In this section we study the forward problem of the Navier-Stokes equation driven by white noise. Section 2.1 describes the forward problem, the Navier-Stokes equation, and rewrites it as an ordinary differential equation in a Hilbert space. In Sect. 2.2 we define the functional setting used throughout the paper. Section 2.3 highlights the solution concept that we use, leading in Sect. 2.4 to proof of the key fact that the solution of the Navier-Stokes equation is continuous as a function of the rough driving of interest and the initial condition. All our theoretical results in this paper are derived in the case of Dirichlet (no flow) boundary conditions. They may be extended to the problem on the periodic torus \(\mathbb {T}^d\), but we present the more complex Dirichlet case only for brevity.
Overview
Let \(D\in \mathbb {R}^2\) be a bounded domain with smooth boundary. We consider in \(D\) the Navier-Stokes equation
$$\begin{aligned} \partial _t u-\nu \Delta u+u\cdot \nabla u&= f-\nabla p,\quad (x,t) \in D \times (0,\infty ) \nonumber \\ \nabla \cdot u&= 0,\quad (x,t) \in D \times (0,\infty )\nonumber \\ u&= 0,\quad (x,t)\in \partial D \times (0,\infty ),\nonumber \\ u&= u_0,\quad (x,t) \in D \times \{0\}. \end{aligned}$$
(1)
We assume that the initial condition \(u_0\) and the forcing \(f(\cdot ,t)\) are divergence-free. We will in particular work with Eq. (3) below, obtained by projecting (1) into the space of divergence free functions—the Leray projector [19]. We denote by \(\mathsf {V}\) the space of all divergence-free smooth functions from \(D\) to \(\mathbb {R}^2\) with compact support, by \(\mathbb {H}\) the closure of \(\mathsf {V}\) in \((L^2(D))^2\), and by \(\mathbb {H}^1\) the closure of \(\mathsf {V}\) in \((H^1(D))^2\). Let \(\mathbb {H}^2=(H^2(D))^2\bigcap \mathbb {H}^1\). The initial condition \(u_0\) is assumed to be in \(\mathbb {H}\). We define the linear Stokes’ operator \(A:\mathbb {H}^2\rightarrow \mathbb {H}\) by \(A u=-\Delta u\) noting that the assumption of compact support means that Dirichlet boundary condition is imposed on the Stokes’ operator \(A\). Since \(A\) is selfadjoint, \(A\) possesses eigenvalues \(0<\lambda _1\le \lambda _2\le \cdots \) with the corresponding eigenvectors \(e_1, e_2, \ldots \in \mathbb {H}^2\).
We denote by \(\langle \cdot ,\cdot \rangle \) the inner product in \(\mathbb {H}\), extended to the dual pairing on \(\mathbb {H}^{-1} \times \mathbb {H}^1\). We then define the bilinear form \(B: \mathbb {H}^1\times \mathbb {H}^1\rightarrow \mathbb {H}^{-1}\)
$$\begin{aligned} \langle B(u,v),z\rangle =\int \limits _Dz(x)\cdot (u(x)\cdot \nabla )v(x)dx \end{aligned}$$
which must hold for all \(z \in \mathbb {H}^1.\) From the incompressibility condition we have, for all \(z \in \mathbb {H}^1\),
$$\begin{aligned} \langle B(u,v),z\rangle =-\langle B(u,z),v\rangle . \end{aligned}$$
(2)
By projecting problem (1) into \(\mathbb {H}\) we may write it as an ordinary differential equation in the form
$$\begin{aligned} du(t)=-\nu A udt-B(u,u)dt+dW(t),\quad u(0)=u_0\in \mathbb {H}, \end{aligned}$$
(3)
where \(dW(t)\) is the projection of the forcing \(f(x,t)dt\) into \(\mathbb {H}\). We will define the solution of this equation pathwise, for suitable \(W\), not necessarily differentiable in time.
