Wellposedness and regularity estimates for stochastic Cahn–Hilliard equation with unbounded noise diffusion

Abstract

In this article, we consider the one dimensional stochastic Cahn–Hilliard equation driven by multiplicative space-time white noise with diffusion coefficient of sublinear growth. By introducing the spectral Galerkin method, we obtain the well-posedness of the approximated equation in finite dimension. Then with help of the semigroup theory and the factorization method, the approximation processes are shown to possess many desirable properties. Further, we show that the approximation process is strongly convergent in a certain Banach space with an explicit algebraic convergence rate. Finally, the global existence and regularity estimates of the unique solution process are proven by means of the strong convergence of the approximation process, which fills a gap on the global existence of the mild solution for stochastic Cahn–Hilliard equation when the diffusion coefficient satisfies a growth condition of order $\alpha \in (\frac{1}{3},1).$

Appendix

Proof of (2.4).

Proof

Thanks to the series expansion $S(t)v=\sum _{k=1}^{\infty }e^{-\lambda _k^2 t}\langle v,e_k\rangle e_k$, using the fact that $|\int _{\mathcal O}e_k(x) dx|\le \frac{C}{k},$ and $\int _{c_0}^{\infty } e^{-x^4t} x^{-1}dx\le C(c_0)<\infty$ for any $c_0>0$, we have that $\Vert S(t)v\Vert _{L^{\infty }}\le C\Vert v\Vert _{L^\infty }$ and $\Vert S(t)v\Vert _{L^{1}}\le C\Vert v\Vert _{L^1}.$ By using the Riesz–Thorin interpolation theorem (see e.g. [21]), it follows that for $q\ge 1,$

$$\begin{aligned} \Vert S(t)v\Vert _{L^q}\le C \Vert v\Vert _{L^q}. \end{aligned}$$

(8.1)

We first show that the contractivity property (2.4) holds for the case $1\le p\le 2.$ When $p=1,$ by using Minkowski’s inequality and Hölder’s inequality,

$$\begin{aligned}&\Vert S(t)v\Vert _{L^\infty }\le \sum _{k=1}^{\infty } e^{-\lambda _k^2t} |\langle v,e_k\rangle | \Vert e_k\Vert _{L^\infty } \le C t^{-\frac{1}{4}}\Vert v\Vert _{L^1}, \end{aligned}$$

where we use the fact that $\sum _{k=1}^{\infty }e^{-\lambda _k^2 t}\le Ct^{-\frac{1}{4}}.$ By (8.1), the above estimate and an interpolation argument, it follows that (2.4) holds for $p=1, q\ge 1$. Similarly, by using Hölder’s inequality, for $p=2,$ we obtain that

$$\begin{aligned} \Vert S(t)v\Vert _{L^\infty }&\le \sum _{k=1}^{\infty } e^{-\lambda _k^2t} |\langle v,e_k\rangle | \Vert e_k\Vert _{L^\infty }\\&\le C\Big (\sum _{k=1}^{\infty } e^{-2\lambda _k^2t}\Big )^{\frac{1}{2}} \sum _{k=1}^{\infty } |\langle v,e_k\rangle |^2\le C t^{-\frac{1}{8} }\Vert v\Vert . \end{aligned}$$

Thus, by (8.1) and interpolation arguments, we also have (2.4) in the case of $p=2, q\ge 2.$ By using the Riesz–Thorin interpolation theorem (see e.g. [21]), we could get (2.4) for $1\le p\le 2,q \ge p.$ To show the case that $q\ge p>2,$ we use the dual form of $L^q$ norm. In fact, according to the self-adjoint property of S(t),  it follows that $\Vert S(t)v\Vert _{L^q}=\sup \limits _{\Vert w\Vert _{L^{q'}}\le 1} \Big |\langle S(t)v,w\rangle \Big |$, where $\frac{1}{q}+\frac{1}{q'}=1.$ Moreover, using Hölder’s inequality and (2.4) for $1\le p'\le 2,q' \le p',$ we obtain that for $\frac{1}{p}+\frac{1}{p'}=1,$

