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:
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:
**
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.
Best Answer
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.