Principal roots and uniform time kernels

Written by GPT-6.1 Sol (OpenAI), Ultra reasoning effort, October 2026. Self-checked by GPT-6.1 Sol (OpenAI). Public domain (CC0).

Time evolution starts with scalar polynomial roots. We bound their displacement from the real principal roots and construct time kernels uniformly even when roots repeat.

Read Primitive factors and moving polynomial roots, Cauchy bounds, root counts and analytic extensions. Fourier transforms, finite spectra and convex separation supplies Schwartz inversion; Banach estimates, quotient spaces and compact parameter arguments supplies the integration estimates. Metric and topological foundations supplies scalar calculus and cutoffs. Cauchy kernels and distributional boundary limits proves the Cauchy–Pompeiu formula; Polynomial and contour interfaces for stable boundary models proves scalar factorization and the exponential series.

For Fourier background, Grubb's author-hosted Chapter 5, Fourier transformation of distributions, gives the differentiation convention used in Exercise 1: Theorem 5.4(2), equation (5.9), printed pp. 5.3–5.4. The root-displacement and uniform time-kernel estimates are proved below. Other general background references are Melrose's Differential Analysis and Hörmander's The Analysis of Linear Partial Differential Operators. The proofs below use the linked prerequisite lessons and the stated planned results.

Sublinear displacement from the real characteristic roots

Lemma 1. Let \(P(\zeta,\tau)\) have total degree \(m\ge1\), with principal part \(F\) hyperbolic with respect to the last frequency vector \(N=(0,\ldots,0,1)\). Every root of \(P(\zeta,\tau)=0\) satisfies \[ |\tau|\le C(1+|\zeta|),\qquad \zeta\in\mathbb C^{n-1}, \tag{1} \] and for real \(\xi\in\mathbb R^{n-1}\) also \[ |\operatorname{Im}\tau| \le C(1+|\xi|)^{1-1/m}. \tag{2} \] The constant depends on the full polynomial; every lower-order complex coefficient is allowed.

Proof. Divide by the nonzero constant \(F(N)\), which does not change any root or the hyperbolicity property. The result is monic in \(\tau\): \[ \begin{gathered} P(\zeta,\tau)=\tau^m+\sum_{j=1}^m a_j(\zeta)\tau^{m-j}, \\ \qquad \deg a_j\le j. \end{gathered} \tag{3} \] There is \(C_1\ge1\) with \(|a_j(\zeta)|\le[C_1(1+|\zeta|)]^j\) for all \(j\): bound the finitely many polynomial coefficients and choose \(C_1\) greater than each resulting coefficient bound's \(j\)-th root. If \(|\tau|>2C_1(1+|\zeta|)\), then \[ \left|\sum_{j=1}^m a_j(\zeta)\tau^{m-j}\right| \le|\tau|^m\sum_{j=1}^m2^{-j}<|\tau|^m. \tag{4} \] This contradicts \(P=0\). Thus (1) holds, even for complex \(\zeta\).

For real \(\xi\), hyperbolicity and scalar factorization give \[ \begin{gathered} F(\xi,\tau)=\prod_{\ell=1}^m(\tau-\lambda_\ell(\xi)), \\ \qquad \lambda_\ell(\xi)\in\mathbb R. \end{gathered} \tag{5} \] Multiplicity is included. Hence \(|F(\xi,\tau)|\ge|\operatorname{Im}\tau|^m\). At a root of \(P\), \(F=-(P-F)\). The latter polynomial has total degree at most \(m-1\), and (1) bounds \(|\tau|\) by a fixed multiple of \(1+|\xi|\). Bounding its finitely many monomials gives \(|P-F|\le C_2(1+|\xi|)^{m-1}\). Taking the \(m\)-th root proves (2). For \(m=1\), the exponent is zero, and the imaginary parts are uniformly bounded. We do not extend the real-principal-root argument to complex \(\xi\). \(\square\)

