A perturbed function has a stationary point, while the unperturbed has not

Question

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)& = & 0\end{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 $x\to \infty$, 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

Answer 1

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.

Answer 2

For the stationary point, you obtain the equations $$ \frac{1}{x}\left(F'(y) + \frac{\epsilon}{x-y}\right) = 0,\\ \frac{1}{x^2}\left(F(y) + \epsilon\frac{y}{x-y}-\epsilon \log(x-y)\right) = 0. $$ 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 $$ \left(\frac{F'(y)}{\epsilon - y F'(y)}\right)^2\left[-F(y) + y F'(y) + \epsilon \log\left(-\frac{\epsilon}{F'(y)}\right)\right] = 0. $$ 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 $$ y = \hat{y} - \frac{\epsilon}{\hat{y} F''(\hat{y})} \log\left(-\frac{\epsilon}{F'(\hat{y})}\right). $$

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 $$ y = y_0 - \frac{\epsilon}{F''(y_0)} \exp\left(-\frac{F(y_0)}{\epsilon}\right), $$ so $$ x = \exp\left(\frac{F(y_0)}{\epsilon}\right) $$ to leading order. The latter confirms your idea that there is a solution of the unperturbed equation `for $x$ at infinity', in some sense.