Extremal curves in the plane which make a line integral stationary

Question

The Question

---

Consider the extremal curves in the plane which make the line integral $$I=\int_P^Q \left(y^2dx+\frac{dy}{dx}dy\right)$$ stationary, where $P=(0,2)$ and $Q=(1,b)\ (b>0)$. To do so, first rewrite the integral in the standard form of the fundamental problem of variational calculus, involving $y(x)$. Then obtain the form of the curve(s).

---

Problem

I have been given the above question, I am posting on here to see if anyone can offer some advice as to where/ how to start with this problem, as I am clueless, Thanks in advance.

Answer 1

You can write the problem as

$$ I = \int_{P}^Q \left(y^2\frac{dx}{dt}dt + \frac{dy/dt}{dx/dt} \frac{dy}{dt}dt \right) = \int_P^Q \left(y^2x' + \frac{y'^2}{x'}\right)dt = \int_P^Q L(x,y,x',y')dt $$

Where $L$ is given by

$$ L = y^2x' + \frac{y'^2}{x'} $$

To solve this problem you can use the Euler-Lagrange equations:

\begin{eqnarray} \frac{d}{dt}\left(\frac{\partial L}{\partial y'}\right) - \frac{\partial L}{\partial y} &=& 0 \\ \frac{d}{dt}\left(\frac{\partial L}{\partial x'}\right) - \frac{\partial L}{\partial x} &=& 0 \\ \end{eqnarray}

For example, the first equation becomes

$$ \frac{d}{dt}\left(2\frac{y'}{x'}\right) - (2yx') = 0 $$

Rearranging we arrive to

$$ y(t) x'(t)^2+\frac{x''(t) y'(t)}{x'(t)}=y''(t) \tag{1} $$

Similarly for the $x$:

$$ x''(t) y'(t)-x'(t) y''(t)+y(t) x'(t)^3=0 \tag{2} $$

Looking at this, probably your best bet is to solve them numerically

Answer 2

$$I=\int_P^Q \left(y^2dx+\frac{dy}{dx} \frac{dy}{dx} dx\right);\quad L= y^2 +y ^{'2} $$ Apply Euler- Lagrage equation in Beltrami $x$ independence form

$$ y^2 +y ^{'2} - y^{'} (2 y^{'}) = const = a^2; \quad y^{'} = \sqrt{y^2-c^2} ;$$

$$ x = \int \frac{dy}{\sqrt{y^2-c^2}} = \cosh^{-1}\frac {y}{c} + h ;$$

$$ y = c \cosh (x-h) $$

which is a catenary equation with arbitrary constants $c,h$.

Say given $b=3$ we have two points of catenary $(0,2), (1,3)$

There are 2 equations to numerically solve for $(c,h)$

$$ 2= c \cosh (h), 3= c \cosh (1-h)$$