ENERGY CONSERVATIVE SOLUTIONS TO A NONLINEAR WAVE SYSTEM OF NEMATIC LIQUID CRYSTALS

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1 ENERGY CONSERVATIVE SOLUTIONS TO A NONLINEAR WAVE SYSTEM OF NEMATIC LIQUID CRYSTALS GENG CHEN, PING ZHANG, AND YUXI ZHENG Abstract. We establish the global existence of solutions to the Cauchy problem for a system of hyperbolic partial differential equations in one space dimension modeling a type of nematic liquid crystals that has equal splay and twist coefficients. Our results have no restrictions on the angles of the director, as we use the director in its natural three-component form, rather than the two-component form of spherical angles.. Introduction In this paper, we consider the global existence of energy conservative weak solutions to the following one-dimensional wave system of liquid crystals tt n x c 2 x n ) = n t 2 + 2c 2 γ) n x 2) n,.) tt n 2 x c 2 x n 2 ) = n t 2 + 2c 2 α) n x 2) n 2, tt n 3 x c 2 x n 3 ) = n t 2 + 2c 2 α) n x 2) n 3, together with initial data.2) n i t=0 = n i0 H, n i ) t t=0 = n i L 2, i =, 2, 3, where c depends on n with c 2 = α + γ α)n 2, α > 0, γ > 0, and n = n, n 2, n 3 ) with.3) n =. Subscripts t or x represent partial derivatives with respect to t or x, and tt = t t. We firstly mention briefly the origin of this system. The mean orientation of the long molecules in a nematic liquid crystal is described by a director field of unit vectors, n S 2, the unit sphere. Associated with the director field n, there is the well-known Oseen-Franck potential energy density W given by.) W n, n) = 2 α n)2 + 2 β n n)2 + 2 γ n n) 2. The positive constants α, β, and γ are elastic constants of the liquid crystal, corresponding to splay, twist, and bend, respectively. For the special case α = β = γ, the potential energy density reduces to W n, n) = 2 α n 2, which is the potential energy density used in harmonic maps into S 2. There are many studies on the constrained elliptic system of equations for n derived through variational principles from the potential.), and on the parabolic flow associated with it, see [2, 5, 6, 8, 9, 7 and references therein. In the regime in which inertia effects dominate viscosity, however, the propagation of the orientation waves in the director field may then be modelled by the least action principle [0, ) { } δ.5) δn 2 tn t n W n, n) dx dt = 0, n n =. R Date: March 9, 202.

2 2 G. CHEN, P. ZHANG, AND Y. ZHENG In the special case α = β = γ, this variational principle.5) yields the equation for harmonic wave maps from +3)-dimensional Minkowski space into S 2, see [,, 2 for example. One may also check [3,, 5, 6 for wave maps in dimension + 2). For planar deformations depending on a single space variable x, i.e, the director field has the special form n = cos ux, t)e x + sin ux, t)e y, where the dependent variable u R measures the angle of the director field to the x- direction, and e x and e y are the coordinate vectors in the x and y directions, respectively, one finds that the variational principle.5) yields.6) u tt c c u x ) x = 0 with the wave speed c given specifically by.7) c 2 u) = γ cos 2 u + α sin 2 u. If we let n be arbitrary in S 2 while maintaining dependence on a single space variable x, i.e.,.8) n = cos u, sin u cos v, sin u sin v) where u, v) are both functions of x, t), we find that the Lagrangian density of.5) is.9) 2 tn t n W n, n) = [ u 2 2 t c 2 u)u 2 x + 2 a2 u) [ vt 2 c 2 2u)vx 2 where.0) c 2 u) = γ cos 2 u + α sin 2 u, c 2 2u) = γ cos 2 u + β sin 2 u, a 2 u) = sin 2 u. The Euler-Lagrange equations are { utt c u)c u)u x ) x = aa vt 2 c 2 2 v2 x) a 2 c 2 c 2 v2 x,.) a 2 v t ) t c 2 2 a2 v x ) x = 0. See Alì and Hunter [ for more details on the derivation of the above system. Under the additional assumption that c u) 0, Zhang and Zheng established the global existence of energy dissipative weak solutions to.6) in [8, 9. Without any additional assumption, Bressan and Zheng established the global existence of energy conservative weak solutions to.6) in [3. In [20, 2, Zhang and Zheng solved the global existence of energy conservative solutions to.) under the un-physical assumption that au) a min > 0 and c 2 < c for [2). The goal of this paper is to get rid of this assumption in [20. Indeed.5) can be equivalently reformulated as { δ δn 2 tn t n W n, x n) + λ } 2 n 2 ) dx dt = 0, R 2 which gives rise to the Euler-Langrage equations.2) tt n i + ni W n, x n) x [ xn i W n, x n) = λn i, for i =, 2, 3. Using n =, multiplying.2) with n i, and summing them up for i from to 3, we obtain 3 {.3) λ = t n i 2 [ + n i ni W n, x n) n i x xni W n, x n) }. i= It is easy to calculate that.2) has the following energy conservation law: [.) t 2 n t 2 + W n, x n) [ 3 x t n i xni W n, x n) = 0. i=

