Jtdylktuy

How can we find some radius of circle with center at origin which contains $xarctan(x)-ax+yarctan(y)-by=0$,

where $pi/2>a>0$ and $pi/2>b>0$.

I'm not sure how can we prove that these inequalities should hold so we can have closed curve: $pi/2>a$ and $pi/2>b$ .

Also very interesting similar equation $xarctan(x)-ax+(-x+y)arctan(-x+y)-b(-x+y)+yarctan(y)-cy=0$.

I need some approximate estimation so I can prove that for some radius this estimation will be correct.

For example I've found such a circle for $a=1.5$ and $b=1.5$:

And for $xarctan(x)-1.5x+(-x+y)arctan(-x+y)-1.5(-x+y)+yarctan(y)-1.5y=0$:

Maybe Lagrange multipliers can help?
I'm also interested in higher dimensions where we can add $z$ and find some radius of a sphere which fully contains $xarctan(x)-ax+yarctan(y)-by+zarctan(z)-cz=0$. But I think it can be done in similar way as for two dimensions.

Definition. Given $ainmathbb R$, define the function $f_a(x)=x(arctan x-a)$.

Definition. Given $a,binmathbb R$, consider the closed set
$$
begin{split}
E_{a,b} &= {(x,y)inmathbb R^2 : x(arctan x-a)+y(arctan y-b)=0} \
&= {(x,y)inmathbb R^2 : f_a(x)+f_b(y)=0}
end{split}
$$

Lemma 1. For every $ainmathbb R$, the set $[0,infty)$ is in the image of $f_a$.

Proof. Assume $ageq0$. Then $f_a(0)=0$ and
$$
lim_{xto-infty}x(arctan x-a)=(-pi/2-a) lim_{xto-infty}x=infty.
$$
Since $f_a$ is continous, these limits imply the thesis. The case $aleq0$ is analogous, with the signs changed. □

Lemma 2. If $|a|>pi/2$, then the function $f_a:mathbb Rtomathbb R$ is surjective.

Proof. Assume $a>pi/2$. We have $pmpi/2-a < 0$, so
$$
lim_{xto+infty} x(arctan x-a)
= (pi/2-a)lim_{xtoinfty} x = -infty
$$
and
$$
lim_{xto-infty} x(arctan x-a)
= (-pi/2-a)lim_{xto-infty} x = infty.
$$
Since $f_a$ is continuous, these limits imply that it is surjective. The case $a<-pi/2$ is analogous. □

Lemma 3. If $|a|<pi/2$, then
$$
lim_{xtopminfty} f_a(x) = infty.
$$

Proof. A direct computation shows
$$
lim_{xtoinfty} f_a(x) = (pi/2-a)lim_{xtoinfty}x = infty
$$
and the same holds for $xto-infty$. □

Proposition. If $|a|>pi/2$. Then $E_{a,b}$ is unbounded.

Proof. Since $f_a$ is surjective by Lemma 2, for every $yinmathbb R$ there exists $xinmathbb R$ such that $f_a(x)=-f_b(y)$. This means that $(x,y)in E_{a,b}$. Therefore the set $E_{a,b}$ is unbounded because we can find points with arbitrary large $y$.

Proposition. If $|a|=pi/2$. Then $E_{a,b}$ is unbounded.

Proof. Assume $a=pi/2$. The other case is analogous. Recall the well known limit
$$
lim_{xtoinfty} f_{pi/2}(x) = lim_{xtoinfty} x(arctan x-pi/2) = -1.
$$
For every $x$ sufficiently large we have that $f_{pi/2}(x)leq0$, so by Lemma 1 there exists $yinmathbb R$ such that $f_b(y)=-f_a(x)in[0,infty)$. Therefore we can find points $(x,y)in E_{pi/2,b}$ with arbitrary large $x$. □

Proposition. If $|a|<pi/2$ and $|b|<pi/2$, then $E_{a,b}$ is bounded.

Proof. By Lemma 3, the function $f_a(x)+f_b(y)$ is coercive, meaning that $f_a(x)+f_b(y)toinfty$ if $|(x,y)|toinfty$, therefore its sublevel sets are bounded. In particular $E_{a,b}$ is bounded as a consequence. □

Corollary. $E_{a,b}$ is bounded if and only if $|a|<pi/2$ and $|b|<pi/2$.

Now, given $a,bin(-pi/2,pi/2)$, how can we find an estimate on $max_{(x,y)in E_{a,b}} x^2+y^2$? We could try the Lagrange multipliers approach again, similarly to what we did here.
The stationary points must satisfy
$$
bigl(f'_a(x), f'_b(y)bigr) = lambda (x, y) qquad text{for some $lambdainmathbb R$},
$$
which is equivalent to
$$
frac1{1+x^2} + frac{arctan x-a}x
= frac{f'_a(x)}{x} = lambda
= frac{f'_b(y)}{y}
=frac1{1+y^2} + frac{arctan y-b}y.
$$

Unfortunately, this time I'm not able to find a closed form solution to the system of equations
$$
left{begin{array}{l}
f_a(x)+f_b(y)=0 , \
frac{f'_a(x)}x=frac{f'_b(y)}y .
end{array}right.
$$

One can of course fall back to numerical solutions. I'm using Mathematica for that.
Here I fix the values $a=3/2$ and $b=5/4$, then I find extremal points numerically both with the built-in NMaximize function and by solving the Lagrange multiplier system with FindRoot. The two solutions are the same, up to machine precision. Then I plot the set $E_{a,b}$ in blue, the root locus of the Lagrange multiplier equation in orange, and the smallest fitting circle in gray.