Forward invariance and Wong–Zakai approximation for stochastic moving boundary problems

Abstract

We discuss a class of stochastic second-order PDEs in one space-dimension with an inner boundary moving according to a possibly nonlinear, Stefan-type condition. We show that proper separation of phases is attained, i.e., the solution remains negative on one side and positive on the other side of the moving interface, when started with the appropriate initial conditions. To extend results from deterministic settings to the stochastic case, we establish a Wong–Zakai-type approximation. After a coordinate transformation, the problems are reformulated and analyzed in terms of stochastic evolution equations on domains of fractional powers of linear operators.

Appendices

A Nemytskii operators on Sobolev spaces

We continue with the analysis on the Sobolev spaces $H^k({\mathbb {R}}_+)$, $k\in {\mathbb {N}}$. In this section we prove some regularity results on the nonlinear Nemytskii operator

$$\begin{aligned} u \mapsto N(u)(x) = \mu (x,u(x)), \end{aligned}$$

where, $\mu :{\mathbb {R}}_+ \times {\mathbb {R}}^d \rightarrow {\mathbb {R}}$ and $x\in {\mathbb {R}}_+$. Note that these operators are well understood but most of the literature focuses on bounded domains, see, e.g., [2, 38, 39]. However, in the case of unbounded domains several additional conditions on $\mu$ are necessary to make them work. First, we state a result on the spaces $H^k$ which guarantees that, under certain assumptions on $\mu$, N will map $H^k$ into $H^k$. For a proof, we refer to [39, Theorem 1], of which it is a special case.

Lemma A.1

For each integer $k\ge 1$, the space $H^k({\mathbb {R}}_+)$ is a Banach algebra. In particular, there exists a constant c such that for all u, $v\in H^k({\mathbb {R}}_+)$ it holds that $uv\in H^k({\mathbb {R}}_+)$ and

$$\begin{aligned} \left||uv\right||_{H^k} \le c \left||u\right||_{H^k}\left||v\right||_{H^k}. \end{aligned}$$

1.1 A.1 Continuity

We now adapt the proof of the continuity result [39, Theorem 2] to our setting, but with some corrections in the proof. For notational reasons, we also introduce the Nemytskii operators

$$\begin{aligned} N_x(u)(x) := \left( \tfrac{\partial }{\partial x}\mu \right) (x,u(x)),\quad N_{y_j}(u)(x):=\left( \tfrac{\partial }{\partial y_j} \mu \right) (x,u(x)),\;j=1,\ldots ,d,\nonumber \\ \end{aligned}$$

(A.1)

for $\quad u\in H^k({\mathbb {R}}_+; {\mathbb {R}}^d),\;x\in {\mathbb {R}}_+$. In order for N to map $H^k$ into $H^k$ again, we need certain growth restrictions, which is not the case on bounded domains. For a multi-index $\alpha$, we denote by $D^\alpha$ the respective partial derivative operator.

Assumption A.2

Assume $\mu \in C^m({\mathbb {R}}_{\ge 0}\times {\mathbb {R}}^d, {\mathbb {R}})$ and

• (a) For each integer l, $0\le l\le m$, there exists an $a_l\in L^2({\mathbb {R}}_+)$ and some $b_l:{\mathbb {R}}^d\rightarrow {\mathbb {R}}_+$ locally bounded, such that

  $$\begin{aligned} \left|D^{(l,0,\ldots ,0)} \mu (x,y)\right| \le b_l(y)\left( a_l(x) + \left|y\right|\right) ,\,\forall \,x\in {\mathbb {R}}_+,\,y\in {\mathbb {R}}^d \end{aligned}$$

• (b) For each multi-index $\alpha$ with $\alpha _1 < \left|\alpha \right| \le m$, the family of functions $(D^\alpha \mu (x,.))_{x\in {\mathbb {R}}_{\ge 0}}$ is equicontinuous and $\sup _{x\in {\mathbb {R}}_+} \left|D^\alpha \mu (x,.)\right|$ is locally bounded.

Assumption A.3

Assume that $\mu \in C^{m}({\mathbb {R}}_{\ge 0} \times {\mathbb {R}}^d, {\mathbb {R}})$ and $D^{\alpha }\mu (x,.)$ is locally Lipschitz for all multi-indices $\alpha$, $\left|\alpha \right|\le m$ with Lipschitz constants uniform in $x\in {\mathbb {R}}_{\ge 0}$, i.e., we assume that for all $r\ge 0$ there exists $L_r\ge 0$ such that

$$\begin{aligned} \left|D^{\alpha }\mu (x,y) - D^{\alpha }\mu (x,z)\right| \le L_r \left|y-z\right|. \end{aligned}$$

holds for all x, y, $z\in {\mathbb {R}}^d$ with $\left|y\right|$, $\left|z\right|\le r$ and $\alpha$, $\left|\alpha \right|\le m$

Remark A.4

If $\mu$ satisfies Assumption A.2 for some integer $m\ge 1$, then $\mu$ satisfies Assumption A.3 for $m-1$.

Remark A.5

Recall the Sobolev imbeddings

$$\begin{aligned} H^{m+1}({\mathbb {R}}_+) \hookrightarrow BUC^m({\mathbb {R}}_+). \end{aligned}$$

As usual $BUC^m({\mathbb {R}}_+)$ is equipped with the $C^m_b$-norm. In the following, we will work with the $BUC^m$ representative of the elements in $H^{m+1}$ without further comment.

Note that Assumption A.2 is stronger than [18, Assumption 6.2] so that we get the following two results from [18, Appendix 1].