3 ENERGY CONSERVATIVE SOLUTIONS TO LIQUID CRYSTALS 3 On the other hand, we use.) in the one-dimensional case to obtain.5) W n, x n) = α 2 xn ) 2 + β 2 [ x n 2 ) 2 + x n 3 ) γ β)n2 n x 2, from which and.3), we infer.6) λ = n t 2 + β + 2γ β)n 2 ) nx 2 + β α)n 2 xn. Combining.2),.5) and.6), we have.7) tt n x [ c 2 x n = { nt 2 + 2c 2 2 γ) n x 2 + 2α β) x n ) 2} n with Similarly, we have.8) c 2 def = α + γ α)n 2 and c 2 2 = β + γ β)n 2. tt n 2 x [ c 2 2 x n 2 = { nt 2 + 2c 2 2 β) n x 2 + β α)n xx n } n2, tt n 3 x [ c 2 2 x n 3 = { nt 2 + 2c 2 2 β) n x 2 + β α)n xx n } n3. In particular, taking α = β in.7) and.8) leads to.). When α = β, the two speeds are equal, so we let.9) c 2 = c 2 = c 2 2 = α + γ α)n 2, and it follows that.20) 0 < min{α, γ} c 2 max{α, γ} < ; max{ c } <, for any n. The energy equation.) becomes.2) 2 [ t nt 2 + c 2 n x 2 [ x c 2 n t n x = 0, where the energy density W in.5) becomes.22) W n, x n) = 2 c2 x n 2. We consider in this paper the global existence of energy-conservative weak solutions of.).3) to its initial value problem with data.2). Since some smooth initial data for equation.6) is known to form singularity later in time, it is easy to see that the same is true for system.) and.). Thus we shall give up classical solutions and consider weak solutions instead. Definition. Weak solution). The vector function nt, x), defined for all t, x) R R, is a weak solution to the Cauchy problem.).3) if it satisfies i): In the t-x plane, the functions n, n 2, n 3 ) are locally Hölder continuous with exponent /2. This solution t n, n 2, n 3 )t, ) is continuously differentiable as a map with values in L p loc, for all p < 2. Moreover, it is Lipschitz continuous with.23) respect to w.r.t.) the L 2 distance, i.e. n i t, ) n i s, ) L 2 L t s, i = 3, for all t, s R. ii): The functions n, n 2, n 3 ) take on the initial conditions in.2) pointwise, while their temporal derivatives hold in L p loc for p [, 2[. iii): The equations.) hold in distributional sense for test functions φ Cc R R).

4 G. CHEN, P. ZHANG, AND Y. ZHENG Our conclusions are as follows. Theorem. Existence). The problem.).3) has a global weak solution defined for all t, x) [0, ) R. We shall use the method of energy-dependent coordinates, used in papers [3 and related Camassa-Holm equation, to dilate the singularity in building the solution. Our constructive procedure yields solutions which depend continuously on the initial data. Moreover, the energy.2) Et) := 2 [ n t 2 + c 2 n x 2 dx remains uniformly bounded by its initial level:.25) E 0 := 2 [ n x) 2 + c 2 n 0 x)) n 0 x)) x 2 dx. More precisely, we have Theorem.2 Continuous dependence). A family of weak solutions to the Cauchy problem.).3) can be constructed with the additional properties: For every t R we have.26) Et) E 0. Moreover, let a sequence of initial conditions satisfy k n i0 ) x n i0 ) L x 2 0, k ni n L i 2 0, i = 3, and n k 0 n 0 uniformly on compact sets, as k. Then we have the convergence of the corresponding solutions n k n, uniformly on bounded subsets of the t-x plane. It appears in.26) that the total energy of our solutions may decrease in time. Yet, we emphasize that our solutions are conservative, in the following sense. Theorem.3 Conservation of energy). There exists a continuous family {µ t ; t R} of positive Radon measures on the real line with the following properties. i): At every time t, we have µ t R) = E 0. ii): For each t, the absolutely continuous part of µ t has density 2 nt 2 + c 2 n x 2) w.r.t. Lebesgue measure. iii): For almost every t R, the singular part of µ t is concentrated on the set where n = 0 or ±, when α γ. In other words, the total energy represented by the measure µ is conserved in time. Occasionally, some of this energy is concentrated on a set of measure zero. At those times τ when this happens, µ τ has a non-trivial singular part and Eτ) < E 0. Condition iii) places some restrictions on the set of such times τ. The paper is organized as follows. In Section 2 we introduce a new set of independent and dependent variables, and derive some identities valid for smooth solutions. We formulate a set of equations in the new variables which is equivalent to.). Remarkably, in the new variables all singularities disappear: Smooth initial data lead to globally smooth solutions. In Section 3, we prove the existence of the solutions in the energy coordinates. In Sections 7, we present proofs to Theorems..3.

5 ENERGY CONSERVATIVE SOLUTIONS TO LIQUID CRYSTALS 5 2. New formulation in energy-dependent coordinates 2.. Energy-dependent coordinates. We denote 2.) R = R, R 2, R 3 ) def = n t + cn x, S = S, S 2, S 3 ) def = n t cn x. Then.) can be reformulated as: t R c x R = c 2 { c 2 γ) R 2 + S 2 ) 23c 2 γ) R S } n 2.2) + c 2c R S )R, t S + c x S = c 2 { c 2 γ) R 2 + S 2 ) 23c 2 γ) R S } n c 2c R S )S, t R 2 c x R 2 = c 2 { c 2 α) R 2 + S 2 ) 23c 2 α) R S } n 2 + c 2c R 2 S 2 )R, t S 2 + c x S 2 = c 2 { c 2 α) R 2 + S 2 ) 23c 2 α) R S } n 2 c 2c R 2 S 2 )S, t R 3 c x R 3 = c 2 { c 2 α) R 2 + S 2 ) 23c 2 α) R S } n 3 + c 2c R 3 S 3 )R, t S 3 + c x S 3 = c 2 { c 2 α) R 2 + S 2 ) 23c 2 α) R S } n 3 n x = R S 2c or n t = R+ S 2. c 2c R 3 S 3 )S, Similar to.), system 2.2) has the following form of energy conservation law: 2.3) t R 2 + S 2) [ x cn ) R 2 S 2 ) = 0. We define the forward and backward characteristics as follows { d ds x± s, t, x) = ±cn s, x ± s, t, x))), 2.) x ± s=t = x. Then we define the coordinate transformation: X def = This implies x 0,t,x) 0 [ + R 2 0, y) dy, and Y def = 0 x + 0,t,x) 2.5) X t cx x = 0, Y t + cy x = 0. Furthermore, for any smooth function f, we obtain by using 2.5) that [ + S 2 0, y) dy. 2.6) f t + cf x = X t + cx x )f X = 2cX x f X f t cf x = Y t cy x )f Y = 2cY x f Y.

