Boundary Integral Operators for the Heat Equation in Time-Dependent Domains

Abstract

This article provides a functional analytical framework for boundary integral equations of the heat equation in time-dependent domains. More specifically, we consider a non-cylindrical domain in space-time that is the $C^2$-diffeomorphic image of a cylinder, i.e., the tensor product of a time interval and a fixed domain in space. On the non-cylindrical domain, we introduce Sobolev spaces, trace lemmata and provide the mapping properties of the layer operators. Here it is critical that the Neumann trace requires a correction term for the normal velocity of the moving boundary. Therefore, one has to analyze the situation carefully.

1 Introduction

Boundary integral equations are a well-known technique to solve elliptic partial differential equations, see for example [16, 17]. For parabolic equations on time-independent, so-called cylindrical domains, Sobolev spaces and the mapping properties of the layer operators for the heat equation are introduced in [3, 15]. To the best of our knowledge, no corresponding theory exists for time-dependent or so-called non-cylindrical domains, or simply, tubes. Some results can be found in e.g. [9, 10]. Therefore, the aim of this article is to close the gaps in the theory also for non-cylindrical domains. To this end, we consider a special class of non-cylindrical domains. We fix a cylindrical domain which serves as a reference domain and define the non-cylindrical domain as the image under a time-dependent $C^2$-diffeomorphism. In this setting, different approaches are possible to establish analogue integral equations and properties to integral operators on a cylindrical domain.

A first approach could be to exploit the fact that the fundamental solution does not depend on the boundary data and is thus defined on the free space $\mathbb {R}\times \mathbb {R}^d$. Therefore, it is the same for a cylindrical and a non-cylindrical domain and allows to state the integral operators in cylindrical and non-cylindrical domains. To establish the mapping properties of the integral operators, one could make use of the equivalence of norms on the tube and on the cylindrical domain by establishing equivalence results of the fundamental solution, evaluated on the tube and on the cylindrical domain. The problem is that the Neumann trace, which will be considered here, contains an additional term involving the normal velocity of the tube. Therefore, one has to come up with a solution how to deal with this.

A second approach could be to map back the partial differential equation from the non-cylindrical domain onto the cylindrical domain. The advantage is that one now considers a cylindrical domain, for which more theory is available. The drawback is that the differential equation in the reference domain is more complicated because of time and space dependent coefficients. Finding a fundamental solution is more difficult and one could for example pursue the parametrix ansatz, taken in [7], and then find the according mapping properties of the respective layer operators.

A third approach considers the partial differential equation on the non-cylindrical domain. Here, the partial differential equation is simple, but the domain is more involved in contrast to the second approach. This approach was used in [18], but without the corresponding Sobolev spaces and mapping properties of the integral operators. Since we already did computations in [1] based on [18], and since [3] provides a self-contained analysis of the mapping properties of the layer operators for the heat equation in a cylindrical domain, we choose this approach for this article.

Although we follow the argumentation line of Costabel [3], we have to modify several proofs for the present non-cylindrical setting. In particular, we have to use the appropriate function spaces and the correct Neumann traces. We would like to emphasize that, once one has the appropriate Neumann trace operator at hand and the mapping properties of the trace operators, the ideas of [3] can basically be followed. We indicate the required adaptations in the article.

The remainder of this article is organized as follows: In Sect. 2, we introduce anisotropic Sobolev spaces on cylindrical and non-cylindrical domain, which are used for the mapping properties. Section 3 is dedicated to the Dirichlet traces and the existence and uniqueness of solutions of the Dirichlet problem. In Sect. 4, we have a look at the appropriate Neumann trace. The main result is presented in Sect. 5, where we establish the mapping properties of the integral operators and the existence and uniqueness of solutions of the Neumann problem. In Sect. 6, we state some concluding remarks.

2 Anisotropic Sobolev Spaces

In order to study the heat equation, we shall introduce appropriate anisotropic Sobolev spaces on cylindrical domains. From these spaces, we will then derive Sobolev spaces on time-dependent domains.

