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Boundedness in a quasilinear parabolic-parabolic chemotaxis system with nonlinear logistic source

Ji LiuJia-Shan Zheng — 2015

Czechoslovak Mathematical Journal

We study a quasilinear parabolic-parabolic chemotaxis system with nonlinear logistic source, under homogeneous Neumann boundary conditions in a smooth bounded domain. By establishing proper a priori estimates we prove that, with both the diffusion function and the chemotaxis sensitivity function being positive, the corresponding initial boundary value problem admits a unique global classical solution which is uniformly bounded. The result of this paper is a generalization of that of Cao (2014).

Some finite generalizations of Euler's pentagonal number theorem

Ji-Cai Liu — 2017

Czechoslovak Mathematical Journal

Euler's pentagonal number theorem was a spectacular achievement at the time of its discovery, and is still considered to be a beautiful result in number theory and combinatorics. In this paper, we obtain three new finite generalizations of Euler's pentagonal number theorem.

Reconstruction algorithms for an inverse medium problem

Ji-Chuan Liu — 2018

Applications of Mathematics

In this paper, we consider a two-dimensional inverse medium problem from noisy observation data. We propose effective reconstruction algorithms to detect the number, the location and the size of the piecewise constant medium within a body, and then we try to recover the unknown shape of inhomogeneous media. This problem is nonlinear and ill-posed, thus we should consider stable and elegant approaches in order to improve the corresponding approximation. We give several examples to show the viability...

Truncations of Gauss' square exponent theorem

Ji-Cai LiuShan-Shan Zhao — 2022

Czechoslovak Mathematical Journal

We establish two truncations of Gauss’ square exponent theorem and a finite extension of Euler’s identity. For instance, we prove that for any positive integer n , k = 0 n ( - 1 ) k 2 n - k k ( q ; q 2 ) n - k q k + 1 2 = k = - n n ( - 1 ) k q k 2 , where n m = k = 1 m 1 - q n - k + 1 1 - q k and ( a ; q ) n = k = 0 n - 1 ( 1 - a q k ) .

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