Poiseuille Flow of Nematic Liquid Crystals via the Full Ericksen–Leslie Model

Abstract

In this paper, we study the Cauchy problem of the Poiseuille flow of the full Ericksen–Leslie model for nematic liquid crystals. The model is a coupled system of a parabolic equation for the velocity and a quasilinear wave equation for the director. For a particular choice of several physical parameter values, we construct solutions with smooth initial data and finite energy that produce, in finite time, cusp singularities—blowups of gradients. The formation of cusp singularity is due to local interactions of wave-like characteristics of solutions, which is different from the mechanism of finite time singularity formations for the parabolic Ericksen–Leslie system. The finite time singularity formation for the physical model might raise some concerns for purposes of applications. This is, however, resolved satisfactorily; more precisely, we are able to establish the global existence of weak solutions that are Hölder continuous and have bounded energy. One major contribution of this paper is our identification of the effect of the flux density of the velocity on the director and the reveal of a singularity cancellation—the flux density remains bounded while its two components approach infinity at formations of cusp singularities.

Appendix A

1.1 A.1. Derivation of System (2.1)

We consider the following form of solutions to the system (1.10)

$$\begin{aligned} \mathbf{u}(x,t)=(0,0,u(x,t))^T\; \text{ and } \; \mathbf{n}(x,t)=\big (\sin \theta (x,t), 0, \cos \theta (x,t)\big )^T. \end{aligned}$$

It is easy to see that $\nabla \cdot \mathbf{u}(x,t)=0$, and $\mathbf{u}\cdot \nabla \mathbf{u}=\mathbf{u}\cdot \nabla \mathbf{n}=\mathbf{u}\cdot \nabla \dot{\mathbf{n}}=0$. Direct computation implies

$$\begin{aligned} D= & {} \frac{1}{2}(\nabla \mathbf{u}+\nabla ^T\mathbf{u})=\frac{1}{2} \left[ \begin{array}{ccc} 0&{}0&{}u_x\\ 0&{}0&{}0\\ u_x&{}0&{}0 \end{array} \right] , \\ \omega= & {} \frac{1}{2}(\nabla \mathbf{u}-\nabla ^T\mathbf{u})=\frac{1}{2} \left[ \begin{array}{ccc} 0&{}0&{}-u_x\\ 0&{}0&{}0\\ u_x&{}0&{}0 \end{array} \right] , \\ \mathbf{n}\otimes \mathbf{n}= & {} \left[ \begin{array}{ccc} \sin ^2\theta &{}0&{}\sin \theta \cos \theta \\ 0&{}0&{}0\\ \sin \theta \cos \theta &{}0&{}\cos ^2\theta \end{array} \right] . \end{aligned}$$

Then

$$\begin{aligned} N= & {} \dot{\mathbf{n}}-\omega \mathbf{n}=\left( \theta _t+\frac{1}{2}u_x\right) \big (\cos \theta , 0,- \sin \theta \big )^T, \\ N\otimes \mathbf{n}= & {} \left( \theta _t+\frac{1}{2}u_x\right) \left[ \begin{array}{ccc} \sin \theta \cos \theta &{}0&{}\cos ^2\theta \\ 0&{}0&{}0\\ -\sin ^2\theta &{}0&{}-\sin \theta \cos \theta \end{array} \right] , \\ \mathbf{n}\otimes N= & {} \left( \theta _t+\frac{1}{2}u_x\right) \left[ \begin{array}{ccc} \sin \theta \cos \theta &{}0&{}-\sin ^2\theta \\ 0&{}0&{}0\\ \cos ^2\theta &{}0&{}-\sin \theta \cos \theta \end{array} \right] , \end{aligned}$$

and also

$$\begin{aligned} D\mathbf{n}= & {} \frac{1}{2}u_x\big (\cos \theta , 0,\sin \theta \big )^T, \\ \mathbf{n}^TD\mathbf{n}=u_x\cos \theta \sin \theta , \\ D\mathbf{n}\otimes \mathbf{n}= & {} \frac{1}{2}u_x \left[ \begin{array}{ccc} \sin \theta \cos \theta &{}0&{}\cos ^2\theta \\ 0&{}0&{}0\\ \sin ^2\theta &{}0&{}\sin \theta \cos \theta \end{array} \right] , \\ \mathbf{n}\otimes D\mathbf{n}= & {} \frac{1}{2}u_x \left[ \begin{array}{ccc} \sin \theta \cos \theta &{}0&{}\sin ^2\theta \\ 0&{}0&{}0\\ \cos ^2\theta &{}0&{}\sin \theta \cos \theta \end{array} \right] . \end{aligned}$$

Hence

$$\begin{aligned} \begin{aligned} \mathbf{g}=&\gamma _1N+\gamma _2D\mathbf{n}\\ =&\gamma _1\theta _t\big (\cos \theta , 0,- \sin \theta \big )^T+\frac{1}{2}u_x\big ((\gamma _1+\gamma _2)\cos \theta , 0,(\gamma _2-\gamma _1) \sin \theta \big )^T. \end{aligned} \end{aligned}$$

