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<g(a)<g(2)=(b-1)(b-2)

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

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