free.}
\end{array}
\right.
$$
\end{theorem}

For our next result, besides the boundary conditions, an integral constraint
is imposed on the set of admissible functions. Such type of problems are known as
isoperimetric problems, and the first example goes back to Dido, Queen of Cartage
in Africa. She was interested if finding the shape of a curve with fixed perimeter,
of maximum possible area. As we know, the solution is given by a circle.
The calculus of variations presents a solution to such problem,
with an integral constraint of type
$$
\int_a^b \sqrt{1+(x'(t))^2}\,dt=\mbox{constant.}
$$
In our case, we consider that the new constraint depends
also on a fractional derivative, and it is of form
\begin{equation}
\label{isope}
G(x):=\int_a^b M\left(t,x(t),\LD x(t)\right)\,dt=K,\quad K\in\mathbb{R},
\end{equation}
where $M:[a,b]\times\mathbb{R}^2\to\mathbb{R}$ is a constant,
such that there exist and are continuous the functions
$\partial_2 M$ and $\partial_3 M$.

\begin{theorem}
Suppose that $\mathcal{J}$ \eqref{funct1}, subject to the constraints \eqref{bound}
and \eqref{isope}, attains a minimum value at $x^\star$.
If $x^\star$ is not a solution for
\begin{equation}
\label{aux8}
\partial_2 M[x]^\alpha(t)+\RRD \left(\partial_3 M[x]^\alpha(t)\right)=0,
\quad \forall t \in[a,b],
\end{equation}
and if there exist and are continuous the functions
$t\mapsto \RRD \left(\partial_{3} L[x^\star]^\alpha(t)\right)$ and $t\mapsto \RRD
\left(\partial_{3} M[x^\star]^\alpha(t)\right)$ on $[a,b]$, then there exists
$\lambda\in\mathbb R$ such that $x^\star$ satisfies
$$
\partial_2 F[x]^\alpha(t)+\RRD \left(\partial_3 F[x]^\alpha(t)\right)=0,
\quad \forall t \in[a,b],
$$
where we define the function $F$ as $F:=L+\lambda M$.
\end{theorem}

The case when $x^\star$ is a solution of \eqref{aux8} can be easily included:

\begin{theorem}
Suppose that $\mathcal{J}$ \eqref{funct1}, subject to the constraints \eqref{bound} and \eqref{isope},
attains a minimum value at $x^\star$. If there exist and are continuous the functions
$t\mapsto \RRD \left(\partial_{3} L[x^\star]^\alpha(t)\right)$ and $t\mapsto \RRD
\left(\partial_{3} M[x^\star]^\alpha(t)\right)$ on $[a,b]$, then there exist
$\lambda_0,\lambda\in\mathbb{R}$, not both zero, such that $x^\star$ satisfies the equation
$$
\partial_2 F[x]^\alpha(t)+\RRD \left(\partial_3 F[x]^\alpha(t)\right)=0,
\quad \forall t \in[a,b],
$$
where we define the function $F$ as $F:=\lambda_0 L+\lambda M$.
\end{theorem}

Next, we present another constrained type problem, but now the restriction is given by
\begin{equation}
\label{isoperi2}g(t,x(t))=0,
\quad \forall t\in[a,b],
\end{equation}
where $x=(x_1,x_2)$ is a vector and
$g:[a,b]\times\mathbb{R}^2\to\mathbb{R}$ is differentiable
with respect to $x_1$ and $x_2$. Also, we have the boundary conditions
\begin{equation}
\label{bound2}
x(a)=x_a \quad \mbox{and} \quad x(b)=x_b,
\quad x_a,x_b\in\mathbb{R}^2.
\end{equation}

\begin{theorem}
Consider the functional
$$
\mathcal{J}(x)=\int_a^b L[x]^\alpha_2(t)\,dt,
$$
where
$$
[x]^\alpha_2(t):=\left(t,x_1(t),x_2(t),\LD x_1(t),\LD x_2(t)\right),
$$
defined on $C^1[a,b]\times C^1[a,b]$, subject to the constraints
\eqref{isoperi2} and \eqref{bound2}. If $\mathcal{J}$ attains an extremum
at $x^\star=(x_1^\star,x_2^\star)$, the maps
$t \mapsto\RRD  \left(\partial_{i+3} L[x^\star]^\alpha_2(t)\right)$,
$i=1,2$, are continuous, and
$\partial_3 g(t,x(t))\not=0$ for all $t\in[a,b]$,
then there exists a continuous function
$\lambda:[a,b]\to\mathbb{R}$ such that
$$
\partial_{i+1}L [x^\star]^\alpha_2(t)+\RD\left( \partial_{i+3}
L[x^\star]^\alpha_2(t)\right)+\lambda \partial_{i+1} g (t,x(t))=0
$$
for all $t\in[a,b]$ and $i=1,2$.
\end{theorem}

