# Calculus of variations question

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1. Jan 13, 2016

### phi1123

Okay, so I've run into a rather weird functional that I am trying to optimize using calculus of variations. It is a functional of three functions of a single variable, with a constraint, but I can't figure out how to set up the Euler-Lagrange equation. The functional in question is (sorry it's kind of messy):
$$J\left[\phi,y,z\right]=\int_0^b\left( \int^b_0\left( \dot{\phi}(t)\frac{(y(s)-y(t))\dot{y}(s)+(z(s)-z(t))\dot{z}(s)}{(y(s)-y(t))^2+(z(s)-z(t))^2} \right)dt + \Lambda\left(a\phi(s)\dot{y}(s)-c\right) \right)ds$$
I'm wondering as to what exactly the best way to deal with the inner integral is. I've considered pulling out the inner integral, and rewriting the whole thing as:
$$J\left[\phi,y,z\right]=\int_0^b \int^b_0\left( \dot{\phi}(t)\frac{(y(s)-y(t))\dot{y}(s)+(z(s)-z(t))\dot{z}(s)}{(y(s)-y(t))^2+(z(s)-z(t))^2} + \Lambda\left(\frac{a}{b}\phi(s)\dot{y}(s)-\frac{c}{b}\right) \right)dsdt$$
and then taking the integrand as my Lagrangian in the Euler-Lagrange equation, but then I seem to run into trouble with having to different independent variables as the argument to the functions $\phi, y,z$. Anyone have any experience with similar problems? Thanks for your help!

2. Jan 15, 2016

### Orodruin

Staff Emeritus
You cannot use the Euler-Lagrange equation because it assumes that the variable is only integrated over a single time. Instead, you may want to review how the EL equation is derived and see if you can do something similar.

3. Jan 20, 2016

### phi1123

So I've been thinking about this problem some more (sorry I didn't reply earlier). Following your suggestion, lets assume only $\phi$ is varied for simplicity. In that case we want to minimize a functional of the form
$$J[\phi]=\int^b_0\int^b_0L(\phi(s),\phi(t),\dot{\phi}(s),\dot{\phi}(t))dsdt$$
Then setting $\frac{d}{d\epsilon}J[\phi+\epsilon\eta]=0$ at $\epsilon=0$, we get
$$\frac{d}{d\epsilon}J[\phi+\epsilon\eta]=0=\int^b_0\int^b_0 \left( \frac{\partial L}{\partial\phi(s)}\eta(s) + \frac{\partial L}{\partial\phi(t)}\eta(t) +\frac{\partial L}{\partial\dot{\phi}(s)}\dot{\eta}(s) + \frac{\partial L}{\partial\dot{\phi}(t)}\dot{\eta}(t) \right)dsdt$$
Separating out the s & t terms, and doing integration by parts
$$0=\int^b_0\int^b_0 \left( \frac{\partial L}{\partial\phi(s)} - \frac{\partial}{\partial s}\left( \frac{\partial L}{\partial\dot{\phi}(s)} \right) \right)\eta(s) dsdt + \int^b_0\int^b_0 \left( \frac{\partial L}{\partial\phi(t)} - \frac{\partial}{\partial t}\left( \frac{\partial L}{\partial\dot{\phi}(t)} \right) \right)\eta(t) dsdt$$
Therefore the differential equations describing the function that maximizes J will be the two equations:
$$\frac{\partial L}{\partial\phi(s)} - \frac{\partial}{\partial s}\left( \frac{\partial L}{\partial\dot{\phi}(s)} \right)=0$$
$$\frac{\partial L}{\partial\phi(t)} - \frac{\partial}{\partial t}\left( \frac{\partial L}{\partial\dot{\phi}(t)} \right)=0$$
I can't think of any way to simplify this further, and I could very well have made some silly mistake, so if you have any thoughts, they'd be appreciated!

One thing that's bugging me about the above is that it doesn't seem to be "symmetrical" with respect to the 2 variables s & t when I apply it to my particular example. I guess this could be that my functional is wrong, but I was kind of expecting the symmetry to fall out of the derivation, since $\phi(s)$ and $\phi(t)$ are really just the same function by a different name.

4. Jan 20, 2016

### Orodruin

Staff Emeritus
I agree up to this point. There is really no need for the integrals to be zero separately (although that would provide a solution). I am also not sure how to continue with this line of argumentation.

5. Jan 21, 2016

### phi1123

Ah, I see your point. Okay suppose we interchange the variables s and t in the second integral, and then exchange the order of integration (not sure if this is justified). Then we can combine the integrals and factor out the $\eta(s)$ such that
$$0=\int\int\left( \frac{\partial}{\partial \phi(s)}\left(L(s,t)+L(t,s)\right) - \frac{\partial}{\partial s}\frac{\partial}{\partial \dot{\phi}(s)}\left(L(s,t)+L(t,s)\right) \right)\eta(s)dsdt$$
And then the integrand would definitely equal 0, so
$$\frac{\partial}{\partial \phi(s)}\left(L(s,t)+L(t,s)\right) - \frac{\partial}{\partial s}\frac{\partial}{\partial \dot{\phi}(s)}\left(L(s,t)+L(t,s)\right)=0$$

(by L(s,t) I mean $L\left(\phi(s),\phi(t),\dot{\phi}(s),\dot{\phi}(t)\right)$, and L(t,s) I mean $L\left(\phi(t),\phi(s),\dot{\phi}(t),\dot{\phi}(s)\right)$, same Lagrangian, but with the positions of the $\phi(s)$ and $\phi(t)$'s switched)

6. Jan 22, 2016

### Orodruin

Staff Emeritus
I do not see anything immediately wrong with doing that. On the other hand I have not thought about it much.