Theorem A.6

If Assumption A.2 holds for some integer $m\ge 1$, then the operator N is continuous from $(H^m({\mathbb {R}}_+))^d$ into $H^m({\mathbb {R}}_+)$.

Theorem A.7

Let $\mu$ satisfy Assumptions A.2 and A.3 for some positive integer m. Then, N is Lipschitz continuous from bounded subsets of $(H^m({\mathbb {R}}_+))^d$ into $H^m({\mathbb {R}}_+)$.

1.2 A.2 Differentiability

We now discuss differentiability of N in Fréchet sense. Here we run into the following problem compared with the literature. To get continuity of the Fréchet derivatives, Valent [38] uses that $H^m$ is a Banach algebra and the Nemytskii operator corresponding to $(\tfrac{\partial }{\partial y_j} \mu )$ maps into $H^m$. On unbounded domains, this would exclude the linear case $\mu (x,y):= y$ which is of particular interest for applications in this work. We resolve this problem in Lemma A.9. Note that multiplication is not only bilinear continuous on $H^k$, but also from $C^k_b\times H^k$ into $H^k$. More precisely, see Lemma B.1, for all $k\ge 0$ there exists $c>0$ such that for all $g\in C^k_b({\mathbb {R}}_+)$, $u\in H^k({\mathbb {R}}_+)$ it holds that

$$\begin{aligned} \left||gu\right||_{H^k} \le c \left||g\right||_{C^k_b} \left||u\right||_{H^k}. \end{aligned}$$

(A.2)

We start with a result on continuity of the nonlinear operators, adapting [39, Theorem 2]. We now write shortly $H^m({\mathbb {R}}_+)^d$ for $H^m({\mathbb {R}}_+;{\mathbb {R}}^d)$.

Theorem A.8

If Assumption A.2.(b) holds for some integer $m\ge 1$, then the operator $N_{y_j}$ is continuous from $H^m({\mathbb {R}}_+)^d$ into $C^{m-1}_b({\mathbb {R}}_+)$ and maps bounded sets into bounded sets.

Proof

We prove the continuity in a similar way as done for Theorem A.6. First, let $m=1$, and $(u_n) \subset H^1({\mathbb {R}}_+)^d$ converging to some $u\in H^1({\mathbb {R}}_+)^d$. Sobolev imbeddings imply that $u_n$, $u \in BUC({\mathbb {R}}_+)^d$, $n\in {\mathbb {N}}$ and $u_n \rightarrow u$ uniformly, as $n\rightarrow \infty$. Thus, $x\mapsto \frac{\partial }{\partial y_i} \mu (x,u(x))$ is continuous and bounded.

Define $R:= \sup _{n\in {\mathbb {N}}} \left||u_n\right||_{\infty }<\infty$ and observe that the family of functions

$$\begin{aligned} y\mapsto \tfrac{\partial }{\partial y_j}\mu (x,y),\quad x\in {\mathbb {R}}_+ \end{aligned}$$

is uniformly equicontinuous on the ${\mathbb {R}}^d$-ball of radius R.

Let $\epsilon >0$ be arbitrary and $\delta = \delta _{\epsilon ,R}>0$ such that for all y, ${\tilde{y}}\in {\mathbb {R}}^d$ with $\left|y\right|$, $\left|{\tilde{y}}\right| \le R$ and $\left|{\tilde{y}}-y\right|<\delta$ it holds that

$$\begin{aligned} \sup _{x\in {\mathbb {R}}_+} \left| \frac{\partial }{\partial y_i} \mu (x,y) - \frac{\partial }{\partial y_i} \mu (x,{\tilde{y}})\right| <\epsilon . \end{aligned}$$

Now, let $N_\delta \in {\mathbb {N}}$ such that for all $n\ge N$ it holds that

$$\begin{aligned} \left||u_n - u\right||_{\infty } <\delta . \end{aligned}$$

Hence, $\sup _{x\in {\mathbb {R}}_+} \left|| \frac{\partial }{\partial y_i} \mu (x,u(x)) - \frac{\partial }{\partial y_i} \mu (x,u_n(x))\right||_{\infty } <\epsilon$ for all $n\ge N_\delta$, and thus

$$\begin{aligned} \left||N_{y_j}(u_n) - N_{y_j} (u)\right||_{\infty } \rightarrow 0, \quad n\rightarrow \infty . \end{aligned}$$

Let $M\subset H^1({\mathbb {R}}_+)$ be bounded and $R>0$ so that M is contained in the radius R ball of $C_b({\mathbb {R}}_+)$. Then, for all $u\in M$,

$$\begin{aligned} \left||N_{y_j}(u)\right||_{\infty } \le \sup _{\left|y\right|< R} \sup _{x\in {\mathbb {R}}_+} \left|\left( \tfrac{\partial }{\partial y_j} \mu \right) \mu (x,y)\right| <\infty . \end{aligned}$$

We finish the proof by induction, so assume the claim holds true for $m\in {\mathbb {N}}$. By induction hypothesis, $N_{y_j}$ is continuous from $H^{m+1}$ into $C^{m-1}_b$, so it remains to show that the same holds true for $\frac{\,{\text {d}}}{\,{\text {d}}x} N_{y_j}$. Chain rule yields

$$\begin{aligned} \tfrac{\,{\text {d}}}{\,{\text {d}}x} N_{y_j}(u)(x) = \frac{\partial ^2}{\partial x \partial y_j} \mu (x,u(x)) + \sum _{i=1}^d \frac{\partial ^2}{\partial y_i \partial y_j} \mu (x,u(x)) \nabla u_i(x), \end{aligned}$$

