Jtdylktuy

I am looking for extrema of the function

$$ B(x,y) = frac{1}{x} Big[ F(y) - epsilon [log(x-y) +1] Big]$$

limited to the the domain $Omega = {y ge 0, x geq y} $

$F$ is a twice differentiable function such that $F(0) = 0$ and having one only stationary point (a maximum) for a $y_0$, $y_0 > 0$.

In general I would like a closed-form solution, but it seems rather unfeasible so I would settle for a characterisation for small $epsilon$.

If $epsilon = 0$, setting the partial derivatives to zero yields

begin{array}{lcl} -frac{1}{x^2}F(y) & = & 0 \ frac{1}{x} F^prime (y)& = & 0end{array}

and it can be concluded that no stationary point exists.
However, taking a point of the form $ (x, y_0) $, the partial derivatives $to 0$ for $xto infty$, and in this sense one could maybe state there is a stationary point at infinity.

On the other hand, for $epsilon neq 0$ a proper stationary point does exist.

I would like to have a characterisation of the stationary point for small $epsilon$, if not a closed form solution, an asymptotic description or so.

For example, if a function $Y(epsilon)$ were to be defined, such that it returns the $y$ coordinate of the stationary point for a certain value of the perturbation parameter $epsilon$,
a "Big O" description of the function $Y$ would be very interesting.

I thought something like perturbation theory could be of assistance, but the problem is that the unperturbed problem has not got a solution.
I tried to handle the stationary point "at infinity" with the coordinate transformation $z = frac{1}{x}$, but with little success.

Ths question is related to System of equations and perturbation methods, which regretfully contained multiple errors in its formulation.

Thanks

For the stationary point, you obtain the equations
$$
frac{1}{x}left(F'(y) + frac{epsilon}{x-y}right) = 0,\
frac{1}{x^2}left(F(y) + epsilonfrac{y}{x-y}-epsilon log(x-y)right) = 0.
$$
You can solve the first equation for $x$ to obtain
$$
x = y - frac{epsilon}{F'(y)}.
$$
Note that the condition $x geq y$ implies $F'(y)<0$. Substituting the above expression for $x$ in the second equation yields
$$
left(frac{F'(y)}{epsilon - y F'(y)}right)^2left[-F(y) + y F'(y) + epsilon logleft(-frac{epsilon}{F'(y)}right)right] = 0.
$$
Now, there are two distinct possibilities to obtain a solution to this equation for small $epsilon$.

First, suppose $F'(y)$ is $mathcal{O}(1)$ for small $epsilon$. Then, the term $epsilon log (-epsilon/F'(y))$ is small in $epsilon$, so the leading order equation to satisfy is $- F(y) + y F'(y) = 0$. Suppose we can find a value for $y := hat{y}$ such that this leading order equation is satisfied (note that, necessarily, $F(hat{y})<0$). Then, we have
$$
x = hat{y} - epsilon frac{1}{F'(hat{y})} + text{higher order terms}
$$
and
$$
y = hat{y} - frac{epsilon}{hat{y} F''(hat{y})} logleft(-frac{epsilon}{F'(hat{y})}right).
$$

However, given your comments, I suspect that the example functions $F$ that you tried numerically are such that we cannot find a solution to $- F(y) + y F'(y) = 0$. So, let's assume that this equation does not have a solution. In that case, we see that the term $epsilon log left(-frac{epsilon}{F'(y)}right)$ must be of the same order as $-F(y) + y F'(y)$. This means in particular that $F'(y)$ must be small. This inspires us to focus on a neighbourhood of the local maximum $y_0$, where $F'(y_0) = 0$. Writing $y = y_0 + eta$, we obtain to leading order (check this!) the equation
$$
-F(y_0) + epsilon log left(-frac{epsilon}{F''(y_0) eta}right) = 0
$$
(note that the argument of the logarithm is positive as $F''(y_0) < 0$ since $y_0$ is a local maximum, and $eta > 0$ since $F'(y) < 0$, i.e. we are at the right side of the local maximum), which we can solve to obtain to leading order
$$
y = y_0 - frac{epsilon}{F''(y_0)} expleft(-frac{F(y_0)}{epsilon}right),
$$
so
$$
x = expleft(frac{F(y_0)}{epsilon}right)
$$
to leading order. The latter confirms your idea that there is a solution of the unperturbed equation `for $x$ at infinity', in some sense.

You have $frac{dB}{dy}=frac{1}{x}(F'(y)+frac{epsilon}{x-y})$ So you can write $x=y-frac{epsilon}{F'(y)}$.

$frac{dB}{dx}=frac{-1}{x^2}(F(y)-epsilon (log{(x-y)}+1)) - frac{epsilon}{x(x-y)}=0$

You can substitute for $x$ in this and get an equation in just $y$ which you will probably have to solve numerically.

Some More Thoughts.

You can make $B$ as large as you like by taking $y$ close to $x$. If you track down a line parallel and close to $y=x$ then there is a maximum I think roughly at the maximum of $F(x)/x$ (??).

Also, consider being on the x-axis, ($y=0$). There is a minimum at $x=1$. Now consider being on a line close to the x-axis $y=y_1$. Then you can approximate for small $y_1$ and find a minimum at $x=exp({frac{F(y_1)}{epsilon})}$. (Note: $F(y_1)$ is small.) I know this is only a minimum for a given fixed $y$, but some further thought may lead you somewhere.