June 14, 2026
Let \(Q\) be a monic polynomial of degree \(M+1\). We study the algebraic system \[\sum_{j\ne i}\frac{1}{x_i-x_j}=Q(x_i),\qquad i=1,\ldots,N,\] for pairwise distinct complex numbers \(x_1,\ldots,x_N\), modulo permutations of these numbers. The case \(M=0\) is, after a translation, the classical Stieltjes system for the zeros of a Hermite polynomial. We prove that, for arbitrary \(Q\), the number of solutions is at most \(\binom{N+M}{N}\), and that the coefficient equations for the associated monic Stieltjes polynomial have total intersection multiplicity exactly \(\binom{N+M}{N}\). Consequently the bound is attained for all \(Q\) in a non-empty Zariski open subset of the affine space of monic polynomials of degree \(M+1\). We also describe the solutions when the coefficient of the linear term of \(Q\) is large: the system splits into \(M+1\) weakly coupled classical Stieltjes systems, one near each zero of \(Q\).
The classical Heine–Stieltjes theory relates equilibrium configurations of movable logarithmic charges to polynomial solutions of second order linear differential equations; see Heine [1], Stieltjes [2], and the modern survey [3]. The Hermite equation is a basic degenerate example of the same general philosophy, and its electrostatic interpretation is discussed, for instance, in Szegő’s book [4] and in [5]. The present note treats a simple polynomial-source version of this degenerate situation.
Fix integers \(N\geq1\) and \(M\geq0\). For a monic polynomial \[\label{eq:source} Q(z)=z^{M+1}+a_Mz^M+\cdots+a_1z+a_0\tag{1}\] of degree \(M+1\), consider \[\label{eq:genStieltjes} \sum_{j\ne i}\frac{1}{x_i-x_j}=Q(x_i),\qquad i=1,\ldots,N,\tag{2}\] with \((x_1,\ldots,x_N)\in\operatorname{Conf}_N(\mathbb{C})\), where \[\operatorname{Conf}_N(\mathbb{C}):=\{(x_1,\ldots,x_N):x_i\ne x_j\text{ for }i\ne j\}.\] Two ordered solutions are considered equivalent if they differ by a permutation of the \(x_i\)’s. Equivalently, an equivalence class is represented by the monic polynomial \[P(z)=\prod_{i=1}^N(z-x_i).\] We shall therefore freely pass between unordered solutions of 2 and monic polynomials \(P\) of degree \(N\).
Theorem 1. For every monic polynomial \(Q\) of degree \(M+1\), the number of inequivalent solutions of 2 is at most \[\binom{N+M}{N}.\] More precisely, the associated coefficient equations for the monic polynomial \(P\) have total intersection multiplicity exactly \(\binom{N+M}{N}\). Hence the same binomial coefficient is the exact number of inequivalent solutions for all \(Q\) in a non-empty Zariski open subset of the affine space of monic polynomials of degree \(M+1\).
The proof has two parts. First, we rewrite 2 as a Heine–Stieltjes divisibility condition and compute the length of the resulting coefficient scheme by a weighted Bézout argument. Second, we exhibit an explicit non-empty open set by analysing the limit in which the coefficient \(a_1\) of the linear term tends to infinity. In this limit the zeros of \(Q\) consist of \(M\) large zeros and one small zero, and each solution of 2 is obtained by distributing the \(N\) movable zeros among these \(M+1\) centres. After the natural rescaling near each centre, one obtains the standard Hermite/Stieltjes system.
Let \(P(z)=\prod_{i=1}^{N}(z-x_i)\) have simple roots. The elementary identity \[\label{eq:logder-id} \frac{P''(x_i)}{2P'(x_i)}=\sum_{j\ne i}\frac{1}{x_i-x_j}\tag{3}\] shows that 2 is equivalent to \[P''(x_i)-2Q(x_i)P'(x_i)=0, \qquad i=1,\ldots,N.\] Thus \(P\) divides \(P''-2QP'\). This gives the following standard form of the Heine–Stieltjes correspondence.
Lemma 1. Let \(Q\) be as in 1 . Equivalence classes of solutions of 2 are in bijection with monic polynomials \(P\) of degree \(N\) for which there exists a polynomial \(V\) of degree at most \(M\) such that \[\label{eq:HS} P''(z)-2Q(z)P'(z)-V(z)P(z)=0.\tag{4}\] For such a polynomial \(P\), all roots of \(P\) are simple.
Proof. If \((x_1,\ldots,x_N)\) is a solution of 2 , then 3 shows that \(P''-2QP'\) vanishes at every root of \(P\). Since the roots are simple, \(P\) divides \(P''-2QP'\). The quotient has degree at most \(M\), because \(Q\) has degree \(M+1\) and \(P\) has degree \(N\).