Function spaces
For any \(s\ge 0\) we define \(\mathbb {H}^s\subset \mathbb {H}\) to be the Hilbert space of functions \(u=\sum _{k=1}^\infty u_ke_k\in \mathbb {H}\) such that
$$\begin{aligned} \sum _{k=1}^\infty \lambda _k^{s}u_k^2<\infty ; \end{aligned}$$
we note that the \(\mathbb {H}^j\) for \(j \in \{0,1,2\}\) coincide with the preceding definitions of these spaces. The space \(\mathbb {H}^s\) is endowed with the inner product
$$\begin{aligned} \langle u,v\rangle _{\mathbb {H}^s}=\sum _{k=1}^\infty \lambda _k^{s}u_kv_k, \end{aligned}$$
for \(u=\sum _{k=1}^\infty u_ke_k\), \(v=\sum _{k=1}^\infty v_ke_k\) in \(\mathbb {H}\). We denote by \(\mathbb {V}\) the particular choice \(s=\frac{1}{2}+\epsilon \), namely \(\mathbb {H}^{\frac{1}{2}+\epsilon }\), for given \(\epsilon >0\). In what follows we will be particularly interested in continuity of the mapping from the forcing \(W\) into linear functionals of the solution of (3). To this end it is helpful to define the Banach space \({\mathbb X}:=C([0,T];\mathbb {V})\) with the norm
$$\begin{aligned} \Vert W\Vert _{{\mathbb X}}=\sup _{t\in (0,T)}\Vert W(t)\Vert _{\mathbb {V}}. \end{aligned}$$
Solution concept
In what follows we define a solution concept for Eq. (3) for each forcing function \(W\) which is continuous, but not necessarily differentiable, in time. We always assume that \(W(0)=0.\) Following Flandoli [6], for each \(W\in {\mathbb X}\) we define the weak solution \(u(\cdot ;W)\in C([0,T];\mathbb {H})\bigcap L^2([0,T];\mathbb {H}^{1/2})\) of (3) as a function that satisfies
$$\begin{aligned} \langle u(t),\phi \rangle +{\nu }\int \limits _0^t\langle u(s),A\phi \rangle ds-\int \limits _0^t\langle {B\bigl (u(s),\phi \bigr ),u(s)}\rangle dx= \langle u_0,\phi \rangle +\langle W(t),\phi \rangle ,\nonumber \\ \end{aligned}$$
(4)
for all \(\phi \in \mathbb {H}^2\) and all \(t\in (0,T)\); note the integration by parts on the Stokes’ operator and the use of (2) to derive this identity from (3). Note further that if \(u\) and \(W\) are sufficiently smooth, (4) is equivalent to (3).
To employ this solution concept we first introduce the concept of a solution of the linear equation
$$\begin{aligned} dz(t)=-\nu A zdt+dW(t),\quad z(0)=0\in \mathbb {H}\end{aligned}$$
(5)
where \(W\) is a deterministic continuous function obtaining values in \(\mathbb {X}\) but not necessarily differentiable. We define a weak solution of this equation as a function \(z\in C([0,T];\mathbb {H})\) such that
$$\begin{aligned} \langle z(t),\phi \rangle +{\nu } \int \limits _0^t\langle z(s),A\phi \rangle ds=\langle W(t),\phi \rangle \end{aligned}$$
(6)
for all \(\phi \in \mathbb {H}^2\).
Then for this function \(z(t)\) we consider the solution \(v\) of the equation
$$\begin{aligned} dv(t)=-\nu A vdt-B(z+v,z+v)dt,\ \ v(0)=u_0\in \mathbb {H}. \end{aligned}$$
(7)
As we will show below, \(z(t)\) possesses sufficiently regularity so (7) possesses a weak solution \(v\). We then deduce that \(u=z+v\) is a weak solution of (3) in the sense of (4). When we wish to emphasize the dependence of \(u\) on \(W\) (and similarly for \(z\) and \(v\)) we write \(u(t;W).\)
We will now show that the function \(z\) defined by
$$\begin{aligned} z(t)&= \int \limits _0^t e^{-\nu A(t-s)}dW(s)\nonumber \\&= W(t)-\int \limits _0^ t\nu A e^{-\nu A(t-s)}W(s)ds \end{aligned}$$
(8)
satisfies the weak formula (6). Let \(w_k=\langle W, e_k \rangle \), that is
$$\begin{aligned} W(t)\,{:=}\,\sum _{k=1}^\infty {w_k}(t)e_k\in {\mathbb X}. \end{aligned}$$
(9)
We then deduce from (8) that
$$\begin{aligned} z(t;W)=W(t)-\sum _{k=1}^\infty \left( \int \limits _0^t w_k(s){\nu }\lambda _ke^{(t-s)(-{\nu }\lambda _k)}ds\right) e_k. \end{aligned}$$
(10)
We have the following regularity property for \(z\):
Lemma 1
For each \(W \in {\mathbb X}\), the function \(z=z(\cdot ;W)\in C([0,T];\mathbb {H}^{1/2})\).