6 6 G. CHEN, P. ZHANG, AND Y. ZHENG Introduce 2.7) p def = + R 2 X x and q def = + S 2 Y x. Then By 2.2), we have t p c x p = 2X x ) [ R t R cn ) x R) X x ) 2 [ t X x c x X x + R 2 ). R tr cn ) xr) { = c 2 c 2 α) R 2 + S 2 ) 23c 2 α) R S } R n + α γ R 2 c 2 + S 2 2R S ) R n + c 2c R R 2 R S). Using n = and c = γ α)n c, so that R n = 0, we further obtain 2.8) R t R cn ) x R) = c c R R 2 S 2 ), which together with 2.5) applied gives t p c x p = c p 2c + R [ R + S 2 ) + S + R 2 ), 2 from which, and ), we infer 2.9) Similarly, p Y = 2c Y x ) p t cp x ) = c c 2 pq R + R 2 + S + S 2 ). t q + c x q = 2 Y x ) [ S t S + cn ) x S) while again thanks to 2.2), we have +Y x ) 2 t Y x + c x Y x ) + S 2 ), S ts cn ) xs) { = c 2 c 2 α) R 2 + S 2 ) 23c 2 α) R S } S n + α γ R 2 c 2 + S 2 2R S ) S n + c 2c S S 2 R S), which along with the fact that S n = 0 leads to 2.0) S t S cn ) x S) = c c S R 2 S 2 ), from which and 2.5), we infer t q + c x q = c q 2c + S [R + S 2 ) S + R 2 ). 2

7 Then applying ) gives rise to 2.) Now we introduce 2.2) Then ENERGY CONSERVATIVE SOLUTIONS TO LIQUID CRYSTALS 7 l = l, l 2, l 3 ) def = h def = q X = q t + cq x ) 2cX x = c c 2 pq R + R S ) 2 + S. 2 + R, h def 2 2 = R + R, m = m, m 2, m 3 ) def = 2 + S 2. t l c x l = + R 2 ) t R c x R ) which together with 2.2) and 2.8) imply S + S 2, and 2 + R 2 ) 2 R [ R t R cn ) x R), t l c x l = + R 2 ) [ c 2 γ c 2 R 2 + S 2 ) 3c2 γ R 2c 2 S n + c 2c + R 2 ) 2 [ R R + S 2 ) S + R 2 ). Then thanks to ), we obtain Y l = q [ c 2 8c 3 γ)h + h 2 2h h 2 ) 23c 2 γ) l m n 2.3) + c c 2 l ql m ). Similarly thanks to 2.2), we have t m + c x m = + S 2 ) t S + c x S ) which together with 2.2) and 2.0) ensure that 2 + S 2 ) 2 S [ S t S + cn ) x S), t m + c x m = + S 2 ) [ c 2 γ c 2 R 2 + S 2 ) 3c2 γ R 2c 2 S n c 2c + S 2 ) 2 [ S R + S 2 ) S + R 2 ), from which and ), we infer X m = p [ c 2 8c 3 γ)h + h 2 2h h 2 ) 23c 2 γ) l m n 2.) c c 2 m pl m ).

8 8 G. CHEN, P. ZHANG, AND Y. ZHENG Following the same line, we deduce from 2.2) and 2.8) that t l 2 c x l 2 = + R 2 ) [ c 2 α c 2 R 2 + S 2 ) 3c2 α R 2c 2 S n 2 + c 2c + R 2 ) 2 [ R R2 + S 2 ) S 2 + R 2 ), and t l 3 c x l 3 = + R 2 ) [ c 2 α c 2 R 2 + S 2 ) 3c2 α R 2c 2 S n 3 + c 2c + R 2 ) 2 [ R R3 + S 2 ) S 3 + R 2 ), which together with ) ensure that 2.5) Y l 2 = q [ c 2 8c 3 α)h + h 2 2h h 2 ) 23c 2 α) l m n 2 + c c 2 l ql 2 m 2 ) and Y l 3 = q [ c 2 8c 3 α)h + h 2 2h h 2 ) 23c 2 α) l m n 3 + c c 2 l ql 3 m 3 ). While we deduce from 2.2) and 2.0) that and t m 2 + c x m 2 = + S 2 ) [ c 2 α c 2 R 2 + S 2 ) 3c2 α R 2c 2 S n 2 c 2c + S 2 ) 2 [ S R2 + S 2 ) S 2 + R 2 ), t m 3 + c x m 3 = + S 2 ) [ c 2 α c 2 R 2 + S 2 ) 3c2 α R 2c 2 S n 3 c 2c + S 2 ) 2 [ S R3 + S 2 ) S 3 + R 2 ), which together with ) imply that 2.6) X m 2 = p [ c 2 8c 3 α)h + h 2 2h h 2 ) 23c 2 α) l m n 2 c c 2 m pl 2 m 2 ) and X m 3 = p [ c 2 8c 3 α)h + h 2 2h h 2 ) 23c 2 α) l m n 3 c c 2 m pl 3 m 3 ).

9 ENERGY CONSERVATIVE SOLUTIONS TO LIQUID CRYSTALS 9 On the other hand, it follows from 2.8) and 2.2) that t h c x h = 2 + R 2 ) 2[ R t R cn ) x R) Then we obtain by using ) that = c 2c + R 2 ) 2 R R 2 S 2 ). 2.7) Y h = c c 2 ql h h 2 ). Similar calculations together with 2.0) yield t h 2 + c x h 2 = c 2c + S 2 ) 2 S R 2 S 2 ), which together with ) imply that 2.8) X h 2 = c c 2 pm h 2 h ). Finally we observe from 2.6) and 2.2) that 2.9) Y n = t n c x n) = 2cX x X n = 2c Y x ) tn + c x n) = q 2c m, p 2c l. In summary, we obtain Y l = q 8c 3 [ c 2 γ)h + h 2 2h h 2 ) 23c 2 γ) l m n 2.20) + c c 2 l ql m ), X m = p 8c 3 [ c 2 γ)h + h 2 2h h 2 ) 23c 2 γ) l m n Y l 2 = c c 2 m pl m ), q 8c 3 [ c 2 α)h + h 2 2h h 2 ) 23c 2 α) l m n 2 + c c 2 l ql 2 m 2 ), X m 2 = p 8c 3 [ c 2 α)h + h 2 2h h 2 ) 23c 2 α) l m n 2 Y l 3 = c c 2 m pl 2 m 2 ), q 8c 3 [ c 2 α)h + h 2 2h h 2 ) 23c 2 α) l m n 3 + c c 2 l ql 3 m 3 ), X m 3 = p 8c 3 [ c 2 α)h + h 2 2h h 2 ) 23c 2 α) l m n 3 c c 2 m pl 3 m 3 ), Y n = q 2c m, or Xn = p 2c l) Y h = c c 2 ql h h 2 ), X h 2 = c c 2 m ph 2 h ), p Y = c c 2 pql m ), q X = c c 2 pql m ).