$$\begin{aligned} \Vert S(t)v\Vert _{L^q}&=\sup \limits _{\Vert w\Vert _{L^{q'}}\le 1} \Big |\langle S(t)v,w\rangle \Big |\\&\le \sup \limits _{\Vert w\Vert _{L^{q'}}\le 1} \Vert v\Vert _{L^p}\Vert S(t)w\Vert _{L^{p'}}\\&\le C \sup \limits _{\Vert w\Vert _{L^{q'}}\le 1} \Vert v\Vert _{L^p}t^{\frac{1}{4}(\frac{1}{q'}-\frac{1}{p'})}\Vert w\Vert _{L^{q'}}\\&\le C t^{\frac{1}{4}(\frac{1}{p}-\frac{1}{q})} \Vert v\Vert _{L^p}, \end{aligned}$$

which completes the proof for any $1\le p\le q\le \infty .$ Since for any $t>0,$ $v\in L^p$, $S(t)v\in E,$ we also have

$$\begin{aligned} \Vert S(t)v\Vert _{E}\le Ct^{-\frac{1}{4p}}\Vert v\Vert _{L^{p}}. \end{aligned}$$

$\square$

The proof of Proposition 5.1in the case that $\gamma \le \frac{1}{2}$:

Following the steps in the proof of Proposition 5.1, it suffices to show that $Y=X-Z$ satisfies

$$\begin{aligned} Y(t)=S(t)X(0)-\int _0^tS(t-s)AF(X(s))ds. \end{aligned}$$

More precisely, we will show that $S(t)X^N(0)$ converges to S(t)X(0) and that $-A\int _0^tS(t-s)P^NF(X^N(s))ds$ converges to $-A\int _0^tS(t-s)F(X(s))ds$. To this end, we need the following convergence result.

Proposition 8.1

Under the condition of Proposition 5.1, $\{X^{N}\}_{N\in \mathbb N^+}$ is a Cauchy sequence in $L^{2p}(\Omega ; L^{\kappa _1}([0,T];E))\cap L^{2p}(\Omega ; L^{\kappa _2}([0,T];L^6))$, where $2\le \kappa _1 <8$ and $2 \le \kappa _2<12.$ Furthermore, there exists $\widetilde{\tau }\in (0,\max (\frac{\gamma }{2},\frac{3}{4}))$ such that

$$\begin{aligned} \Vert X^N-X^M\Vert _{L^{2p}(\Omega ; L^{\kappa _1}([0,T];E))}+\Vert X^N-X^M\Vert _{L^{2p}(\Omega ; L^{\kappa _2}([0,T];L^6))}\le C \lambda _N^{-\widetilde{\tau }}, \end{aligned}$$

where $M\ge N.$

Proof

The proof of the convergence in $L^{2p}(\Omega ; L^{\kappa _1}([0,T];E))$ and $L^{2p}(\Omega ;$ $L^{\kappa _2} ([0,T];L^6))$ are similar. We only present the details on the convergence in $L^{2p}(\Omega ; L^{\kappa _1}([0,T];E))$. Since $X^N=Y^N+Z^N$, it suffices to show the convergence of $Y^N$ and $Z^N$ respectively. Let $M\ge N$. Then by using the arguments in the proof of Lemma 3.2, we have that for $\frac{3}{8}-\frac{\alpha }{4}>\alpha _1>\frac{1}{8}+\frac{1}{p}$ with a large enough $p\ge 1,$ and a small enough $\epsilon >0$,

$$\begin{aligned}&\mathbb E\Big [\sup _{t\in [0,T]}\Vert Z^N(t)-Z^M(t)\Vert _E^{2p}\Big ] \\&\le C_p\mathbb E\Big [\sup _{t\in [0,T]}\Vert Z^N(t)-P^NZ^M(t)\Vert _E^{2p}\Big ] +C_p\mathbb E\Big [\sup _{t\in [0,T]}\Vert (I-P^N)Z^M(t)\Vert _E^{2p}\Big ]\\&\le C(p,T) \int _0^T \mathbb E\Big [\Big (\int _0^{{t}} ({t}-r)^{-2\alpha _1-\frac{1}{4}}\Vert X^N(r)-X^M(r)\Vert ^{2}dr\Big )^{p}\Big ] dt\\&\quad +C(p,T) \int _0^T \mathbb E\Big [\Big (\int _0^{{t}} ({t}-r)^{-2\alpha _1-\frac{1}{4}-\epsilon } \lambda _N^{-2\epsilon } (1+\Vert X^M(r)\Vert ^{2})dr\Big )^{p}\Big ]dt, \end{aligned}$$