Proof
We first show that for each \(t, z(t;W)\) as defined in (10) belongs to \(\mathbb {H}^{1/2}\). Fixing an integer \(M>0\), using inequality \(a^{1-\epsilon /2}e^{-a}<c\) for all \(a>0\) for an appropriate constant \(c\), we have
$$\begin{aligned}&\sum _{k=1}^M\lambda _k^{1/2}\left( \int \limits _0^t\nu \lambda _ke^{(t-s)(-\nu \lambda _k)}w_k(s)ds\right) ^{2}\\&\quad \le \sum _{k=1}^M\lambda _k^{1/2}\left( \int \limits _0^t{c\over (t-s)^{1-\epsilon /2}}\lambda _k^{\epsilon /2}|w_k(s)|dx\right) ^{2}. \end{aligned}$$
Therefore,
$$\begin{aligned} \left\| \sum _{k=1}^M\int \limits _0^t\nu \lambda _ke^{(t-s)(-\nu \lambda _k)}w_k(s)e_kds\right\| _{\mathbb {H}^{1/2}}&\le \left\| \sum _{k=1}^M\int \limits _0^t{c\over (t-s)^{1-\epsilon /2}}\lambda _k^{\epsilon /2}|w_k(s)|e_kds\right\| _{\mathbb {H}^{1/2}}\\&\le \int \limits _0^t{c\over (t-s)^{1-\epsilon /2}}\left\| \sum _{k=1}^M\lambda _k^{\epsilon /2}|w_k(s)|e_k\right\| _{\mathbb {H}^{1/2}}ds\\&\le \max _{s\in (0,T)}\Vert W(s)\Vert _{\mathbb {H}^{1/2+\epsilon }}\int \limits _0^t{c\over (t-s)^{1-\epsilon /2}}ds, \end{aligned}$$
which is uniformly bounded for all \(M\). Therefore,
$$\begin{aligned} \sum _{k=1}^{\infty }\left( \int \limits _0^tw_k(s)\nu \lambda _ke^{(t-s)(-\nu \lambda _k)}ds\right) e_k\in \mathbb {H}^{1/2}. \end{aligned}$$
It follows from (10) that, since \(W \in {\mathbb X}\), for each \(t\), \(z(t;W)\in \mathbb {H}^{1/2}\) as required. Furthermore, for all \(t\in (0,T)\)
$$\begin{aligned} \Vert z(t;W)\Vert _{\mathbb {H}^{1/2}}\le c\Vert W\Vert _{{\mathbb X}}. \end{aligned}$$
(11)
Now we turn to the continuity in time. Arguing similarly, we have that
$$\begin{aligned}&\left\| \sum _{k=M}^\infty \left( \int \limits _0^tw_k(s)\nu \lambda _ke^{(t-s)(-\nu \lambda _k)}ds\right) e_k\right\| _{\mathbb {H}^{1/2}}\\&\quad \le \int \limits _0^t{c\over (t-s)^{1-\epsilon /2}}\left\| \sum _{k=M}^\infty w_k(s)e_k\right\| _{\mathbb {H}^{1/2+\epsilon }}ds\\&\quad \le \left( \int \limits _0^t{c\over (t-s)^{(1-\epsilon /2)^p}}ds\right) ^{1/p}\left( \int \limits _0^t\left\| \sum _{k=M}^\infty w_k(s)e_k\right\| ^q_{\mathbb {H}^{1/2+\epsilon }}ds \right) ^{1/q}, \end{aligned}$$
for all \(p,q>0\) such that \(1/p+1/q=1\). From the Lebesgue dominated convergence theorem,
$$\begin{aligned} \lim _{M\rightarrow \infty }\int \limits _0^{t}\left\| \sum _{k=M}^\infty w_k(s)e_k\right\| ^q_{\mathbb {H}^{1/2+\epsilon }}ds=0; \end{aligned}$$
and when \(p\) sufficiently close to 1,
$$\begin{aligned} \int \limits _0^t{c\over (t-s)^{(1-\epsilon /2)p}}ds \end{aligned}$$
is finite. We then deduce that
$$\begin{aligned} \lim _{M\rightarrow \infty }\left\| \sum _{k=M}^\infty \left( \int \limits _0^tw_k(s)\nu \lambda _ke^{(t-s)(-\nu \lambda _k)}ds\right) e_k\right\| _\mathbb {H}^{1/2}=0, \end{aligned}$$
uniformly for all \(t\).
