Ngô Quốc Anh

January 20, 2011

On an algebraic identity araising in the Toda systems SU(3)

Filed under: Uncategorized — Ngô Quốc Anh @ 21:42

Recently, in their paper published in J. Diffierential Equations [here], H. Ohtsuka and T. Suzuki prove the following

Suppose we have real numbers a and b satisfying $a^2-ab+b^2=4\pi(a+b)$.

If we assume $\min\{a,b\} \geqslant 4\pi$,

then we can prove $\max\{a,b\}\geqslant 8\pi$.

Furthermore, if we know that both $2a-b>4\pi,\quad 2b-a>4\pi$,

then we can further show that $\max\{a,b\}\leqslant 4\left( 1+\frac{2}{\sqrt{3}}\right)\pi \approx 8.61\pi$.

In this note, we show that the restriction $2a-b>4\pi,\quad 2b-a>4\pi$,

can be relaxed. By a scaling $a=4\pi a_1$, $b=4 \pi b_1$ the above question becomes

Suppose we have real numbers a and b satisfying $a^2-ab+b^2=4\pi(a+b)$.

If we assume $\min\{a,b\} \geqslant 1$,

then we can prove $\displaystyle 2 \leqslant \max\{a,b\}\leqslant 1+\frac{2}{\sqrt{3}} \approx 2.31$.

Following is a proof provided by Phan Thanh Nam from the University of Copenhagen. Indeed, for the first part, by contradiction and WLOG we can assume that $1\leqslant b \leqslant a < 2$.

For the upper bound, let us consider the following auxiliary function $f(x)=a^2-ax+x^2-(a+x)$

with $x \in [1,a]$. Since $f'(x)=2x-a-1$

we know that $f$ achives its minimum value at $x = \frac{a+1}{2}$. Therefore, $\displaystyle 0 =f(b)\geqslant f\left(\frac{a+1}{2}\right)=\frac{3a^2-6a-1}{4}$.

Consequently, $a$ is bounded from above by the maximum solution of the following algebraic equation $3a^2-6a-1=0$, i.e., $1+\frac{2}{\sqrt{3}}$. The equality occurs if and only if $b=\frac{a+1}{2}=1+\frac{1}{\sqrt{3}}$.

For the second part, assuming $a<2$. The case of $b=1$ is trivial since $a=2$. Thus it suffices to consider the case of $b>1$. Let $g(x)=x^2-ab+b^2-(x+b)$

with $2>x\geqslant b>1$. Since $g'(x)=2x-b-1>0$

function $g$ is strictly monotone increasing. Consequently, $0

which gives a contradiction to the fact that $1. Thus $a \geqslant 2$. The equality occurs if and only if $b=1$.