Since the last term in Oseen-Frank energy density is null Lagrangian term, without loss of generalization, we only compute the first three terms. The Oseen-Frank energy density of this case will be

$$\begin{aligned} W(\mathbf{n},\nabla \mathbf{n})=K_1(\mathbf{n}^1_x)^2+K_3\big ((\mathbf{n}^1\mathbf{n}^1_x)^2+(\mathbf{n}^1\mathbf{n}^3_x)^2\big )=\frac{1}{2}\theta ^2_x\big (K_1\cos ^2\theta +K_3\sin ^2\theta \big ), \end{aligned}$$

where $\mathbf{n}^i$ is the i-th component of $\mathbf{n}$. Thus

$$\begin{aligned} \frac{\partial W}{\partial \mathbf{n}}= & {} \big (K_3\theta _x^2\sin \theta , 0,0\big )^T, \\ \frac{\partial W}{\partial \nabla \mathbf{n}}= & {} \left[ \begin{array}{ccc} K_1\theta _x\cos \theta +K_3\theta _x\sin ^2\theta \cos \theta &{}0&{}0\\ 0&{}0&{}0\\ -K_3\theta _x\sin ^3\theta &{}0&{}0 \end{array} \right] . \end{aligned}$$

The Lagrangian constant is

$$\begin{aligned} \begin{aligned} \gamma =&\frac{\partial W}{\partial \mathbf{n}}\cdot \mathbf{n}+\gamma _2D\mathbf{n}\cdot \mathbf{n}-\nabla \cdot \left( \frac{\partial W}{\partial \nabla \mathbf{n}}\right) \cdot \mathbf{n}-|\mathbf{n}_t|^2\\ =&(K_1+2K_3)\theta _x^2\sin ^2\theta -K_1\theta _{xx}\sin \theta \cos \theta +\gamma _2u_x\sin \theta \cos \theta -|\theta _t|^2. \end{aligned} \end{aligned}$$

We are ready to derive the system (2.1). We first work on the equation of $\theta$. By the third equation of (1.10), we have

$$\begin{aligned} \begin{aligned} \mathbf{n}_{tt}=&\theta _{tt}\big (\cos \theta ,0,-\sin \theta \big )^T+|\theta _t|^2\big (-\sin \theta ,0,-\cos \theta \big )^T\\ =&\gamma \mathbf{n}-\frac{\partial W}{\partial \mathbf{n}}-\mathbf{g}+\nabla \cdot \left( \frac{\partial W}{\partial \nabla \mathbf{n}}\right) \\ =&-|\theta _t|^2\big (\sin \theta ,0,\cos \theta \big )^T-\gamma _1\theta _t\big (\cos \theta , 0,- \sin \theta \big )^T\\&+u_x{{\mathbf {T}}_1}+K_1{\mathbf {T}}_2+K_3{\mathbf {T}}_3. \end{aligned} \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} (\theta _{tt}+\gamma _1\theta _t)\big (\cos \theta ,0,-\sin \theta \big )^T =u_x{{\mathbf {T}}_1}+K_1{\mathbf {T}}_2+K_3{\mathbf {T}}_3. \end{aligned} \end{aligned}$$

(A.1)

Here the vector ${\mathbf {T}}_1$ is given by

$$\begin{aligned} {\mathbf {T}}_1=&\gamma _2\sin \theta \cos \theta \big (\sin \theta , 0,\cos \theta \big )^T-\frac{1}{2}\big ((\gamma _1+\gamma _2)\cos \theta , 0,(\gamma _2-\gamma _1) \sin \theta \big )^T\nonumber \\ =&-\frac{\gamma _1}{2}\big (\cos \theta , 0,-\sin \theta \big )^T\nonumber \\&+\frac{\gamma _2}{2}\big (2\sin ^2\theta \cos \theta -\cos \theta , 0,2\sin \theta \cos ^2\theta -\sin \theta \big )^T\nonumber \\ =&\left( -\frac{\gamma _1}{2}-\frac{\gamma _2}{2}\cos (2\theta )\right) \big (\cos \theta , 0,-\sin \theta \big )^T. \end{aligned}$$

(A.2)

The nonzero components of vector ${\mathbf {T}}_2$ is given by

$$\begin{aligned} {\mathbf {T}}_2^1=&\theta _x^2\sin ^3\theta -\theta _{xx}\sin ^2\theta \cos \theta +(\theta _x\cos \theta )_x\\ =&-\sin ^2\theta (\theta _x\cos \theta )_x+(\theta _x\cos \theta )_x =\cos ^2\theta (\theta _x\cos \theta )_x,\\ {\mathbf {T}}_2^3=&\theta _x^2\sin ^2\theta \cos \theta -\theta _{xx}\sin \theta \cos ^2\theta =-\sin \theta \cos \theta (\theta _x\cos \theta )_x, \end{aligned}$$

so the vector ${\mathbf {T}}_2$ is

$$\begin{aligned} \begin{aligned} {\mathbf {T}}_2= \cos \theta (\theta _x\cos \theta )_x\big (\cos \theta , 0,-\sin \theta \big )^T. \end{aligned} \end{aligned}$$

(A.3)

Similarly, the vector ${\mathbf {T}}_3$ is

$$\begin{aligned} \begin{aligned} {\mathbf {T}}_3= \sin \theta (\theta _x\sin \theta )_x\big (\cos \theta , 0,-\sin \theta \big )^T. \end{aligned} \end{aligned}$$

(A.4)

Plugging (A.2), (A.3) and (A.4) into (A.1), we obtain

$$\begin{aligned} \theta _{tt}+\gamma _1\theta _t=K_1\cos \theta (\theta _x\cos \theta )_x +K_3\sin \theta (\theta _x\sin \theta )_x+\left( -\frac{\gamma _1}{2} -\frac{\gamma _2}{2}\cos (2\theta )\right) u_x, \end{aligned}$$

which is exactly the second equation in (2.1).

