# What am I doing wrong here? Trying to show functional solution gives catenary

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

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

$$1+(z’)^2-(z’)^2=0.$$

$$1=0.$$

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.

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.