2.1 Anisotropic Sobolev Spaces on Cylindrical Domains

Let $\Omega _0 \subset \mathbb {R}^d$, $d \ge 2$, be a Lipschitz domain in the spatial variable with boundary $\Gamma _0 \mathrel {\mathrel {\mathop :}=}\partial \Omega _0$ and let $0<T<\infty$. Then, the product set $Q_0 \mathrel {\mathrel {\mathop :}=}(0,T) \times \Omega _0 \subset \mathbb {R}^{d+1}$ forms a time-space cylinder with the lateral boundary $\Sigma _0\mathrel {\mathrel {\mathop :}=}(0,T)\times \Gamma _0$. The appropriate function spaces for parabolic problems in time-invariant domains, i.e. in cylindrical domains, are the anisotropic Sobolev spaces defined by

$$\begin{aligned} H^{r,s} (Q_0) \mathrel {\mathrel {\mathop :}=}L^2\big ((0,T); H^r (\Omega _0)\big ) \cap H^s\big ((0,T); L^2(\Omega _0)\big ) \end{aligned}$$

for $r,s \in \mathbb {R}_{\ge 0}$, see, e.g., [2, 3, 12]. The corresponding boundary spaces are

$$\begin{aligned} H^{r,s} (\Sigma _0) \mathrel {\mathrel {\mathop :}=}L^2\big ((0,T); H^r (\Gamma _0)\big ) \cap H^s\big ((0,T); L^2(\Gamma _0)\big ) \end{aligned}$$

Note that these spaces are well-defined for $r \le 1$ (while $s\ge 0$ is arbitrary) if $\Gamma _0$ is Lipschitz.

Remark 2.1

The space $H^{r,s} (Q_0)$ consists of all functions $u \in L^2 (Q_0)$, where the $L^2 (Q_0)$-norm of the partial derivatives $\partial _{\mathbf {x}}^{{\varvec{\alpha }}} \partial _t^\beta u (t, \mathbf {x})$ is finite for all $|{\varvec{\alpha }}| \le \lambda r$, $\beta \le (1-\lambda ) s$, and $\lambda \in [0,1]$.

With these definitions at hand, we can moreover define spaces for functions with zero initial condition by setting

$$\begin{aligned} H^{r,s}_{;0,}(Q_0) \mathrel {\mathrel {\mathop :}=}L^2 \big ( (0,T); H^r (\Omega _0)\big ) \cap H_{0,}^{s} \big ( (0,T); L^2 (\Omega _0) \big ) , \end{aligned}$$

where

$$\begin{aligned}&H_{0,}^s \big ( (0,T); L^2 (\Omega _0) \big ) \mathrel {\mathrel {\mathop :}=}\Big \lbrace u = \tilde{u}|_{(0,T)}:\\&\tilde{u} \in H^s \big ( (-\infty ,T); L^2 (\Omega _0) \big ):\tilde{u} (t) = 0 \text { for } t<0 \Big \rbrace . \end{aligned}$$

Note that we adopted the notation from [4, 5]. In addition, we can define functions which vanish at $t=T$ by setting

$$\begin{aligned} H^{r,s}_{;,0}(Q_0) \mathrel {\mathrel {\mathop :}=}L^2 \big ( (0,T); H^r (\Omega _0)\big ) \cap H_{,0}^{s} \big ( (0,T); L^2 (\Omega _0) \big ) , \end{aligned}$$

where in complete analogy

$$\begin{aligned}&H_{,0}^s \big ( (0,T); L^2 (\Omega _0) \big ) \mathrel {\mathrel {\mathop :}=}\Big \lbrace u = \tilde{u}|_{(0,T)} :\\&\tilde{u} \in H^s \big ( (0,\infty ); L^2 (\Omega _0) \big ):\tilde{u} (t) = 0 \text { for } t>T \Big \rbrace . \end{aligned}$$

As in the elliptic case, we can also include (spatial) zero boundary conditions into the function spaces by setting

$$\begin{aligned} \begin{aligned} H^{r,s}_{0;0,} (Q_0) \mathrel {\mathrel {\mathop :}=}L^2 \big ( (0,T); H^r_0 (\Omega _0 ) \big ) \cap H_{0,}^s \big ( (0,T); L^2(\Omega _0) \big ), \\ H^{r,s}_{0;,0} (Q_0) \mathrel {\mathrel {\mathop :}=}L^2 \big ( (0,T); H^r_0 (\Omega _0 ) \big ) \cap H_{,0}^s \big ( (0,T); L^2(\Omega _0) \big ), \end{aligned} \end{aligned}$$