For the equation of u, direct computation gives

$$\begin{aligned} \nabla \cdot \left( \frac{\partial W}{\partial \nabla \mathbf{n}}\otimes \nabla \mathbf{n}\right) =\big ((K_1\theta _x^2\cos ^2\theta )_x+(K_3\theta _x^2\sin ^2\theta )_x, 0,0\big )^T \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \nabla \cdot \sigma =&\alpha _1\big ((u_x\sin ^3\theta \cos \theta )_x,0,(u_x\sin ^2\theta \cos ^2\theta )_x\big )^T\\&+\alpha _2\left( \big (\big (\theta _t+\frac{1}{2}u_x\big )\sin \theta \cos \theta \big )_x,0, -\big (\big (\theta _t+\frac{1}{2}u_x\big )\sin ^2\theta \big )_x\right) ^T\\&+\alpha _3\left( \big ( \big (\theta _t+\frac{1}{2}u_x\big )\sin \theta \cos \theta \big )_x,0, \big (\big (\theta _t+\frac{1}{2}u_x\big )\cos ^2\theta \big )_x\right) ^T\\&+\frac{\alpha _4}{2}\big (0,0, u_{xx}\big )^T +\frac{\alpha _5}{2}\big ((u_x\sin \theta \cos \theta )_x,0,(u_x\sin ^2\theta )_x\big )^T\\&+\frac{\alpha _6}{2}\big ((u_x\sin \theta \cos \theta )_x,0,(u_x\cos ^2\theta )_x\big )^T. \end{aligned} \end{aligned}$$

Therefore the first equation of system (1.10) can be written into following three equations:

$$\begin{aligned} P_x=&K_1(\theta _x^2\cos ^2\theta )_x+K_3(\theta _x^2\sin ^2\theta )_x+\alpha _1(u_x\sin ^3\theta \cos \theta )_x\nonumber \\&+\frac{1}{2}\left( \alpha _2+\alpha _3+\alpha _5+\alpha _6\right) (u_x\sin \theta \cos \theta )_x+(\alpha _2+\alpha _3)\big (\theta _t\sin \theta \cos \theta \big )_x, \end{aligned}$$

(A.5)

$$\begin{aligned} P_y=&0, \end{aligned}$$

(A.6)

$$\begin{aligned} u_t+P_z=&\frac{\alpha _4}{2}u_{xx} +\alpha _1(u_x\sin ^2\theta \cos ^2\theta )_x -\alpha _2\big (\big (\theta _t+\frac{1}{2}u_x\big )\sin ^2\theta \big )_x\nonumber \\&+\alpha _3\big (\big (\theta _t+\frac{1}{2}u_x\big )\cos ^2\theta \big )_x +\frac{\alpha _5}{2}(u_x\sin ^2\theta )_x +\frac{\alpha _6}{2}(u_x\cos ^2\theta )_x. \end{aligned}$$

(A.7)

By these equations, one can obtain that $P_z=a$ for some constant a. The right hand side of (A.7) can be rewritten as $(g(\theta )u_x+h(\theta )\theta _t)_x$ where $g(\theta )$ and $h(\theta )$ is defined as (2.2). Therefore, we obtain the first equation of (2.1).

1.2 A.2. Derivation of System (5.9)

We will in fact derive the semilinear system in XY-coordinates for (2.1) with $\nu =\rho =1$ and $a=0$. Recall, from (5.5) and (5.6), that we have introduced

$$\begin{aligned} w=2\arctan R,\quad z=2\arctan S, \quad p=\frac{1+R^2}{X_x},\quad q=-\frac{1+S^2}{Y_x}. \end{aligned}$$

It is easy to have that

$$\begin{aligned} R= & {} \tan \frac{w}{2},\quad \frac{R}{1+R^2}=\frac{1}{2}\sin w, \quad \frac{1}{1+R^2}=\cos ^2\frac{w}{2}, \\ S= & {} \tan \frac{z}{2}, \quad \frac{S}{1+S^2}=\frac{1}{2}\sin z, \quad \frac{1}{1+S^2}=\cos ^2\frac{z}{2}. \end{aligned}$$

By (5.4) and (4.2), we have

$$\begin{aligned} \theta _X=\frac{\sin w}{4c}p,\quad \theta _Y=\frac{\sin z}{4c}q. \end{aligned}$$

Denote

$$\begin{aligned} U=h(\theta )u_x. \end{aligned}$$

By (4.2), we have

$$\begin{aligned} \left\{ \begin{array}{rcl} (S^2)_t+c(\theta )(S^2)_x &{}=&{} \frac{c'}{2c}(S^3-R^2S)-\gamma _1(RS+S^2)-2SU,\\ (R^2)_t-c(\theta )(R^2)_x &{}=&{} \frac{c'}{2c} (R^3-RS^2)-\gamma _1(RS+R^2)-2RU, \end{array}\right. \end{aligned}$$

