$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$

**Definition 1.**
*Functional* is a map from some space of functions (or subset in the space of functions) $\mathsf{H}$ to $\mathbb{R}$ (or $\mathbb{C}$):
\begin{equation}
\Phi: \mathsf{H}\ni u\to \Phi[u]\in \mathbb{R}.
\label{eq-10.1.1}
\end{equation}

**Remark 1.**
Important that we consider a whole function as an argument, not its value at some particular point!

**Example 1.**

- On the space $C(I)$ of continuous functions on the closed interval $I$ consider functional $\Phi[u]=u(a)$ where $a\in I$ (value at the point);
- On $C(I)$ consider functionals $\Phi[u]=\max_{x\in I} u(x)$, $\Phi[u]=\min_{x\in I} u(x)$ and $\Phi[u]=\max_{x\in I} |u(x)|$, $\Phi[u]=\min_{x\in I} |u(x)|$;
- Consider $\Phi[u]=\int_I f(x) u(x)\,dx$ where $f(x)$ is some fixed function.
- On the space $C^1(I)$ of continuous and continuously differentiable functions on the closed interval $I$ consider functional $\Phi[u]=u'(a)$.

**Definition 2.**

- Sum of functionals $\Phi_1+\Phi_2$ is defined as $(\Phi_1+\Phi_2)[u]=\Phi_1[u]+\Phi_2[u]$;
- Product of functional by a number: $\lambda \Phi$ is defined as $(\lambda \Phi)[u]=\lambda (\Phi[u])$;
- Function of functionals: $F(\Phi_1,\ldots,\Phi_s)$ is defined as $F(\Phi_1,\ldots,\Phi_s)[u]=F (\Phi_1 [u],\ldots,\Phi_s[u])$.

**Definition 3.**
Functional $\Phi[u]$ is called *linear* if
\begin{gather}
\Phi [u+v]= \Phi[u]+\Phi [v],\label{eq-10.1.2}\\
\Phi [\lambda u]= \lambda \Phi[u] \label{eq-10.1.3}
\end{gather}
for all functions $u$ and scalars $\lambda$.

**Remark 2.**
Linear functionals will be crucial in the definition of *distributions* later.

**Exercise 1.**
Which functionals of Example 1. are linear?

We start from the classical variational problems: a single real valued function $q(t)$ of $t\in [t_0,t_1]$, then consider vector-valued function. This would lead us to ODEs (or their systems), and rightfully belongs to advanced ODE course.

Let us consider functional \begin{equation} S[q]= \int_{I} L(q(t),\dot{q}(t),t)\,dt \label{eq-10.1.4} \end{equation} where in traditions of Lagrangian mechanics we interpret $t\in I=[t_0,t_1]$ as a time, $q(t)$ as a coordinate, and $\dot{q}(t):=q'_t(t)$ as a velocity.

Let us consider $q+\delta q$ where $\delta q $ is a "small" function. We do not formalize this notion, just $\delta q=\varepsilon \varphi$ with fixed $\varphi$ and $\varepsilon\to 0$ is considered to be small. We call $\delta q$ *variation* of $q$ and important is that we change a function as a whole object. Let us consider

\begin{multline} \delta S:=S[q+\delta q]-S[q]= \int_I\Bigl(L(q+\delta q,\dot{q} + \delta \dot{q},t) -L(q,\dot{q},t)\Bigr)\,dt\\ \approx \int_I \Bigl(\frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot{q}}\delta \dot{q}\Bigr)\,dt\qquad \label{eq-10.1.5} \end{multline} where we calculated the linear part of expression in the parenthesis; if $\delta q=\varepsilon \varphi$ and all functions are sufficiently smooth then $\approx$ would mean "equal modulo $o(\varepsilon)$ as $\varepsilon\to 0$".

**Definition 4.**

- Function $L$ we call
*Lagrangian*. - The right-hand expression of (\ref{eq-10.1.5}) which is a linear functional with respect to $\delta q$ we call
*variation of functional $S$*.

**Assumption 1.**
All functions are sufficiently smooth.

Under this assumption, we can integrate the right-hand expression of (\ref{eq-10.1.5}) by parts: \begin{gather} \delta S:= \int_I \Bigl(\frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot{q}}\delta \dot{q}\Bigr)\,dt = \int_I\Bigl(\frac{\partial L}{\partial q}+ \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} \Bigr)\delta q \,dt + \Bigl( \frac{\partial L}{\partial \dot{q}} \Bigr)\delta q \Bigr|_{t=t_0}^{t=t_1}.\qquad \label{eq-10.1.6} \end{gather}

**Definition 5.**
If $\delta S=0$ for all *admissible variations* $\delta q$ we call $q$ a *stationary point* or *extremal* of functional $S$.

**Remark 3.**

- We consider $q$ as a point in the functional space.
- In this definition we did not specify which variations are admissible. Let us consider as admissible all variations which are $0$ at both ends of $I$: \begin{equation} \delta q (t_0)=\delta q(t_1)=0. \label{eq-10.1.7} \end{equation} We will consider different admissible variations later.