where the spaces include zero initial and end conditions, respectively. On the boundary, we introduce

$$\begin{aligned} \begin{aligned} H^{r,s}_{;0,} (\Sigma _0)&\mathrel {\mathrel {\mathop :}=}L^2 \big ( (0,T); H^r (\Gamma _0) \big ) \cap H^{s}_{0,} \big ( (0,T); L^2(\Gamma _0) \big ), \\ H^{r,s}_{;,0} (\Sigma _0)&\mathrel {\mathrel {\mathop :}=}L^2 \big ( (0,T); H^r (\Gamma _0) \big ) \cap H^{s}_{,0} \big ( (0,T); L^2(\Gamma _0) \big ). \end{aligned} \end{aligned}$$

These spaces are the closures of $H^{r,s}(\Sigma _0)$ for zero start and end condition, respectively, compare [4, Section 2.3].

By duality we have

$$\begin{aligned} H^{-r, -s}_{;0,} (Q_0) = \big [H_{0;,0}^{r,s} (Q_0)\big ]' \quad \text {for } r - \frac{1}{2} \notin \mathbb {Z} \end{aligned}$$

according to [3]. The anisotropic Sobolev spaces on the boundary with negative smoothness index are defined by

$$\begin{aligned} H^{-r,-s}_{;,0} (\Sigma _0)&\mathrel {\mathrel {\mathop :}=}\big [ H^{r,s}_{;0,} (\Sigma _0) \big ]',\\ H^{-r,-s}_{;0,} (\Sigma _0)&\mathrel {\mathrel {\mathop :}=}\big [ H^{r,s}_{;,0} (\Sigma _0) \big ]',\\ \widetilde{H}^{-r,-s} (\Sigma _0)&\mathrel {\mathrel {\mathop :}=}\big [ H^{r,s} (\Sigma _0) \big ]', \end{aligned}$$

see [4, Section 2.3]. Moreover, according to [4, Remark 2.1], for $r \ge 0$ and $0 \le s < \frac{1}{2}$ it holds $H^{r,s}(\Sigma _0) = H^{r,s}_{;0,}(\Sigma _0) = H^{r,s}_{;,0}(\Sigma _0)$ and, therefore, the above introduced dual spaces are equal and we simply write $H^{-r,-s}(\Sigma _0)$.

Remark 2.2

We would like to clarify the intuition behind the slightly cumbersome notation. In $H^{r,s}_{0;,} (Q_0)$, a zero before the semicolon indicates a zero boundary condition in space. After the semicolon, a zero initial condition can be indicated by writing a zero between the semicolon and the comma. Whereas, a present zero after the comma stands for a zero end condition. Thus, this notation allows to see the spatial and temporal boundary condition at one glance.

2.2 Anisotropic Sobolev Spaces on Non-Cylindrical Domains

Having at hand the Sobolev spaces defined on cylindrical domains, we can also introduce Sobolev spaces on non-cylindrical domains. Non-cylindrical domains consist of a spatial domain, which we denote by $\Omega _t$. The subscript t indicates that the spatial domain might differ for every point of time. To obtain a non-cylindrical domain $Q_T$ we set

$$\begin{aligned} Q_T \mathrel {\mathrel {\mathop :}=}\bigcup _{0< t< T} \big ( \lbrace t \rbrace \times \Omega _t \big ). \end{aligned}$$

This domain has a lateral boundary $\Sigma _T$ defined by

$$\begin{aligned} \Sigma _T \mathrel {\mathrel {\mathop :}=}\underset{0< t < T}{\bigcup } \big ( \{t \} \times \Gamma _t \big ), \end{aligned}$$

where $\Gamma _t\mathrel {\mathrel {\mathop :}=}\partial \Omega _t$.

For every point of time t, we assume to have a smooth diffeomorphism $\varvec{\kappa }$, which maps the initial domain $\Omega _0$ onto the time-dependent domain $\Omega _t$. In accordance with [14], we write

$$\begin{aligned} \varvec{\kappa }:[0, T ] \times \mathbb {R}^d \rightarrow \mathbb {R}^d, \quad (t,\mathbf {x}) \mapsto \varvec{\kappa }(t, \mathbf {x}) \end{aligned}$$

(2.1)

to emphasize the dependence of the mapping $\varvec{\kappa }$ on the time, where we have $\varvec{\kappa }(t,\Omega _0) = \Omega _t$. Especially, $\Omega _t$ is also a Lipschitz domain for all $t\in [0,T]$.