Infinite horizon problems are an important field of research, that deal
with phenomena that spread along time \cite{MR3046396,MR3031162,MR2966852}.
In this case, the cost functional
is evaluated in an infinite interval $[a,\infty[$:
\begin{equation}
\label{funct2}
\mathcal{J}(x):=\int_a^\infty L[x]^\alpha\,dt,
\end{equation}
defined on a set of functions with fixed initial conditions: $x(a)=x_a$.
Since we are dealing with improper integrals, some attention to what
is a minimizer is needed. We say that $x^\star$ is a local minimizer
for $\mathcal{J}$ as in \eqref{funct2} if there exists some $ \epsilon>0$ such that,
for all  $x$, if $\|x^\star-x\|<\epsilon$, then
$$
\lim_{T\to\infty}\inf_{b \geq T}\int_{a}^{b}[L[x^\star]^\alpha(t)
-L[x]^\alpha(t)] \, dt
\leq 0.
$$
Also, let us define the functions
\begin{align*}
A(\epsilon, b) &:= \int_a^{b} \frac{L[x^\star+\epsilon v]^\alpha(t)
- L[x^\star]^\alpha(t)}{\epsilon}dt;\\
V(\epsilon, T) &:= \inf_{b \geq T}\int_a^{b} \left[L[x^\star+\epsilon v]^\alpha(t)
-L[x^\star]^\alpha(t)\right]dt;\\
W(\epsilon)&:= \lim_{T\to\infty} V(\epsilon,T),
\end{align*}
where $v \in C^1[a,\infty[$ is a function and $\epsilon$ a real.

\begin{theorem}
Let $x^\star$ be a local minimal for $\mathcal{J}$
as in \eqref{funct2}. Suppose that:
\begin{enumerate}
\item $\displaystyle \lim_{\epsilon \to 0} \frac{V(\epsilon, T) }{\epsilon}$ exists for all $T$;
\item $\displaystyle \lim_{T\to\infty}\frac{V(\epsilon, T) }{\epsilon}$ exists uniformly for all $\epsilon$;
\item For every $T> a$ and $\epsilon\not=0$, there exists a sequence
$\left(A(\epsilon, b_n)\right)_{n \in \mathbb{N}}$ such that
$$
\lim_{n \to \infty} A(\epsilon, b_n)= \inf_{b\geq T} A(\epsilon, b)
$$
uniformly for $\epsilon$.
\end{enumerate}
If there exists and is continuous the function $t\mapsto \RRD \left(
\partial_3 L[x^\star]^\alpha(t)\right)$ on $[a,b]$, for all $b>a$, then
$$
\partial_2 L[x^\star]^\alpha(t)+\RRD \left(\partial_3 L[x^\star]^\alpha(t)\right)=0,
$$
for all $b> a$. Also, we have
$$
\lim_{T\to\infty}\inf_{b \geq T}{I_{b-}^{1-\a}}\left(\partial_3
L[x^\star]^\alpha(t)\right)=0 \quad \mbox{at} \quad t=b.
$$
\end{theorem}

So far, we considered variational problems with order $\alpha\in]0,1[$.
Now we proceed by extending them to functionals depending on higher-order derivatives.
To that purpose, consider the functional
\begin{equation}
\label{funHig}
\mathcal{J}(x):=
\int_a^b L\left(t,x(t),{^CD_{a+}^{\a_1}} x(t),\ldots, {^CD_{a+}^{\a_m}} x(t)\right)\,dt,
\end{equation}
defined on $C^m[a,b]$, such that $x^{(i)}(a)$ and $x^{(i)}(b)$ are fixed reals
for $i\in\{0,1,\ldots,m-1\}$. Here, $m$ is a positive integer, $\a_i\in(i-1,i)$,
for all $i\in\{1,\ldots,m\}$, and $L:[a,b]\times\mathbb R^{m+1}\to\mathbb{R}$
is differentiable with respect to the $i$th variable, for $i\in\{2,3,\ldots,m+1\}$.

\begin{theorem}
Let $x^\star$ be a minimizer of $\mathcal{J}$ \eqref{funHig}, and suppose that for all
$i\in\{1,\ldots,m\}$, there exist and are continuous the functions $t\mapsto
{D_{b-}^{\a_i}}\left(\partial_{i+2} L[x^\star]^\alpha_m(t)\right)$ on $[a,b]$. Then,
$$
\partial_2 L[x^\star]^\alpha_m(t)+\sum_{i=1}^m{D_{b-}^{\a_i}}\left(
\partial_{i+2} L[x^\star]^\alpha_m(t)\right)=0
$$
for all $t\in[a,b]$, where
$[x^\star]^\alpha_m(t):=\left(t,x(t),{^CD_{a+}^{\a_1}} x^\star(t),
\ldots, {^CD_{a+}^{\a_m}} x^\star(t)\right)$.
\end{theorem}

After solving the presented Euler--Lagrange equations, we still need to verify
if we are in presence of a minimizer of the functional or not. We recall that
the Euler--Lagrange equation is just a necessary optimality condition of first order, and thus
its solutions may not be a solution for the problem. For the question of existence
of solutions, we refer to \cite{MR3071538,MR3200762}.
One possible way to check if we have a candidate for minimizer or maximizer
is to apply the Legendre condition, which is a second order
necessary optimality condition \cite{MR3784384,MR3223604}.
Under suitable convexity of the Lagrangian, sufficient
conditions for global minimizer hold \cite{MR2736825}.

