I am trying to minimise the functional

$$V[z]= \int_{-a}^{a} z \sqrt{1+(z’)^2}dx.$$

Note $f(z, z’, x)= z \sqrt{1+(z’)^2} $ hasn’t got any dependence on $x$, so

$$f-z’ \frac{\partial f}{\partial z’}=0$$

at the extremum.

$$\frac{\partial f}{\partial z’}=z \frac{1}{2} (1+(z’)^2)^{-1/2} 2z’$$

So the equation becomes

$$z \left( \sqrt{1+(z’)^2} - (z’)^2/\sqrt{1+(z’)^2} \right)=0.$$



I’ve obviously got to be doing something wrong here, but I cannot figure out what.

I’m expecting a solution of the form $z-z_0=B \cosh (x/B)$ since the functional describes the potential energy of a uniform rope in a gravitational field.


1 Answer 1


Your reduction of the E-L equation is incorrect. The difference should be constant, but that constant need not be zero.

$$ f - z'\partial_{z'} f = c $$

Substitute the given values

$$ z\sqrt{1+z'^2} - zz'^2/\sqrt{1+z'^2} = c$$

Rearrange to get

$$ z = c\sqrt{1+z'^2} $$

This looks like it will give you the $\cosh$ you're looking for.


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