where we have used the following estimate

$$\begin{aligned} \Vert S(\frac{t-r}{2})(I-P^N) u\Vert&\le \Vert S(\frac{t-r}{2})A^{\epsilon }\Vert \Vert (I-P^N) A^{-\epsilon } u\Vert \\&\le C(t-r)^{-\frac{\epsilon }{2}} \lambda _N^{-\epsilon }\Vert u\Vert . \end{aligned}$$

Lemma 3.3 and Theorem 4.1 lead to

$$\begin{aligned}&\mathbb E\Big [\sup _{t\in [0,T]}\Vert Z^N(t)-Z^M(t)\Vert _E^{2p}\Big ] \le C(T,X_0,p) (\lambda _{N}^{-\tau p}+\lambda _N^{-2\epsilon p}) \end{aligned}$$

(8.2)

where $\tau <\gamma .$ Since $C([0,T];E)\subset L^{\kappa _1}([0,T];E)$, it remains to prove the convergence $Y^N$. The mild form $Y^N-Y^M$, together with (2.4), (2.3) and (4.2), yields that for any $\gamma _2<\frac{3}{4},$

$$\begin{aligned}&\Vert Y^N(t)-Y^M(t)\Vert _E \\&\le \Vert S(t)(I-P^N)X^M(0)\Vert _E +\Big \Vert \int _0^t S(t-s)P^MA(I-P^N)F(X^M(s))ds\Big \Vert _E\\&\quad + \Big \Vert \int _0^t S(t-s) P^N A(F(X^M(s))-F(X^N(s)))ds \Big \Vert _E\\&\le C t^{-\frac{1}{8}} \lambda _N^{-\frac{\gamma }{2}} \Vert X^M(0)\Vert _{\mathbb H^{\gamma }}\\&\quad +C \int _0^t (t-s)^{-\frac{5}{8}-\frac{\gamma _2}{2}} \lambda _N^{-\gamma _2} (1+ \Vert X^M(s)\Vert _{L^6}^3)ds\\&\quad +C \int _0^t (t-s)^{-\frac{5}{8}}\big (1+\Vert X^N(s)\Vert _E^2+\Vert X^M(s)\Vert _E^2\big )ds \sup _{s\in [0,t]}\Vert X^N(s)-X^M(s)\Vert . \end{aligned}$$

Using Proposition 3.2 and Theorem 4.1, we obtain that for $\gamma _2<\frac{3}{4},$ $\tau <\gamma ,$

$$\begin{aligned}&\mathbb E\Big [ \Big (\int _0^T\Vert Y^N(t)-Y^M(t)\Vert _E^{\kappa _1}dt\Big )^{\frac{2p}{\kappa _1}}\Big ]\\&\le C(\lambda _N^{-2\gamma _2 p}+\lambda _N^{-\tau p}) + C \lambda _N^{- \gamma p} \Vert X^M(0)\Vert _{\mathbb H^{\gamma }}^{2p} \mathbb E\Big [\Big (\int _0^T t^{-\frac{1}{8} \kappa _1} dt\Big )^{2p}\Big ]\\&\le C(\lambda _N^{-2\gamma _2 p}+\lambda _N^{-\tau p}+\lambda _N^{-\gamma p}). \end{aligned}$$

Thus, $\{Y^N\}_{N\in \mathbb N^+}$ is a Cauchy sequence in $L^{2p}(\Omega ; L^{\kappa _1}([0,T];E))$ and $\widetilde{\tau }\le \max (\epsilon ,\frac{\tau }{2}, \frac{3}{4})$ This, together with the convergence of $Z^N$ in $L^{2p}(\Omega ; L^{\kappa _1}([0,T];E))$, implies that $\{X^N\}_{N\in \mathbb N^+}$ forms a Cauchy sequence in $L^{2p}(\Omega ; L^{\kappa _1}([0,T];E))$. $\square$

