Find the infimum of $J(\phi) = \int_{\phi}^{} e^x dl$ Dear members of community! I have a trouble, with my attempts to understand provided in solution manual for such task.
Task:
So let the standard Cartesian coordinate system $(x,y)$ be given on the plane $\mathbb{R}^2$. Let's assume that $$dl \overset{\operatorname{def}}{=} \sqrt{(dx)^2 + (dy)^2}.$$
Let $\Omega$ the set of $C^{(1)}$ smooth paths on plane $\mathbb{R}^2$ with initial point $O(0;0)$ and ending at a point $A(2;4)$.
Consider a functional $J: \Omega \rightarrow \mathbb{R}$ such that
$$
J(\phi) = \int_{\phi}^{} e^x dl, \quad \forall \phi \in \Omega.
$$
Find the $\underset{\phi \in \Omega} \inf J(\phi)$.
Provided solution: (manual solution)
The Euler-Lagrange equation for this functional is given by:
$$
\frac{d}{dt} \frac{\partial L}{\partial \phi'(t)} - \frac{\partial L}{\partial \phi(t)} = 0,
$$
where $L(\phi(t),\phi'(t)) = e^{\phi(t)} \sqrt{\phi'(t)^2 + 1}$
Solving the Euler-Lagrange equation subject to the boundary conditions $\phi(0) = 0$ and $\phi(1) = 2$ will give us the path $\phi(t) = 2t$ that minimizes the functional $J(\phi)$.
After we get:
$$
\underset{\phi \in \Omega} \inf J(\phi) = \int_{0}^{1} e^{2t} \sqrt{2^2 + 1} dt = e^2 \sqrt{5}.
$$
My trouble:
And to be honest, I did not quite understand how the Euler-Lagrange equations are essentially solved in this case, i.e. the minimality of this functional is proved, and why the Lagrangian is written in this case in this way $L(\phi(t),\phi'(t)) = e^{\phi(t)} \sqrt{\phi'(t)^2 + 1}$. Is it possible to explain in more detail how the minimality of the functional is proved in this case, because the lower bound is then found?
UPD:
Having returned to this problem 5 times already, I still understand that this solution is more than strange and not entirely correct. Therefore, after reading one more book on the calculus of variations, I came to the conclusion that we should try again to do the same steps using the parametrization of the curve.
My attempt solution:
First, I need to express $J(\phi)$ in terms of the parameterization of $\phi$. Let $\phi(t) = (x(t),y(t))$ be a parameterization of a path $\phi$ with $t \in [0,1]$ such that $\phi(0) = O$ and $\phi(1) = A$. Then, we have
\begin{align*}
J(\phi) &= \int_{\phi} e^x dl \
= \int_0^1 e^{x(t)} \sqrt{(dx/dt)^2 + (dy/dt)^2} dt \
= \int_0^1 e^{x(t)} \sqrt{x'(t)^2 + y'(t)^2} dt.
\end{align*}
Next, we write the Euler-Lagrange equation for $J(\phi)$. The Euler-Lagrange equation for a functional of the form $J(\phi) = \int_a^b L(\phi,\phi',t) dt$ is
\begin{align*}
\frac{d}{dt} \frac{\partial L}{\partial \phi'} - \frac{\partial L}{\partial \phi} &= 0,
\end{align*}
where $L(\phi,\phi',t)$ is the Lagrangian, and $\phi'$ denotes the derivative of $\phi$ with respect to $t$.
For our functional $J(\phi)$, we have $L(\phi,\phi',t) = e^{x(t)} \sqrt{x'(t)^2 + y'(t)^2}$, and so the Euler-Lagrange equation becomes
\begin{align*}
\frac{d}{dt} \frac{\partial L}{\partial \phi'} - \frac{\partial L}{\partial \phi} &= 0 \
\frac{d}{dt} \left( \frac{e^{x(t)} x'(t)}{\sqrt{x'(t)^2 + y'(t)^2}} \right) - e^{x(t)} \frac{y'(t)^2}{\sqrt{x'(t)^2 + y'(t)^2}^3} = 0.
\end{align*}
After simplify this equation by multiplying both sides by $\sqrt{x'(t)^2 + y'(t)^2}^3$ and using the chain rule to get
\begin{align*}
\frac{d}{dt} \left( e^{x(t)} x'(t) \sqrt{x'(t)^2 + y'(t)^2}^2 \right) - e^{x(t)} x'(t)^2 - e^{x(t)} y'(t)^2 &= 0 \\
\frac{d}{dt} \left( e^{x(t)} x'(t) \sqrt{x'(t)^2 + y'(t)^2}^2 \right) &= e^{x(t)} \left( x'(t)^2 + y'(t)^2 \right).
\end{align*}
(use Maple with subs for evaluate and check correct equation)
Integrate both sides of the equation with respect to $t$ from $0$ to $1$ and use the boundary conditions $\phi(0) = O$ and $\phi(1) = A$ to obtain
\begin{align*}
&\left. e^{x(t)} x'(t) \sqrt{x'(t)^2 + y'(t)^2}^2 \right|_{0}^1 = \int_0^1 e^{x(t)} \left( x'(t)^2 + y'(t)^2 \right) dt \\
\Rightarrow \quad & e^{1} 2^2 - e^{0} 0^2 = \int_0^1 e^{x(t)} \left( x'(t)^2 + y'(t)^2 \right) dt \\
\Rightarrow \quad & 4e = \int_0^1 e^{x(t)} \left( x'(t)^2 + y'(t)^2 \right) dt.
\end{align*}
We see that the integral on the right-hand side of the equation is the same as $J(\phi)$, up to a constant factor. Therefore, to find the infimum of $J(\phi)$ over all $\phi \in \Omega$, it suffices to find the infimum of the integral $\int_0^1 e^{x(t)} \left( x'(t)^2 + y'(t)^2 \right) dt$ over all $C^{(1)}$ smooth paths $\phi$ with initial point $O$ and ending point $A$.
Note that the integrand $e^{x(t)} \left( x'(t)^2 + y'(t)^2 \right)$ is non-negative, and so the infimum is achieved when the integrand is zero, that is, when $x'(t) = y'(t) = 0$ for all $t$. In other words, the infimum is achieved when the path $\phi$ is constant and equal to the point $(2,4)$. Therefore, the infimum of $J(\phi)$ over all $\phi \in \Omega$ is
\begin{align*}
\inf_{\phi \in \Omega} J(\phi) &= \inf_{\phi \in \Omega} \int_{\phi} e^x dl \ = \inf_{\phi \in \Omega} \int_0^1 e^{x(t)} \sqrt{x'(t)^2 + y'(t)^2} dt = \inf_{\phi \in \Omega} \int_0^1 e^{x(t)} \cdot 0 dt = 0,
\end{align*}
where the infimum is taken over all $C^{(1)}$ smooth paths $\phi$ with initial point $O$ and ending point $A$.
$0$ and the previously received answer differ very much, in the end, if you follow the rules for searching for functionals, then you still get $0$? Right? Or am I wrong somewhere?
 A: *

