Finding the minimum of an integral with Lagrange multipliers

[dRic2]

Homework Statement

I have to minimize the integral
$$\int_{\tau_1}^{\tau_2} y \sqrt{{x'}^2+{y'}^2} d \tau$$
with the isoperimetrical condition
$$\int_{\tau_1}^{\tau_2} \sqrt{{x'}^2+{y'}^2} = c$$
where $c$ is a constant.

Relevant Equations

Euler-Lagrange equations.

Using Lagrange multiplier $\lambda$ (only one is needed) the integral to minimize becomes
$$\int_{\tau_1}^{\tau_2} (y + \lambda) \sqrt{{x'}^2+{y'}^2} d \tau = \int_{\tau_1}^{\tau_2} F(x, x', y, y', \lambda, \tau) d\tau $$
Using E-L equations:
$$\frac {\partial F}{\partial x} - \frac d {d \tau} \frac {\partial F}{\partial x'} = 0 - \frac d {d \tau} \left( (y+\lambda) \frac {x'}{\sqrt{{x'}^2+{y'}^2}} \right) = 0$$
$$\frac {\partial F}{\partial y} - \frac d {d \tau} \frac {\partial F}{\partial y'} = \sqrt{{x'}^2+{y'}^2} - \frac d {d \tau} \left( (y+\lambda) \frac {y'}{\sqrt{{x'}^2+{y'}^2}} \right) = 0$$
$$\frac {\partial F}{\partial \lambda} - \frac d {d \tau} \frac {\partial F}{\partial \lambda'} = \sqrt{{x'}^2+{y'}^2} - 0 = 0$$
If I sum the equation for $y$ and $\lambda$ I get:
$$\frac d {d \tau} \left( (y+\lambda) \frac {y'}{\sqrt{{x'}^2+{y'}^2}} \right) = 0$$
Integrating the first equation I get:
$$(y+\lambda) \frac {x'}{\sqrt{{x'}^2+{y'}^2}} = \text{const.} \rightarrow y+\lambda = \frac {\sqrt{{x'}^2+{y'}^2}} {x'}$$
Putting together these last two equations I get:
$$\frac d {d \tau} \left( \frac {y'}{x'} \right) = \frac d {d \tau} \left( \frac {dy}{dx} \right) = 0$$
which is wrong!

I think I'm missing something fundamental because it is a pretty straightforward exercise.

 

[Orodruin]

For isoperimetric constraints, $\lambda$ is just a number, not a function to be varied with respect to.

I suggest you use the x-coordinate rather than path length as your curve parameter and apply the Beltrami identity.

 

[dRic2]

For isoperimetric constraints, λλ\lambda is just a number, not a function to be varied with emrespect to.

Dumb me... Thank you! ( btw I noticed that in my original post I wrote $x$ instead of $\lambda$... nice deduction!)

I suggest you use the x-coordinate rather than path length as your curve parameter and apply the Beltrami identity.

I just need to show that this variational problem is equal to the differential equation:
$$c' \frac d {d \tau} \left( \frac {dy}{dx} \right) = 1$$
where $c'$ is an other constant.
Putting the first equation into the second and integrating:
$$\text{const} \frac {y'}{x'} = \int_{\tau_1}^{\tau_2} \sqrt{{x'}^2+{y'}^2} d \tau = c$$

So, I'd say the two problems are equivalent if and only if $\frac {\tau_2 - \tau_1} {c'} = \frac {c} {\text{const}}$. What do you think ?