Graduate Maximization problem using Euler Lagrange

[petterson]

Hi,

I'm trying to solve the following problem

$\max_{f(x)} \int_{f^{-1}(0)}^0 (kx- \int_0^x f(u)du) f'(x) dx$.

I have only little experience with calculus of variations - the problem resembles something like

$I(x) = \int_0^1 F(t, x(t), x'(t),x''(t))dt$

but I don't know about the boundary $f^{-1}(0)$.

Is this there a way to solve this with a Euler Lagrange equation?

Thanks very much for your help!

 

[Stephen Tashi]

Hi,

I'm trying to solve the following problem

$\max_{f(x)} \int_{f^{-1}(0)}^0 (kx- \int_0^x f(u)du) f'(x) dx$.

If this comes from an applied math problem, is there a reason to think that integral has an upper bound?

Must a solution have $f^{-1}(0) < 0$ or do we also consider the case $f^{-1}(0) > 0$?

but I don't know about the boundary $f^{-1}(0)$.

Can you solve the problem with the constraint that $f^{-1}(0)$ is equal to a given constant $C$ ?

Let $f_C$ be a solution to Maximize $I_C(f) = \int_{f^{-1}(0)=C}^ 0 [ (kx - \int_0^x f(u)du) f'(x) ] dx$ then the set of solutions is a family of functions parameterized by $C$. You might be able to solve the further problem of maximizing $I_C(f_C)$ with respect to $C$.

 

[petterson]

If this comes from an applied math problem, is there a reason to think that integral has an upper bound?

Must a solution have $f^{-1}(0) < 0$ or do we also consider the case $f^{-1}(0) > 0$?

Can you solve the problem with the constraint that $f^{-1}(0)$ is equal to a given constant $C$ ?

Let $f_C$ be a solution to Maximize $I_C(f) = \int_{f^{-1}(0)=C}^ 0 [ (kx - \int_0^x f(u)du) f'(x) ] dx$ then the set of solutions is a family of functions parameterized by $C$. You might be able to solve the further problem of maximizing $I_C(f_C)$ with respect to $C$.

We consider only the case $f^{-1}(0) > 0$. Typically, the solution would be such that $f'(x) < 0$. The problem is also such that $k$ is always an upper bound on the integral.

If I choose $f^{-1}(0)$ equal to a given constant $C$ , I think I can find a stationary point with the Lagrange equation. It gives me $f'(x) = 0$, which I know should be part of a valid solution, but I can't find out anything about the constant $C$ then, because the integrand is zero.

 

[Stephen Tashi]

We consider only the case $f^{-1}(0) > 0$.

In that case, it's easier for me to visualize the problem as:

$I(f) = \int_0^{f^{-1}0} [ \int_0^x f(u)du - kx) f'(x)] dx$
Find $M =$max$_f \ I(f)$

Typically, the solution would be such that $f'(x) < 0$. The problem is also such that $k$ is always an upper bound on the integral.

Are we insisting on the constraint $f'(x) < 0$? $\$ If not, I don't understand why there is any upperbound on $I(f)$. To make $I(f)$ big, use a function $f$ that is big at $x = 0$ and increases more steeply than the graph of $y = kx$.

Edit: No, the above idea isn't correct. $f$ must eventually cross the positive x-axis in order for $f^{-1}(0)$ to be positive. So $f$ can't increase indefinitely.

 

[petterson]

In that case, it's easier for me to visualize the problem as:

$I(f) = \int_0^{f^{-1}0} [ \int_0^x f(u)du - kx) f'(x)] dx$
Find $M =$max$_f \ I(f)$

Are we insisting on the constraint $f'(x) < 0$? $\$ If not, I don't understand why there is any upperbound on $I(f)$. To make $I(f)$ big, use a function $f$ that is big at $x = 0$ and increases more steeply than the graph of $y = kx$.

Edit: No, the above idea isn't correct. $f$ must eventually cross the positive x-axis in order for $f^{-1}(0)$ to be positive. So $f$ can't increase indefinitely.