(A.8)

so

$$\begin{aligned} \left\{ \begin{array}{rcl} (S^2)_X &{}=&{} \frac{1}{2c}p\frac{1}{1+R^2}\Big \{\frac{c'}{2c}(S^3-R^2S)-\gamma _1(RS+S^2)-2SU\Big \},\\ (R^2)_Y &{}=&{} \frac{1}{2c}q\frac{1}{1+S^2}\Big \{\frac{c'}{2c} (R^3-RS^2)-\gamma _1(RS+R^2)-2RU\Big \}. \end{array}\right. \end{aligned}$$

(A.9)

Hence,

$$\begin{aligned} \begin{aligned} z_X=&\textstyle \frac{1}{1+S^2}\frac{1}{S}(S^2)_X \\ =&\textstyle \frac{1}{2c}p\frac{1}{(1+R^2)(1+S^2)}\Big \{\frac{c'}{2c}(S^2-R^2)-\gamma _1(R+S)-2U\Big \} \\ =&\textstyle p \Big \{\frac{c'}{4c^2}\Big (\cos ^2\frac{w}{2}-\cos ^2\frac{z}{2}\Big )-\frac{\gamma _1}{4c}\Big (\sin w\cos ^2\frac{z}{2}+\sin z\cos ^2\frac{w}{2}\Big )\\&-\frac{1}{c}\cos ^2\frac{z}{2}\cos ^2\frac{w}{2}U\Big \} . \end{aligned} \end{aligned}$$

Similarly,

$$\begin{aligned} \textstyle w_Y= & {} q \Big \{\frac{c'}{4c^2}\Big (\cos ^2\frac{z}{2}-\cos ^2\frac{w}{2}\Big )-\frac{\gamma _1}{4c} \Big (\sin w\cos ^2\frac{z}{2}+\sin z\cos ^2\frac{w}{2}\Big )\\&-\frac{1}{c}\cos ^2\frac{z}{2}\cos ^2\frac{w}{2}U\Big \}. \end{aligned}$$

On the other hand, using $X_t-cX_x=0$, we have

$$\begin{aligned} X_{tx}-cX_{xx}=\frac{c'}{2c}(R-S)X_x. \end{aligned}$$

Then

$$\begin{aligned} p_Y =&\textstyle \frac{1}{X_x}(R^2)_Y-\frac{q\cos ^2{\frac{w}{2}}}{2c}\frac{1+R^2}{X_x^2}(X_{xt}-cX_{xx}) \\ =&\textstyle \frac{pq}{2c}\frac{1}{(1+R^2)(1+S^2)}\Big \{\frac{c'}{2c} (R^3-RS^2)-\gamma _1(RS+R^2)-2RU\Big \} -\frac{pq}{2c}\frac{c'}{2c}\frac{R-S}{1+S^2}\\ =&\textstyle \frac{pq}{2c}\frac{1}{(1+R^2)(1+S^2)}\Big \{\frac{c'}{2c} (-R-RS^2)-\gamma _1(RS+R^2)-2RU\Big \} +\frac{pq}{2c}\frac{c'}{2c}\frac{S}{1+S^2}\\ =&\textstyle pq\frac{c'}{8c^2}(\sin z-\sin w) -\gamma _1\frac{pq}{2c}\Big [\frac{1}{4}\sin w\sin z+\sin ^2 \frac{w}{2}\cos ^2\frac{z}{2}\Big ]\\&-\frac{pq}{2c}U\sin w\cos ^2\frac{z}{2}. \end{aligned}$$

Similarly, we have

$$\begin{aligned} q_X= & {} \textstyle pq\frac{c'}{8c^2}(\sin w-\sin z) -\gamma _1\frac{pq}{2c}\Big [\frac{1}{4}\sin w\sin z+\sin ^2 \frac{z}{2}\cos ^2\frac{w}{2}\Big ]\\&-\frac{pq}{2c}U\sin z\cos ^2\frac{w}{2}. \end{aligned}$$

In summary, we have the following system of equations:

$$\begin{aligned} \begin{aligned} \theta _X\textstyle =&\frac{\sin w}{4c}p,\qquad \theta _Y=\frac{\sin z}{4c}q\\ z_X&\textstyle =p \Big \{\frac{c'}{4c^2}\Big (\cos ^2\frac{w}{2}-\cos ^2\frac{z}{2}\Big ) -\frac{\gamma _1}{4c}\Big (\sin w\cos ^2\frac{z}{2}+\sin z\cos ^2\frac{w}{2}\Big )\\&-\frac{1}{c}\cos ^2\frac{z}{2}\cos ^2\frac{w}{2}U\Big \} \\ w_Y\textstyle =&q \Big \{\frac{c'}{4c^2}\Big (\cos ^2\frac{z}{2}-\cos ^2\frac{w}{2}\Big ) -\frac{\gamma _1}{4c}\Big (\sin w\cos ^2\frac{z}{2}+\sin z\cos ^2\frac{w}{2}\Big )\\&-\frac{1}{c}\cos ^2\frac{z}{2}\cos ^2\frac{w}{2}U\Big \} \\ p_Y\textstyle =&pq\frac{c'}{8c^2}(\sin z-\sin w) -\gamma _1\frac{pq}{2c}\Big [\frac{1}{4}\sin w\sin z+\sin ^2 \frac{w}{2}\cos ^2\frac{z}{2}\Big ]\\&-\frac{pq}{2c}U\sin w\cos ^2\frac{z}{2}\\ q_X\textstyle =&pq\frac{c'}{8c^2}(\sin w-\sin z) -\gamma _1\frac{pq}{2c}\Big [\frac{1}{4}\sin w\sin z+\sin ^2 \frac{z}{2}\cos ^2\frac{w}{2}\Big ] \\&-\frac{pq}{2c}U\sin z\cos ^2\frac{w}{2}, \end{aligned} \end{aligned}$$