For the Legendre condition, we have to assume that the Lagrange function $L$ is such that
its second order partial derivatives $\partial^2_{ij}L$, with $i,j\in\{2,3\}$, exist and are continuous.

\begin{theorem}
Suppose that $x^\star$ is a local minimum for $\mathcal{J}$ as in \eqref{funct1}. Then,
$$
\partial^2_{33}L[x^\star]^\alpha(t)\geq0, \quad \forall  t \in[a,b].
$$
\end{theorem}

For our next result, we recall that a function $L:[a,b]\times\mathbb{R}^2\to\mathbb{R}$
is convex with respect to the second and third variables if
$$
L(t,x+v,y+w)-L(t,x,y)\geq \partial_2 L(t,x,y)v+\partial_3 L(t,x,y)w
$$
for all $t\in[a,b]$ and $x,y,v,w\in\mathbb{R}$.

\begin{theorem}
If $L$ is convex and if $x^\star$ satisfies the Euler--Lagrange equation \eqref{ELeq},
then $x^\star$ minimizes $\mathcal{J}$ \eqref{funct1}
when restricted to the boundary conditions \eqref{bound}.
\end{theorem}

\section{Direct methods}
\label{sec:04}

There are several different numerical approaches and methods
to solve fractional differential equations
\cite{MR3513174,MR3157930,MR3294734}.
Different discretizations of fractional derivatives are possible,
but many of them do not preserve the fundamental properties
of the systems, such as stability \cite{MR2656354,MyID:397}.
For numerical calculations, a simple and powerful method preserving stability
is obtained from Gr\"{u}nwald--Letnikov discretizations.
For a detailed numerical treatment of fractional differential equations,
based on Gr\"unwald--Letnikov fractional derivatives, we refer
the interested reader to \cite{Ort,MR2824679}. Here we just mention
that both explicit and implicit methods are possible:
Theorem~5.1 of \cite{MR2824679} shows that the explicit and implicit
Gr\"unwald--Letnikov methods are asymptotically stable; while
Theorem~5.2 of \cite{MR2824679} gives conditions on the step size
for the explicit method to be absolute stable, asserting that
the implicit method is always absolute stable,
without any step size restriction.
It is also worth to underline the good convergence properties
and error behavior of Gr\"{u}nwald--Letnikov methods \cite{MR2824679}.

We start this section by recalling the (left) Gr\"{u}nwald--Letnikov fractional
derivative of a function $x$. Let $\alpha>0$ be a real.
The Gr\"{u}nwald--Letnikov fractional derivative of order $\alpha$
is defined by
$$
{_a^{GL}D_x^\alpha}x(t):=\lim_{h\rightarrow 0^+}
\frac{1}{h^\alpha}\sum_{k=0}^\infty(-1)^k\binom{\alpha}{k}x(t-kh),
$$
where $\binom{\alpha}{k}$ stands for the generalization of binomial coefficients
to real numbers. As usual, we will adopt the notation
$$
(w_k^\alpha):= (-1)^k\binom{\alpha}{k}.
$$
This fractional derivative is particularly useful to approximate
the Caputo derivative. The method is now briefly explained.