Conversely, suppose that 4 holds. It remains first to observe that \(P\) cannot have a multiple root. If \(P(z)=(z-\xi)^m\widetilde{P}(z)\) with \(m\ge2\) and \(\widetilde{P}(\xi)\ne0\), then \(P''\) has order \(m-2\) at \(\xi\) with leading coefficient \(m(m-1)\widetilde{P}(\xi)\), while \(QP'\) has order at least \(m-1\) and \(VP\) has order at least \(m\). The coefficient of \((z-\xi)^{m-2}\) in the left-hand side of 4 is therefore non-zero, a contradiction.
Thus the roots of \(P\) are simple. Substituting any root \(\xi\) of \(P\) in 4 gives \(P''(\xi)-2Q(\xi)P'(\xi)=0\), and 3 gives 2 . ◻
Let \[P(z)=z^N+c_1z^{N-1}+\cdots+c_N.\] Divide \(P''-2QP'\) by \(P\) and denote by \[\label{eq:remainder} \operatorname{Rem}_Q(P)=r_1(c)z^{N-1}+r_2(c)z^{N-2}+\cdots+r_N(c)\tag{5}\] the remainder, where \(c=(c_1,\ldots,c_N)\). By Lemma 1, solutions of 2 are exactly the common zeros of \[\label{eq:coefficient-system} r_1(c)=\cdots=r_N(c)=0.\tag{6}\]
Assign the weights \[\operatorname{wt}(z)=1, \qquad \operatorname{wt}(c_j)=j,\quad j=1,\ldots,N.\] Then \(P(z)=z^N+c_1z^{N-1}+\cdots+c_N\) is weighted homogeneous of total weight \(N\). For a polynomial \(f(c)\), write \(\operatorname{in}_{\!w}(f)\) for its highest weighted homogeneous part.
Lemma 2. For \(j=1,\ldots,N\), the polynomial \(r_j\) has weighted degree at most \(M+j\). Its highest weighted homogeneous part is the coefficient of \(z^{N-j}\) in the remainder of \(-2z^{M+1}P'(z)\) modulo \(P(z)\). These highest parts have no common zero in \(\mathbb{C}^N\setminus\{0\}\).
Proof. If a weighted homogeneous polynomial \(G(z,c)\) has total weight \(N+d\), then the remainder of \(G\) modulo the weighted homogeneous monic polynomial \(P\) has the form \[h_1(c)z^{N-1}+\cdots+h_N(c), \qquad \operatorname{wt}(h_j)=d+j.\] Indeed, Euclidean division by the monic polynomial \(P\) is compatible with the weight filtration. The term \(-2z^{M+1}P'\) has total weight \(N+M\). The terms coming from \(P''\) and from the lower degree part of \(Q\) have strictly smaller weight. This proves the first two assertions.
Assume now that all highest parts vanish at some \(c\in\mathbb{C}^N\). Then the remainder of \(z^{M+1}P'\) modulo \(P\) is zero, i.e. \[\label{eq:leading-divisibility} P\mid z^{M+1}P'.\tag{7}\] If \(P\) has a non-zero root \(\alpha\) of multiplicity \(m\ge1\), then the right-hand side of 7 has multiplicity \(m-1\) at \(\alpha\), which is impossible. Therefore every root of \(P\) is zero, so \(P=z^N\) and \(c_1=\cdots=c_N=0\). Hence the highest parts have no common zero away from the origin. ◻
We shall use the following weighted form of Bézout’s theorem. The proof is included to make clear that the multiplicity counted below is the affine length of the coefficient scheme.
Lemma 3 (weighted Bézout). Let \(f_1,\ldots,f_N\in\mathbb{C}[c_1,\ldots,c_N]\), with \(\operatorname{wt}(c_j)=w_j>0\), have weighted degrees at most \(d_1,\ldots,d_N\). Suppose that their highest weighted homogeneous parts \(g_i:=\operatorname{in}_{\!w}(f_i)\) have no common zero in \(\mathbb{C}^N\setminus\{0\}\). Then the common zero scheme of \(f_1,\ldots,f_N\) in \(\mathbb{C}^N\) is finite and has length \[\frac{d_1\cdots d_N}{w_1\cdots w_N}.\]
Proof. The polynomials \(g_1,\ldots,g_N\) form a weighted homogeneous system of parameters in the polynomial ring \(S=\mathbb{C}[c_1,\ldots,c_N]\). Since \(S\) is Cohen–Macaulay, they form a regular sequence. Hence \[\operatorname{Hilb}_{S/(g_1,\ldots,g_N)}(t) =\frac{\prod_{i=1}^N(1-t^{d_i})}{\prod_{j=1}^N(1-t^{w_j})},\] and the length of \(S/(g_1,\ldots,g_N)\) is the value at \(t=1\) after cancelling the zero of numerator and denominator, namely \(\prod_i d_i/\prod_j w_j\).