where recall that $U=h(\theta )u_x.$

Denote $J=g(\theta )u_x+h(\theta )\theta _t$. Then

$$\begin{aligned} U=\frac{h(\theta )}{g(\theta )}(J-h(\theta )\theta _t). \end{aligned}$$

In terms of the variable J, the above system is

$$\begin{aligned} \theta _X=&\frac{\sin w}{4c}p,\quad \theta _Y=\frac{\sin z}{4c}q \nonumber \\ \textstyle z_X=&p \Big \{\frac{c'}{4c^2}\Big (\cos ^2\frac{w}{2}-\cos ^2\frac{z}{2}\Big ) +\frac{\frac{h^2(\theta )}{g(\theta )}-\gamma _1}{4c} \Big (\sin w\cos ^2\frac{z}{2}+\sin z\cos ^2\frac{w}{2}\Big )\nonumber \\&-\frac{h(\theta )}{cg(\theta )}J\cos ^2\frac{z}{2}\cos ^2\frac{w}{2} \Big \} \nonumber \\ \textstyle w_Y=&q \Big \{\frac{c'}{4c^2} \Big (\cos ^2\frac{z}{2}-\cos ^2\frac{w}{2}\Big ) +\frac{\frac{h^2(\theta )}{g(\theta )}-\gamma _1}{4c} \Big (\sin w\cos ^2\frac{z}{2}+\sin z\cos ^2\frac{w}{2}\Big )\nonumber \\&-\frac{h(\theta )}{cg(\theta )}J\cos ^2\frac{z}{2}\cos ^2\frac{w}{2}\Big \} \nonumber \\ \textstyle p_Y=&pq\Big \{\frac{c'}{8c^2}(\sin z-\sin w) +\frac{\frac{h^2(\theta )}{g(\theta )}-\gamma _1}{2c}\Big (\frac{1}{4}\sin w\sin z+\sin ^2 \frac{w}{2}\cos ^2\frac{z}{2}\Big )\nonumber \\&-\frac{h(\theta )}{2cg(\theta )}J\sin w\cos ^2\frac{z}{2}\Big \}\nonumber \\ \textstyle q_X=&pq\Big \{\frac{c'}{8c^2}(\sin w-\sin z) +\frac{\frac{h^2(\theta )}{g(\theta )}-\gamma _1}{2c}\Big (\frac{1}{4}\sin w\sin z+\sin ^2 \frac{z}{2}\cos ^2\frac{w}{2}\Big )\nonumber \\&-\frac{h(\theta )}{2cg(\theta )}J\sin z\cos ^2\frac{w}{2}\Big \}, \end{aligned}$$

(A.10)

where the coefficient $\frac{h^2(\theta )}{g(\theta )}-\gamma _1<0$ by (2.11).

For the special case where $\gamma _1=2$, $\gamma _2=0$ and $g(\theta )=h(\theta )=1$, system (A.10) reduces to system (5.9).

1.3 A.3. Hölder Continuity of ${\mathcal {M}}(J)$ in §6.3

First, we consider three types of integrals in the definition of J in (6.31):

$$\begin{aligned} L_i(x,t):=&\int _{\mathbb {R}}H(x-y,t)f(y)\,\mathrm{d}y \\ L_{ii}(x,t):=&\int _0^t\int _{\mathbb R}H(x-y,t-s)g(y,s)\,\mathrm{d}y\mathrm{d}s\\ L_{iii}(x,t):=&\int _0^t\int _{\mathbb R}H_x(x-y,t-s)f(y,s)\,\mathrm{d}y\mathrm{d}s. \end{aligned}$$

We first prove the following lemma, working generally:

Lemma A.1

If $f(x,t)\in L^{\infty }([0,T], L^2(\mathbb {R}))$, $g(x,t)\in L^{\infty }([0,T], L^1(\mathbb {R}))$ or $g(x,t)\in L^{\infty }([0,T], L^2(\mathbb {R}))$ we have $L_j(x,t)$ is Hölder continuous w.r.t x and t, for all $j=i,ii,iii$, with exponent $\beta \in (0,\frac{1}{4})$.