(A.3)

for $u\in H^{m+1}({\mathbb {R}}_+)^d\hookrightarrow BUC^m({\mathbb {R}}_+)^d$. The function ${\bar{\mu }}$ defined as

$$\begin{aligned} {\bar{\mu }}(x,y,z) := \frac{\partial }{\partial x} \mu (x,y) + \sum _{i=1}^d \frac{\partial }{\partial y_i} \mu (x,y) z_i, \end{aligned}$$

for $x\in {\mathbb {R}}_{\ge 0}$, $(y,z)\in {\mathbb {R}}^{d+d}$, satisfies Assumption A.2.(b) for m. Hence, by induction hypothesis, the Nemytskii operators corresponding to the $y_j$ (and $z_i$)-derivatives of ${\bar{\mu }}$ are continuous and map bounded sets into bounded sets, from $H^{m}({\mathbb {R}}_+)^{d+d}$ into $C^{m-1}_b({\mathbb {R}}_{\ge 0})^{d+d}$. Since the map $u\mapsto \nabla u_i$ is linear continuous from $H^{m+1}({\mathbb {R}}_+)^d$ into $H^{m}({\mathbb {R}}_+)$, we get the properties for $\frac{\,{\text {d}}}{\,{\text {d}}x} N_{y_j}$. $\square$

In the following, we write for $j=1,\ldots ,d$, $u\in H^m({\mathbb {R}}_+)^d$, $v\in H^m({\mathbb {R}}_+)$,

$$\begin{aligned} {\widetilde{N}}_{y_j}(u,v) := N_{y_j}(u) v = (\tfrac{\partial }{\partial y_j} \mu )(.,u(.)) v(.). \end{aligned}$$

Lemma A.9

Let $m\ge 1$ and $\mu$ fulfilling Assumption A.2.(b) for $m+1$, then, the mapping

$$\begin{aligned} \varPhi _j: \,u\mapsto {\widetilde{N}}_{y_j}(u,.) \end{aligned}$$

is continuous from $H^m({\mathbb {R}}_+)^d$ into ${\mathscr {L}}(H^m({\mathbb {R}}_+))$, for all $j= 1,\ldots ,d$. Moreover, $\varPhi _i$ maps bounded sets into bounded sets.

Proof

Note that ${\tilde{\mu }}(x,y,z) := \mu (x,y)z$ fulfills Assumption A.2.(a) for m so that Theorem A.6 yields continuity of

$$\begin{aligned} H^m({\mathbb {R}}_+)^{d+1} \ni (u,v)\mapsto {\widetilde{N}}_{y_j}(u,v) \in H^m({\mathbb {R}}_+). \end{aligned}$$

Of course, ${\widetilde{N}}_{y_j}$ is linear in its second argument so that ${\widetilde{N}}_{y_j}(u,.)\in {\mathscr {L}}(H^m({\mathbb {R}}_+))$, for each $u\in H^m({\mathbb {R}}_+)^d$. It remains to prove continuity in the uniform operator topology for which we proceed by induction, again.

• Step I. With $m=1$ let $(u^{(n)})\subset H^1({\mathbb {R}}_+)^d$, $u\in H^1({\mathbb {R}}_+)^d$ and $v\in H^1({\mathbb {R}}_+)$. First note that by Theorem A.8, $N_{y_j}$ is continuous from $H^1$ into $C_b$, so that

  $$\begin{aligned} \left||{\widetilde{N}}_{y_j}(u^{(n)},v) - {\widetilde{N}}_{y_j}(u,v)\right||_{L^2} \le c \left||N_{y_j} (u^{(n)})- N_{y_j}(u^{(n)})\right||_{C_b} \left||v\right||_{L^2}, \end{aligned}$$

  converges to 0, as $n\rightarrow \infty$, uniformly in v. Similar we get uniform $L^2$-convergence for $N_{y_j}(u^{(n)})\tfrac{\partial }{\partial x}v$ and for the operator

  $$\begin{aligned} (u,v) \mapsto (\tfrac{\partial ^2}{\partial x\partial y_j} \mu )(x,u(x))v(x). \end{aligned}$$

  Moreover, for all i, $j\in \{1,\ldots ,d\}$, by Sobolev imbeddings

  $$\begin{aligned}&\int _{{\mathbb {R}}_+} \left|\tfrac{\partial ^2}{\partial y_j\partial y_i} \mu (x,u^{(n)}(x)) \nabla u_i^{(n)}(x) - \tfrac{\partial ^2}{\partial y_j\partial y_i}\mu (x,u(x))\nabla u_i(x)\right|^2 \left|v(x)\right|^2 \,{\text {d}}x \le \nonumber \\&\le K \left||v\right||_{H^1}^2 \int _{{\mathbb {R}}_+} \left|\tfrac{\partial }{\partial y_j\partial y_i} \mu (x,u^{(n)}(x)) \nabla u_i^{(n)}(x) - \tfrac{\partial }{\partial y_j\partial y_i}\mu (x,u(x))\nabla u_i(x)\right|^2 \,{\text {d}}x .\nonumber \\ \end{aligned}$$

  (A.4)

  Note that $\tfrac{\partial }{\partial y_i}\mu$ fulfills Assumption A.2.(b) and recall that multiplication is continuous from $C_b\times L^2$ into $L^2$. Hence, the integral converges to 0, as $n\rightarrow \infty$ by application of Theorem A.8 to the corresponding Nemytskii operator, and so we conclude continuity of $\varPhi$. Using almost the same estimates and applying the corresponding of part Theorem A.8, we get that $\varPhi _i$ maps bounded sets into bounded sets again.