It remains to pass from the highest parts \(g_i\) to \(f_i\). We use the following elementary filtered-graded fact. If \(A\) is a filtered algebra, \(f\in A\) has initial form \(g\in\operatorname{gr}A\), and \(g\) is a non-zero-divisor on \(\operatorname{gr}A\), then \(\operatorname{gr}(A/(f))\simeq (\operatorname{gr}A)/(g)\); this follows by comparing highest weighted terms. Applying this fact successively to the regular sequence \(g_1,\ldots,g_N\) shows that \(f_1,\ldots,f_N\) is a regular sequence, that \(S/(f_1,\ldots,f_N)\) is finite-dimensional, and that its associated graded algebra is \(S/(g_1,\ldots,g_N)\). Thus the two quotients have the same length.
Equivalently, one may homogenize the equations with a new variable \(\tau\) of weight \(1\) and apply Bézout in the weighted projective space \(\mathbb{P}(w_1,\ldots,w_N,1)\); see, for example, [6]. The hypothesis precisely says that the homogenized system has no solutions on the hyperplane \(\tau=0\). ◻
Proposition 2. For every monic polynomial \(Q\) of degree \(M+1\), the coefficient system 6 has finitely many zeros and total intersection multiplicity \[\binom{N+M}{N}.\]
Proof. Apply Lemma 3 to \(r_1,\ldots,r_N\). By Lemma 2, the weights are \(w_j=j\) and the degrees are \(d_j=M+j\). Therefore the total multiplicity is \[\prod_{j=1}^{N}\frac{M+j}{j} =\binom{N+M}{N}.\] ◻
Let \(\mathsf H_n\) denote the monic Hermite polynomial, normalized by \[\label{eq:monic-Hermite} \mathsf H_n(z)=2^{-n}(-1)^n e^{z^2}\frac{d^n}{dz^n}e^{-z^2}.\tag{8}\] It satisfies \[\label{eq:Hermite-ODE} \mathsf H_n''(z)-2z\mathsf H_n'(z)+2n\mathsf H_n(z)=0.\tag{9}\] Its zeros are real, simple, and symmetric with respect to the origin. We write them as \[u_1^{(n)}<\cdots<u_n^{(n)}, \qquad u_i^{(n)}=-u_{n+1-i}^{(n)}.\] For \(n=0\) the set of zeros is empty.
Lemma 4. The standard Stieltjes system \[\label{eq:standard-Stieltjes} \sum_{j\ne i}\frac{1}{t_i-t_j}=t_i, \qquad i=1,\ldots,n,\tag{10}\] has, modulo permutations, the unique solution given by the zeros of \(\mathsf H_n\).
Proof. If \(R(z)=\prod_{i=1}^n(z-t_i)\) corresponds to a solution, then \(R''-2zR'\) is divisible by \(R\). Since \(R\) is monic of degree \(n\), comparison of leading terms gives \[R''-2zR'+2nR=0.\] The monic polynomial solution of 9 is unique by the usual coefficient recursion; hence \(R=\mathsf H_n\). ◻
Lemma 5. Let \(J^{(n)}\) be the Jacobian matrix of the map \[F_i(t)=t_i-\sum_{j\ne i}\frac{1}{t_i-t_j}, \qquad i=1,\ldots,n,\] at the Hermite zero configuration \(u^{(n)}=(u_1^{(n)},\ldots,u_n^{(n)})\). Then the eigenvalues of \(J^{(n)}\) are \(1,2,\ldots,n\). In particular, \(J^{(n)}\) is invertible.
Proof. For \(m=0,\ldots,n-1\), consider \[R_\varepsilon(z)=\mathsf H_n(z)+\varepsilon\mathsf H_m(z)\] and let \(t_i(\varepsilon)\) be the root of \(R_\varepsilon\) near \(u_i^{(n)}\). Differentiating \(R_\varepsilon(t_i(\varepsilon))=0\) at \(\varepsilon=0\) gives \[t_i'(0)=-\frac{\mathsf H_m(u_i^{(n)})}{\mathsf H_n'(u_i^{(n)})}.\] Using 9 for \(\mathsf H_n\) and \(\mathsf H_m\), one obtains, at the roots of \(R_\varepsilon\), \[F_i(t(\varepsilon)) =-(n-m)\varepsilon\, \frac{\mathsf H_m(t_i(\varepsilon))}{R_\varepsilon'(t_i(\varepsilon))}.\] Differentiating at \(\varepsilon=0\) gives \[J^{(n)}t'(0)=(n-m)t'(0).\] Thus the eigenvalues are \(n,n-1,\ldots,1\), and the corresponding eigenvectors are linearly independent because the eigenvalues are distinct. ◻
For later use, note also the translated form of the case \(M=0\). If \(Q(z)=z+a_0\), then the unique solution of 2 is \[x_i=u_i^{(N)}-a_0, \qquad i=1,\ldots,N,\] up to permutation, and it is reduced by Lemma 5.