Fixing \(t\in (0,T)\) we show that
$$\begin{aligned} \lim _{t'\rightarrow t}\Vert z(t;W)-z({t';W})\Vert _{\mathbb {H}^{1/2}}=0. \end{aligned}$$
We have
$$\begin{aligned}&\Vert z(t;W)-z(t';W)\Vert _{\mathbb {H}^{1/2}}\le \Vert W(t)-W(t')\Vert _{\mathbb {H}^{1/2}}\\&\quad + \left\| \sum _{k=1}^{M-1}\left( \int \limits _0^tw_k(s)\nu \lambda _ke^{(t-s)(-\nu \lambda _k)}ds-\int \limits _0^{t'}w_k(s)\nu \lambda _ke^{(t'-s)(-\nu \lambda _k)}ds\right) e_k\right\| _{\mathbb {H}^{1/2}}\\&\quad +\left\| \sum _{k=M}^\infty \left( \int \limits _0^tw_k(s)\nu \lambda _ke^{(t-s)(-\nu \lambda _k)}ds\right) e_k\right\| _{\mathbb {H}^{1/2}}\\&\quad +\left\| \sum _{k=M}^\infty \left( \int \limits _0^{t'}w_k(s)\nu \lambda _ke^{(t'-s)(-\nu \lambda _k)}ds\right) e_k\right\| _{\mathbb {H}^{1/2}}. \end{aligned}$$
For \(\delta >0\), when \(M\) is sufficiently large, the argument above shows that
$$\begin{aligned}&\left\| \sum _{k=M}^\infty \left( \int \limits _0^tw_k(s)\nu \lambda _ke^{(t-s)(-\nu \lambda _k)}ds\right) e_k\right\| _{\mathbb {H}^{1/2}}\\&+\left\| \sum _{k=M}^\infty \left( \int \limits _0^{t'}w_k(s)\nu \lambda _ke^{(t'-s)(-\nu \lambda _k)}ds\right) e_k\right\| _{\mathbb {H}^{1/2}}<\delta /3. \end{aligned}$$
Furthermore, when \(|t'-t|\) is sufficiently small,
$$\begin{aligned} \left\| \sum _{k=1}^{M-1}\left( \int \limits _0^tw_k(s)\nu \lambda _ke^{(t-s)(-\nu \lambda _k)}ds-\int \limits _0^{t'}w_k(s)\nu \lambda _ke^{(t'-s)(-\nu \lambda _k)}ds\right) e_k\right\| _{\mathbb {H}^{1/2}}<\delta /3. \end{aligned}$$
Finally, since \(W \in {\mathbb X}\), for \(|t'-t|\) is sufficiently small we have
$$\begin{aligned} \Vert W(t)-W(t')\Vert _{\mathbb {H}^{1/2}} <\delta /3 \end{aligned}$$
Thus when \(|t'-t|\) is sufficiently small, \(\Vert z(t;W)-z(t';W)\Vert _{\mathbb {H}^{1/2}}<\delta \). The conclusion follows.\(\square \)
Having established regularity, we now show that \(z\) is indeed a weak solution of (5).
Lemma 2
For each \(\phi \in \mathbb {H}^2\), \(z(t)=z(t;W)\) satisfies (6).
Proof
It is sufficient to show this for \(\phi =e_k\). We have
$$\begin{aligned} \int \limits _0^t\langle z(s),Ae_k\rangle ds&= \int \limits _0^t\langle W(s),Ae_k\rangle ds-\int \limits _0^t\int \limits _0^s {w_k(\tau )\nu }\lambda _k^2e^{(s-\tau )(- {\nu }\lambda _k)}d\tau ds\\&= \lambda _k\int \limits _0^t {w_k}(s)ds-{\nu }\lambda _k^2\int \limits _0^t {w_k}(\tau )\Bigl (\int \limits _\tau ^te^{(s-\tau )(-{\nu }\lambda _k)} ds\Bigr )d\tau \\&= \lambda _k\int \limits _0^t{w_k}(s)ds-\lambda _k\int \limits _0^t{w_k}(\tau )d\tau +\lambda _k\int \limits _0^t{w_k}(\tau )e^{(t-\tau )(-{\nu }\lambda _k)}d\tau \\&= \lambda _k\int \limits _0^t{w_k}(\tau )e^{(t-\tau )(-{\nu }\lambda _k)}d\tau . \end{aligned}$$
On the other hand,
$$\begin{aligned} \langle z(t),e_k\rangle =\langle W(t),e_k\rangle -{\nu } \lambda _k\int \limits _0^t{w_k}(s)e^{(t-s)(-\nu \lambda _k)}ds. \end{aligned}$$
The result then follows.\(\square \)
We now turn to the following result, which concerns \(v\) and is established on page 416 of [6], given the properties of \(z(\cdot ;W)\) established in the preceding two lemmas.