• Step II: By induction hypothesis, the Lemma holds true for $m\in {\mathbb {N}}$ fixed, so assume that $\mu$ fulfills Assumption A.2.(b) for $m+2$. Then, $\varPhi _j$ is continuous from $H^{m+1}({\mathbb {R}}_+)^d$ into ${\mathscr {L}}(H^m({\mathbb {R}}_+))$. Let $(u^{(n)})\subset H^{m+1}({\mathbb {R}}_+)^d$ converging to $u\in H^{m+1}({\mathbb {R}}_+)^d$. Note that

  $$\begin{aligned} \left||\varPhi (u^{(n)}) - \varPhi (u)\right||_{{\mathscr {L}}(H^{m+1})}^2\le & {} \sup _{w\in H^{m}} \frac{\left||\varPhi (u^{(n)})w - \varPhi (u)w\right||_{H^m}^2}{\left||w\right||_{H^m}^2}\nonumber \\&+ \sup _{v\in H^{m+1}}\frac{\left||\frac{\,{\text {d}}^{m+1}}{\,{\text {d}}x^{m+1}} (\varPhi (u^{(n)})v - \varPhi (u)v)\right||_{L^2}^2}{\left||v\right||_{H^{m+1}}^2}.\nonumber \\ \end{aligned}$$

  (A.5)

  The first term vanishes as $n\rightarrow \infty$ by induction hypothesis. For all $v\in H^{m+1}$, we can write the latter one can be estimated by

  $$\begin{aligned} \frac{\,{\text {d}}^{m+1}}{\,{\text {d}}x^{m+1}} (\varPhi (u^{(n)})v - \varPhi (u)v)&= \frac{\,{\text {d}}^{m}}{\,{\text {d}}x^{m}} [(\tfrac{\partial }{\partial x}N_{y_j}(u^{(n)}) - \tfrac{\partial }{\partial x}N_{y_j}(u)) v] \\&\qquad + \frac{\,{\text {d}}^{m}}{\,{\text {d}}x^{m}} [(N_{y_j}(u^{(n)}) - N_{y_j}(u))w], \end{aligned}$$

  for $w:= \tfrac{\partial }{\partial x}v \in H^m$. By induction hypothesis, the latter term converges to 0 in $L^2$, uniformly over all $w\in H^m({\mathbb {R}}_+)$. For the first summand, we observe that $D_x \mu (x,y)$ fulfills Assumption A.2.(b) for $m+1$ so that the induction hypothesis applied on

  $$\begin{aligned} \varPsi (u)v := (\tfrac{\partial ^2}{\partial x \partial y_j} \mu )(.,u(.)) v \end{aligned}$$

  yields $L^2$ convergence. Plugging in into (A.5) finally yields the convergence uniform in ${\mathscr {L}}(H^{m+1})$. With the same decomposition, we deduce from induction hypothesis that $\varPhi _i$ maps bounded sets from $H^{m+1}$ into bounded sets of ${\mathscr {L}}(H^{m+1})$.

$\square$

Based on the continuity in the uniform topology, we are now able to prove Fréchet differentiability.

Theorem A.10

If Assumption A.2 holds for some integer $m+1$, $m\ge 1$, then the operator N defined above is in $C^1(H^m({\mathbb {R}}_+)^d, H^m({\mathbb {R}}_+))$ with derivative

$$\begin{aligned} DN(u)v = \sum _{j=1}^d N_{y_j}(u) v_j,\quad u,v\in H^m({\mathbb {R}}_+)^d. \end{aligned}$$

Proof

From Theorem A.6, we already know that N maps $H^m({\mathbb {R}}_+;{\mathbb {R}}^d)$ continuously into $H^m({\mathbb {R}}_+)$. Moreover, Lemma A.9 tells us that DN, defined as above, is continuous from $H^m({\mathbb {R}}_+)^d$ into ${\mathscr {L}}(H^m({\mathbb {R}}_+))$. Thus, it remains to verify that DN is at least the Gâteaux derivative of N, i.e., that for any u, $v\in H^{m +1}({\mathbb {R}}_+;{\mathbb {R}}^d)$,

$$\begin{aligned} \frac{1}{\epsilon } \left||N(u+\epsilon v) - N(u) - \epsilon \sum _{j=1}^d N_{y_j}(u)v_j\right||_{H^m} \longrightarrow 0,\;\quad \text { as }\epsilon \rightarrow 0. \end{aligned}$$

(A.6)

By fundamental theorem of calculus, we get for fixed u, $v\in H^{m}$, and all $x\in {\mathbb {R}}_+$,

$$\begin{aligned} \begin{aligned} N(u+\epsilon v)(x) - N(u)(x)&= \int _0^1 \sum _{j=1}^d \frac{\partial }{\partial y_j}\mu (x,u(x) + t\epsilon v(x)) \epsilon v_j(x)\,{\text {d}}t \\&= \epsilon \sum _{j=1}^d \int _0^1 {\widetilde{N}}_{y_j}(u+t\epsilon v,v_j)(x) \,{\text {d}}t \end{aligned} \end{aligned}$$

(A.7)

The map $t\mapsto {\widetilde{N}}_{y_i}(u+t\epsilon v, v_i)$ is continuous from [0, 1] into $H^{m}({\mathbb {R}}_+)$ by Theorem A.6. Therefore, the integral in Eq. (A.7) can be considered as an Bochner integral and (A.6) follows from Lemma A.9 and the estimate

$$\begin{aligned}&\left||N(u+\epsilon v) - N(u) - \epsilon \sum _{j=1}^d N_{y_j}v_j\right||_{H^m}\\&\quad \le \left|\epsilon \right|\sum _{j=1}^d \int _0^1 \left||\left( N_{y_j}(u+t\epsilon v) - N_{y_j}(u)\right) v_j\right||_{H^m} \,{\text {d}}t. \end{aligned}$$