In this section assume \(M\ge1\). Fix all coefficients of \(Q\) except the linear one, and write them as \(a_0,a_2,\ldots,a_M\); if \(M=1\), this means only \(a_0\) is fixed. Put \[\label{eq:Qnu} Q_\nu(z)=z^{M+1}+\sum_{\ell=2}^{M}a_\ell z^\ell-\nu^{2M}z+a_0.\tag{11}\] Thus \(a_1=-\nu^{2M}\), and the limit \(|a_1|\to\infty\) is studied through \(\nu\to\infty\).
Lemma 6. For \(|\nu|\) sufficiently large, the polynomial \(Q_\nu\) has \(M\) simple large zeros \[\alpha_k(\nu)=\omega^k\nu^2+O(1), \qquad k=0,\ldots,M-1, \qquad \omega=e^{2\pi i/M},\] and one simple small zero \[\gamma(\nu)=a_0\nu^{-2M}+O(\nu^{-4M}).\] More precisely, \(\nu^{-2}\alpha_k(\nu)\) and \(\nu^{2M}\gamma(\nu)\) are analytic functions of \(\nu^{-1}\) near \(\nu^{-1}=0\), after the choice of the branch of \(\nu\). Moreover \[\label{eq:derivatives-centres} Q_\nu'(\alpha_k(\nu))=M\nu^{2M}\bigl(1+O(\nu^{-2})\bigr), \qquad Q_\nu'(\gamma(\nu))=-\nu^{2M}\bigl(1+O(\nu^{-2})\bigr).\tag{12}\]
Proof. For the large zeros set \(z=\nu^2y\). Dividing \(Q_\nu(\nu^2y)\) by \(\nu^{2M+2}\) gives \[y^{M+1}-y+O(\nu^{-2})\] uniformly for \(y\) in compact sets. The non-zero roots of \(y^{M+1}-y\) are \(1,\omega,\ldots,\omega^{M-1}\), and the derivative of \(y^{M+1}-y\) at each of these roots is \(M\). The analytic implicit function theorem gives the \(M\) large roots. For the small zero set \(z=\nu^{-2M}y\); then \[Q_\nu(\nu^{-2M}y)=a_0-y+O(\nu^{-4M}),\] which gives the stated small root. The derivative estimates follow by substitution in \(Q_\nu'\). ◻
Choose analytic square roots and define \[\label{eq:local-scales} s_k(\nu):=Q_\nu'(\alpha_k(\nu))^{-1/2}, \qquad s_O(\nu):=Q_\nu'(\gamma(\nu))^{-1/2},\tag{13}\] with asymptotics \[\label{eq:scale-asymptotics} s_k(\nu)=M^{-1/2}\nu^{-M}(1+O(\nu^{-2})), \qquad s_O(\nu)=i\nu^{-M}(1+O(\nu^{-2})).\tag{14}\] The sign choices are immaterial, since the Hermite zero configuration is invariant under \(t\mapsto -t\) up to permutation.
A weak composition of \(N\) into \(M+1\) parts will be written \[\lambda=(\lambda_0,\ldots,\lambda_{M-1},\lambda_O), \qquad |\lambda|=\sum_{k=0}^{M-1}\lambda_k+\lambda_O=N.\] There are \(\binom{N+M}{N}\) such compositions. Given such a \(\lambda\), we place \(\lambda_k\) variables near \(\alpha_k\) and \(\lambda_O\) variables near \(\gamma\): \[\begin{align} \label{eq:cluster-coordinates} x_{k,i}&=\alpha_k(\nu)+s_k(\nu)t_{k,i}, && i=1,\ldots,\lambda_k, \quad k=0,\ldots,M-1,\\ x_{O,i}&=\gamma(\nu)+s_O(\nu)t_{O,i}, && i=1,\ldots,\lambda_O.\nonumber \end{align}\tag{15}\] Empty blocks are omitted.