Given an interval $[a,b]$ and a fixed integer $N$, let $t_j:=a+ jh$,
$j = 0,1,\ldots,N$ and $h>0$, be a partition of the interval $[a,b]$. Then,
$$
\LD x(t_j)= \frac{1}{h^\alpha}\sum_{k=0}^j(w_k^\alpha) x(t_{j-k})
-\frac{x(a)}{\Gamma(1-\alpha)}(t_j-a)^{-\alpha}+O(h).
$$
Thus, truncated Gr\"{u}nwald--Letnikov fractional derivatives are first-order
approximations of the Caputo fractional derivatives. The approximation
used for the Caputo derivative is then
$$
\LD x(t_j)\approx  \frac{1}{h^\alpha}\sum_{k=0}^j(w_k^\alpha) x(t_{j-k})
-\frac{x(a)}{\Gamma(1-\alpha)}(t_j-a)^{-\alpha}:= \tilde{D} x(t_j).
$$
The next step is to discretize the functional \eqref{funct1}.
For simplicity, let $h=(b-a)/N$ and let us consider
the grid $t_j=a+jh$, $j=0,1,\ldots, N$. Then,
\begin{equation}
\label{disFuncl}
\begin{array}{ll}
\mathcal{J}(x)&=\displaystyle \sum_{k=1}^N\int_{t_{k-1}}^{t_k} L(t,x(t),\LD x(t))\,dt\\
&\approx \displaystyle \sum_{k=1}^N h L(t_k,x(t_k),\LD x(t_k))\\
&\approx \displaystyle \sum_{k=1}^N h L(t_k,x(t_k),\tilde{D} x(t_k))\,dt.
\end{array}
\end{equation}
The right hand side of \eqref{disFuncl} can be regarded as a function
$\Psi$ of $N-1$ unknowns:
$$
\Psi(x_1,x_2,\ldots,x_{N-1})=\sum_{k=1}^N h L(t_k,x(t_k),\tilde{D} x(t_k)).
$$
To find an extremum for $\Psi$, one has to solve the following system
of algebraic equations:
$$
\frac{\partial \Psi}{\partial x_i}=0,\qquad i=1,\ldots,N-1.
$$
Suppose that $h$ goes to zero. The solution obtained by this method converges
to a function $x^\star$. Then, $x^\star$ is a solution of
the Euler--Lagrange equation \eqref{ELeq} (for details,
see \cite[Theorem~8.1]{MR3443073}).

\begin{example}
\label{ex01}
Consider the functional
$$
\mathcal{J}(x)=\int_0^{10}\left({^CD_{0+}^{0.5}} x(t)-\frac{2}{\Gamma(3/2)}t^{1.5}\right)^2\,dt
$$
subject to the boundary conditions $x(0)=0$ and $x(10)=100$. Since ${^CD_{0+}^{0.5}}t^2
= \frac{2}{\Gamma(3/2)}t^{1.5}$, and $\mathcal{J}$ is nonnegative, we conclude that the function
$x(t)=t^2$ is a minimizer of the functional. If we discretize the functional, for different values
of $n$, we obtain several numerical approximations of $x$. In Table~\ref{tab1} we present the error
of each $n$, where the error is given by the maximum of the absolute value
of the difference between $x$ and the numerical approximation.

\begin{table}[h]
\centering
\caption{Errors for the numerical solution of problem of Example~\ref{ex01}.} \label{tab1}
\begin{tabular}{lcccc}\hline
$n$ & 10 & 50 & 100 & 200 \\  \hline
Error  & 0.425189 & 0.107622 & 0.056370 & 0.029215\\ \hline
\end{tabular}
\end{table}
\end{example}

\begin{example}
\label{ex02}
For our second example, we consider a problem where we do not
know its exact solution. Let
$$
\mathcal{J}(x)=\int_0^{1}\left(x(t)\left({^CD_{0+}^{0.5}} x(t)\right)^2
-\sin(x(t))\right)^2\,dt
$$
subject to the boundary conditions $x(0)=0$ and $x(1)=1$.
In Figure~\ref{fig1}, we present the results for $n=100$.

\begin{figure}[ht]
\centering
\includegraphics[width=6cm]{n100}\\
\caption{Plot of the numerical solution of problem of Example~\ref{ex02}.}\label{fig1}
\end{figure}
\end{example}

\begin{acknowledgement}
the authors are grateful to the Research and Development
Unit UID/MAT/04106/2013 (CIDMA) and to two anonymous referees
for several constructive comments.
\end{acknowledgement}


TITLE: Why this function is continuous?
QUESTION [2 upvotes]: Let $(\Omega,\Sigma,\mu)$ be a sample space and let $L^2= \lbrace f:\Omega \rightarrow R / \int f^2d\mu <\infty  \rbrace$ be a Hilbert space.
Let $L_n=L^2\times L^2 \times .... \times L^2$ ($n$ times) be a vector space with inner product $\langle X,Y \rangle = \sum_{i=1}^{n} \int X_iY_id\mu$ where $X=(X_1,...,X_n)$ and $Y=(Y_1,...,Y_n)$.
Also consider:

$X \geq 0$ if and only if $X_i \geq 0$, $\forall \: i \in \lbrace 1,..,n \rbrace$.
$X = Y$ if and only if $X_i = Y_i$ almost everywhere, $\forall \: i \in \lbrace 1,..,n \rbrace$.

If $F:L_n \rightarrow R$ is a linear functional and it's positive (i.e. $F(X)\geq0$ for $X\geq0$) then $F$ is continuous. (i.e. $\Vert X_n - X \Vert \rightarrow 0$ imply $F(X_n) \rightarrow F(X)$)

REPLY [0 votes]: The following works in the case $n=1$. The general case easily follows.
Assume that $F$ is not continuous. Then for every $k \in \Bbb{N}$, there is some $f_k \in L^2$ with $\Vert f_k \Vert_2 \leq \frac{1}{2^k}$ and $F(f_k) \geq 2^k$ (show this!).
Using e.g. monotone convergence, one can show (do it!) $g := \sum_{\ell = 1}^{\infty} |f_\ell| \in L^2$ with $\Vert g \Vert_2 \leq 1$.
But now $0 \leq g \pm f_k$ for every $k$ and thus $0 \leq F(g \pm f_k)$ for all $k$.
Why does this yield a contradiction?