Since $S(t)X^N(0)$ is convergent to S(t)X(0), we only need to estimate

$$\begin{aligned} Err_1(t):= \Big \Vert \int _0^t S(t-s)A(F(X(s))-P^NF(X^N(s)))ds\Big \Vert . \end{aligned}$$

Applying (2.3) and Theorem 4.1, we obtain that for $\gamma _3\in (0,\frac{1}{2}),$

$$\begin{aligned} Err_1(t)&\le \Big \Vert \int _0^t S(t-s)A(I-P^N) F(X(s)ds\Big \Vert \\&\quad + \Big \Vert \int _0^t S(t-s) AP^N( F(X^N(s))- F(X(s)) )ds \Big \Vert \\&\le C\int _0^t (t-s)^{-\frac{1}{2} -\gamma _3} \lambda _N^{-2\gamma _3}(1+\Vert X(s)\Vert ^3_{L^6})ds\\&\quad +C \int _0^t (t-s)^{-\frac{1}{2}}(1+\Vert X^N(s)\Vert ^{2}_E+\Vert X(s)\Vert ^2_E) \Vert X^N(s)-X(s)\Vert ds. \end{aligned}$$

Hölder’s inequality, together with Proposition 8.1, yields that for $\tau <\gamma$, $\gamma _3\in (0,\frac{1}{2}),$ $3l<12,$ $(\frac{1}{2} +\gamma _3)\frac{l}{l-1}<1$ and $2l_1<8$, $\frac{l_1}{2(l_1-1)}<1,$

$$\begin{aligned}&\mathbb E\Big [ \sup _{t\in [0,T]}\Vert Err_1(t)\Vert ^{2p}\Big ]\\&\le C \mathbb E\Big [ \sup _{t\in [0,T]} \Big (\int _0^t (t-s)^{-\frac{1}{2} -\gamma _3} \lambda _N^{-2\gamma _3}(1+\Vert X(s)\Vert ^3_{L^6})ds\Big )^{2p} \Big ]\\&\quad +C \mathbb E\Big [\sup _{t\in [0,T]}\Big (\int _0^t (t-s)^{-\frac{1}{2}} (1+\Vert X^N(s)\Vert ^{2}_E+\Vert X(s)\Vert ^2_E) ds\Big )^{2p} \sup _{t\in [0,T]}\Vert X^N(s)-X(s)\Vert ^{2p}\Big ]\\&\le C\lambda _N^{{-4\gamma _3}p} \mathbb E\Big [ \sup _{t\in [0,T]} \Big (\int _0^t (t-s)^{-\frac{1}{2} -\gamma _3} (1+\Vert X(s)\Vert ^3_{L^6})ds\Big )^{2p} \Big ]\\&\quad +C \lambda _N^{-\tau p} \Big (\mathbb E\Big [\sup _{t\in [0,T]}\Big ( \int _0^t (t-s)^{-\frac{1}{2}}(1+\Vert X^N(s)\Vert ^{2}_E+\Vert X(s)\Vert ^2_E) ds\Big )^{4p}\Big ]\Big )^{\frac{1}{2}} \\&\le C\lambda _N^{-{4\gamma _3}p} \Big (\mathbb E\Big [ \sup _{t\in [0,T]} \Big (\int _0^t (t-s)^{(-\frac{1}{2} -\gamma _3)\frac{l}{l-1}}\Big )^{4p\frac{l-1}{l}} \Big (\int _0^T (1+\Vert X(s)\Vert ^{3l}_{L^6})ds\Big )^{\frac{4p}{l}} \Big ]\Big )^{\frac{1}{2}}\\&\quad +C \lambda _N^{-\tau p}\sup _{t\in [0,T]}\Big ( \int _0^t (t-s)^{-\frac{l_1}{2(l_1-1)}} ds \Big )^{2p\frac{l_1-1}{l_1}} \Big (\mathbb E\Big [\Big (\int _0^T (1+\Vert X^N(s)\Vert ^{2 l_1}_E+\Vert X(s)\Vert ^{2 l_1}_E) ds\Big )^{\frac{4p}{l_1}}\Big ]\Big )^{\frac{1}{2}} \\&\le C(\lambda _N^{-{4\gamma _3}p}+\lambda _N^{-\tau p}), \end{aligned}$$

which implies that Y satisfies $Y(t)=S(t)X(0)-\int _0^tS(t-s)AF(X(s))ds$. This, together with the convergence of $Z^N$ to Z, shows that X is the global mild solution.