Proof

We only provide the proof of Hölder continuity w.r.t x for $L_{iii}$, the rest cases can be proved similarly. For any $x_1<x_2$, it is sufficient to show that

$$\begin{aligned}&\left| \int _0^t\int _{\mathbb {R}} \frac{H_y(x_2-y,t-s)-H_y(x_1-y,t-s)}{(x_2-x_1)^\beta }f(y,s) \mathrm{d}y\mathrm{d}s\right| \nonumber \\&\quad \leqq \hbox {Constant}\, \cdot \, t^{\frac{1}{4}-\frac{\beta }{2}} \end{aligned}$$

(A.11)

for some $\beta \in (0,1)$. Since

$$\begin{aligned}&H_x(x,t)\\&\quad =-\frac{1}{4\sqrt{\pi }}\frac{x}{t^{3/2}}\exp \left( -\frac{x^2}{4t}\right) ,\\&H_x(x_2-y,t-s)-H_x(x_1-y,t-s)\\&\quad =-\frac{1}{4\sqrt{\pi }}\frac{1}{(t-s)^{3/2}} \left\{ (x_2-x_1)\exp \left( -\frac{(x_2-y)^2}{4(t-s)}\right) \right. \\&\qquad -\left. (x_1-y)\left[ \exp \left( -\frac{(x_2-y)^2}{4(t-s)}\right) -\exp \left( -\frac{(x_1-y)^2}{4(t-s)}\right) \right] \right\} \\&\quad :=I_1+I_2. \end{aligned}$$

The integral related to the first term $I_1$ can be estimated as follows:

$$\begin{aligned}&\int _0^t\int _{\mathbb {R}}\frac{1}{(t-s)^{3/2}}(x_2-x_1)^{1-\beta }\exp \left( -\frac{(x_2-y)^2}{4(t-s)}\right) |f(y,s)|\,\mathrm{d}y\mathrm{d}s \nonumber \\&\quad \leqq C\int _0^t\int _{\mathbb {R}}\frac{1}{(t-s)^{3/2}}\left( |x_2-y|^{1-\beta }+|x_1-y|^{1-d}\right) \nonumber \\&\qquad \exp \left( -\frac{(x_2-y)^2}{4(t-s)}\right) |f(y,s)|\,\mathrm{d}y\mathrm{d}s. \end{aligned}$$

(A.12)

By the same argument in (3.9), and choosing $0<\beta <\frac{1}{8}$, the first term in (A.12) should be controlled by

$$\begin{aligned}&\int _0^t\int _{\mathbb {R}}\frac{1}{(t-s)^{3/2}}|x_2-y|^{1-\beta }\exp \left( -\frac{(x_2-y)^2}{4(t-s)}\right) |f(y,s)|\,\mathrm{d}y\mathrm{d}s\nonumber \\&\quad \leqq Ct^{\frac{1}{4}-\frac{\beta }{2}}\Vert f\Vert _{L^{\infty }((0,T), L^2(\mathbb {R}))}. \end{aligned}$$

(A.13)

For the second term in (A.12),

$$\begin{aligned}&\int _0^t\int _{\mathbb {R}}\frac{1}{(t-s)^{3/2}}|x_1-y|^{1-\beta }\exp \left( -\frac{(x_2-y)^2}{4(t-s)}\right) |f(y,s)|\,\mathrm{d}y\mathrm{d}s\nonumber \\&\quad \leqq C\left( \int _0^t\int _{\mathbb {R}}\frac{|x_1-y|^{2(1-\beta )}}{(t-s)^{9/4-\beta /2}} \exp \left( -\frac{(x_2-y)^2}{2(t-s)}\right) \,\mathrm{d}y\mathrm{d}s\right) ^{\frac{1}{2}}\nonumber \\&\qquad \left( \int _0^t\int _{\mathbb {R}}\frac{|f(y,s)|^2}{(t-s)^{3/4+\beta /2}}\,\mathrm{d}y\mathrm{d}s\right) ^{\frac{1}{2}}\nonumber \\&\quad \leqq C\left( \int _0^t\int _{\mathbb {R}}\frac{|\sqrt{t-s}u|^{2(1-\beta )}}{(t-s)^{7/4-\beta /2}}\exp \left( -\frac{u^2}{2}-\frac{x_2-x_1}{\sqrt{t-s}}u\right) \,\mathrm{d}u\mathrm{d}s \right) ^{\frac{1}{2}}\nonumber \\&\qquad \left( \int _0^t\int _{\mathbb {R}}\frac{|f(y,s)|^2}{(t-s)^{3/4+\beta /2}}\,\mathrm{d}y\mathrm{d}s\right) ^{\frac{1}{2}}\nonumber \\&\quad \leqq C\left( \int _0^t\int _{\mathbb {R}}\frac{|u|^{2(1-\beta )}}{(t-s)^{3/4+\beta /2}} \exp \left( -u\big (\frac{u}{2}+\frac{x_2-x_1}{\sqrt{t-s}}\big )\right) \,\mathrm{d}u\mathrm{d}s \right) ^{\frac{1}{2}}\nonumber \\&\qquad \left( \int _0^t\int _{\mathbb {R}}\frac{|f(y,s)|^2}{(t-s)^{3/4+\beta /2}}\,\mathrm{d}y\mathrm{d}s\right) ^{\frac{1}{2}}\nonumber \\&\quad \leqq Ct^{\frac{1}{4}-\frac{\beta }{2}}\Vert f\Vert _{L^{\infty }((0,T), L^2(\mathbb {R}))}, \end{aligned}$$

