Sentence examples for have a singularity from inspiring English sources

A workaround for this problem is to rewrite this ratio using a function that does not have a singularity.

It can happen that even when the Aij are smooth functions of t or constant, the solution X may have a singularity or even infinitely many singularities.

The interesting point is that the weak singularity of restoring force (frac{alpha(t)}{x^{mu}}) at (x=0) is allowed and f may have a singularity at (x=0).

Analytic continuation of the function (G x,t,lambda)) with respect to λ out of sectors (S^) and (S^) does not have a singularity on the line (operatorname{Re}(lambdaomega_{0})=0}.

The interesting point is that the weak singularity of restoring force term (frac{alpha (t)}{x^{mu}}) at (x=0) is allowed and f may have a singularity at (x=0).

Consider the Schrödinger operator, either in a bounded domain (Xni 0) or in (mathbb {R}^d) like in Example 3.2(i) and assume that (g^{jk}(x)) and V x) have a singularity at 0 satisfying there (3.14)–(3.16) with (gamma (x)=|x|) and (rho (x)=|x|^m) with (m<-1).

This paper investigates the singular differential equation, having a singularity at.

The paper studies the singular differential equation, which has a singularity at.

As far as we assume a non-singular potential ((V(Phi )0), the vacuum solution has a singularity at the origin.

Wang and Ma [14] first studied the resonant singular equation x"+frac{1}{4}n^{2}x+g(x)=p t), (1.2) where g has a singularity and satisfies lim_{xrightarrowinfty}g(x)=g(+infty).

In 2015, Wang and Ma [12] investigated the following singular Rayleigh equation: x"+fbigl t,x'bigr)+g(x)=p t), where g had a singularity at the origin, i.e., (lim_{xto +infty} g(x)=+infty).