TITLE: find the upper and lower bound for a finite sum
QUESTION [0 upvotes]: Find upper and lower bound for the following finite sum
$1/(1 + 1^3)+1/(1 +2^3)+1/(1 + 3^3) + ··· + 1/(1 + n^3)$
My attempt:
$1/(1 + 1^3)+1/(1 +2^3)+1/(1 + 3^3) + ··· + 1/(1 + n^3)$ = $\sum_{i=1}^n 1/(1+i^3)$ = $\int_1^n$1/$(1+i^3)$di = $\int_1^n1/(1+x^3)$dx
But now I'm stuck.Is my attempt correct?

REPLY [1 votes]: No, because$$\sum_{i=1}^n\frac1{1+i^3}\not=\int_1^n\frac1{1+x^3}\,\mathrm dx.$$However,$$\int_i^{i+1}\frac1{1+x^3}\,\mathrm dx\leqslant\frac1{1+i^3}\leqslant\int_{i-1}^i\frac1{1+x^3}\,\mathrm dx,$$and you can use this to solve your problem.


TITLE: Find the singularities and residues that correspond. Show details
QUESTION [0 upvotes]: Find all the singularities in the finite plane and the corresponding residues. Show the details.

$$f(z) = \frac{1}{1-e^z}$$
So I'm trying to find the singularities. So the singularities are located when the denominator = 0 and that occurs when $e^z = 1$ which occurs when $z = 2 \pi in$ when $n \in \mathbb{R}$
Now what? How can I find the residues?
so the residue formula that I know is:
$$\operatorname{Res}_{z = z_0}f(z) = \lim_{z \to z_0} (z-z_0) f(z)$$
That's what I'm trying to use. But how do I determine this when $z_0$ isn't fixed?

REPLY [1 votes]: For each $n\in\Bbb Z$, you have\begin{align}\operatorname{res}_{z=2\pi in}\frac1{1-e^z}&=\lim_{z\to2\pi in}\frac{z-2\pi in}{1-e^z}\\&=-\frac1{\lim_{z\to2\pi in}\frac{e^z-e^{2\pi in}}{z-2\pi in}}\\&=-\frac1{\exp'(2\pi in)}\\&=-\frac1{e^{2\pi in}}\\&=-1.\end{align}


TITLE: Saddle point, point of inflection, extremum, stationary point
QUESTION [0 upvotes]: What is the difference between a point of inflection and a saddle point?
What is the difference between an extremum and a stationary point?

REPLY [1 votes]: A point of inflexion is a point where the tangent line exists and crosses the curve.  So the context is the graph of a 1-dimensional curve in 2 dimensions.  A saddle point is a point on a surface (so the context is a two dimensional surface in 3 dimensions.) where the tangent plane is horizontal, but the point is neither a max or a min.
A stationary point is a point where the derivative exists and is zero.  It may or may not be an extremum.  An extremum is either a local max or local min.  For instance the function $f(x) = |x|$ has a local min at $x=0$, but the derivative doesn't exist, and therefore it is not a stationary point.


TITLE: Minimum $9a+25b+49c$ when $ab+bc+ca+abc=4$
QUESTION [1 upvotes]: If $a,b,c\ge 0$ such that $ab+bc+ca+abc=4$, find the minimum value of $9a+25b+49c$.
I know that $a+b+c\ge 3$, but I don't this is good to use here. So I tried with Lagrange multipliers:
$$L(x,y,z)=9a+25b+49c+\lambda(ab+bc+ca+abc-4)$$
With the partial derivative I found:
$$\frac{b+c+bc}{9}=\frac{a+c+ac}{25}=\frac{a+b+ab}{49}$$
and with $ab+bc+ca+abc=4$, I found minimum $59$ at $(3,1,1/7)$. My question is, can it be done with traditional ways? I tried to prove $ab+bc+ca+abc\le 4$ when $9a+25b+49c=59$ (with idea from this question: Minimum value when $abc+ab+4bc+9ca=144$), but I got lost after expanding.