(A.14)

where $x_1-y=u\sqrt{t-s}$, and

$$\begin{aligned} \begin{aligned}&\exp \left( -\frac{(x_2-y)^2}{2(t-s)}\right) =\exp \left( -\frac{(x_2-x_1+x_1-y)^2}{2(t-s)}\right) \\&\quad =\exp \left( -\frac{(x_2-x_1)^2}{2(t-s)}\right) \cdot \exp \left( -\frac{2(x_2-x_1)(x_1-y)}{2(t-s)}\right) \cdot \exp \left( -\frac{(x_1-y)^2}{2(t-s)}\right) . \end{aligned} \end{aligned}$$

Putting (A.13) and (A.14) into (A.12), it holds that

$$\begin{aligned} \begin{aligned}&\int _0^t\int _{\mathbb {R}}\frac{1}{(t-s)^{3/2}}(x_2-x_1)^{1-\beta }\exp \left( -\frac{(x_2-y)^2}{4(t-s)}\right) |f(y,s)|\,\mathrm{d}y\mathrm{d}s\\&\quad \leqq Ct^{\frac{1}{4}-\frac{\beta }{2}}\Vert f\Vert _{L^{\infty }((0,T), L^2(\mathbb {R}))}. \end{aligned} \end{aligned}$$

(A.15)

On the other hand, by the mean value theorem, there exists a $\xi \in (x_1,x_2)$ such that

$$\begin{aligned} \begin{aligned} e^{-\frac{(x_2-y)^2}{4(t-s)}}-e^{-\frac{(x_1-y)^2}{4(t-s)}} =&-e^{-\frac{(\xi -y)^2}{4(t-s)}} \frac{\xi -y}{2(t-s)}(x_2-x_1)\\ \leqq&-e^{-\frac{(x_1-y)^2}{4(t-s)}}\frac{x_1-y}{2(t-s)}(x_2-x_1). \end{aligned} \end{aligned}$$

Hence, for the term related to $I_2$, it holds that

$$\begin{aligned}&\int _0^t\int _{\mathbb {R}}\frac{1}{(t-s)^{3/2}}\frac{(x_1-y)^2}{t-s}(x_2-x_1)^{1-\beta }\exp \left( -\frac{(x_1-y)^2}{4(t-s)}\right) |f(y,s)|\,\mathrm{d}y\mathrm{d}s \nonumber \\&\quad \leqq C\int _0^t\int _{\mathbb {R}}\frac{1}{(t-s)^{3/2}}\frac{(x_1-y)^2}{t-s}\left( |x_2-y|^{1-\beta } +|x_1-y|^{1-\beta }\right) \nonumber \\&\qquad \exp \left( -\frac{(x_1-y)^2}{4(t-s)}\right) |f(y,s)|\,\mathrm{d}y\mathrm{d}s. \end{aligned}$$

(A.16)

The second term is similar to (3.9) and (A.13):

$$\begin{aligned}&\int _0^t\int _{\mathbb {R}}\frac{1}{(t-s)^{5/2}}|x_1-y|^{3-\beta }\exp \left( -\frac{(x_1-y)^2}{4(t-s)}\right) |f(y,s)|\,\mathrm{d}y\mathrm{d}s\nonumber \\&\quad \leqq Ct^{\frac{1}{4}-\frac{\beta }{2}}\Vert f\Vert _{L^{\infty }((0,T), L^2(\mathbb {R}))}. \end{aligned}$$

(A.17)

For the first term, one has

$$\begin{aligned}&\int _0^t\int _{\mathbb {R}}\frac{1}{(t-s)^{3/2}}\frac{(x_1-y)^2}{t-s}|x_2-y|^{1-\beta }\exp \left( -\frac{(x_1-y)^2}{4(t-s)}\right) f(y,s)\,\mathrm{d}y\mathrm{d}s\nonumber \\&\quad \leqq C\left( \int _0^t\int _{\mathbb {R}}\frac{|x_2-y|^{2(1-\beta )}}{(t-s)^{9/4-d/2}}\frac{(x_1-y)^4}{(t-s)^2}\exp \left( -\frac{(x_1-y)^2}{2(t-s)}\right) \,\mathrm{d}y\mathrm{d}s\right) ^{\frac{1}{2}}\nonumber \\&\qquad \cdot \left( \int _0^t\int _{\mathbb {R}}\frac{|f(y,s)|^2}{(t-s)^{3/4+\beta /2}}\,\mathrm{d}y\mathrm{d}s\right) ^{\frac{1}{2}}. \end{aligned}$$

(A.18)

Let $x_2-y=\sqrt{t-s}u$. Then it holds that

$$\begin{aligned}&\int _0^t\int _{\mathbb {R}}\frac{|u|^{2(1-d)}}{(t-s)^{3/4+\beta /2}} \frac{(x_1-x_2+\sqrt{t-s}u)^4}{(t-s)^2}\nonumber \\&\qquad \exp \left( -\frac{(x_1-x_2+\sqrt{t-s}u)^2}{2(t-s)}\right) \,\mathrm{d}u\mathrm{d}s\nonumber \\&\quad \leqq C\int _0^t\int _{\mathbb {R}}\frac{|u|^{2(1-\beta )}}{(t-s)^{3/4+\beta /2}}\frac{(x_1-x_2)^4 +(\sqrt{t-s}u)^4}{(t-s)^2}\nonumber \\&\qquad \exp \left( -\frac{(x_1-x_2+\sqrt{t-s}u)^2}{2(t-s)}\right) \,\mathrm{d}u\mathrm{d}s. \end{aligned}$$