Sorry I wasn't quite clear with $f'(x)$ and the upper bound - from the setup of the problem it is a given that $k$ is an upper bound on the integral. Typically we find $f'(x) < 0$ (in related problems) - it would be nicer not to impose it as a constraints but we may easily do so if it makes it mathematically more tractable.
It would probably be fine to write the problem as
$I(f) = \int_0^{f^{-1}0} [ kx - \int_0^x f(u)du )|f'(x)|] dx$
Find $M =$max$_f \ I(f)$

 

[Stephen Tashi]

from the setup of the problem it is a given that $k$ is an upper bound on the integral.

Do you mean $I(f) \le k$? - or is k a bound for $\int_0^x f(u)du$ ?

It would probably be fine to write the problem as
$I(f) = \int_0^{f^{-1}0} [ kx - \int_0^x f(u)du )|f'(x)|] dx$

In that case, do we use the constraint $f(x) \ge 0$?

 

[petterson]

Do you mean $I(f) \le k$? - or is k a bound for $\int_0^x f(u)du$ ?

In that case, do we use the constraint $f(x) \ge 0$?

Yes I mean $I(f) \le k$. In the original problem $x$ is between 0 and 1. The boundary $f^{-1}(0)$ comes from a substitution. We also restrict to $f(x) \ge 0$. I'm sorry for all the confusion (I thought there might be more of a standard technique that could help me) . Below is the original problem with all constraints :

Assuming $x \in [0,1], a \in [0,1], k > 0$

$$\max_{f} I(f) = \int_{0}^{f(0)} (k f^{-1}(a) - \int_0^{f^{-1}(a)} f(t)dt) da \\

\text{s.t. } f(x) \geq 0, f'(x) \leq 0$$

 

[Stephen Tashi]

I don't see how to formulate that problem as a typical Calculus Of Variations problem.

I have some vague thoughts about approaching it in the spirit of the Calculus Of Variations (- or perhaps "dynamic programming").

Suppose we only consider functions that are piecewise linear. Let [0,1] on the x-axis be partition into given intervals so that on $[x_i, x_{i+1}]$ , $f(x) = A_i x + B_i$. Assume $f_m(x)$ is a particular function that maximizes $I(f)$. Can we develop simultaneous equations that would allow us to solve for the $f_m(x_i)$ ?

We get some equations by requiring that each linear segment of the graph of $f_m$ meets the next linear segment at a common point: $A_i x_{i+1} + B_i = A_{i+1} x_{i+1} + B_{i+1}$. ( Each of the $A_i$ and $B_i$ can be expressed in terms of values of $f_m$.)

It seems possible to get other equations by using the idea that the value of $f_m$ at the point $x_{i+1}$ must be the optimum value given we knew the values of $f_m$ at all other $x_j$.

Momentarily take the viewpoint that the $f(x_j)$ are known except that we don't know $f( x_{i+1})$ (i.e. We don't know that to do about $f_m$ between $(x_i,f(x_i))$ and $(x_{i+2} f( x_{i+2})$ ). Solve the problem of maximizing $I(f_m)$ with respect to chosing $f_m(x_{i+1})$.
We can consider $I(f)$ as the difference of two integrals. They will be evaluated as finite sums.

Pretending we know ordered pairs $(x_j, f_m(x_j))$ of $f_m$ implies we also know ordered pairs $(f_m(x_j),x_j)$ of $f_m^{-1}$.

For $I_1(f) = \int_0^{f(0}{k f^{-1}(a)} da$ , the formula for $I_1(f)$ will be a summation of areas of trapezoids. We are pretending these areas are known except for the two trapezoids with common vertex $(x_{i+1}, f_m(x_{i+1}))$

For $I_2(f) = \int_0^{f(0)}\ [ \int_0^{f^{-1}(a)} f(t)dt ]\ da$ , thinking about it makes my head spin!

My intuition is that $I_2(f)$ can be expressed as a function of the unknown $f_m(x_{i+1})$ and the other known $f_m(x_j)$ in an explicit manner.

Assuming it can, solving the 1-variable optimization problem

Max$_{f_m(x_i+1)} (I_1(f_m) - I_2(f_m))$

should give an equation relating $f_m(x_{i+1})$ to the other values of $f_m$.

 

[petterson]

Thanks very much for this idea, Stephen - I'll have to think about this for a little bit.