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In the catenary variational problem we have to find the curve that results from letting a weighted string with fixed length hang between two endpoints, and waiting until it minimizes its potential energy:

enter image description here

The constrained lagrangian for this problem is: $$L(y(x),y'(x))=(y(x) - \lambda)\cdot \sqrt{1+y'(x)^2}$$

What is wrong with my approach?

We have to solve the Euler-lagrange Equation, which is:

$$L_y=\frac d{dx}L_{y'}$$

So we find the derivatives of $L$: $$L_y=\sqrt{1+y'(x)^2}$$ $$L_{y'}= (y(x)-\lambda)\frac {y'(x)}{\sqrt{1+y'(x)^2}}$$ $$\frac d {dx} L_{y'}= \frac {y'(x)^2}{\sqrt{1+y'(x)^2}}+ y''(x)\cdot (y(x)-\lambda)\frac {1}{({1+y'(x)^2})^{3/2}}$$

This results in the Euler-Lagrange equation: $$\frac {y'(x)^2}{\sqrt{1+y'(x)^2}}+ y''(x)\cdot (y(x)-\lambda)\frac {1}{({1+y'(x)^2})^{3/2}}=\sqrt{1+y'(x)^2}$$

What surprised me is that according to this website, the Euler-lagrange equation looks very different:

enter image description here

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There is not even a $y''(x)$ in their Euler-Lagrange equation. What am I doing wrong?

ps. here is another question that I posted where I seem to be making a very similar mistake, so I must fundamentally misunderstand something about the calculus of variations and the Euler-Lagrange Equation.


EDIT: (based on @user121049's comment): I've tried doing what you said, and it gets me to the same result, so this solves my question. (But I'm still not sure about the final solution) :

$$y''\frac {(y-\lambda)} {1+y'^2}=1 \implies \int y'' \cdot \frac {y'}{1+y'^2}dx=\int \frac {y'}{y-\lambda}dx$$ Which implies $$\int \cdot \frac {y'}{1+y'^2}d(y')=\int \frac {1}{y-\lambda}dy\implies \ln|1+y'^2|=2 \ln(|y-\lambda|^2)+C\implies$$

$$1+y'^2=c_!\cdot(y-\lambda)^2$$

This Answers my question.

Now to solve it further: substitute $z(x)=y(x)-\lambda$ to get $z^2=c_1(1+z')\implies z'=\sqrt{\frac {z^2 -c_1}{c_1}}\implies c_1\int \frac 1 {\sqrt {z^2-a}}dz=x+c_2\implies \ln|\sqrt {z^2-c_1}+x|=\frac x c_1 +c_2\implies$ $$y(x)=z(x)+\lambda =\sqrt {\frac {c_2}{x^2}e^{\frac x {c_1} } +c_1}+\lambda$$

Is this correct? It doesn't seem like it describes the problem.

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  • $\begingroup$ They have already integrated once. Your equations look right. Multiple through by sqrt term to simplify. You can then rejig it to look like $y'' f(y')=y'/(y-\lambda)$. This you can integrate. $\endgroup$ – user121049 Aug 30 '17 at 19:39
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The clue is that your Lagrangian is independent of $x$. For any Lagrangian function $L\equiv L(\,y(x),y'(x)\,)$:

$$\frac{dL}{dx} \,=\, \frac{\partial L}{\partial y}\,y' + \frac{\partial L}{\partial y'} y'' \quad\Longrightarrow\quad \frac{\partial L}{\partial y}\,y' \,=\, \frac{dL}{dx} - \frac{\partial L}{\partial y'} y''. \qquad\qquad(*)$$

From the Euler-Lagrange equations:

$$\begin{aligned} \frac{\partial L}{\partial y} - \frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right) \;=\; 0\quad&\Longrightarrow \quad y'\frac{\partial L}{\partial y} - y'\frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right) \;=\; 0 \\[0.2cm] \quad&\Longrightarrow\quad \frac{dL}{dx} - \frac{\partial L}{\partial y'} y'' - y'\frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right) \;=\; 0 \\[0.2cm] \quad&\Longrightarrow\quad \frac{dL}{dx} - \frac{d}{dx}\left[y'\frac{\partial L}{\partial y'}\right] \;=\; 0\\[0.2cm] \quad &\Longrightarrow\quad \frac{d}{dx}\left[ L - y'\frac{\partial L}{\partial y'}\right] \;=\; 0 \end{aligned}$$

where on the second line we substituted in $(*)$. Integrating this, as @user121049 already mentioned, gives you

$$L - y'\frac{\partial L}{\partial y'} \;=\; a$$

for some constant $a$. This is the Beltrami identity; you should be able to use this derivation to help with your problem.

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  • $\begingroup$ What do you mean by $\frac{\partial L}{dy}$ in the Euler-Lagrange equation? shouldn't this be $\frac{\partial L}{\partial y'}$? $\endgroup$ – user56834 Aug 31 '17 at 5:08
  • $\begingroup$ To be clear, I actually have never before seen someone write a partial operator $\partial$ in the numerator but a $d$ operator in the denominator. Does this have a specific meaning or is it just a typo? $\endgroup$ – user56834 Aug 31 '17 at 5:44
  • $\begingroup$ Yes - you're right on both counts! The primes have been added as has $d\rightarrow\partial$, so thanks for the corrections. My math gets sloppy late at night! $\endgroup$ – AloneAndConfused Aug 31 '17 at 6:37
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    $\begingroup$ I now understand it, thank you :) $\endgroup$ – user56834 Aug 31 '17 at 7:39

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