Lemma 3
For each \(W\in {\mathbb X}\), problem (7) has a unique solution \(v\) in the function space \(C(0,T;\mathbb {H})\bigcap L^2(0,T;\mathbb {H}^1)\).
We then have the following existence and uniqueness result for the Navier-Stokes Eq. (3), more precisely for the weak form (4), driven by rough additive forcing [6]:
Proposition 1
For each \(W\in {\mathbb X}\), problem (4) has a unique solution \(u\in C(0,T;\mathbb {H})\bigcap L^2(0,T;\mathbb {H}^{1/2})\) such that \(u-z\in L^2(0,T;\mathbb {H}^1)\).
Proof
A solution \(u\) for (4) can be taken as
$$\begin{aligned} u(t;W)=z(t;W)+v(t;W). \end{aligned}$$
(12)
From the regularity properties of \(z\) and \(v\) in Lemmas 1 and 3, we deduce that \(u\in C(0,T;\mathbb {H})\bigcap L^2(0,T;\mathbb {H}^{1/2})\). Assume that \(\bar{u}(t;W)\) is another solution of (4). Then \(\bar{v}(t;W)=\bar{u}(t;W)-z(t;W)\) is a solution in \(C(0,T;\mathbb {H})\bigcap L^2(0,T;\mathbb {H}^1)\) of (7). However, (7) has a unique solution in \(C(0,T;\mathbb {H})\bigcap L^2(0,T;\mathbb {H}^1)\). Thus \(\bar{v}=v\).\(\square \)
Continuity of the forward map
The purpose of this subsection is to establish continuity of the forward map from \(W\) into the weak solution \(u\) of (3), as defined in (4), at time \(t>0.\) In fact we prove continuity of the forward map from \((u_0,W)\) into \(u\) and for this it is useful to define the space \({\mathcal H}=\mathbb {H}\times {\mathbb X}\) and denote the solution \(u\) by \(u(t;u_0,W)\).
Theorem 1
For each \(t>0\), the solution \(u(t;\cdot ,\cdot )\) of (3) is a continuous map from \({\mathcal H}\) into \(\mathbb {H}\).
Proof
First we fix the initial condition and just write \(u(t;W)\) for simplicity. We consider Eq. (3) with driving \(W \in \mathbb {X}\) given by (9) and by \(W' \in \mathbb {X}\) defined by
$$\begin{aligned} W'(s)=\sum _{k=1}^\infty {w_{k}}'(s)e_{k}\in {\mathbb X}. \end{aligned}$$
We will prove that, for \(W,W'\) from a bounded set in \(\mathbb {X}\), there is \(c={c}(T)>0\), such that
$$\begin{aligned} \sup _{t\in (0,T)}\Vert z(t;W)-z(t;W')\Vert _{\mathbb {H}^{1/2}}\le c \Vert W-W'\Vert _{{\mathbb X}} \end{aligned}$$
(13)
and, for each \(t \in (0,T)\),
$$\begin{aligned} \Vert v(t;W)-v(t;W')\Vert _{\mathbb {H}}^2\le c\sup _{s\in (0,T)}\Vert z(s;W)-z(s;W')\Vert _{L^4(D)}^2. \end{aligned}$$
(14)
This suffices to prove the desired result since Sobolev embedding yields, from (14),
$$\begin{aligned} \Vert v(t;W)-v(t;W')\Vert _{\mathbb {H}}^2\le c\sup _{s\in (0,T)}\Vert z(s;W)-z(s;W')\Vert _{\mathbb {H}^{\frac{1}{2}}}^2. \end{aligned}$$
(15)
Since \(u=z+v\) we deduce from (13) and (15) that \(u\) as a map from \({\mathbb X}\) to \(\mathbb {H}\) is continuous.