$\square$

If $\mu \in C^2$, then we define for $u\in H^m({\mathbb {R}}_+)^d$, v, $w\in H^m({\mathbb {R}}_+)$, $x\in {\mathbb {R}}_+$,

$$\begin{aligned} N_{y_i,y_j} (u)(x)&:= \frac{\partial ^2}{\partial y_i \partial y_j} \mu (x, u(x)). \end{aligned}$$

Theorem A.11

Assume that $\mu$ satisfies Assumption A.2 for $m+2$, $m\ge 1$, then the Nemytskii operator $N: H^m({\mathbb {R}}_+)^d \rightarrow H^m({\mathbb {R}}_+)$ is of class $C^2$ with second derivative

$$\begin{aligned} D^2 N(u)[v,w] = \sum _{i=1}^d \sum _{j=1}^d N_{y_i,y_j}(u) v_j w_i , \end{aligned}$$

for u, v, $w\in H^m({\mathbb {R}}_+)^d$.

Proof

By the previous theorem, N is of class $C^1$, so we have to show the same for the map

$$\begin{aligned} DN: H^m({\mathbb {R}}_+) \rightarrow {\mathscr {L}}(H^m({\mathbb {R}}_+)^d; H^m({\mathbb {R}}_+)), \quad DN(u) := \left( v\mapsto \sum _{j=1}^d {\widetilde{N}}_{y_j}(u,v_j)\right) \end{aligned}$$

Since $H^m$ is a Banach algebra, we get for u, ${\bar{u}} \in H^m({\mathbb {R}}_+)^d$,

$$\begin{aligned}&\left||D^2N(u) - D^2N({\bar{u}})\right||_{{\mathscr {L}}(H^m({\mathbb {R}}_+)^d,H^m({\mathbb {R}}_+)^d;H^m)}\\&\quad \le \sum _{i,j=1}^d\sup _{\left||v\right||_{}= 1}\sup _{\left||w\right||_{}= 1}\left|| \left( N_{y_i,y_j}(u) - N_{y_i,y_j} ({\bar{u}})\right) v_j w_i\right||_{H^m}\\&\quad \le c \sum _{i,j=1}^d\sup _{\left||v\right||_{}= 1}\left||N_{y_i,y_j}(u)v - N_{y_i,y_j} ({\bar{u}})v\right||_{H^m}. \end{aligned}$$

Now, we apply Lemma A.9 to $\frac{\partial }{\partial y_i} \mu (x,u(x))$, $i=1,\ldots ,d$, which indeed fulfill Assumption A.2.(b) for $m+1$. This yields continuity of $D^2N$.

To finish the proof, it again suffices to show differentiability in Gâteaux sense. Fix u, $w\in H^m({\mathbb {R}}_+)^d$ and let $\epsilon >0$. As in the proof of Theorem A.10, cf. (A.7), we get by fundamental theorem of calculus, for all $v\in H^m({\mathbb {R}}_+)^d$

$$\begin{aligned}&DN(u+\epsilon w)v - DN(u)v - D^2N(u)(v,w) \\&\quad = \epsilon \sum _{i,j=1}^d \int _0^1 \left( N_{y_i,y_j}(u+t\epsilon w) - N_{y_i,y_j}(u)\right) w_iv_j \,{\text {d}}t \\&\quad = \epsilon \int _0^1 D^2 N(u+t\epsilon w)(v,w) - D^2N(u)(v,w)\,{\text {d}}t. \end{aligned}$$

From the first part of this proof, we know that $\epsilon \mapsto D^2N(u+ \epsilon w)$ is uniformly continuous from [0, 1] into the space of continuous bilinear operators. Hence, the right-hand side is in $o(\epsilon )$, uniformly in $v\in H^m$. $\square$

The following conclusion is a combination of Theorem A.8 and the representations of DN and $D^2N$.

Corollary A.12

Under the assumptions of, respectively, Theorems A.10 and A.11, the maps

$$\begin{aligned} DN: H^m({\mathbb {R}}_+)^d \rightarrow {\mathscr {L}}(H^m({\mathbb {R}}_+)^d;H^m({\mathbb {R}}_+)) \end{aligned}$$

and

$$\begin{aligned} D^2N : H^m({\mathbb {R}}_+)^d \rightarrow {\mathscr {L}}(H^m({\mathbb {R}}_+)^d, H^m({\mathbb {R}}_+)^d;H^m({\mathbb {R}}_+) \end{aligned}$$

map bounded sets into bounded sets.

B The noise operator

We will now study the operator-valued map ${\mathcal {C}}$, defined previously by

$$\begin{aligned} ({\mathcal {C}}(u)w)(x) = \begin{pmatrix} \sigma _+(x,u_1(x)) (T_\zeta w)(u_3+x)\\ \sigma _-(-x,u_2(x)) (T_\zeta w)(u_3-x) \\ 0 \end{pmatrix} \end{aligned}$$

for $u\in {\mathcal {D}}({\mathcal {C}})\subset L^2({\mathbb {R}}_+)\oplus L^2({\mathbb {R}}_+) \oplus {\mathbb {R}}$, $w\in L^2({\mathbb {R}})=: U$ and $x\in {\mathbb {R}}$. We can reduce the problem to the operator

$$\begin{aligned} \varPsi : (u,x_*)\mapsto \sigma (.,u(.)) T_\zeta (.+x_*) \end{aligned}$$