A contour that remains away from every root

Lemma 2. Let \(p(\tau)=\tau^m+a_1\tau^{m-1}+\cdots+a_m\) be a monic scalar polynomial, \(m\ge1\). Suppose each of its roots satisfies \(|\tau|\le A\) and \(|\operatorname{Im}\tau|\le B\), where \(A,B\ge0\). For \(0\le k<m\), the Cauchy problem \[ \begin{gathered} p(D_t)v_k=0,\\ \qquad D_t^jv_k(0)=\delta_{jk}\\ \quad(0\le j<m) \end{gathered} \tag{6} \] has a unique smooth solution on the real line, and, for every integer \(j\ge0\), \[ |D_t^jv_k(t)| \le2^m(A+1)^{j+m-k}e^{(B+1)|t|}. \tag{7} \] Repeated roots are permitted.

Proof. Put \(R=A+1\), \(h=B+1\), and let \(\Omega\) be the intersection of the disk \(|\tau|<R\) with the strip \(|\operatorname{Im}\tau|<h\), with positively oriented boundary. All roots are inside. At a boundary point on the circle, its distance from every root is at least \(R-A=1\); at a boundary point on a horizontal edge, its imaginary-coordinate distance is at least \(h-B=1\). Thus \[ |p(\tau)|=\prod_{\ell=1}^m|\tau-\tau_\ell|\ge1, \qquad \tau\in\partial\Omega. \tag{8} \]

Its perimeter is at most \(2\pi R\). If \(h\ge R\), it is the circle. Otherwise put \(u=h/R\in(0,1)\). The two circular side arcs have total length \(4R\arcsin u\), and the two horizontal chords have total length \(4R\sqrt{1-u^2}\). Their sum is \(4R(\arcsin u+\sqrt{1-u^2})\). The expression in parentheses has derivative \((1-u)/\sqrt{1-u^2}\ge0\) and value \(\pi/2\) at \(u=1\), proving the bound. This retains the constant used below, including the corners where arcs meet chords.

Let \(q_k\) be the polynomial part of \(p(\tau)\tau^{-1-k}\). Its degree is \(m-1-k\). Define \[ v_k(t)=\frac1{2\pi i} \int_{\partial\Omega}e^{it\tau}\frac{q_k(\tau)}{p(\tau)}\,d\tau. \tag{9} \] The contour is compact and avoids all poles, so differentiation of every time order is legitimate. Applying \(p(D_t)\) cancels its denominator; the remaining entire function \(e^{it\tau}q_k(\tau)\) has zero contour integral by Cauchy's formula. Thus the equation holds.

For the initial conditions, \[ \begin{gathered} \frac{q_k(\tau)}{p(\tau)} =\tau^{-1-k}+O(\tau^{-m-1})\\ \quad\text{at infinity}. \end{gathered} \tag{10} \] Indeed, subtract the negative-power part of \(p(\tau)\tau^{-1-k}\); its highest power is at most \(-1\), and division by the monic degree-\(m\) polynomial gives the displayed remainder. For \(j<m\), the coefficient of \(\tau^{-1}\) in \(\tau^jq_k/p\) is precisely \(\delta_{jk}\). Its integral over our boundary equals the integral over a large positive circle: the rational function has no poles between those contours, so the annular Cauchy formula applies. Expanding its convergent Laurent series on that circle selects exactly the coefficient of \(\tau^{-1}\). In (9), \(D_t^je^{it\tau}=\tau^je^{it\tau}\); hence all the initial values in (6) follow.

The scalar coefficients obey \(|a_\ell|\le\binom m\ell A^\ell\), by their elementary-symmetric root formula. On \(|\tau|\le R\) this gives \[ \begin{gathered} |q_k(\tau)| \\ \le R^{-1-k}\sum_{\ell=0}^m\binom m\ell A^\ell R^{m-\ell} \\ =R^{-1-k}(R+A)^m \\ \le2^mR^{m-1-k}. \end{gathered} \tag{11} \] Here \(a_0=1\); extending the sum through \(m\) only increases its nonnegative bound. On our contour, \(|\tau|^j\le R^j\) and \(|e^{it\tau}|\le e^{h|t|}\). Combine (8), (11), the perimeter bound and the prefactor \(1/(2\pi)\) in (9). The result is exactly (7).