To prove (13) we note that
$$\begin{aligned}&\Vert z(t;W)-z(t;W')\Vert _{\mathbb {H}^{\frac{1}{2}}} \le \Vert W(t)-W'(t)\Vert _{\mathbb {H}^{\frac{1}{2}}}\nonumber \\&\quad + \left\| \int \limits _0^t \nu Ae^{-\nu A(t-s)}\left( W(s)-W'(s)\right) ds\right\| _{\mathbb {H}^{\frac{1}{2}}} \end{aligned}$$
so that
$$\begin{aligned}&\sup _{t \in (0,T)}\Vert z(t;W)-z(t;W')\Vert _{\mathbb {H}^{\frac{1}{2}}} \le \Vert W-W'\Vert _{{\mathbb X}}\\&\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad +\sup _{t \in (0,T)}\left\| \int \limits _0^t \nu Ae^{-\nu A(t-s)}\left( W(s)-W'(s)\right) ds\right\| _{\mathbb {H}^{\frac{1}{2}}}. \end{aligned}$$
Thus it suffices to consider the last term on the right hand side. We have
$$\begin{aligned}&\left\| \int \limits _0^t Ae^{-\nu A(t-s)}\left( W(s)-W'(s)\right) ds\right\| _{\mathbb {H}^{\frac{1}{2}}}^2\\&\qquad =\left\| \sum _{k=1}^{\infty }\int \limits _0^t\lambda _ke^{(t-s)(-\nu \lambda _k)}(w'_k(s)-w_k(s))e_kds\right\| _{\mathbb {H}^{1/2}}^2\\&\qquad =\sum _{k=1}^\infty \lambda _k^{1/2}\left( \int \limits _0^t\lambda _ke^{(t-s)(-{\nu }\lambda _k)}(w_k'(s)-w_k(s))ds\right) ^{2}\\&\qquad \le \sum _{k=1}^\infty \lambda _k^{1/2}\left( \int \limits _0^t\lambda _ke^{(t-s)(-{\nu }\lambda _k)}|w_k'(s)-w_k(s)|ds\right) ^{2}\\&\qquad \le \sum _{k=1}^\infty \lambda _k^{1/2}\left( \int \limits _0^t{c\over (t-s)^{1-\epsilon /2}}\lambda _k^{\epsilon /2}|w_k'(s)-w_k(s)|ds\right) ^{2} \end{aligned}$$
where we have used the fact that \(a^{1-\epsilon /2}e^{-a}<c\) for all \(a>0\) for an appropriate constant \(c\). From this, we deduce that
$$\begin{aligned}&\left\| \int \limits _0^t Ae^{-\nu A(t-s)}\bigl (W(s)-W'(s)\bigr )ds\right\| _{\mathbb {H}^{\frac{1}{2}}}\\&\qquad \le \left\| \sum _{k=1}^\infty \int \limits _0^t{c\over (t-s)^{1-\epsilon /2}}\lambda _k^{\epsilon /2}|w_k'(s)-w_k(s)|e_kds\right\| _{\mathbb {H}^{1/2}}\\&\qquad \le \int \limits _0^t{c\over (t-s)^{1-\epsilon /2}}\left\| \sum _{k=1}^\infty \lambda _k^{\epsilon /2}|w_k'(s)-w_k(s)|e_k\right\| _{\mathbb {H}^{1/2}}ds\\&\qquad \le \int \limits _0^t{c\over (t-s)^{1-\epsilon /2}}ds\sup _{s\in (0,T)}\left\| \sum _{k=1}^\infty \lambda _k^{\epsilon /2}|w_k'(s)-w_k(s)|e_k\right\| _{\mathbb {H}^{1/2}}\\&\qquad \le c\sup _{s\in (0,T)}\Vert W'(s)-W(s)\Vert _{\mathbb {V}}. \end{aligned}$$
Therefore (13) holds.
We now prove (14). We will use the following estimate for the solution \(v\) of (7) which is proved in Flandoli [6], page 412, by means of a Gronwall argument:
$$\begin{aligned} \sup _{s\in (0,T)} \Vert v(s)\Vert ^2_{\mathbb {H}}+\int \limits _0^T\Vert v(s)\Vert ^2_{\mathbb {H}^1}\le C\left( T,\sup _{s\in (0,T)}\Vert z(s)\Vert _{L^4(D)}\right) . \end{aligned}$$
(16)
We show that the map \(C([0,T];L^4(D))\ni z(\cdot ;W)\mapsto v(\cdot ;W)\in \mathbb {H}\) is continuous.