10 0 G. CHEN, P. ZHANG, AND Y. ZHENG 2.2. Consistency of variables of 2.20). The various variables introduced for 2.20) are consistent, following from the proposition below. Proposition 2.. For smooth enough data, there hold the following conservative quantities: 2.2) l nx, Y ) = m nx, Y ) = 0, nx, Y ) = and lx, Y ) 2 + h 2 X, Y ) = h X, Y ), mx, Y ) 2 + h 2 2X, Y ) = h 2 X, Y ), X, Y, as long as l nx, ϕx)) = m nx, ϕx)) = 0, nx, ϕx)) = and lx, ϕx)) 2 + h 2 X, ϕx)) = h X, ϕx)), mx, ϕx)) 2 + h 2 2X, ϕx)) = h 2 X, ϕx)), X. Here the function ϕ represents the initial curve, see 3.2) from the next section. Proof. We first deduce from 2.20) that [ Y l 2 + h 2 q [ h = c 2 c 3 α)h + h 2 2h h 2 ) 23c 2 α) l m l n + α γ)n c 3 q[ h + h 2 2h h 2 ) 2 l m l + c c 2 ql [ 2 l 2 l m) + 2h h h 2 ) h h 2 ), from which and the fact that c = γ α)n c, we deduce that [ Y l 2 + h 2 q [ h = c 2 c 3 α)h + h 2 2h h 2 ) 2.22) 23c 2 α) l m c l n + 2c 2 ql [ l 2 + h 2 h. Similarly it follows from 2.20) that Y [ l n = Y l n + l Y n = q [ c 2 8c 3 α)h + h 2 2h h 2 ) 23c 2 α) l m n 2 ) q + 8c 3 h + h 2 2h h 2 ) [ c 2 α) + α γ)n 2 q c 3 l m [ 3c 2 α + α γ)n 2 q + c 2 ql [ l n m n + q 2c l m, which along with the fact that c 2 = α + γ α)n 2 leads to Y [ l n = q [ c 2 8c 3 α)h + h 2 2h h 2 ) 23c 2 α) l m n 2 ) 2.23) + c c 2 ql [ l n m n.

11 ENERGY CONSERVATIVE SOLUTIONS TO LIQUID CRYSTALS On the other hand, observe that Y n = 2c m is consistent with Xn = again thanks to 2.20), one has on the one hand X [ Y n = c 8c 3 pql + m ) m + q 2c X m, and on the other hand Y [ X n = c 8c 3 pql + m ) l + p 2c Y l, q p 2c l. Indeed which along with the l equations and m equations of 2.20) shows that X [ Y n = Y [ X n. So we can also use the equation X n = p 2c l, from which and the m equation of 2.20), we obtain X [ m n = X m n + m X l = p [ c 2 8c 3 α)h + h 2 2h h 2 ) 23c 2 α) l m n 2 ) p [ + c 2 8c 3 α)h + h 2 2h h 2 ) 23c 2 α) l m + α γ [ 8c 3 pn2 h + h 2 2h h 2 ) 2 l m + c c 2 pm [ m n l n + p 2c l m, which gives rise to Y [ m n = p [ c 2 8c 3 α)h + h 2 2h h 2 ) 23c 2 α) l m n 2 ) 2.2) c c 2 pm [ l n m n. Next it is easy to observe that 2.25) Y [ n 2 = q m n. 2c Finally, to control the evolution of m 2 + h 2 2 h 2, we obtain by applying 2.20) that [ X m 2 + h 2 p [ 2 h 2 = c 2 c 3 α)h + h 2 2h h 2 ) 23c 2 α) l m m n + α γ c 3 p[ h + h 2 2h h 2 ) 2 l m m n + c 2c 2 pm [ m 2 l m) + h 2 2 )h 2 h ), which simplifies to [ X m 2 + h 2 p [ 2 h 2 = c 2 c 3 α)h + h 2 2h h 2 ) 2.26) 23c 2 α) l m m n + c 2c 3 pm [ m 2 + h 2 2 h 2. Summing up 2.22) to 2.26) gives rise to 2.2). This completes the proof of the proposition.

12 2 G. CHEN, P. ZHANG, AND Y. ZHENG 3. Solutions in the energy coordinates Using 2.6) by letting f = t or x, we obtain the equations 3.) t X = ph 2c, t Y = qh 2 2c, x X = ph 2, x Y = qh 2 2. Only one of the two equations of t X and t Y is needed for recovering t, and the two equations are consistent since t XY = t Y X. The same is true for x. The initial line t = 0 in the x, t) plane is transformed to a parametric curve 3.2) γ : Y = ϕx) in the X, Y ) plane, where Y = ϕx) if and only if there is an x such that X = x 0 [ + R 2 0, y) dy, 3.3) Y = 0 x [ + S 2 0, y) dy. We point out that the curve is non-characteristic. We introduce 3.) E 0 := [ R 2 0, y) + S 2 0, y) dy <. It equals to the number in.25). The two functions X = Xx), Y = Y x) from 3.3) are well-defined and absolutely continuous, provided that.2) is satisfied. Clearly, X is strictly increasing while Y is strictly decreasing. Therefore, the map X ϕx) is continuous and strictly decreasing. From 3.) it follows 3.5) X + ϕx) E 0. As t, x) ranges over the domain [0, ) R, the corresponding variables X, Y ) range over the set 3.6) Ω + := { X, Y ) ; Y ϕx) }. Along the curve γ := { X, Y ) ; Y = ϕx) } R 2 parametrized by x Xx), Y x) ), we can thus assign the boundary data l, m, h, h 2, p, q, n) L defined by their definition evaluated at the initial data.2), i.e., 3.7) where n = n 0 x), p =, q =, l = R0, x) h, m = S0, x) h2, h = h 2 = R0, x) = n x) + cn 0 x))n 0x), + R 2 0, x), and + S 2 0, x). S0, x) = n x) cn 0 x))n 0x). We consider solutions to the boundary value problem 2.20)3.7).3). Theorem 3.. The problem 2.20)3.7).2).3) has a unique global solution defined for all X, Y ) R 2.