REPLY [0 votes]: A similar idea as the accepted answer after rewriting the condition as:
$$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1$$
and using Cauchy-Schwarz:
$$
\begin{aligned}
\frac{a+1}{a+2}&=\frac{1}{b+2}+\frac{1}{c+2}\\
&=5\cdot\frac{1}{5(b+2)}+7\cdot\frac{1}{7(c+2)}\\
&\geq \frac{144}{25b+49c+148}
\end{aligned}
$$
Therefore
$$25b+49c \geq \frac{144}{a+1}-4$$
and using AM-GM:
$$
\begin{aligned}
9a+25b+49c &\geq 9a+\frac{144}{a+1}-4\\
&=9(a+1)+\frac{144}{a+1}-13\\
& \geq 2\sqrt{9(a+1)\cdot \frac{144}{a+1}}-13\\
& = 59
\end{aligned}
$$


TITLE: Continuous function existence
QUESTION [1 upvotes]: Prove that there is at most one continuous function on $[0,2]$ that satisfies:
$$v(x)=f(x)+\int_0^2 e^{-(x-y)^2} \cos(0.3v(y)) \, dy$$
I don't know how to estimate this integral...

REPLY [1 votes]: If $v_1$ and $v_2$ are two solutions,
$$v_1(x) - v_2(x) = \int_0^2 e^{-(x-y)^2} (\cos(0.3 v_1(y)) - \cos(0.3 v_2(y)))\ dy$$
Now $\cos(.3 v_1(y)) - \cos(.3 v_2(y) = -.3 \sin(.3 s) (v_1(y) - v_2(y))$ for some $s$ between $v_1(y)$ and $v_2(y)$. Only quite obvious estimates of
$\exp(-(x-y)^2)$ and $|\sin(s)|$ are needed ...
you should get $|v_1(x) - v_2(x)| \le c \max_y |v_1(y) - v_2(y)|$ with $c < 1$.


TITLE: Isomorphic relation between dihedral groups
QUESTION [2 upvotes]: Theorem
Let $G,H$ be abelian groups such that $Dih(G)\cong Dih(H)$
If $G$ is finitely generated, then $G\cong H$.

I'm curious whether "finitely generated" hypothesis can be removed.
If it's not true in general, what would be a counterexample?
=====
EDIT:
The below statement is equivalent to the above question.

Let $G,H$ be abelian groups such that every element of $G,H$ is of order 2.
If $G\times Z_2 \cong H\times Z_2$ (direct product), then $G\cong H$.

Is this statement true?

REPLY [1 votes]: For your second question, there is no need to assume that $G,H$ are abelian, because all groups in which every element has order $2$ are abelian! Such groups are vector spaces over the field of order $2$, so the groups $G$ and $H$ in your question are subspaces of $G \times C_2$, and they clearly have the same dimension, so they are isomorphic.


TITLE: Eisenstein Series and cusps of $\Gamma_{1}(N)$
QUESTION [11 upvotes]: Let $q = e^{2\pi i\tau}$, $\operatorname{Im}\tau > 0$ and let $$G(\tau) = 1 - 24\sum_{n = 1}^{\infty}\frac{nq^{n}}{1 - q^{n}}$$ be the weight 2 Eisenstein series for $\Gamma(1)$. Consider the inequivalent cusps of $\Gamma_{1}(N)$. For each cusp, there is an Eisenstein series of weight 2 associated to this cusp. How does one relate this Eisenstein series to a linear combination of $G(k\tau)$'s (for some integers $k$)?
In particular, the case I am considering is when $N = 7$. The inequivalent cusps are $0$, $2/7$, $1/3$, $3/7$, $1/2$, and $\infty$. Consider the cusp $2/7$ and denote by $E_{2/7}$ the associated Eisenstein series. I would like to write $E_{2/7}$ as a finite sum of $G(k\tau)$ for some integers $k$. One of the first issues I'm having is how would I know the $q$-expansions of Eisenstein series of $E_{2/7}$ (which is related to the issue that I can't seem to find a good reference for the definition of a weight 2 Eisenstein series associated to a cusp)? Second, would finding the desired linear combination of $G(k\tau)$'s be just an exercise in matching up finitely many coefficients in the $q$-expansion?

REPLY [8 votes]: The question has some implicit hypotheses, possibly not clear to the questioner, and this implicitness and ambiguities about it complicate matters. First, the more natural descriptions of Eisenstein series for GL(2) of weights k>2 are not of the form in the question, but are $\sum_{c,d} 1/(cz+d)^k$. This does not converge for $k=2$, so $k=2$ has to be approached more delicately (via an analytic continuation, Hecke summation, producing _something_like_ the expression in the question). However, at level one, that is, for $SL_2(\mathbb Z)$, there is no truly-holomorphic Eisenstein series of weight $2$. The analytic continuation has an extra term, which one may discard, but then destroying the literal automorphy condition. Maybe that doesn't matter, but one should be careful about "understandings".
Thus, depending what one means, wants, or needs, while at higher levels the meromorphic continuation can produce holomorphic modular forms at level 2. (This positive outcome always occurs for Hilbert modular forms, that is, for ground fields totally real anything other than $\mathbb Q$.) First, whatever description one chooses for "Eisenstein series" (attached to cusps?), a suitable weighted average of the level-7 (for example) such should be level-one. A literal notion of holomorphy tells us there is no level-one, weight-two such. Thus, there are (at most) six linearly independent Eisenstein series at that level, so not quite possible to "attach" one to each cusp. It is not hard to say more.
Then there is the further issue of expressing various Eisenstein series in terms of each other, by the group action. At square-free level, the underlying (!) representation theory is simpler (Iwahori-fixed vectors in principal series are well understood, at least up to a very useful point.)
But, at this point, without knowing more precisely what the questioner wants, or may discover is wanted, there are too many things that can be said to know which to choose to say. :)