For \(W\) and \(W'\) in \({\mathbb X}\), define \(v=v(t;W), v'=v(t;W'), z=z(t;W), z'=z(t;W'), e=v-v'\) and \(\delta =z-z'\). Then we have
$$\begin{aligned} {de\over dt}+\nu Ae+ B\bigl (v+z,v+z\bigr )-B\bigl (v'+z',v'+z'\bigr )=0. \end{aligned}$$
(17)
From this, we have
$$\begin{aligned}&{1\over 2}{d\Vert e\Vert _{\mathbb {H}}^2\over dt}+\nu \Vert e\Vert _{\mathbb {H}^1}^2=-\bigl \langle B\bigl (v+z,v+z\bigr ),e\bigr \rangle \\&\quad +\,\big \langle B\bigl (v'+z',v'+z'\bigr ),e\big \rangle . \end{aligned}$$
From (2) we obtain
$$\begin{aligned}&{1\over 2}{d\Vert e\Vert _{\mathbb {H}}^2\over dt}+\nu \Vert e\Vert _{\mathbb {H}^1}^2= +\bigl \langle B\bigl (v+z,e\bigr ),v+z\bigr \rangle \\&\quad -\,\big \langle B\bigl (v'+z',e\bigr ),v'+z'\big \rangle =\bigl \langle B\bigl (v+z,e\bigr ),v+z-v'-z'\bigr \rangle \\&\quad -\,\big \langle B\bigl (v'+z'-v-z,e\bigr ),v'+z'\big \rangle =\bigl \langle B\bigl (v+z,e\bigr ),e+\delta \bigr \rangle \\&\quad +\,\big \langle B\bigl (e+\delta ,e\bigr ),v'+z'\big \rangle \le (\Vert e\Vert _{L^4(D)}+\Vert \delta \Vert _{L^4(D)}) (\Vert v\Vert _{L^4(D)}+\Vert z\Vert _{L^4(D)}\\&\quad +\,\Vert v'\Vert _{L^4(D)}+\Vert z'\Vert _{L^4(D)})\Vert e\Vert _{\mathbb {H}^1}. \end{aligned}$$
We now use the following interpolation inequality
$$\begin{aligned} \Vert e\Vert _{L^4(D)}\le c_0\Vert e\Vert _{\mathbb {H}^1}^{1/2}\Vert e\Vert _{\mathbb {H}}^{1/2}, \end{aligned}$$
(18)
which holds for all two dimensional domains \(D\) with constant \(c_0\) depending only on \(D\); see Flandoli [6]. Using this we obtain
$$\begin{aligned}&{1\over 2}{d\Vert e\Vert ^2_{\mathbb {H}}\over dt}+\nu \Vert e\Vert ^2_{\mathbb {H}^1}\le c_1\left( \Vert e\Vert _{\mathbb {H}^1}^{3/2}\Vert e\Vert _{\mathbb {H}}^{1/2}+\Vert \delta \Vert _{L^4(D)}\Vert e\Vert _{\mathbb {H}^1}\right) \\&\quad \cdot \left( \Vert v\Vert _{L^4(D)}+\Vert v'\Vert _{L^4(D)}+\Vert z\Vert _{L^4(D)}+\Vert z'\Vert _{L^4(D)}\right) \end{aligned}$$
for a positive constant \(c_1\). From the Young inequality, we have
$$\begin{aligned}&\Vert e\Vert ^{3/2}_{\mathbb {H}^1}\Vert e\Vert _{\mathbb {H}}^{1/2}\left( \Vert v\Vert _{L^4(D)}+\Vert v'\Vert _{L^4(D)}+\Vert z\Vert _{L^4(D)}+\Vert z'\Vert _{L^4(D)}\right) \\&\quad \le \frac{3}{4}c_2^{4/3}\Vert e\Vert _{\mathbb {H}^1}^2+\frac{1}{4c_2^4}\Vert e\Vert _{\mathbb {H}}^2\left( \Vert v\Vert _{L^4(D)}+\Vert v'\Vert _{L^4(D)}+\Vert z\Vert _{L^4(D)}+\Vert z'\Vert _{L^4(D)}\right) ^4 \end{aligned}$$
and
$$\begin{aligned}&\Vert \delta \Vert _{L^4(D)}\Vert e\Vert _{\mathbb {H}^1}\left( \Vert v\Vert _{L^4(D)}+\Vert v'\Vert _{L^4(D)}+\Vert z\Vert _{L^4(D)}+\Vert z'\Vert _{L^4(D)}\right) \\&\qquad \le \frac{c_3^2}{2}\Vert e\Vert _{\mathbb {H}^1}^2+\frac{1}{2c_3^2}\Vert \delta \Vert ^2_{L^4(D)}\left( \Vert v\Vert _{L^4(D)}+\Vert v'\Vert _{L^4(D)}+\Vert z\Vert _{L^4(D)}+\Vert z'\Vert _{L^4(D)}\right) ^2 \end{aligned}$$
