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

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

• maybe try rewriting in $u,v=1/x,1/y$ and then the optimal $u,v$ might be $O(\epsilon)$ – user619894 Jan 15 '19 at 12:17

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.

• Dear Frits, thanks a lot for this. In the last days i luckily came to follow exactly the same path, but of course i did know already, as you cleverly supposed, that the equation $-F+yF^\prime=0$ has no solutions. Thank you ever so much, great support. – An aedonist Jan 17 '19 at 9:50

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.

• thanks for your answer. I have already tried your suggestion, but it is not a numerical solution i sm looking for ( i have computed it already anyhow, that is actually what made me sure one only ststionary point exists). Thanks anyhow – An aedonist Jan 8 '19 at 18:52
• many thanks for your edits. I will ponder on the first part: the second bit i have already thought about in the past. Albeit intetesting in itself, the focus os on the stationary point, which for small $\epsilon$ does not occur for small $y$ – An aedonist Jan 9 '19 at 15:37