(B.1)

for $\sigma : {\mathbb {R}}^2\rightarrow {\mathbb {R}}$, and $\zeta$ and an integral kernel $\zeta :{\mathbb {R}}^2\rightarrow {\mathbb {R}}$, which we aim to take values in spaces of Hilbert–Schmidt operators like ${\mathscr {L}}_2(U; L^2({\mathbb {R}}_+))$. As above, we write

$$\begin{aligned} T_\zeta :w\mapsto \int _{\mathbb {R}}\zeta (x,y)w(y) \,{\text {d}}y \end{aligned}$$

and define the Nemytskii operator

$$\begin{aligned} N_\sigma :u\mapsto \sigma (.,u(.)). \end{aligned}$$

Naturally, it will make sense to separate the study of $\varPsi$ into the operators $N_\sigma$ and $T_\zeta$. Recall that we have discussed the Nemytskii operators $N_\sigma$ in Appendix A.

1.1 B.1 The Hilbert–Schmidt property

Note that on $L^2(D)$, for a domain $D\subset {\mathbb {R}}^d$, $d\in {\mathbb {N}}$, every Hilbert–Schmidt operator is of the form $T_\kappa$, for an integral kernel $\kappa$ satisfying

$$\begin{aligned} \int _D\int _D \left|\kappa (x,y)\right|^2 \,{\text {d}}x\,{\text {d}}y<\infty , \end{aligned}$$

(B.2)

see, e.g., [8, Section XI.6]. When D has infinite Lebesgue measure, this condition is obviously violated for convolution kernels $\kappa (x,y) = \kappa (x-y)$, in which have been interested in Example 1.4 for instance. Hence, $T_\zeta$ itself will typically not be Hilbert–Schmidt on the spaces of interest. We skip the proofs in the following three lemmas since they will be the same as the proofs of, respectively, Lemmas 7.1, 7.2 and 7.4 in [18].

Lemma B.1

For any integer $n\ge 0$, multiplication is bilinear continuous from $H^n({\mathbb {R}}_+) \times C_b^n({\mathbb {R}}_{\ge 0})$ into $H^n({\mathbb {R}}_+)$.

The lemma is the first step in the direction to separate our discussion of $\varPsi$ into the operators $N_\sigma$ and $T_\zeta$. Provided that $\zeta$ is sufficiently nice, $T_\zeta$ will indeed map into the space of bounded and uniformly continuous functions.

Assumption B.2

Let $n\ge 1$, $\zeta (.,y) \in C^{n+1}({\mathbb {R}})$ for all $y\in {\mathbb {R}}$ and $\tfrac{\partial ^{i}}{\partial x^i}\zeta (x,.)\in L^2({\mathbb {R}})$ for all $x\in {\mathbb {R}}$, $i\in \{0,\dots , n+1\}$. Moreover,

$$\begin{aligned} \sup _{x\in {\mathbb {R}}} \left||\tfrac{\partial ^{i}}{\partial x^i} \zeta (x,.)\right||_{L^2({\mathbb {R}})} <\infty ,\quad i=0,1,\ldots ,n+1. \end{aligned}$$

(B.3)

In the following, we use the notation $\zeta ^{(i)}:=\tfrac{\partial ^{i}}{\partial x^i} \zeta$.

Remark B.3

For convolution kernels $\zeta (x,y):= \zeta (x-y)$, this assumption is satisfied when $\zeta \in H^{n+1}({\mathbb {R}})\cap C^{n+1}({\mathbb {R}})$.

Lemma B.4

Let Assumption B.2 be fulfilled for $n\in {\mathbb {N}}$. Then, $T_\zeta$ maps U into $BUC^n({\mathbb {R}})$. Moreover, $T_\zeta w$ and its first n derivatives are Lipschitz continuous for all $w\in U$ and it holds that $T_\zeta \in {\mathscr {L}}(U; BUC^2({\mathbb {R}}))$.

Lemma B.5

Let $n\in {\mathbb {N}}$ and Assumption B.2 be satisfied. For $u\in H^n({\mathbb {R}}_+)$ and $x_*\in {\mathbb {R}}$ it holds that

$$\begin{aligned} \left||u \cdot T_\zeta (.+x_*)))\right||_{{\mathscr {L}}_2(U; H^n({\mathbb {R}}_+))} \le K \left||u\right||_{H^n({\mathbb {R}}_+)} \sup _{x\in {\mathbb {R}}}\sum _{i=0}^n\left||\zeta ^{(i)}(x,.)\right||_{L^2({\mathbb {R}})} \end{aligned}$$

For application in Sect. 4, we need to deal ${\mathcal {C}}$ on the domain of the Dirichlet Laplacian. In fact, Assumption $({\hbox {Noise}}_0)$ and Lemma B.1 ensure $N_\sigma (u)\in H^2({\mathbb {R}}_+)\cap H^1_0({\mathbb {R}}_+)$ for all $u\in H^2({\mathbb {R}}_+) \cap H^1_0({\mathbb {R}}_+)$.

1.2 B.2 Lipschitz continuity and differentiability

In order to apply the results, let us introduce the translation group $(\theta _x)_{x\in {\mathbb {R}}}$ which is strongly continuous on $BUC({\mathbb {R}})$.

Remark B.6

By the structure of the direct sum of Hilbert spaces, the following two results directly extend to ${\mathcal {C}}$ as a mapping from $H^n({\mathbb {R}}_+)\oplus H^n({\mathbb {R}}_+) \oplus {\mathbb {R}}$ into ${\mathscr {L}}_2(U; H^n({\mathbb {R}}_+) \oplus H^n({\mathbb {R}}_+) \oplus {\mathbb {R}})$.