Exponential integrability of the mild solution

Corollary 8.1

Let $X_0\in \mathbb H^{\gamma }, \gamma \in (0,\frac{3}{2}).$ There exist $\beta >0$, $c>0$ such that for $t\in [0,T]$,

$$\begin{aligned}&\mathbb E\Big [\exp \Big (\frac{1}{2}e^{-\beta t}\Vert X(t)\Vert _{\mathbb H^{-1}}^2+c\int _0^t e^{-\beta s}\Vert X(s)\Vert _{L^4}^4ds+c\int _0^te^{-\beta s}\Vert \nabla X(s)\Vert ^2ds\Big )\Big ]\nonumber \\&\le C(X_0,T). \end{aligned}$$

(8.3)

Proof

From the Gagliardo–Nirenberg inequality

$$\begin{aligned} \Vert u\Vert _{L^4}\le C\Vert \nabla u\Vert _{L^2}^{\frac{1}{4}}\Vert u\Vert ^{\frac{3}{4}}+C\Vert u\Vert , \end{aligned}$$

and Young’s inequality, it follows that

$$\begin{aligned}&\int _0^t\Vert X^N(s)-X(s)\Vert ^4_{L^4}ds\\&\le C\int _0^t\Vert \nabla (X^N(s)-X(s))\Vert ^2ds +C\int _0^t (1+\Vert X^N(s)-X(s)\Vert ^6)ds. \end{aligned}$$

We first claim that

$$\begin{aligned} \lim _{N\rightarrow \infty }\Vert X^N-X\Vert _{L^{2}(\Omega ; L^{2}([0,t];\mathbb H^1))}+\lim _{N\rightarrow \infty }\Vert X^N-X\Vert _{C([0,t];L^{6}(\Omega ; \mathbb H))}=0. \end{aligned}$$

(8.4)

The estimate of the second term is similar to that of Proposition 8.1. To show the convergence of $\Vert X^N-X\Vert _{L^{2}(\Omega ; L^{2}([0,t];\mathbb H^1))}$, one needs to recall that from the proof of Proposition 4.1 and taking $p=1,$ it holds that

$$\begin{aligned}&\mathbb E[\Vert X^N(t)-P^NX^M(t)\Vert _{\mathbb H^{-1}}^{2}]+2\mathbb E[\int _0^t\Vert X^N(s)-P^NX^{M}(s)\Vert _{\mathbb H^1}^2ds]\\&\le C(T,X_0,p) \lambda _{N}^{-\gamma }. \end{aligned}$$

For convenience, we assume that $\sup \limits _{N\in \mathbb N^+}\Vert X^N_0\Vert _E\le C(X_0)$ or $\gamma >\frac{1}{2}$ here. Otherwise, one needs to deal with the singularity appearing in each integral term via a tedious and technical calculus. Next, it remains to estimate $\mathbb E[\int _0^t\Vert (I-P^N)X^{M}(s)\Vert _{\mathbb H^1}^2ds].$ Indeed, the a priori estimates of $X^M$ and $Z^M$, $\sum _{i=1}^{\infty } e^{-\lambda _i^2t} \le Ct^{-\frac{1}{4}}$, (2.3) and (5.4) yield that

