Nonisothermal nematic liquid crystal flows with th

Authors: Eduard Feireisl Giulio Schimperna Elisabetta Rocca Arghir Zarnescu

Publish Date: 2014/04/21

Volume: 194, Issue: 5, Pages: 1269-1299

PDF Link

Abstract

In this paper we prove the existence of globalintime weak solutions for an evolutionary PDE system modelling nonisothermal Landau–de Gennes nematic liquid crystal LC flows in three dimensions of space In our model the incompressible Navier–Stokes system for the macroscopic velocity mathbfu is coupled to a nonlinear convective parabolic equation describing the evolution of the Qtensor mathbb Q namely a tensorvalued variable representing the normalized secondorder moments of the probability distribution function of the LC molecules The effects of the absolute temperature vartheta are prescribed in the form of an energy balance identity complemented with a global entropy production inequality Compared to previous contributions we can consider here the physically realistic singular configuration potential f introduced by Ball and Majumdar This potential gives rise to severe mathematical difficulties since it introduces in the Qtensor equation a term that is at the same time singular in mathbb Q and degenerate in vartheta To treat it a careful analysis of the properties of f particularly of its blowup rate is carried outIn order to apply the previous argument let us observe that the function gamma m p m2+q m2 of variable pq=p 1p 2p 3q 1q 2q 3in mathbb S2times mathbb S2 attains its maximum value at two points that depend on the maximum element in the set gamma 1gamma 2gamma 3 Let us assume without the loss of generality that gamma 2gamma 3 gamma 1 Then the maximum of the function gamma m p m2+q m2 is attained at two points pp with pin 100100 so in order to apply Laplace’s method we need to split mathbb S2times mathbb S2 into two subdomains Let us denote mathbb S E=pin mathbb S2 pcdot 1000 We then apply the previously mentioned Laplace’s method on each of the sets mathcal E=mathbb S Etimes mathbb S E and mathcal V=mathbb S2times mathbb S2setminus mathcal E chosing h=rho and fp 1p 2p 3q 1q 2q 3=gamma 1gamma i p i2+q i2 note that we can multiply both denominator and numerator in mathcal I ij by erho gamma 1

Keywords: