# Existence of solution to linear fractional equation

We consider the equation

$$\sum_{j=1}^n \frac{\lambda_j}{x-x_j} =i$$

where $$\lambda_j>0$$ and $$x_j$$ are real distinct numbers.

I want to show that if $$\lambda_k$$ is small compared to the distance of all $$x_j$$ from $$x_k$$ then there exists a solution $$x\approx x_k- i y_k$$ to this equation in the neighbourhood of $$x_k.$$

Heuristic argument:

Let $$x=x_k - i y_k$$ by multiplying the equation with $$(x-x_k),$$ we find

$$\lambda_k - \sum_{j=1, j \neq k}^n \frac{i y_k\lambda_j}{x_k-x_j-i y_k} =y_k.$$

Now, if $$x_k-x_j$$ is large, then the sum is small and we can choose $$y_k\approx \lambda_k.$$

However, this argument is (obviously) non-rigorous.

Can we make it rigorous?

• $y_k = 0 + \lambda_k + _ \lambda_k + \cdots$ – AHusain Mar 22 '19 at 0:22
• This is correct (and trivial). When all $\lambda_j$ are small the LHS is small everywhere except small neighborhoods of $x_k$. To each of these neighborhoods apply Rouche's theorem. – Alexandre Eremenko Mar 22 '19 at 1:59
• @AlexandreEremenko sorry, but for the sake of understanding. How is this theorem applied now? What are the holomorphic functions $f,g$ one would consider?- If you think it is off-topic I will close the question also, but I would appreciate an answer very much. Thank you for your time. – user121558 Mar 22 '19 at 2:44
• @AlexandreEremenko I now received two answers and both of them seem to rely not only on distances $x_j-x_k$ but on all the parameters $\lambda_j$ as well. That's nice, but not what I asked for. Was the way you intended to apply Rouch\'e's theorem able to avoid that problem? – user121558 Mar 23 '19 at 22:24
• @J.Doe: You should mark Alexandre Eremenko's answer as the accepted one then. – Willie Wong Mar 28 '19 at 1:30

Here's a "real" method proof:

Without loss of generality, assume $$k = 1$$. You are equivalently looking for the roots of

$$P(x, \eta) = \eta \prod_{j > 1} (x - x_j) + \sum_{\ell > 1} \lambda_\ell \prod_{j \neq \ell} (x - x_j) - i \prod_{j} (x - x_j)$$

where we wrote $$\eta$$ for $$\lambda_1$$. (We consider $$\lambda_\ell$$ and all $$x_j$$ to be fixed.)

Now allow $$\eta$$ to be a parameter in $$\mathbb{C} \cong \mathbb{R}^2$$.

For $$\eta = 0$$, by inspection $$P(x_1, 0) = 0$$. The derivative

$$\partial_\eta P(x_1,\eta) = \prod_{j > 1} (x_1-x_j) \cdot \mathrm{Id}$$

is non-critical. So by implicit function theorem for all sufficiently small $$\eta$$ there exists a solution close to $$x_1$$.

• Smallness bound of your $\eta$ depends on $x_j$ and the rest of $\lambda_j$. Was not the question about $\eta$ depending on $x_j$ only (but not on the rest of $\lambda_j$? – Alexandre Eremenko Mar 22 '19 at 21:27
• @AlexandreEremenko: I dunno. If the OP can clarify that'd be great! – Willie Wong Mar 25 '19 at 7:42
• see my answer. I made the estimate independent on the rest of $\lambda_j$. – Alexandre Eremenko Mar 25 '19 at 12:03
• @AlexandreEremenko: thanks! That's a very nice answer. I +1'd it. – Willie Wong Mar 28 '19 at 1:29

Expanding the comment by Alexandre Eremenko (and assuming $$\lambda_j$$ small for $$j=k$$ only).

Let us rewrite the equation in the form $$\underbrace{\prod_{j=1}^n (x-x_j) + i \sum_{j\in\{1,\ldots,n\} \setminus\{k\}} \lambda_j \prod_{s\in\{1,\ldots,n\} \setminus \{j\}} (x-x_s)}_{f(x)} + \underbrace{i \lambda_k \prod_{s\in\{1,\ldots,n\} \setminus \{k\}} (x-x_s)}_{g(x)} = 0.$$

Choose $$r>0$$ sufficiently small so that $$f(x) \ne 0$$ for $$x\in\partial B_r(x_k)$$. Then $$\min_{x\in \partial B_r(x_k)} |f(x)| = m > 0$$. On the other hand if $$\lambda_k$$ is sufficiently small then $$|g(x)| < m$$ for all $$x \in \partial B_r(x_k)$$. Since $$f(x_k)=0$$ it remains to apply Rouché's theorem.

• This is not wgat the question asks for, it only assumes $\lambda_k$ to be small. – user69109 Mar 22 '19 at 10:32
• @Tokoyo good point, I've updated the answer to handle the case when $\lambda_j$ is small only for $j=k$. – Skeeve Mar 22 '19 at 15:06

Here is an estimate that does not depend on $$\lambda_j,j\geq 2$$.

Lemma. Consider an equation $$\epsilon+az+(b+\phi(z))z^2=0,$$ where $$\epsilon>0$$, $$|a|\geq 1$$ and $$b$$, are complex constants, and $$\phi$$ is analytic, $$|\phi|<|b|/10.$$ Then there exists a solution $$|z|\leq 10\epsilon$$.

Proof. If $$|b|\leq |a|/(5\epsilon)$$, apply Rouche to $$|z|\leq2\epsilon/|a|$$. The dominant term is $$az$$. If $$b>|a|/(5\epsilon)$$, apply Rouche to $$|z|\leq10\epsilon$$. The dominant term is $$(b+\phi)|z^2|$$ now.

To apply this lemma, write your equation $$\frac{\epsilon}{z}+\sum_2^n\frac{\lambda_k}{z-z_k}=i$$ which we bring to the form given in the lemma by writing the sum minus $$i$$ in the form $$a+(b+\phi(z))z$$ and mulpiplying on $$z$$. So $$a=-\sum_2^n\frac{\lambda_k}{z_k}-i,$$ (we have $$|a|\geq 1$$ because the sum is real). $$b=-\sum_2^n\frac{\lambda_k}{z_k^2},$$ and $$\phi$$ is what remains. I used the identity $$\frac{1}{z-w}=-\frac{1}{w}-\frac{z}{w^2}-\frac{z^2/w^3}{1-z/w},$$ in which I set $$w=z_k$$ and then add.

Now estimate $$\phi$$, assuming that $$|z|/|z_k|<1/20$$ for all $$k\geq 2$$. $$|\phi(z)|=\sum_2^n\lambda_k\left|\frac{z/z_k^3}{1-z/z_k}\right|< \frac{1}{10}\sum_2^n\lambda_k/|z_k^2|<|b|/10.$$ It is important here that all $$\lambda_j$$ are positive.