The domains $\Omega _t$ each have a spatial normal $\mathbf {n}_t$, which we will also denote by $\mathbf {n}$ if it is clear from the context. Besides having a spatial normal, we also have a time-space normal ${\varvec{\nu }}$. We can write the time-space normal as

$$\begin{aligned} {\varvec{\nu }}= \frac{1}{\sqrt{1+v_{{\varvec{\nu }}}^2}} \begin{bmatrix} v_{{\varvec{\nu }}} \\ \mathbf {n}\end{bmatrix} \end{aligned}$$

(2.2)

for some appropriate $v_{{\varvec{\nu }}} \in \mathbb {R}$. According to [6], it holds $v_{{\varvec{\nu }}}= - \langle \mathbf {V}, \mathbf {n}\rangle$ for the vector field $\mathbf {V}$, which deforms the cylinder $Q_0$ into the tube $Q_T$ and for which the relation $\mathbf {V}= \partial _t {\varvec{\kappa }}\circ {\varvec{\kappa }}^{-1}$ holds.

We introduce the non-cylindrical analogues of the Sobolev spaces by setting

$$\begin{aligned}H^{r,s}(Q_T) \mathrel {\mathrel {\mathop :}=}\big \lbrace v \in L^2(Q_T):v \circ {\varvec{\kappa }}\in H^{r,s} (Q_0) \big \rbrace \end{aligned}$$

where the composition with $\varvec{\kappa }$ only acts on the spatial component. Due to the chain rule, $v\circ \varvec{\kappa }$ and v have the same Sobolev regularity, provided that the mapping $\varvec{\kappa }$ is smooth enough, see for example [13, Theorem 3.23] for the elliptic case. For what follows, we assume that that ${\varvec{\kappa }}\in C^2\big ([0,T]\times \mathbb {R}^d\big )$ satisfies

$$\begin{aligned} \Vert \varvec{\kappa }(t, \mathbf {x}) \Vert _{C^2([0,T]\times \mathbb {R}^d ; \mathbb {R}^d)}, \Vert \varvec{\kappa }(t, \mathbf {x})^{-1} \Vert _{C^2([0,T] \times \mathbb {R}^d ; \mathbb {R}^d)} \le C_{{\varvec{\kappa }}} \end{aligned}$$

(2.3)

for some constant $C_{{\varvec{\kappa }}} \in (0, \infty )$ as in [8, pg. 826]. We define the norm of $H^{r,s}(Q_T)$ as $\Vert u\Vert _{H^{r,s}(Q_T)}\mathrel {\mathrel {\mathop :}=}\Vert u\circ \kappa \Vert _{H^{r,s}(Q_0)}$ for $r,s \ge 0$. Notice that the Sobolev spaces on the boundary are defined in a similar manner.

Remark 2.3

(i) The space $H^{r,s}(Q_T)$ contains all functions such that $u \circ {\varvec{\kappa }}\in H^{r,s}(Q_0)$. This means that $\partial _{\mathbf {x}}^{{\varvec{\alpha }}} \partial _t^\beta (u \circ {\varvec{\kappa }}) \in L^2 (Q_0)$ for all $|{\varvec{\alpha }}| \le \lambda r$, $\beta \le (1-\lambda ) s$, and $\lambda \in [0,1]$. According to (2.3), the partial derivatives $\partial ^{{\varvec{\alpha }}}_\mathbf {x}\partial _t^\beta {\varvec{\kappa }}$ exist and are uniformly bounded for all $|{\varvec{\alpha }}|+\beta \le 2$.