Remark B.7

Note that for $x\in {\mathbb {R}}$

$$\begin{aligned} \theta _x\circ T_{\zeta } = T_{\zeta _x}, \end{aligned}$$

where $\zeta _x := \zeta (x+.,.)$ satisfies Assumption B.2, whenever $\zeta$ does.

We impose the following conditions on $\sigma$.

Assumption B.8

Let $n\ge 1$ and assume that $\sigma \in C^n({\mathbb {R}}^2;{\mathbb {R}})$ satisfies

• (i) For every multi-index $I = (i,j) \in {\mathbb {N}}^2$ with $\left|I\right| \le n$, there exist $a_I\in L^2({\mathbb {R}}_+)$ and $b_I \in L^\infty _{loc}({\mathbb {R}}, {\mathbb {R}}_+)$ such that

  $$\begin{aligned} \left|\tfrac{\partial ^{\left|I\right|}}{\partial x^{i}\partial y^{j}} \sigma (x,y)\right| \le {\left\{ \begin{array}{ll} b_I(y)\left( a_I(|x|) +\left|y\right|\right) , &{} j=0, \\ b_I (y), &{} j\ne 0. \end{array}\right. } \end{aligned}$$

• (ii) $\sigma$ and its partial derivatives (in x and y) are locally Lipschitz with Lipschitz constants independent of $x\in {\mathbb {R}}$.

Theorem B.9

Let $n\in {\mathbb {N}}$ and assume that Assumption B.2 is fulfilled for $n+1$ and, respectively, B.8 for $n+2$. Then, $\varPsi$ is of class $C^2$ from $H^n({\mathbb {R}}_+)\oplus {\mathbb {R}}$ into ${\mathscr {L}}_2:= {\mathscr {L}}_2(U; H^n({\mathbb {R}}_+))$, with derivatives

$$\begin{aligned} D\varPsi (u,x)(v,y)&= DN_\sigma (u)v \cdot \theta _{x}T_\zeta + y N_\sigma (u)\cdot \theta _{x} T_{\zeta '} \\&=\left( w\mapsto \tfrac{\partial }{\partial y}\sigma (.,u) v T_\zeta w(.+x) + y \sigma (.,u) T_{\zeta '} w(.+x)\right) ,\\ D^2\varPsi (u)[(v,y),({\bar{v}},{\bar{y}})]&= D^2N_\sigma (u)[v,{\bar{v}}] \cdot \theta _{x}T_\zeta + y DN_\sigma (u){\bar{v}} \cdot \theta _{x} T_{\zeta '}\\&\qquad + {\bar{y}} DN_\sigma (u)v \cdot \theta _{x}T_{\zeta '} + y{\bar{y}} N_\sigma (u)\cdot \theta _{x} T_{\zeta ''}. \end{aligned}$$

Moreover, $\varPsi$, $D\varPsi$, and $D^2\varPsi$ map bounded sets into bounded sets.

Remark B.10

For $n=2$ and under the additional assumption that $\sigma (0,0) = 0$, it holds that $\varPsi (u,x)\in H^2\cap H^1_0({\mathbb {R}}_+)$, when $u\in H^2\cap H^1_0({\mathbb {R}}_+)$. This even translates to $D\varPsi$ and $D^2\varPsi$.

Proof

Let u, $v\in H^n({\mathbb {R}}_+)$, x, $y\in {\mathbb {R}}$ and $\epsilon >0$, then with Lemma B.5,

$$\begin{aligned}&\left||\varPsi ((u,x)+\epsilon (v,y))-\varPsi (u,x) - \epsilon D\varPsi (u,x)(v,y)\right||_{{\mathscr {L}}_2(U;H^n)} \nonumber \\&\quad \le K_{\zeta } \left||N_\sigma (u+v) - N_\sigma (u) - DN_\sigma (u)\epsilon v\right||_{H^n} \nonumber \\&\qquad + K \left||N_\sigma (u) + \epsilon DN_\sigma (u)v\right||_{H^n} \sup _{z\in {\mathbb {R}}} \sum _{i= 0}^n \left||\zeta ^{(i)}_{\epsilon y}(z,.)\right||_{L^2({\mathbb {R}})} \end{aligned}$$

(B.4)

In fact, the first term is in $o(\epsilon )$ because of differentiability of $N_\sigma$ we get from Theorem A.10. For the second summand, we have defined

$$\begin{aligned} \zeta _{z}(x,y) := \zeta (z + x, y) - \zeta (x,y) - z\tfrac{\partial }{\partial x}\zeta (x,y),\; x,y,z\in {\mathbb {R}}. \end{aligned}$$

Thus,

$$\begin{aligned} \sup _{z\in {\mathbb {R}}} \left|| \zeta _{\epsilon y}(z,.)\right||_{L^2}^2= & {} \sup _{z\in {\mathbb {R}}}\int _{\mathbb {R}}\left|\zeta (z+\epsilon y, \xi ) - \zeta (z,\xi ) - \epsilon y \zeta '(z,\xi )\right|^2 \,{\text {d}}\xi \nonumber \\\le & {} \left|\epsilon y\right|^2\sup _{z\in {\mathbb {R}}}\int _{\mathbb {R}}\int _0^1 \left|\zeta '(z+\alpha \epsilon y,\xi ) - \zeta '(z,\xi )\right|^2\,{\text {d}}\alpha \,{\text {d}}\xi .\nonumber \\ \end{aligned}$$

(B.5)

