# Variational calculus, bending a stick and stationary states

We have a horizontal stick, one of its ends is on the wall, and we can apply a force to the other end. We assume that anything that we can do will leave this in the same plane. Our question is to understand when do we get the stationary states.

1. Actually I haven't quite understood the formulation, is the solution is the shape of a curve plus a force? How would you interpret this?

2. Then, one thing that I was told is that the stationary state is when the derivative of the potential energy of a system equals zero. Can you explain what kind of derivative should that be?

3. If we think of a stick as a graph of the function $u(x)$ then the potential energy in a point depends on a sign of the curvature in that point, so it depends on a square of the curvature which is $(\frac{u''}{(1+u'^2)^{3/2}})^2$. Then I was told that with some assumptions we can think that it's equivalent to $u''^2$, but when can we assume that and why? I don't see that too.

4. So the total potential energy was $\int \limits_0^1{(a(u'')^2-\lambda(u')^2)\ dx}$ where $\lambda$ is the applied force, and $a -$ some constant. I don't understand this too, why is $\lambda$ multiplied by the derivative, and also it should depend on the direction of a force, isn't that true?

Not meant to confuse you guys with stupid questions, it's really my first time with this kind of problems (and also the first time in here =) and I don't really know physics as well so I can't yet understand explanations even like those in wiki =(

I'll try to get you started. It would be helpful if you could post the text (if any) you're working with. Stationary state: net force is zero which means the restoring (bending) force of the stick equals the externally applied force, plus gravity if that's not to be ignored. I suspect they mean the derivative with respect to time is zero, as you'd hope for a system where nothing is moving. I don't know what the function $u(x)$ looks like, but apparently $u^{'2}$ can be shown to be small relative to 1, so the denominator is essentially equal to 1. You need to demonstrate this in your actual problem, either by expanding the denominator and evaluating the terms, or by other variational analysis. The potential energy depends on the amount of bend, not the direction, at least assuming the stick is uniform and isotropic, so the direction of bend doesn't enter into the integral. As to why each term is what it is, I recommend you dig back into the textbook or class notes.
• Thank you for an answer! Some things became more clear but why do we multiply the force by the square of $u'$? Unfortunately I was given this task on the practice, so have no textbook or notes on this problem, only lecture notes which are yet all about the mathematical basis And still don't understand why can we assume that $1+u'^2$ equals 1 if $u'$ is small but still not zero?.. don't get this idea :( Mar 10 '15 at 12:26