$$\begin{aligned}&\mathbb E\Big [\int _0^t\Vert (I-P^N)X^{M}(s)\Vert _{\mathbb H^1}^2ds\Big ]\\&\le \mathbb E\Big [\int _0^t\Vert (I-P^N) S(s)X^M(0)\Vert _{\mathbb H^1}^2ds \Big ]\\&\quad +\mathbb E\Big [\int _0^t\Vert (I-P^N) \int _0^s S(s-r)A F(X^M(r))dr\Vert _{\mathbb H^1}^2ds \Big ] \\&\quad +\mathbb E\Big [\int _0^t\Vert (I-P^N) Z^M(s)\Vert _{\mathbb H^1}^2ds \Big ]\\&\le C \lambda _N^{-2\epsilon }\Big (\int _0^t s^{-\frac{1}{2}-\epsilon }ds+\int _0^t(\int _0^s (s-r)^{-\frac{3}{4}-\epsilon }dr)^2ds+\int _0^t\int _0^s (s-r)^{-\frac{3}{4}-\epsilon }drds\Big )\\&\le C \lambda _N^{-2\epsilon }. \end{aligned}$$

The above claim (8.4) implies that

$$\begin{aligned} \lim _{N\rightarrow \infty }\Vert X^N-X\Vert _{L^{4}(\Omega ; L^{4}([0,t];L^4))}=0. \end{aligned}$$

Take a subsequence $X^{N_k}$ of $X^N$ such that

$$\begin{aligned} X^{N_k} \rightarrow X \; \text {in} \; C([0,T];\mathbb H^{-1})\cap L^2([0,T];\mathbb H^1) \cap L^4([0,T];L^4), \text {a.s.} \end{aligned}$$

Thus, to prove (8.3), by using Fatou’s lemma, it suffices to show the uniform boundedness of the exponential moment for $X^N$, i.e.,

$$\begin{aligned} \mathbb E\Big [\exp \Big (\frac{1}{2}e^{-\beta t}\Vert X^N(t)\Vert _{\mathbb H^{-1}}^2+c\int _0^t e^{-\beta s}\Vert X^N(s)\Vert _{L^4}^4ds+c\int _0^te^{-\beta s}\Vert \nabla X^N(s)\Vert ^2ds\Big )\Big ] \end{aligned}$$

since the terms inside the expectation converges to those of X,  a.s.

Denote $\mu (x)=-A^2x-AP^NF(x)$ and $\sigma (x)=P^NG(x)I_{\mathbb H}$ and $U(x)=\frac{1}{2}\Vert x\Vert _{\mathbb H^{-1}}^2$, where $x\in P^N(\mathbb H)$. Using (2.5), the Lipschitz continuity of G, and applying Hölder’s and Young’s inequality, we get for a small $\epsilon >0,$

$$\begin{aligned}&\langle DU(x),\mu (x)\rangle +\frac{1}{2}\text {tr}[D^2U(x)\sigma (x)\sigma ^*(x)]+\frac{1}{2} \Vert \sigma (x)^*DU(x)\Vert ^2\\&=\langle x, -A^2x{-A}F(x)\rangle _{\mathbb H^{-1}} +\frac{1}{2}\sum _{i\in \mathbb N^+} \Vert P^N(G(x)e_i)\Vert _{\mathbb H^{-1}}^2 +\frac{1}{2} \sum _{i\in \mathbb N^+}\langle x,G(x)e_i\rangle _{\mathbb H^{-1}}^2\\&\le -(1-\epsilon )\Vert \nabla x\Vert ^2-(4c_4-\epsilon )\Vert x\Vert _{L^4}^4+\epsilon \Vert x\Vert _{\mathbb H^{-1}}^2+C(\epsilon ). \end{aligned}$$

Using the exponential integrability lemma in [8] and taking $\beta =\epsilon$, we have

$$\begin{aligned}&\mathbb E\Big [\exp \Big (e^{-\beta t}\frac{1}{2}\Vert X^N(t)\Vert _{\mathbb H^{-1}}^2+(4c_4-\epsilon )\int _0^t e^{-\beta s}\Vert X^N(s)\Vert _{L^4}^4ds\\&\quad +(1-\epsilon )\int _0^te^{-\beta s}\Vert \nabla X^N(s)\Vert ^2ds\Big )\Big ]\le C(X_0,T,\epsilon ), \end{aligned}$$

which completes the proof. $\square$