Using fundamental theorem of calculus again, we obtain

$$\begin{aligned} \sup _{z\in {\mathbb {R}}}\int _{\mathbb {R}}\int _0^1 \left|\zeta '(z+\alpha \epsilon y, \xi ) - \zeta '(z,\xi )\right|^2\,{\text {d}}\alpha \,{\text {d}}\xi \le \left|\epsilon y\right|^2 \sup _{z\in {\mathbb {R}}} \left||\zeta ''(z,.)\right||_{L^2}, \end{aligned}$$

which goes to 0, as $\epsilon \rightarrow 0$. Using that (B.3) holds for $i=0,\ldots ,n+2$, the same calculation can be done for $\zeta ^{(i)}$, $i=1,\dots , n$ which then shows that $D\varPsi$ is at least the Gâteaux derivative of $\varPsi$. To finish the proof, it is now enough to show that

$$\begin{aligned} D\varPsi : H^n\oplus {\mathbb {R}}\rightarrow {\mathscr {L}}(H^n\oplus {\mathbb {R}};{\mathscr {L}}_2(U;H^n)) \end{aligned}$$

is Gâteaux differentiable, and

$$\begin{aligned} D^2\varPsi : H^2\oplus {\mathbb {R}}\rightarrow {\mathscr {L}}(H^n\oplus {\mathbb {R}}, H^n\oplus {\mathbb {R}}; {\mathscr {L}}_2(U;H^n)) \end{aligned}$$

is continuous. Let us start with the latter claim and show continuity of each summand separately. To this end, we first decompose as above

$$\begin{aligned} D^2\varPsi (u,x)[(v,y),({\bar{v}},{\bar{y}})] = \sum _{k=1}^4 R_k(u,v,{\bar{v}},x,y,{\bar{y}}). \end{aligned}$$

Consider u, ${\tilde{u}}$, v, ${\bar{v}}\in H^n$, $x,{\tilde{x}},y,\bar{y}\in {\mathbb {R}}$. Because $N_\sigma \in C^2$ by Theorem A.11, we get

$$\begin{aligned}&\left||R_1(u,v,{\bar{v}},x,y,{\bar{y}}) - R_1({\tilde{u}},{\tilde{x}},v,{\bar{v}})\right||_{L_2(U;H^n)} \nonumber \\&\quad \le \left||\left( D^2N_\sigma (u)[v,{\bar{v}}] - D^2N_\sigma ( {\tilde{u}})[v,{\bar{v}}]\right) \cdot \theta _x(T_\zeta (.))\right||_{{\mathscr {L}}_2} \nonumber \\&\qquad + \left||D^2N_\sigma ({\tilde{u}})[v,{\bar{v}}] \cdot \left( \theta _x(T_\zeta (.)) - \theta _{{\tilde{x}}}(T_\zeta (.))\right) \right||_{{\mathscr {L}}_2}. \end{aligned}$$

(B.6)

Applying Lemma B.5, we see that both terms go to 0, as $\left||u-{\tilde{u}}\right||_{H^n} + \left|x-{\tilde{x}}\right|$ does, and that the convergence is uniformly in v, ${\bar{v}}\in H^n$ with norm smaller than 1. Indeed, for the first term this is continuity of $D^2N$, the second term can be estimated by

$$\begin{aligned}&\left||D^2N_\sigma ({\tilde{u}})[v,{\bar{v}}] \cdot \left( \theta _x(T_\zeta (.)) - \theta _{{\tilde{x}}}(T_\zeta (.))\right) \right||_{{\mathscr {L}}_2(U;H^n)} \nonumber \\&\quad \le K \left||D^2N_\sigma ({\tilde{u}})\right||_{{\mathscr {L}}(H^n)}\left||v\right||_{H^n} \sup _{z\in {\mathbb {R}}} \sum _{i=0}^n \left||\zeta ^{(i)}_{x}(z,.) - \zeta ^{(i)}_{{\tilde{x}}}(z,.)\right||_{L^2({\mathbb {R}})}. \end{aligned}$$

(B.7)

Convergence of the right-hand side follows with the same procedure as in (B.5). For $R_2$ and $R_3$, we use continuity of $DN_\sigma$, for $R_4$ continuity of $N_\sigma$ itself. With almost the same estimates, we observe that $D^2\varPsi$ maps bounded sets into bounded sets. In fact, this property is inherited by N, DN and $D^2N$, see Corollary A.12.

It remains to show that $D^2\varPsi$ is indeed the derivative of $D\varPsi$. The derivative of the second summand can be computed in the same way as $D\varPsi$ itself has been computed. For the first summand, we have to be slightly more careful, but note that by Lemma B.5

$$\begin{aligned}&\sup _{\left||v\right||_{}\le 1}\left||DN_\sigma (u)v \cdot (\theta _{x+\epsilon y}T_\zeta - \theta _x T_\zeta - \epsilon y T_{\zeta '})\right||_{{\mathscr {L}}_2(U;H^n)} \\&\quad \le K \left||DN_\sigma (u)\right||_{{\mathscr {L}}(H^n({\mathbb {R}}))} \sup _{z\in {\mathbb {R}}}\sum _{i=0}^n \left||\zeta ^{(i)}_{\epsilon y}(z,.)\right||_{L^2({\mathbb {R}})} \end{aligned}$$

which is in $o(\epsilon )$ thanks to (B.5). The remaining estimates follow in the same way: First apply Lemma B.5, but then use that $N_\sigma$ is of class $C^2$. Hence, $D^2\varPsi$ is the Gâteaux derivative of $D\varPsi$. By continuity of $D^2\varPsi$, the differentiability also holds true in Fréchet sense. $\square$