Behaviour of roots of family of polynomials

Question

Let $$ P(x)=(1+\frac{w^3}{12})x^4 + w^2 \sqrt{u}x^3+(wu-\frac{2}{u}(1+\frac{w^3}{12}) )x^2-\frac{w^2}{\sqrt{u}}x+\frac{1}{u^2}(1+\frac{w^3}{12}) $$ a polynomial in the variable $x$, where $w,u>0$. If $w$ is fixed, How I can study the behaviour of the roots in the cases when $u\to 0$ and $u\to \infty$?

Answer 1

You can combine the obviously related terms to get $$ P(x)=\left(1+\frac{w^3}{12}\right)\left(x^2-\frac1u\right)^2+w\sqrt{u}x\left(wx^2+\sqrt{u}x-\frac{w}{u}\right) $$ With $y=x-\frac1{xu}$ and for $xu\ne 0$ the equation $P(x)=0$ is equivalent to $$ 0=Q(y)=\left(1+\frac{w^3}{12}\right)y^2+w^2\sqrt{u}y+wu $$ This means that you can actually compute the roots via a nested sequence of two quadratic equations.

Answer 2

The trick is to "clear denominators" to avoid infinities so that you can use the continuity of roots of homogeneous polynomials with respect to the coefficients. In particular, let us assume that $1+w^3/12\neq 0$ and consider the roots of $P(x)$ when $u$ is close to but not equal to $0$. These are the same as the roots of $$u^2P(x)=u^2vx^4 + w^2u^{5/2}x^3+(wu^3-2uv)x^2-w^2u^{3/2}x+v$$ where $v=1+w^3/12$. Now unfortunately, we cannot just directly use the continuity of roots of polynomials of the same degree with respect to the coefficients here, since the leading coefficient $u^2v$ is going to $0$ so the degree decreases in the limit. The trick is to homogenize: let $x=s/t$ and consider the homogeneous polynomial $$Q(s,t)=t^4u^2P(x)=u^2vs^4+ w^2u^{5/2}s^3t+(wu^3-2uv)s^2t^2-w^2u^{3/2}st^3+vt^4.$$ There is then a theorem that the roots of a nonzero homegeneous polynomial in projective space vary continuously with the coefficients. So, as $u\to 0$, the roots of $Q(s,t)$ approach the roots of $$vt^4$$ since all the other coefficients have factors of $u$ and thus go to $0$. But those roots are all just $t=0$, which is the point at $\infty$ in projective space. In other words, writing $x=s/t$, all the roots for $x$ approach $\infty$ as $u\to 0$.

(In this case, we could see that more directly by just observing that if $x$ is bounded in size and $u$ is very close to $0$, then the final term $\frac{1}{u^2}(1+\frac{w^3}{12})$ will be much larger than all the other terms of $P(x)$ and so $P(x)$ cannot be $0$. This means that as $u\to 0$, $P(x)$ cannot have any roots inside larger and larger balls, so that the roots must all approach $\infty$.)

Similar analyses can be done for the case where $1+w^3/12=0$ or $u\to\infty$, to find that some of the roots approach $\infty$ and others approach $0$, depending on which coefficient of $P(x)$ grows fastest in the limit (and thus survives in the limit after we clear denominators so that the others go to $0$).

Answer 3

Let $u = U^2$, then the question is about finding the Puiseux series solution for a polynomial equation. If we draw the Newton polygon, we can see that the terms corresponding to the small $U$ and large $x$ expansion are $$v U^4 x^4 - 2 v U^2 x^2 + v, \\ x_{1, 2} \sim -\frac 1 U, \; x_{3, 4} \sim \frac 1 U, \quad U \to 0,$$ where $v = w^3 + 12$. The terms corresponding to large $U$ and small $x$ are $$12 w U^6 x^2 - 12 w^2 U^3 x + v, \\ x_{1, 2} \sim \frac {3 w^2 \pm \sqrt D} {6 w U^3}, \quad U \to \infty,$$ $D = 6 w (w^3 - 6)$. Large $U$ and large $x$ gives $$v U^4 x^4 + 12 w^2 U^5 x^3 + 12 w U^6 x^2, \\ x_{3, 4} \sim \frac {2 (-3 w^2 \pm \sqrt D) U} v, \quad U \to \infty.$$