Minimization with cos

1. Mar 12, 2015

dirk_mec1

I'm trying to find a function for x in [0, L] that minimizes this:

$$\int_0^{L} A \phi(x) \frac{ d \phi(x) }{dx} + B cos(\phi(x))\ d\mbox{x}$$

For real (given) positve numbers A and B.

with
$\phi(0) = 0$
$\phi(x)$ is an increasing positve function.

Can somebody point me in the right direction?

Last edited: Mar 12, 2015
2. Mar 12, 2015

jfizzix

Sounds like a calculus of variations problem.

I'm no expert, but the principal thing to do would be to find the Euler-Lagrange equation for $\phi(x)$. The solution (or solutions) to that equation will be the function which minimizes that integral.

I don't exactly know how to do this, but if you look at the Wikipedia article on the calculus of variations, and go down to the section on the Euler-Lagrange equation, you should find some useful tips there.

3. Mar 12, 2015

dirk_mec1

Yes I get the zero function then but that is not what I am looking for.

4. Mar 12, 2015

jfizzix

the Euler-Lagrange equation only has zero as a solution?

5. Mar 12, 2015

dirk_mec1

Yes I get the zero function then but that is not what I am looking for.

$$\frac { \partial L } { \partial \phi } = A \frac{d \phi}{dx} -C sin ( \phi(x))$$

$$\frac{d}{dx } \left( \frac { \partial L } { \partial \phi' } \right) = A \frac{d \phi}{dx}$$

$$\frac { \partial L } { \partial \phi } - \frac{d}{dx } \left( \frac { \partial L } { \partial \phi' } \right) = C sin ( \phi(x)) = 0$$

6. Mar 12, 2015

jfizzix

well that sure is interesting...

How do you know that the zero function isn't actually the solution you're looking for?

7. Mar 12, 2015

jfizzix

Also, since you're integrand doesn't depend explicitly on $x$, you can try the simplified Euler Lagrange equation instead:
$L-\frac{\partial L}{\partial \phi}\frac{d \phi}{d x} = const$.
Here, that would translate to
$A \phi \phi' + B Cos(\phi) -(A \phi' - B Sin(\phi))\phi'= const$
or rather
$A (\phi-\phi') \phi' + B \big(Cos(\phi) -\frac{d}{dx}( Cos(\phi)\big)= const$

I don't know how to solve this off the top of my head, but this equation probably admits more than zero as a solution.

8. Mar 12, 2015

dirk_mec1

Yes, you are right but I do not see how this is going to help me. This might be easier than an integral but this a non linear DE and I do not think an analytical solution exists.

9. Mar 12, 2015

pasmith

Since $\phi\phi' = \frac12 (\phi^2)'$ you are trying to minimize $$\frac{A\phi(L)^2}{2} + \int_0^L B\cos(\phi(x))\,dx.$$ The Euler-Lagrange equation is then $$B\sin(\phi) = 0$$ so that indeed $\phi(x) = 0$.

Do you have reason to believe that any other (differentiable) solutions exist? Also, although $\phi(x) = 0$ is not a strictly positive strictly increasing function, it is a positive increasing function.

(For weak solutions, you can take $$\phi(x) = \begin{cases} 0 & 0 \leq x < x_0 \\ (2n+1)\pi & x_0 \leq x \leq L \end{cases}$$ for some positive integer $n$ and if $0 < 2L - \frac{A(2n+1)^2\pi^2}{2B} < L$ then for any $0 < x_0 < 2L - \frac{A(2n+1)^2\pi^2}{2B}$ you have $$\frac{A\phi(L)^2}{2} + \int_0^L B \cos(\phi(x))\,dx = \frac{A(2n+1)^2\pi^2}{2} - B(L - x_0) < BL,$$ the last being the value attained when $\phi$ is zero.)

10. Mar 13, 2015

dirk_mec1

I made a mistake with the model. I will create a new thread.

Last edited: Mar 13, 2015