13 ENERGY CONSERVATIVE SOLUTIONS TO LIQUID CRYSTALS 3 Sketch of Proof. In the following, we shall construct the solution on the domain Ω + where Y ϕx). On the complementary set Ω where Y < ϕx), the solution can be constructed similarly. Observing that all equations in 2.20) have a locally Lipschitz continuous right hand side, the construction of a local solution as fixed point of a suitable integral transformation is straightforward. To make sure that this solution is actually defined on the whole domain Ω +, one must establish a priori bounds, showing that the solution remains bounded on bounded sets. By 2.2), we have 3.8) h h ) 0, h 2 h 2 ) 0. Thus h, h 2 are bounded between zero and one, and both l and m are uniformly bounded. By the p and q equations in 2.20), we have 3.9) p Y + q X = 0 which implies that X ϕ Y ) px, Y )dx + Y ϕx) qx, Y )dy = X ϕ Y ) + Y ϕx) where ϕ denotes the inverse of ϕ, following an integration over the characteristic triangle with vertex X, Y ). Thus, by the energy assumption 3.), we find 3.0) X ϕ Y ) px, Y )dx + Y ϕx) qx, Y ) dy 2 X + Y + E 0 ). Integrating the p equation in 2.20) vertically and use the bound on q from 3.0), we find { } Y c px, Y ) = exp ϕx) c 2 qx, Y ) l + m ) dy 3.) } exp {C 0 YϕX) qx, Y ) dy exp{2c 0 X + Y + E 0 )}. Here C 0 represents a finite number. Similarly, we have 3.2) qx, Y ) exp{2c 0 X + Y + E 0 )}. Relying on the local bounds 3.)3.2), the local solution to 2.20)3.7).2).3) can be extended to the entire plane. One may consult paper [3 for details. This completes the sketch of the proof. Let us state a useful consequence of the above construction for future reference. Corollary 3.. If the initial data n 0, n ) are smooth, the solution U := n, p, q, l, m, h, h 2 ) of 2.20)3.7).2).3) is a smooth function of the variables X, Y ). Moreover, assume that a sequence of smooth functions n i 0, ni ) i satisfies n i 0 n 0, n i 0) x n 0 ) x, n i n uniformly on compact subsets of R. Then one has the convergence of the corresponding solutions: n i, p i, q i, l i, m i, h i, h i 2) U uniformly on bounded subsets of the X-Y plane.

14 G. CHEN, P. ZHANG, AND Y. ZHENG. Inverse transformation By expressing the solution nx, Y ) in terms of the original variables t, x), we shall recover a solution of the Cauchy problem.).3). This will provide a proof of Theorem.. We integrate 3.) with data t = 0, x = x on γ to find t = tx, Y ), x = xx, Y ), which exist for all X, Y ) in R 2. We need the inverse functions X = Xt, x), Y = Y t, x). The inverse functions do not exist as a one-to-one correspondence between t, x) in R 2 and X, Y ) in R 2. There may be a nontrivial set of points in the X, Y ) plane that maps to a single point t, x). To investigate it, we find the partial derivatives of the inverse mapping, valid at points where h 0, h 2 0,.) X t = c ph, Y t = c qh 2, X x = ph, Y x = qh 2. Thus 2.5) holds and so does 2.6) for our solution. As a preliminary, we examine the regularity of the solution U = n, p, q, l,...) constructed in the previous section. Since the initial data n 0 ) x, n etc. are only assumed to be in L 2, the functions U may well be discontinuous. More precisely, on bounded subsets of the X-Y plane, the solutions satisfy the following: -: The functions l, l 2,, l 3, h, p are Lipschitz continuous w.r.t. Y, measurable w.r.t. X. -: The functions m, m 2, m 3, h 2, q are Lipschitz continuous w.r.t. X, measurable w.r.t. Y. -: The vector field n is Lipschitz continuous w.r.t. both X and Y. In order to define n as a vector field of the original variables t, x, we should formally invert the map X, Y ) t, x) and write nt, x) = n Xt, x), Y t, x) ). The fact that the above map may not be one-to-one does not cause any real difficulty. Indeed, given t, x ), we can choose an arbitrary point X, Y ) such that tx, Y ) = t, xx, Y ) = x, and define nt, x ) = nx, Y ). To prove that the values of n do not depend on the choice of X, Y ), we proceed as follows. Assume that there are two distinct points such that tx, Y ) = tx 2, Y 2 ) = t, xx, Y ) = xx 2, Y 2 ) = x. We consider two cases: Case : X X 2, Y Y 2. Consider the set Γ x := {X, Y ) ; xx, Y ) x } and call Γ x its boundary. By 3.), x is increasing in X and decreasing in Y. Hence, this boundary can be represented as the graph of a Lipschitz continuous function: X Y = φx + Y ). We now construct the Lipschitz continuous curve γ consisting of -: a horizontal segment joining X, Y ) with a point A = X A, Y A ) on Γ x, with Y A = Y, -: a portion of the boundary Γ x, -: a vertical segment joining X 2, Y 2 ) to a point B = X B, Y B ) on Γ x, with X B = X 2. Observe that the map X, Y ) t, x) is constant along γ. By 3.) this implies h = 0 on the horizontal segment, h 2 = 0 on the vertical segment, and h = h 2 = 0 on the portion of the boundary Γ x. When either h = 0 or h 2 = 0 or both, we have from the conserved quantities 2.22) that l = 0 or m = 0 or both, correspondingly. Upon examining the derivatives of n in 2.20), we have the same pattern of vanishing property. Thus, along the path from A to B, the values of the components of n remain constant, proving our claim. Case 2: X X 2, Y Y 2. In this case, we consider the set { Γ t := X, Y ) ; tx, Y ) t },