(A.19)

It is easy to see that

$$\begin{aligned} \begin{aligned}&\exp \left( -\frac{(x_1-x_2+\sqrt{t-s}u)^2}{2(t-s)}\right) \\&\quad =\exp \left( -\frac{(x_1-x_2)^2}{2(t-s)}\right) \exp \left( -\frac{(x_1-x_2)u}{\sqrt{t-s}}\right) \exp \left( -\frac{u^2}{2}\right) \\&\quad \leqq \exp \left( -\frac{u^2}{2}-\frac{(x_1-x_2)u}{\sqrt{t-s}}\right) . \end{aligned} \end{aligned}$$

Hence,

$$\begin{aligned}&\int _0^t\int _{\mathbb {R}}\frac{|u|^{2(1-\beta )}}{(t-s)^{3/4+\beta /2}}u^4\exp \left( -\frac{(x_1-x_2+\sqrt{t-s}u)^2}{2(t-s)}\right) \,\mathrm{d}u\mathrm{d}s\nonumber \\&\quad \leqq \int _0^t\int _{\mathbb {R}}\frac{|u|^{2(1-\beta )}}{(t-s)^{3/4+\beta /2}}u^4 \exp \left( -\frac{u^2}{2}-\frac{(x_1-x_2)u}{\sqrt{t-s}}\right) \,\mathrm{d}u\mathrm{d}s\leqq Ct^{\frac{1}{4}-\frac{\beta }{2}},\nonumber \\ \end{aligned}$$

(A.20)

and

$$\begin{aligned} \begin{aligned}&\int _0^t\int _{\mathbb {R}}\frac{|u|^{2(1-\beta )}}{(t-s)^{3/4+\beta /2}}\frac{(x_1-x_2)^4}{(t-s)^2}\exp \left( -\frac{(x_1-x_2+\sqrt{t-s}u)^2}{2(t-s)}\right) \,\mathrm{d}u\mathrm{d}s\\&\quad =\int _0^t\int _{\mathbb {R}}\frac{|u|^{2(1-\beta )}}{(t-s)^{3/4+\beta /2}} \frac{(x_1-x_2)^4}{(t-s)^2} \exp \left( -\frac{(x_1-x_2)^2}{2(t-s)}\right) \\&\qquad \exp \left( -\frac{u^2}{2}-\frac{(x_1-x_2)u}{\sqrt{t-s}}\right) \,\mathrm{d}u\mathrm{d}s\\&\quad \leqq \int _0^t\int _{\mathbb {R}}\frac{|u|^{2(1-\beta )}}{(t-s)^{3/4+\beta /2}}u^4 \exp \left( -\frac{u^2}{2}-\frac{(x_1-x_2)u}{\sqrt{t-s}}\right) \,\mathrm{d}u\mathrm{d}s\leqq Ct^{\frac{1}{4}-\frac{\beta }{2}}. \end{aligned} \end{aligned}$$

(A.21)

Putting (A.20) and (A.21) into (A.19), we obtain

$$\begin{aligned} \begin{aligned}&\int _0^t\int _{\mathbb {R}}\frac{|u|^{2(1-d)}}{(t-s)^{3/4+\beta /2}} \frac{(x_1-x_2+\sqrt{t-s}u)^4}{(t-s)^2}\\&\qquad \exp \left( -\frac{(x_1-x_2+\sqrt{t-s}u)^2}{2(t-s)}\right) \,\mathrm{d}u\mathrm{d}s\\&\quad \leqq Ct^{\frac{1}{4}-\frac{\beta }{2}}. \end{aligned} \end{aligned}$$

(A.22)

Combining (A.22) with (A.17) and (A.18), we have

$$\begin{aligned}&\int _0^t\int _{\mathbb {R}}\frac{1}{(t-s)^{3/2}}\frac{(x_1-y)^2}{t-s}(x_2-x_1)^{1-\beta }\exp \left( -\frac{(x_1-y)^2}{4(t-s)}\right) |f(y,s)|\,\mathrm{d}y\mathrm{d}s\nonumber \\&\quad \leqq Ct^{\frac{1}{4}-\frac{\beta }{2}}\Vert f\Vert _{L^{\infty }((0,T), L^2(\mathbb {R}))}. \end{aligned}$$

(A.23)

It is easy to see that (A.15) and (A.23) imply (A.24).

Similarly, we can show the Hölder continuity of for $L_{iii}$ in t , and

$$\begin{aligned} \left| \int _0^t\int _{\mathbb {R}} \frac{H_y(x_2-y,t-s)-H_y(x_1-y,t-s)}{(t_2-t_1)^\beta }f(y,s) \mathrm{d}y\mathrm{d}s\right| \leqq C t^{\frac{1}{4}-\beta }.\nonumber \\ \end{aligned}$$

(A.24)

The proof of Hölder continuity for other terms is similar, and all bounds include a factor $t^{\frac{1}{4}-\beta }$. $\square$