*OP's functional is
$$ J[x,y] ~:=~\int_{(0,0)}^{(2,4)}\!\mathrm{d}t~L, \qquad
L~=~e^x\sqrt{\dot{x}^2+\dot{y}^2},\tag{A}$$
where dot means differentiation wrt. $t$.


*$J$ is the arclength for the metric tensor
$$\mathrm{d}s^2~=~e^{2x}\left(\mathrm{d}x^2+ \mathrm{d}y^2\right),
\tag{B}$$
which in physics is known as a Liouville wall. Intuitively, it is more and more expensive in terms of arclength to move further and further into the positive $x$-plane. So the problem is to determine the geodesics.


*Note that the functional $J$ is worldline reparametrization invariant. Assuming the solution $t\mapsto y(t)$ is injective, we may assume an
analogue of the static gauge
$$y(t)~=~t.\tag{C}$$


*Then  OP's functional becomes
$$ J[x] ~:=~\int_0^4\mathrm{d}t~L, \qquad
L~=~e^x\sqrt{\dot{x}^2+1},\tag{D}$$
with Dirichlet boundary conditions (DBC)
$$ x(0)~=~0, \qquad x(4)~=~2.\tag{E}$$


*Since $t$ does not appear explicitly in the Lagrangian $L$, we know
from Noether's theorem
that the energy
$$ -e^{x_0}~:=~E~:=~\dot{x}\frac{\partial
L}{\partial\dot{x}}-L~=~-\frac{e^x}{\sqrt{\dot{x}^2+1}}\tag{F}$$
is a negative constant.


*This leads to a 1st order ODE:
$$ \dot{x}~=~\pm\sqrt{e^{2(x-x_0)}-1},\qquad x~\geq~x_0.\tag{G}$$
Keeping in mind the static gauge (C), the solutions are of the form
$$ \pm(y-y_0)~=~\arctan\sqrt{e^{2(x-x_0)}-1},\tag{H}$$
or
$$ \cos^{-2}(y-y_0)~=~\tan^2(y-y_0)+1~=~e^{2(x-x_0)},\tag{I}$$
so that the geodesics are of the form
$$ x(t)-x_0~=~-\ln|\cos(y-y_0)|,\tag{J}$$
see e.g. Fig. 1.

$\uparrow$ Fig. 1. A typical geodesic in the $xy$ plane.


*We conclude from eq. (J) that the $y$-interval length for a geodesic must be less than $\pi$. But the DBC calls for length=$4$. So OP's endpoints $O$ and $A$ are not connected via a geodesic.


*By continuity a minimal curve $O\to A$ must pass through a point $C(x_1,2)$ for some $x_1\in[-\infty,\infty[$. Assume that $x_1>-\infty$ is finite. Then the 2 curves $O\to C$ and $C\to A$ have $y$-interval length$=2$ [which is less than $\pi$ so that they are geodesics] but it is more than $\pi/2$. We conclude from eq. (J) that $\dot{x}=0$ at some internal points, say $D$ and $B$ on each of the 2 geodesic branches $O\to C$ and $C\to A$, i.e. the $x$-coordinates at $B$ and $D$ are smaller than at $C$. We can lower the arclength by straightning the curve $D\to B$ [and in this way move further away from the Liouville wall] i.e. lowering $x_1$. We leave it to the reader to turn this handwaving argument into a proof that $x_1=-\infty$.


*This suggests that a minimal curve would consist of piecewise straight lines
$$ (0,0) \to (-\infty,0) \to (-\infty,4) \to (2,4), \tag{K}$$
corresponding to the arclength
$$ J[x,y]~=~\int_{-\infty}^0\mathrm{d}x~e^x + 0 +\int_{-\infty}^2\mathrm{d}x~e^x~=~1+e^2. \tag{L}$$