15 ENERGY CONSERVATIVE SOLUTIONS TO LIQUID CRYSTALS 5 and construct a curve γ connecting X, Y ) with X 2, Y 2 ) similarly as in case. Details are entirely similar to Case. We now prove that the function nt, x) = n Xt, x), Y t, x) ) thus obtained are Hölder continuous on bounded sets. Toward this goal, consider any characteristic curve, say t x + t), with dx + /dt = c. By construction, this is parametrized by the function X tx, Y ), xx, Y ) ), for some fixed Y. Using the chain rule and the inverse mapping formulas.), we obtain Thus we have n t + cn x = n X X t + cx x ) + n Y Y t + cy x ) = 2cX x n X. τ 0 nt + cn x 2 dt = Xτ X 0 2cX x n X ) 2 2X t ) dx.2) = X τ X 0 2c ph p l 2c )2 ph 2c dx = X τ X 0 p l 2 2c h dx X τ X 0 p 2c dx C τ, for some constant C τ depending only on τ. Notice we have used l 2 h, which follows from 2.2). Similarly, integrating along any backward characteristics t x t) we obtain τ.3) n t cn x 2 dt Cτ. 0 Since the speed of characteristics is ±c, and c is uniformly positive and bounded, the bounds.2)-.3) imply that the function n = nt, x) is Hölder continuous with exponent /2. In turn, this implies that all characteristic curves are C with Hölder continuous derivative. In addition, from.2)-.3) it follows that R, S at 2.) are square integrable on bounded subsets of the t-x plane. However, we should check the consistency that R, S at 2.) are indeed the same as recovered from 2.2). Let us check only one of them, R = l/h. We find n t + cn x = 2cX x n X = 2c ph p l 2c = l h. Finally, we prove that n satisfies the equations of system.) in distributional sense, according to iii) of Definition.. We note that [ 2 φt n t φ x c 2 n x dxdt.) = φ t [ nt + cn x ) + n t cn x ) cφ x [ nt + cn x ) n t cn x ) dxdt = [ φ t cφ x nt + cn x ) dxdt + [ φ t + cφ x nt cn x ) dxdt By 2.6), this is equal to = [ φ t cφ x R dxdt + [ φt + cφ x S dxdt..5) Using the Jacobian.6) [ 2cYx φ Y R + 2cXx φ X S dxdt. x, t) X, Y ) = pqh h 2 2c

16 6 G. CHEN, P. ZHANG, AND Y. ZHENG derived from 3.), and the inverse.), we see that it is equal to [ 2c RφY + 2c pqh h 2 SφX dxdy qh 2 ph 2c.7) = [ph RφY + qh 2 SφX dxdy = [p lφy + q mφ X dxdy = [ p l)y q m) X φ dxdy. Thanks to 2.), 2.2) and 2.20), we have that the first component of the integrand equals to φ [ pl ) Y qm ) X = φ pq [ c 2 c 3 γ)h + h 2 2h h 2 ) 23c 2 γ) l m n = φ pqh h 2 2c = φ pqh h 2 2c = φ pqh h 2 2c = 2φ pqh h 2 2c [c 2 γ 2c 2 + 2) 3c2 γ l m n h 2 h c 2 h h 2 [c 2 γ 2c 2 R 2 + S 2 ) 3c2 γ R S n c 2 [c 2 γ c 2 n t 2 + c 2 n x 2 ) 3c2 γ c 2 n t 2 c 2 n x 2 ) n. nt 2 + 2c 2 γ) n x 2) n, which implies that the first equation in.) holds in integral form. Similarly, n satisfies the second and third equations of system.) in distributional sense. 5. Upper bound on energy We convert the energy conservation.2) formally to the X, Y ) plane to look for conservation of energy. Recall that the energy conservation law.2) can be written as 2.3) in terms of R and S. By the variables 2.2), we rewrite the equation 2.3) as It yields a closed form 5.) which, using the formula 5.2) can be written as 5.3) h + h 2 2 ) t [ c h h 2 ) x = 0. + ) [ c dx + h h 2 2 ) dt h h 2 dt = t X dx + t Y dy = ph 2c dx + qh 2 2c dy dx = x X dx + x Y dy = ph 2 dx qh 2 2 dy p h ) dx q h 2) dy.

17 ENERGY CONSERVATIVE SOLUTIONS TO LIQUID CRYSTALS 7 This is the energy form in the X, Y ) plane. The solutions n = nx, Y ) constructed in Section 2 are conservative, in the sense that the integral of the form 5.3) along every Lipschitz continuous, closed curve in the X-Y plane is zero. Y A D Ω B γ τ O X C γ 0 Figure. Energy conservation We use 5.3) to establish that the energy of our solution is bounded, i.e., the energy inequality.26). Fix τ > 0, and r >>. Define the set { } 5.) Ω := X, Y ) ; 0 tx, Y ) τ, X r, Y r. See Figure, where segment AD is where Y = r while segment BC is where X = r. By construction, the map X, Y ) t, x) will act as follows: A τ, a), B τ, b), C 0, c), D 0, d), for some a < b and d < c. Integrating the -form 5.3) along the boundary of Ω we obtain p h ) dx q h 2) dy 5.5) AB = DC CB p h ) dx q h 2) dy DA dy q h 2 ) p h ) dx p h ) DC dx q h 2) dy = [ c d 2 n t 2 0, x) + c 2 n 0, x) ) n x 2 0, x) dx. On the other hand, we use 5.2) to compute b [ n t 2 τ, x) + c 2 n τ, x) ) n x 2 τ, x) dx a 2 5.6) p h ) q h 2 ) = dx dy E 0. AB {h 2 0} AB {h 0} Notice that the last relation in 5.5) is satisfied as an equality, because at time t = 0, along the curve γ 0 the variables h, h 2 never assume the value zero. The proof for the case τ < 0 is similar. Letting r + in 5.), one has a, b +. Therefore 5.5) and 5.6) together imply Et) E 0, proving.26).