TITLE: Can the Casimir effect also generate a repulsive force at large distances?
QUESTION [1 upvotes]: I've just seen a documentary where they state that particles pop into existence and then annihilate each other all the time in empty space. The experiment discussed that proved this was that two metal plates placed very, very near each other would be pushed together.
It occurs to me that if you placed them very, very far apart, such that more of the universe was between them than was outside them, wouldn't they be pushed apart for the same reasons? Could this be what causes the universe to expand? It seems to me that if there's any pushing at all, there has to be expansion.

REPLY [0 votes]: This is the experiment you cited:  Two uncharged parallel plates separated by a tiny distance, on the order of the wavelength of a single photon, will limit the photons popping in and out of existence between them to those whose wavelengths can fit a whole number of times between the plates.
As the plates draw closer together, one should find that the photons popping in and out of existence between them consist only of those with such whole wavelengths, and should exclude photons of all other wavelengths.  This should reduce the total vacuum energy between the plates, as compared to the vacuum energy in space around the plates, which implies that the plates are drawn together by the reduced vacuum energy.  The attraction is called the Casimir effect, as Alpha001 notes in his comment.
Another way of looking at the effect is as though the two metal plates are connected by a stretched spring (a sea of virtual photons), and move together as the stretched spring loses energy (limitation of the virtual photons) and contracts.  This metaphor was devised by Stephen Reucroft and John Swain.
Moving up in scale to the entire cosmos, quantum field theory views space to be filled with virtual energy that fluctuates in and out of existence.  But unless there are boundaries that limit the virtual energy to certain whole wavelengths in a bounded area, there will be no Casimir effect.
A problem with your interesting suggestion is that it is not believed that the Universe is finite in the sense of having a boundary, and is not divisible into one area containing "more" Universe, and another containing "less" Universe, which sum to the entire Universe, as Countto10 points out in his comment.
Another problem is that even if there were evidence that the Universe were finite and divisible into two parts that sum to the whole, General Relativity provides that the difference in vacuum energy between the two parts of the Universe might cause recognizable curvature, which has not been found.


TITLE: using Schur's complement and Young's inequality to reduce matrix algebraic equation to LMI
QUESTION [2 upvotes]: This questions concerns the practical implementation of schur's complement and Young's inequality.
Consider the following
\begin{align}
\begin{pmatrix} \begin{pmatrix}
\mathbb{A}_{\mathbb{Z}} & \mathcal{O}\\
\mathcal{O} & \mathbb{A}_{\mathbb{P}}
\end{pmatrix} + \begin{pmatrix}
\mathbb{Z} & \mathcal{O}\\
\mathcal{O} & \mathcal{I}_{n}
\end{pmatrix} \begin{pmatrix}
\mathbb{Z} & \mathcal{O}\\
\mathcal{O} & \mathcal{I}_{n}
\end{pmatrix} & \begin{pmatrix}
D_{11} & D_{12}\\
D_{21} & D_{22}
\end{pmatrix}\\
\begin{pmatrix}
D_{11} & D_{12}\\
D_{21} & D_{22}
\end{pmatrix}^T & -\mu \mathbb{I}_{q}
\end{pmatrix} + \underbrace{\begin{pmatrix}
\overline{B} \overline{K}\\
\mathcal{O}\\
\mathcal{O}
\end{pmatrix}}_{V^T} \underbrace{\begin{pmatrix}
\mathcal{O} & \tilde{B} & \mathcal{O}
\end{pmatrix}}_U + \begin{pmatrix}
\mathcal{O}\\
\tilde{B}^T\\
\mathcal{O}
\end{pmatrix} \begin{pmatrix}
\left(\overline{B} \overline{K} \right)^T & \mathcal{O} & \mathcal{O}
\end{pmatrix} \leq 0
\end{align}
Now to avoid any bilinear coupling between terms, the young's inequality is used, as given by
\begin{equation}
X^T \hspace{1mm} Y + Y^T \hspace{1mm} X \leq \dfrac{1}{2} \hspace{1mm} \left( X + S \hspace{1mm} Y \right)^T \hspace{1mm} S^{-1} \hspace{1mm} \left( X + S \hspace{1mm} Y \right)
\end{equation}
where $S$ is a symmetric positive definite matrix. To apply the inequality, we use $S = \dfrac{1}{2} \mathbb{I}_n$ for the term $\begin{pmatrix}
\mathbb{Z} & \mathcal{O}\\
\mathcal{O} & \mathcal{I}_{n}
\end{pmatrix} \begin{pmatrix}
\mathbb{Z} & \mathcal{O}\\
\mathcal{O} & \mathcal{I}_{n}
\end{pmatrix}$ and $S = \epsilon \mathbb{Z}$ for the term $V^TU+U^TV$. Now, the intent is to use the young's inequality and convert the above algebraic equation in LMI using Schur's complement. I am unsure of the proper way and cannot reduce the equation to the requisite LMI. Given: $\mathbb{Z}$ is a $n \times n$ symmetric positive definite matrix.
Can someone walk me through the subsequent steps?