(ii) Consider a function $u\in L^2(Q_T)$ with partial derivatives $\partial _\mathbf{x}^{{\varvec{\alpha }}}\partial _t^\beta u\in L^2(Q_T)$ for all $|{\varvec{\alpha }}| \le \lambda r$, $\beta \le (1-\lambda ) s$, and $\lambda \in [0,1]$. When computing the time-derivative of $u\circ {\varvec{\kappa }}$, we obtain also a spatial derivative as the following shows

$$\begin{aligned} (\partial _t u)\circ {\varvec{\kappa }}= \partial _t(u \circ {\varvec{\kappa }}) -\big \langle ({\text {D}}{\varvec{\kappa }})^{-\intercal } \nabla (u \circ {\varvec{\kappa }}) , \partial _t {\varvec{\kappa }}\big \rangle . \end{aligned}$$

(2.4)

Hence, it holds $u\in H^{r,s}(Q_T)$ only if $r\ge s$ since the temporal derivative $\partial _t^\beta (u\circ \kappa )$ involves also spatial partial derivatives $\partial _\mathbf{x}^{{\varvec{\alpha }}} u$ up to the order $|{\varvec{\alpha }}|=\beta$ besides the temporal derivative $\partial _t^\beta u$.

(iii) Due to the uniformity condition (2.3), we have as in [8]

$$\begin{aligned} 0< \underline{\sigma } \le \min \lbrace \sigma ({\text {D}}\!\varvec{\kappa }) \rbrace \le \max \lbrace \sigma ({\text {D}}\!\varvec{\kappa }) \rbrace \le \overline{\sigma } < \infty , \end{aligned}$$

where ${\text {D}}\!\varvec{\kappa }$ denotes the Jacobian of $\varvec{\kappa }$ and $\sigma ({\text {D}}\!\varvec{\kappa })$ denotes its singular values. Especially, as in [8, Remark 1, pg. 827], we may assume $\det ({\text {D}}\!\varvec{\kappa })$ to be positive. We can thus define the dual of $H^{r,s}_{0;,0}(Q_T)$ in two different ways, namely by using either the norm

$$\begin{aligned} \Vert u \Vert _{H^{-r,-s}_{;0,} (Q_T)} = \sup _{\widetilde{v} \in H^{r,s}_{0;,0} (Q_0)} \int _{Q_0} (u \circ {\varvec{\kappa }}) \widetilde{v} \, \mathrm {d}(\mathbf {x}, t) \end{aligned}$$

or the norm

$$\begin{aligned} {\left| \left| \left| u \right| \right| \right| }_{H^{-r, -s}_{;0,} (Q_T)} = \sup _{v \in H^{r,s}_{0;,0}(Q_T)} \int _{Q_T} u v \,\mathrm {d}(\mathbf {x}, t). \end{aligned}$$

However, we show that these norms are equivalent and so the spaces. On one hand, we have

$$\begin{aligned} {\left| \left| \left| u \right| \right| \right| }_{H^{-r, -s}_{;0,} (Q_T)}&= \sup _{v \in H^{r,s}_{0;,0} (Q_T)} \frac{\int _{Q_0} (u \circ {\varvec{\kappa }}) (v \circ {\varvec{\kappa }}) \det ( {\text {D}}\!{\varvec{\kappa }}) \, \mathrm {d}(\mathbf {x}, t)}{\Vert v \Vert _{H^{r,s}_{0;,0} (Q_T)}} \\&\le \Vert u \Vert _{H^{-r, -s}_{;0,} (Q_T)} \sup _{v \in H^{r,s}_{0;,0}(Q_T)} \frac{ \big \Vert (v \circ {\varvec{\kappa }}) \det ({\text {D}}\!{\varvec{\kappa }}) \big \Vert _{H^{r,s}_{0;,0} (Q_0)}}{\Vert v \circ {\varvec{\kappa }}\Vert _{H^{r,s}_{0;,0} (Q_0)}}\\&\lesssim \Vert u \Vert _{H^{-r, -s}_{;0,} (Q_T)}, \end{aligned}$$

where we used the definition of the norm on $H^{r,s}_{0;,0} (Q_T)$ for $s, r \ge 0$ and that the pointwise multiplication with a smooth function is a continuous operation. On the other hand, we likewise find