18 8 G. CHEN, P. ZHANG, AND Y. ZHENG 6. Regularity of trajectories 6.. Lipschitz continuity. In this part, we first prove the Lipschitz continuity of the map t n, n 2, n 3 )t, ) in the L 2 distance, stated in.23). For any h > 0, we have Thus nt + h, x) nt, x) = h 6.) n i t + h, x) n i t, x) L 2 h 0 0 n t t + τh, x) dτ. n it t + τh, ) L 2 dτ h 2E 0, i = Continuity of derivatives. We prove the continuity of functions t n t, n 2t, n 3t )t, ) and t n x, n x, n x )t, ), as functions with values in L p, p < 2. To keep tradition, we use the exponent p here at the expense of repeating one of our primary variables.) This will complete the proof of Theorem.. We first consider the case where the initial data n 0 ) x and n are smooth with compact support. In this case, the solution n = nx, Y ) remains smooth on the entire X-Y plane. Fix a time τ. We claim that d 6.2) nt, ) dt = n t τ, ) t=τ where 6.3) n t τ, x) := n X X t + n Y Y t = p l 2c c + q m ph 2c c qh 2 = l 2h + m 2h 2. Notice that 6.3) defines the values of n t τ, ) at almost every point x R. By the inequality.26), we obtain 6.) n t τ, x) 2 dx 2 Eτ) 2 E 0. To prove 6.2), we consider the set 6.5) Γ τ := {X, Y ) tx, Y ) τ}, R and let γ τ be its boundary. Let ε > 0 be given. There exist finitely many disjoint intervals [a i, b i R, i =,..., N, with the following property. Call A i, B i the points on γ τ such that xa i ) = a i, xb i ) = b i. Then one has 6.6) min { h P ), h 2 P ) } < 2ε at every point P on γ τ contained in one of the arcs A i B i, while 6.7) h P ) > ε, h 2 P ) > ε, for every point P along γ τ, not contained in any of the arcs A i B i. Call J := i N [a i, b i, J = R \J, and notice that, as a function of the original variables, n = nt, x) is smooth in a neighborhood of the set {τ} J. Using Minkowski s inequality and the differentiability of n on J, we can write, for i = 3, lim h 0 h R n i τ + h, x) n i τ, x) h n it τ, x) p ) /p dx 6.8) lim h 0 h J n i τ + h, x) n i τ, x) p ) /p dx + J nit τ, x) p dx ) /p.

19 ENERGY CONSERVATIVE SOLUTIONS TO LIQUID CRYSTALS 9 We now provide an estimate on the measure of the bad set J: meas J) = J dx = ph i A i B i 2 dx qh 2 2 dy 6.9) 2ε p h ) 2ε i A i B i 2 dx q h 2) 2 dy ε p h ) 2ε γ τ dx q h 2) dy ε 2ε E 0. Notice that dt = 0 on γ τ, so the two parts of the integral are actually equal. Now choose q = 2/2 p) so that p 2 + q =. Using Hölder s inequality with conjugate exponents 2/p and q, and recalling 6.), we obtain for any i = 3 J n i τ + h, x) n i τ, x) p dx ) meas J) /q J n i τ + h, x) n i τ, x) 2 p/2 dx Therefore, 6.0) J meas J) /q n i τ + h, ) n i τ, ) ) 2 p/2 L 2 meas J) /q h 2[ ) p/2 2E 0. lim sup h 0 h ε [ J n i τ + h, x) n i τ, x) p dx 2ε E 0 /pq [2E /2. 0 In a similar way we estimate n it τ, x) p dx [ meas J) /q 6.) J J ) /p n it τ, x) dx) 2 p/2, n it τ, x) p dx) /p meas J) /pq [2E 0 p/2. Since ε > 0 is arbitrary, from 6.8), 6.0) and 6.) we conclude 6.2) lim n i τ + h, x) n i τ, x) h n it τ, x) p /p dx) = 0. h 0 h R The proof of continuity of the map t n it is similar. Fix ε > 0. Consider the intervals [a i, b i as before. Since n is smooth on a neighborhood of {τ} J, it suffices to estimate lim sup nit h 0 τ + h, x) n it τ, x) p dx lim sup h 0 J n it τ + h, x) n it τ, x) p dx [ /q lim sup h 0 meas J) J n it τ + h, x) n it τ, x) 2 dx [ lim sup ε h 0 2ε E /q nit 0 τ + h, ) L 2 + nit τ, ) ) p L 2 [ ε 2ε E 0 /q [ E 0 p. ) p/2

20 20 G. CHEN, P. ZHANG, AND Y. ZHENG Since ε > 0 is arbitrary, this proves continuity. To extend the result to general initial data, such that n 0 ) x, n = n t t=0 L 2, we use Corollary 3. and consider a sequence of smooth initial data, with n ν 0 ) x, n ν C c, with n ν 0 n 0 uniformly, n ν 0 ) x n 0 ) x almost everywhere and in L 2, n ν n almost everywhere and in L 2. The continuity of the function t n x t, ) as maps with values in L p, p < 2, is proved in an entirely similar way. 7. Energy conservation This part is devoted to the proof of Theorem.3, stating that, in some sense, the total energy of the solution remains constant in time. A key tool in our analysis is the wave interaction potential, defined in the smooth case by Λt) := 6 x>y R 2 t, x) S 2 t, y) dxdy. To define it in the general case, for any fixed time τ, we let µ τ = µ τ + µ + τ be the positive measure on the real line defined as follows. In the smooth case, 7.) µ τ To define µ ± τ a, b) ) = b a R 2 τ, x) dx, µ + τ a, b) ) = in the general case, let γ τ be the boundary of the set 7.2) Γ τ := {X, Y ) tx, Y ) τ}, b a S 2 τ, x) dx. and let γ τ be its boundary. Given any open interval a, b), let A = X A, Y A ) and B = X B, Y B ) be the points on γ τ such that Then xa) = a, X P Y P X A Y A for every point P γ τ with xp ) a, xb) = b, X P Y P X B Y B for every point P γ τ with xp ) b. 7.3) µ τ a, b) ) = µ τ a, b) ) + µ + τ a, b) ), where 7.) µ ) τ a, b) := AB p h ) dx µ + τ a, b) ) := AB q h 2 ) It is clear that µ τ, µ + τ are bounded, nonnegative measures, and µ τ R) = E 0, for all τ. The wave interaction potential is defined as 7.5) Λt) := µ t µ + t ){ x, y) ; x > y }. Lemma 7. Bounded variation). The map t Λt) has locally bounded variation; i.e., there exists a one-sided Lipschitz constant L 0 such that Λt) Λs) L 0 t s), t > s > 0. Before we prove the lemma, let us first give a formal argument, valid when the solution nt, x) remains smooth. Thanks to 2.8) and 2.0), we obtain 7.6) t R 2 x c R 2 ) =A, t S 2 + x c S 2 ) = A, dy.