In this framework \begin{equation} \delta S= \int_I\Bigl(\frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} \Bigr)\delta q \,dt . \label{eq-10.1.8} \end{equation}

**Lemma 1.**
Let $f$ be a continuous function in $I$.
If $\int_I f(t)\varphi(t)\,dt=0$ for all $\varphi$ such that
$\varphi(t_0)=\varphi(t_1)=0$ then $f=0$ in $I$.

*Proof.* Indeed, let us assume that $f(\bar{t})> 0$ at some point
$\bar{t}\in I$ (case $f(\bar{t})< 0$ is analyzed in the same way). Then $f(t)>0$ in some vicinity $\mathcal{V}$ of $\bar{t}$. Consider function $\varphi(x)$ which is $0$ outside of $\mathcal{V}$, $\varphi\ge 0$ in $\mathcal{V}$ and $\varphi(\bar{t})>0$. Then $f(t)\varphi(t)$ has the same properties and $\int_I f(t)\varphi(t)\, dt>0$. Contradiction!

As a corollary we arrive to

**Theorem 2.**
Let us consider a functional (\ref{eq-10.1.4}) and consider as admissible all $\delta u$ satisfying (\ref{eq-10.1.7}). Then $u$ is a stationary point of $\Phi$ if and only if it satisfies *Euler-Lagrange equation*
\begin{equation}
\frac{\delta S}{\delta q}:=
\frac{\partial L}{\partial q} - \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}}\right) =0.
\label{eq-10.1.9}
\end{equation}

**Remark 4.**

Equation (\ref{eq-10.1.9}) is the 2nd order ODE. Indeed, \begin{gather*} \frac{d}{dt} \frac{\partial L}{\partial \dot{q}}= \left(\frac{\partial ^2L}{\partial \dot{q}\partial t} +\frac{\partial ^2L}{\partial \dot{q}\partial q}\dot{q} + \frac{\partial ^2L}{\partial \dot{q}^2}\ddot{q}\right). \end{gather*}

If $L_{q}=0$ then it is integrates to \begin{equation} \frac{\partial L}{\partial \dot{q}}=C. \label{eq-10.1.10} \end{equation}

The following equality holds: \begin{equation} \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}}\dot{q}-L\right)=-\frac{\partial L}{\partial t}. \label{eq-10.1.11} \end{equation} The proof will be provided for vector-valued $\mathbf{q}(t)$.

In particular, if $\frac{\partial L}{\partial t}=0$ ($L$ does not depend explicitly on $t$), then \begin{equation} \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}}\dot{q}-L\right)=0\implies \frac{\partial L}{\partial \dot{q}}\dot{q}-L=C. \label{eq-10.1.12} \end{equation}

**Definition 6.**
If $S[q]\ge S[q+\delta q]$ for all small admissible variations
$\delta q$ we call $q$ a *local maximum* of functional $S$. If
$S[q]\le S[q+\delta q]$ for all small admissible variations
$\delta q$ we call $q$ a *local minimum* of functional $S$.

Here again we do not specify what is *small admissible variation*.

**Theorem 3.**
If $q$ is a *local extremum* (that means either local minimum or maximum) of $S$ and variation exists, then $q$ is a stationary point.

*Proof.*
Consider case of minimum. Let $\delta q =\varepsilon \varphi$. Then $S[q+\delta q]- S [q]=\varepsilon (\delta S)(\varphi) +o(\varepsilon)$. If $\pm \delta S > 0$ then choosing
$\mp \varepsilon < 0$ we make
$\varepsilon (\delta S)(\varphi)\le -2\sigma \varepsilon$ with some $\sigma>0$. Meanwhile for sufficiently small $\varepsilon$ "$o(\varepsilon)$" is much smaller and $S [q+\delta q]- S [q]\le -\sigma \varepsilon<0$ and $q$ is not a local minimum.

**Remark 5.**

- We used that $\varepsilon$ can take both signs. If $\varepsilon $ can be only positive or negative, then conclusion is wrong.
- We consider neither sufficient conditions of extremums nor
*second variations*(similar to second differentials). In some cases they will be obvious.

**Example 2.**
Find the line of the fastest descent from fixed point $A$ to fixed point $B$.

*Solution.*
Assuming that the speed at $A(0,0)$ is $0$, we conclude that the speed at point $(x,y)$ is $\sqrt{2gy}$ and therefore the time of the descent from $A$ to $B(a, -h)$
\begin{gather*}
T[y]=\int_0^a \frac {\sqrt{1+y'^2}\,dx}{\sqrt{2gy}}.
\end{gather*}

This equation has a solution
\begin{align*}
x= &r\bigl(\varphi - \sin(\varphi)\bigr),\\
y=&r\bigl(1-\cos(\varphi))
\end{align*}
with $r=D/2$.

So, *brachistochrone is an inverted cycloid*-the curve traced by a point on a circle as it rolls along a straight line without slipping.