# A Function Optimization Problem

1. Jun 26, 2015

### saminator910

Can anyone tell me straightforward information about a way to maximize a certain functional $I[f]=\displaystyle\int_{X} L(f,x)dx$ such that the integral is bounded, $T≥\displaystyle\int_{X}f(x)h(x)dx$. I really know a minimal amount about functional analysis and calculus of variations, but I've looked at things like Hamiltonians and I don't know if they apply to this problem. Intuitively I see this as a problem that would be solved with Lagrange multipliers if it we weren't talking about functions and functionals, ie $\vec{f}\cdot \vec{h}=T$ is a linear constraint, $∇I(x_{1},...,x_{n})=\lambda \vec{h}$. We would then solve systems of equations $\lambda h_{i}=\frac{\partial I}{\partial x_{i}}$. Any help is greatly appreciated.

2. Jun 27, 2015

### Orodruin

Staff Emeritus
How would you find the minimum of any (continuous and differentiable) function f(x) subject to the condition $x^2 \leq 1$? Can you do something similar for functionals?

3. Jun 27, 2015

### saminator910

My guess would be a functional derivative equal to 0, but all the problems I've looked at don't have a constraint.

4. Jun 27, 2015

### Orodruin

Staff Emeritus
Well, for a function you would put the derivative equal to zero. When considering functionals you would put the functional derivative equal to zero. The question is what you would do with the constraints? Is the derivative equal to zero anywhere if the function is $f(x) = x$? Where is the corresponding minumum?

5. Jun 27, 2015

### saminator910

So, the extreme values are at the boundary?

6. Jun 27, 2015

### Orodruin

Staff Emeritus
Not necessarily, but they might be. Hence, you need to minimise the functional both in the interior as well as on the boundary. The minimum value is the minimal value of these.

7. Jun 27, 2015

### saminator910

Ok, so from what I've read I set $\frac{\delta I}{\delta f(x)}=\frac{\partial L}{\partial f}=0$, the derivative is in this specific form because $L$ is not a function of $f'$. Mathematically I don't know how to incorporate the constraint. So, if I don't arrive at a max that satisfies the constraint, then I check the boundary cases?

8. Jun 27, 2015

### Orodruin

Staff Emeritus
You need to check the boundary in all cases. Even if you find a maximum inside the region, it might be a local maximum. You need to check that any maximum you find (a) is inside the region and (b) is a maximum and not another kind of extremum. In addition, you need to check the boundary of the region of functions which fulfill your criterion. The boundary is given by "integral = T". Do you know how to use Lagrange multipliers? The problem constrained to the boundary is exactly equivalent to extremisation with constraints in a finite dimensional vector space.

9. Jun 27, 2015

### Orodruin

Staff Emeritus
You need to check the boundary in all cases. Even if you find a maximum inside the region, it might be a local maximum. You need to check that any maximum you find (a) is inside the region and (b) is a maximum and not another kind of extremum. In addition, you need to check the boundary of the region of functions which fulfill your criterion. The boundary is given by "integral = T". Do you know how to use Lagrange multipliers? The problem constrained to the boundary is exactly equivalent to extremisation with constraints in a finite dimensional vector space.

10. Jun 27, 2015

### saminator910

Yeah, take a look at the bottom of my original question, I thought about using Lagrange multipliers, I just have no idea about how to get rid of the lambda? once I set the functional derivative of the constraint equal to the functional derivative of the boundary multiplied by lambda, and get a solution in terms of lambda, what do I do next?

11. Jun 27, 2015

### Orodruin

Staff Emeritus
Just the same as you would do in a finite vector space. You fix lambda such that the constraint is satisfied.

It strikes me now that you may not have to check the interior separately. You will find a lambda dependent value and find the maximum value of the functional such that the constraint is fulfilled. In other words, you will have reduced the problem to maximising a function of lambda subject to some constraints.

12. Jun 27, 2015

### saminator910

Ok, that makes sense. Now what I'm realizing is my functionals and constraint are such that the functional for the region will be maximized on the boundary.

13. Jun 27, 2015

### Orodruin

Staff Emeritus
This actually does not matter much. The method just described will reduce the problem to maximising a function of lambda subject to constraints, and we already know how to do that for functions.

14. Jun 27, 2015

### saminator910

So, I have an expression $0=A (f,x,\lambda)$ after setting $\frac{\delta I}{\delta f}=\lambda \frac{\delta J}{\delta f}$ (where J is the functional that is set equal to constraint T), so I maximize what w.r.t. what? I really appreciate the help, I'm back from college for the summer so I don't have easy access to teachers.

edit:

Ok, so I have found $f(x,\lambda)$, so now I just find $f$ such that $T=\displaystyle \int_{X} f(x,\lambda)h(x)dx$

Last edited: Jun 27, 2015
15. Jun 28, 2015

### Orodruin

Staff Emeritus
Oncle you have f(x,lambda), you can insert it into the functionals to define $I(\lambda) = I[f(*,\lambda)]$ and $J(\lambda) = J[f(*,\lambda)]$. Your problem is now to maximise the function $I(\lambda)$ under the constraint $J(\lambda) \leq T$.