21 ENERGY CONSERVATIVE SOLUTIONS TO LIQUID CRYSTALS 2 where 7.7) A = c R 2 S S 2c 2 ) R. Then 7.8) d dt [6Λt) 2C L 2c R 2 S 2 dx + S 2 + R 2) dx A dx R 2 S 2 c dx + E 0 2c R 2 S R S 2 dx. L For each ε > 0 we have R ε /2 + ε /2 R 2. Choosing ε > 0 such that c C L E 0 2 ε, 2c we thus obtain d dt [6Λt) C L L R 2 S 2 dx + 6 E2 0 c ε 2c L 6 E2 0 c ε 2c Hence, the map t Λt) has bounded variation on any bounded interval. It can be discontinuous, with downward jumps. Proof. We reproduce the formal argument in terms of the variables X, Y. In particular, we reproduce 7.8) in integral form in the X, Y ) plane. By 2.2), for any ε > 0, there exists a constant κ ε such that Hence m ε h 2 ) + κ ε h 2, l ε h ) + κ ε h. 7.9) h )m h 2 )l ε h ) h 2 ) + κ ε [ h )h 2 + h 2 )h. Fix 0 s < t. Consider the sets Γ s, Γ t as in 7.2) and define Γ st := Γ t \Γ s. Recall that We can write 7.0) dx dt = pqh h 2 2c dx dy, R 2 = h h, S 2 = h 2 h 2. t R 2 + S 2 s dxdt = t s)e 0 = Γ st ) h h + h 2 h 2 pqh h 2 2c dxdy. The first identity holds only for smooth solutions, but the second one is always valid. Recalling 5.) and 7.), and then using 7.9)-7.0), we obtain h Λt) Λs) p h 2 q dxdy Γ st c + E 0 Γst 6c 2 pq [ h )m h 2 )l dxdy h ) h 2 ) pq dxdy 6 Γ st c + E 0 Γst 6c 2 pq {κ ε [ h )h 2 + h 2 )h +ε h ) h 2 )} dxdy κt s), L.

22 22 G. CHEN, P. ZHANG, AND Y. ZHENG for a suitable constant κ. This proves the lemma. To prove Theorem.3, consider the three sets { Ω := X, Y ) ; h X, Y ) = 0, h 2 X, Y ) 0, Ω 2 := Ω 3 := { X, Y ) ; h 2 X, Y ) = 0, h X, Y ) 0, { X, Y ) ; h X, Y ) = 0, h 2 X, Y ) = 0, c 2 n X, Y )) γ ) n X, Y ) 0}, c 2 n X, Y )) γ ) n X, Y ) 0}, c 2 n X, Y )) γ ) n X, Y ) 0}. Here c 2 γ ) n 0 is equivalent to n ± or 0, since we assume that α γ. From the equations 2.20), it follows that 7.) meas Ω ) = meas Ω 2 ) = 0. Indeed, Y l 0 on Ω since h = 0 implies l = 0 by 2.2). Similarly X m 0 on Ω 2 since h 2 = 0 implies m = 0. And, for any i = 3, l i 0 implies h 0, m i 0 implies h 2 0, by 2.2). Let Ω 3 be the set of Lebesgue points of Ω 3. We now show that 7.2) meas {tx, Y ) ; X, Y ) Ω 3 } ) = 0. To prove 7.2), fix any P = X, Y ) Ω 3 and let τ = tp ). We claim that 7.3) lim sup h,k 0+ Λτ h) Λτ + k) h + k = + Q σ P * σ + t = t t = t Figure 2. Lebesgue point By assumption, for any ε > 0 arbitrarily small we can find δ > 0 with the following property. For any square Q centered at P with side of length l < δ, there exists a vertical segment σ and a horizontal segment σ, as in Figure 2, such that 7.) meas Ω 3 σ ) ε)l, meas Ω 3 σ ) ε)l, Call { t + := max tx, Y ) ; X, Y ) σ σ }, { t := min tx, Y ) ; X, Y ) σ σ }.

23 ENERGY CONSERVATIVE SOLUTIONS TO LIQUID CRYSTALS 23 Since h = h 2 = 0 at nearly all points close to P, we can assume that the endpoints of the two segments σ, σ are all in Ω 3. By integrating the equation for l from 2.20), we obtain q [ c 2 σ 8c 3 γ)h + h 2 2h h 2 ) 23c 2 γ) l m n 7.5) + c c 2 l ql m )dy = 0. Notice that h, l are Lipschitz in Y and h = 0 implies l = 0, and they are zero on σ on a set with measure greater than ε)l. So we obtain 7.6) h f + l f 2 + l 2 f 3 + l 3 f ) dy = O)εl) 2 σ for any bounded functions f f. Thus we obtain from 7.5)7.6) that c 2 γ ) n 7.7) 8c 3 qh 2 dy = O)εl) 2. By 5.2), we have σ σ t Y dy = σ qh 2 2c dy. Assume without loss of generality that c 2 γ ) n > C 3 > 0, at the point P. By 7.7), we obtain t Y dy = O)εl) 2. σ Similarly we can estimate the growth of t in the X-direction. Combining them, we obtain 7.8) t + t O)εl) 2. On the other hand, 7.9) Λt ) Λt + ) C ε) 2 l 2 C 2 t + t ) for some constant C > 0, C 2 > 0. Since ε > 0 is arbitrary, this implies 7.3). Recalling that the map t Λ has bounded variation, from 7.3) it follows 7.2). We now observe that the singular part of µ τ is nontrivial only if the set { P γτ ; h P ) = 0 or h 2 P ) = 0 } has positive -dimensional measure. By the previous analysis, restricted to the region where c 2 γ ) n 0, i.e. n ± or 0, this can happen only for a set of times having zero measure. Acknowledgments. Part of this work was done when Zheng was visiting Academy of Mathematics and System Sciences, he would like to thank the financial support from Center of Interdisciplinary Sciences from AMSS. P. Zhang is partially supported by NSF of China under Grant 020 and , the one hundred talents plan from Chinese Academy of Sciences under Grant GJHZ and innovation grant from National Center for Mathematics and Interdisciplinary Sciences. Additionally, Yuxi Zheng is partially supported by NSF DMS

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