Finally, uniqueness does not require distinct roots. The vector of the first \(m\) \(D_t\)-jets of a difference of two solutions satisfies a constant companion system \(\partial_tV=iCV\) with \(V(0)=0\). Integration on an interval of length \(\ell\) gives \(\sup|V|\le\|C\|\ell\sup|V|\). Choose \(\|C\|\ell<1\), or any length if \(C=0\); the supremum must be zero. Iterate in both directions on the real line. The difference is zero everywhere. Existence was already supplied by (9). \(\square\)

Exercises with complete solutions

Exercise 1 — basic: the sideways heat scale. Regard \(-\partial_x^2+\partial_t\) as a second-order equation with Cauchy normal in the \(x\)-direction. Compute the time-kernel roots after Fourier transformation in \(t\), and verify the exponent in Lemma 1.

Solution. With \(D=-i\partial\), the symbol is \(P(\xi,\tau)=\tau^2+i\xi\), where \(\xi\) is the frequency of \(t\) and \(\tau\) that of \(x\). Its principal part is \(\tau^2\), hyperbolic with respect to the last vector, since its repeated characteristic root is real. For real \(\xi\ne0\), the roots satisfy \(\tau^2=-i\xi\); therefore \[ |\tau|=|\xi|^{1/2},\qquad |\operatorname{Im}\tau|=(|\xi|/2)^{1/2}. \tag{12} \] At \(\xi=0\) both roots are zero. The growth exponent \(1-1/m=1/2\) is attained. This identifies the root growth that the full endpoint small-Gevrey order \(\delta=2\) must control; it is not yet a proof of the full Cauchy existence theorem.

Exercise 2 — intermediate: repeated roots create no denominator loss. For \(p(\tau)=(\tau-iB)^m\), \(B\ge0\), find \(v_{m-1}\) in (6) and check the first \(m\) jets.

Solution. The function \[ v_{m-1}(t)=\frac{(it)^{m-1}}{(m-1)!}e^{-Bt} \tag{13} \] has \((D_t-iB)^m v_{m-1}=0\): multiplying by \(e^{Bt}\) conjugates \(D_t-iB\) to \(D_t\), which annihilates the degree-\((m-1)\) polynomial after \(m\) derivatives. Every derivative of order less than \(m-1\) at zero is zero. At order \(m-1\), its ordinary derivative is \(i^{m-1}\); multiplying by \((-i)^{m-1}\) gives one. Thus all prescribed \(D_t\)-jets agree, and uniqueness identifies it with the contour kernel. The polynomial factor records the root multiplicity, while Lemma 2 remains valid with \(A=B\) and no inverse root gap.

Exercise 3 — advanced: the general displacement exponent is sharp. For \(m\ge2\), consider \(P(\xi,\tau)=(\tau-\xi)^m+i\xi^{m-1}\). Show that its principal part is hyperbolic and that an imaginary root displacement of order \(\xi^{1-1/m}\) occurs as real \(\xi\to+\infty\).

Solution. The principal part is \((\tau-\xi)^m\), with the repeated real root \(\tau=\xi\) for every real \(\xi\). Its value at the last frequency vector is 1. For \(\xi>0\), choose any \(c\) with \(c^m=-i\); then \[ \tau=\xi+c\xi^{(m-1)/m} \tag{14} \] is an exact root. At least one such \(c\), and in fact every real candidate fails, has nonzero imaginary part: a real number's integer power is real, whereas \(-i\) is not. Hence \(|\operatorname{Im}\tau|=|\operatorname{Im}c|\xi^{1-1/m}\). The exponent of Lemma 1 cannot be reduced uniformly over this class of full polynomials. The lower-order term and its complex coefficient are retained, rather than replaced by a condition on the full polynomial's hyperbolicity.

References