REPLY [2 votes]: The only term that is not linear in the unknown variables is the quadratic term in $\mathbb{Z}$, so it might be easier to replace that with another variable and add another inequality
$$
\mathbb{Z}^2 \leq \mathbb{X}, \tag{1}
$$
such that the initial in equality can be written as
\begin{align}
\begin{pmatrix} \begin{pmatrix}
\mathbb{A}_{\mathbb{Z}} & \mathcal{O}\\
\mathcal{O} & \mathbb{A}_{\mathbb{P}}
\end{pmatrix} + \begin{pmatrix}
\mathbb{X} & \mathcal{O}\\
\mathcal{O} & \mathcal{I}_{n}^2
\end{pmatrix} & \begin{pmatrix}
D_{11} & D_{12}\\
D_{21} & D_{22}
\end{pmatrix}\\
\begin{pmatrix}
D_{11} & D_{12}\\
D_{21} & D_{22}
\end{pmatrix}^\top & -\mu\,\mathbb{I}_{q}
\end{pmatrix} + \begin{pmatrix}
\overline{B} \overline{K}\\
\mathcal{O}\\
\mathcal{O}
\end{pmatrix} \begin{pmatrix}
\mathcal{O} & \tilde{B} & \mathcal{O}
\end{pmatrix} + \begin{pmatrix}
\mathcal{O}\\
\tilde{B}^\top\\
\mathcal{O}
\end{pmatrix} \begin{pmatrix}
\left(\overline{B} \overline{K} \right)^\top & \mathcal{O} & \mathcal{O}
\end{pmatrix} \leq 0, \tag{2}
\end{align}
which is linear in all variables. One can then use Schur's complement on $(1)$ which allows us to rewrite it also as an LMI
$$
\begin{pmatrix}
I & \mathbb{Z} \\
\mathbb{Z} & \mathbb{X}
\end{pmatrix} \geq 0, \tag{3}
$$
with $I$ an $n$ by $n$ identity matrix, $\mathbb{Z} = \mathbb{Z}^\top > 0$ and $\mathbb{X} = \mathbb{X}^\top > 0$.
So the combination of $(2)$ and $(3)$ should be equivalent to your initial inequality and are all LMI's.


TITLE: Problem leading simple equations
QUESTION [1 upvotes]: A sum of Rs. 8.85 is made up of 124 coins which are either 10 paisa coins or 5 paisa coins ; how many coins are there each
Note : Rs. 1 = 100 paisas

REPLY [2 votes]: 1 RS = 100 paisas, so 8.85 RS = 885 paisas
885 paisas is made up of 124 coins, either 10 paisa coins or 5 paisa coins
Let x = the number of 10 paisa coins and y =  the number of 5 paisa coins
then $124 = x + y$, so  $x = 124 - y$
885 paisas = the number of 10 paisa coins * 10 (what they are worth)  + the number of 5 paisa coins * 5 (what they are worth)
$
\begin{align}
885 &= 10x + 5y \\
 &= 10 (124 - y) + 5y  \quad\quad(\text{ substituting in }x = 124 - y )\\
 &= 1240 - 10y + 5y \\
 &= 1240 - 5y \\
-355 &= -5y \\
71 &= y
\end{align}
$
Since:  $124 - y = x$
$\begin{align}
124 - 71 &= x \\
53 &= x
\end{align}
$
So there are 53 10-paisa coins and 71 5-paisa coins.
Check:
$\begin{align}
10x + 5y &= 10(53) + 5(71) \\
 &= 530 + 355 \\
 &= 885 \quad \checkmark
\end{align}
$

REPLY [1 votes]: Let $P_{10}$ represent a 10 paisa coin and $P_5$ represent a 5 paisa coin.
Then the two equations you want to work with are:
$$\begin{align}
P_5+P_{10} &=124 \\
0.05P_5+0.10P_{10} &=8.85
\end{align}
$$
To solve by substitution:
$$\begin{align}
P_5 &=124-P_{10} \\
0.05(124-P_{10})+0.10P_{10} &=8.85 \\
6.2-0.05P_{10}+0.10P_{10} &=8.85 \\
0.05P_{10} &=2.65 \\
P_{10} &=53 \\
\end{align}
$$
Substitute to solve for $P_5 = 124-53 = 71$