Proposition 3. Fix \(a_0,a_2,\ldots,a_M\) and a weak composition \(\lambda\) of \(N\) into \(M+1\) parts. For \(|\nu|\) sufficiently large, the Stieltjes system 2 with source \(Q_\nu\) has a unique solution, up to permutations inside the blocks of 15 , of the form \[\label{eq:cluster-asymptotics} t_{k,i}=u_i^{(\lambda_k)}+O(\nu^{-1}), \qquad t_{O,i}=u_i^{(\lambda_O)}+O(\nu^{-1}),\qquad{(1)}\] where \(u_i^{(r)}\) are the zeros of the monic Hermite polynomial \(\mathsf H_r\). The corresponding solution is reduced as a zero of the coefficient system 6 .
Proof. We prove the assertion for ordered variables inside each non-empty block; the quotient by the block permutation group is then immediate.
Consider a block centred at a zero \(\rho_b\in\{\alpha_0,\ldots,\alpha_{M-1},\gamma\}\), with scale \(s_b=Q_\nu'(\rho_b)^{-1/2}\). In that block write \(x_{b,i}=\rho_b+s_bt_{b,i}\). Since \(Q_\nu(\rho_b)=0\) and \(s_b^2Q_\nu'(\rho_b)=1\), Taylor’s formula gives, uniformly for \(t\) in compact sets, \[\label{eq:source-local-expansion} s_b Q_\nu(\rho_b+s_bt)=t+O(\nu^{-1}).\tag{16}\] In fact the error is smaller than \(O(\nu^{-1})\), but this bound is sufficient. For two variables in the same block, \[\frac{s_b}{x_{b,i}-x_{b,j}} =\frac{1}{t_{b,i}-t_{b,j}}.\] For variables in distinct blocks the centres differ by order \(\nu^2\), while \(s_b=O(\nu^{-M})\); hence every cross-interaction, after multiplication by \(s_b\), is \(O(\nu^{-M-2})=O(\nu^{-1})\).
Multiplying the Stieltjes equation for \(x_{b,i}\) by \(s_b\) therefore gives a system analytic in the local variables and in \(\nu^{-1}\) near \(\nu^{-1}=0\). Its limiting system is the product, over all blocks, of \[\label{eq:block-limit} t_{b,i}-\sum_{j\ne i}\frac{1}{t_{b,i}-t_{b,j}}=0.\tag{17}\] By Lemma 4, the limiting solution in the \(b\)-th block is the Hermite zero set of size equal to that block. By Lemma 5, the Jacobian of the limiting product system is block diagonal with invertible blocks. The analytic implicit function theorem therefore gives a unique analytic branch with the asymptotics ?? .
The same Jacobian argument shows that the obtained zero is reduced. Passing from ordered block variables to the coefficient system does not introduce an extra multiplicity, because the roots in the limiting Hermite configurations are distinct and the passage from unordered roots to coefficients is locally biholomorphic away from the big diagonal. ◻
Corollary 1. For every fixed \(a_0,a_2,\ldots,a_M\) and all sufficiently large \(|a_1|\), with \(a_1=-\nu^{2M}\), the system 2 has exactly \(\binom{N+M}{N}\) inequivalent solutions. All of them are reduced.
Proof. Proposition 3 constructs one reduced solution for each weak composition of \(N\) into \(M+1\) parts. These solutions are distinct because they have different cluster multiplicities around the zeros of \(Q_\nu\). There are \(\binom{N+M}{N}\) such compositions. Proposition 2 says that the total multiplicity of the whole coefficient scheme is exactly the same number, so no further solutions are possible. ◻
Proof of Theorem 1. The upper bound and the total multiplicity statement are precisely Proposition 2, together with Lemma 1. For \(M=0\) the translated Hermite solution described above is reduced. For \(M\ge1\), Corollary 1 gives at least one monic source \(Q\) for which all zeros of the coefficient system are reduced and their number is \(\binom{N+M}{N}\). Since reducedness of the fibres of a finite algebraic family is a Zariski open condition, equality holds on a non-empty Zariski open subset of the affine coefficient space of monic sources \(Q\). ◻
Remark 4. The proof of Proposition 3 explains why the binomial coefficient appears geometrically: in the large-\(|a_1|\) regime, the \(N\) roots of the Stieltjes polynomial choose one of the \(M\) large zeros of \(Q\) or the one small zero of \(Q\), and inside each chosen cluster the only possible limiting configuration is the Hermite configuration of the corresponding size.
D. Masoero is partially supported by FCT Grant “The Nonlinear Stokes Phenomenon. A unifying perspective on Integrable Models, Enumerative Geometry, and Special Functions”, 2021.00091.CEECIND.