for all positive constants \(c_2\) and \(c_3\). Choosing \(c_2\) and \(c_3\) so that \(c_1(3c_2^{4/3}/4+c_3^2/2)=\nu \), we deduce that there is a positive constant \(c\) such that
$$\begin{aligned} {1\over 2}{d\Vert e\Vert _{\mathbb {H}}^2\over dt}+\nu \Vert e\Vert _{\mathbb {H}^1}^2\le \nu \Vert e\Vert _{\mathbb {H}^1}^2 +c\Vert e\Vert _{\mathbb {H}}^2\cdot I_4 +c\Vert \delta \Vert _{L^4(D)}^2\cdot I_2 \end{aligned}$$
(19)
where we have defined
$$\begin{aligned} I_2&=\Vert v\Vert _{L^4(D)}^2+\Vert v'\Vert _{L^4(D)}^2+\Vert z\Vert _{L^4(D)}^2+\Vert z'\Vert _{L^4(D)}^2\\ I_4&=\Vert v\Vert _{L^4(D)}^4+\Vert v'\Vert _{L^4(D)}^4+\Vert z\Vert _{L^4(D)}^4+\Vert z'\Vert _{L^4(D)}^4. \end{aligned}$$
From Gronwall’s inequality, we have
$$\begin{aligned} \Vert e(t)\Vert _{\mathbb {H}}^2\le c\int \limits _0^t\bigg (e^{\int \limits _s^t I_4(s')ds'} \bigg ) \Vert \delta (s)\Vert _{L^4(D)}^2 I_2(s) ds. \end{aligned}$$
(20)
Applying the interpolation inequality (18) to \(v(s';W)\), we have that
$$\begin{aligned} \int \limits _0^T\Vert v(s';W)\Vert _{L^4(D)}^4 ds'\le c\sup _{s'\in (0,T)}\Vert v(s';W)\Vert _{\mathbb {H}}^2\int \limits _0^T\Vert v(s';W)\Vert _{\mathbb {H}^1}^2ds', \end{aligned}$$
which is bounded uniformly when \(W\) belongs to a bounded subset of \({\mathbb X}\) due to (16). Using this estimate, and a similar estimate on \(v'\), together with (11) and Sobolev embedding of \(\mathbb {H}^{\frac{1}{2}}\) into \(L^4(D)\), we deduce that
$$\begin{aligned} \Vert e(t)\Vert _{\mathbb {H}}^2\le c\sup _{0\le s\le T}\Vert \delta (s)\Vert _{L^4(D)}^2. \end{aligned}$$
We now extend to include continuity with respect to the initial condition. We show that \(u(\cdot ,t;u_0,W)\) is a continuous map from \({\mathcal H}\) to \(\mathbb {H}\). For \(W\in {\mathbb X}\) and \(u_0\in \mathbb {H}\), we consider the following equation:
$$\begin{aligned} \frac{dv}{dt}+A v+B(v+z,v+z)=0, \quad v(0)=u_0. \end{aligned}$$
(21)
We denote the solution by \(v(t)=v(t;u_0,W)\) to emphasize the dependence on initial condition and forcing which is important here. For \((u_0,W)\in {\mathcal H}\) and \((u_0',W')\in {\mathcal H}\), from (19) and Gronwall’s inequality, we deduce that
$$\begin{aligned}&\Vert v(t;u_0,W)-v(t;u_0',W')\Vert ^2_{\mathbb {H}}\le \Vert u_0-u_0'\Vert _{\mathbb {H}}^2e^{\int \limits _0^t I_4(s'))ds'}\\&\quad +\, c\int \limits _0^t\bigg (e^{\int _s^t I_4(s')ds'} \cdot \Vert z(s;W)-z(s;W')\Vert _{L^4(D)}^2. I_2(s)\bigg )ds. \end{aligned}$$
We then deduce that
$$\begin{aligned} \Vert v(t;u_0,W)-v(t;u_0',W')\Vert ^2_{\mathbb {H}}&\le c\Vert u_0-u_0'\Vert ^2_{\mathbb {H}}+c\sup _{0\le s\le T}\Vert z(s;W)-z(s;W')\Vert _{L^4(D)}^2\\&\le c\Vert u_0-u_0'\Vert ^2_{\mathbb {H}} \quad +c\sup _{t\in (0,T)}\Vert W(t)-W'(t)\Vert _{\mathbb {H}^{1/2+\epsilon }}. \end{aligned}$$
This gives the desired continuity of the forward map.\(\square \)