$$\begin{aligned} \Vert u \Vert _{H^{-r,-s}_{;0,} (Q_T)}&= \sup _{\widetilde{v} \in H^{r,s}_{0;,0}(Q_0)} \frac{\int _{Q_T} u (\widetilde{v} \circ {\varvec{\kappa }}^{-1}) \det ({\text {D}}\!{\varvec{\kappa }}^{-1}) \, \mathrm {d}(\mathbf {x}, t)}{\Vert \widetilde{v} \Vert _{H^{r,s}_{0;,0} (Q_0)} }\\&\le {\left| \left| \left| u \right| \right| \right| }_{H^{-r,-s}_{;0,} (Q_T)} \sup _{\widetilde{v} \in H^{r,s}_{0;,0} (Q_0)} \frac{\big \Vert (\widetilde{v} \circ {\varvec{\kappa }}^{-1})\det ({\text {D}}\!{\varvec{\kappa }}^{-1}) \big \Vert _{H^{r,s}_{0;,0}(Q_T)}}{\Vert \widetilde{v} \Vert _{H^{r,s}_{0;,0} (Q_0)}} \\&\lesssim {\left| \left| \left| u \right| \right| \right| }_{H^{-r, -s}_{;0,} (Q_T)}. \end{aligned}$$

Hence, both duality pairings result in the same dual spaces and we can say that $H_{0;,0}^{r,s} (Q_T)$ and $H^{-r, -s}_{;0,} (Q_T)$ are indeed dual, as likewise for the other pairings.

Finally, let the space $\mathcal {V}(Q_T)$ consist of all the functions v with $v\circ {\varvec{\kappa }}\in \mathcal {V} (Q_0)$ and

$$\begin{aligned} \mathcal {V}(Q_0)\mathrel {\mathrel {\mathop :}=}\big \{u \in L^2\big ((0,T); H^1(\Omega _0)\big ) :\partial _t u \in L^2\big ((0,T); H^{-1} (\Omega _0)\big )\big \}. \end{aligned}$$

The norm on this space is given by

$$\begin{aligned} \Vert u \Vert _{\mathcal {V}(Q_0)}^2 \mathrel {\mathrel {\mathop :}=}\Vert u \Vert _{H^{1,0} (Q_0)}^2 + \Vert \partial _t u \Vert _{H^{-1,0} (Q_0)}^2. \end{aligned}$$

Note that the space $\mathcal {V}(Q_0)$ is a dense subspace of $H_{;,}^{1,\frac{1}{2}} (Q_0)$, which follows according to [3, Formula (2.2)] from the interpolation result

$$\begin{aligned} L^2(I; X) \cap H^1 (I;Y) \subset H^{\frac{1}{2}} \big ( I; [X,Y]_{\frac{1}{2}} \big ) \cap C \big ( \bar{I}; [X,Y]_{\frac{1}{2}} \big ) \end{aligned}$$

(2.5)

for $X \subset Y$ being Hilbert spaces.

3 Dirichlet Problem

3.1 Dirichlet Trace Operator on Cylindrical Domains

We first introduce the notion of traces with respect to cylindrical domains. According to [4, Section 2.3], we can define the (interior) Dirichlet trace for a function $u \in C^1(\overline{Q}_0)$ as

$$\begin{aligned} \gamma _0 u (t, \mathbf {x}) \mathrel {\mathrel {\mathop :}=}\lim _{\Omega _0 \ni \mathbf {y}\rightarrow \mathbf {x}\in \Gamma _0} u (t, \mathbf {y}) \quad \text {for } (t, \mathbf {x}) \in \Sigma _0. \end{aligned}$$

We thus have $\gamma _0 u = u|_{\Sigma _0}$. We can introduce a similar operator on anisotropic Sobolev spaces, see the following lemma, being along the lines of [11, Theorem 2.1]. It has been proven for $\Gamma _0 \in C^\infty$, but it is also true for a Lipschitz boundary in accordance with [3, pg. 504ff].

Lemma 3.1

The map $\gamma _0 :H^{1,\frac{1}{2}} (Q_0) \rightarrow H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _0)$ is linear and continuous.

We find the following statement in [3, Lemma 2.4], which holds in the case of a Lipschitz domain.

Lemma 3.2

The Dirichlet trace operator $\gamma _0$ is continuous and surjective as an operator from $H^{1,\frac{1}{2}}_{;0,} (Q_0)$ to $H^{\frac{1}{2},\frac{1}{4}} (\Sigma _0)$.

According to [4, Theorem 2.4], there exists also an extension operator. The extension operator is a right inverse to the surjective Dirichlet trace operator $\gamma _0$ and, thus, extends a function defined only on the boundary to the space (see also [5, pg. 12] and [3, Definition 2.17]).

Lemma 3.3

The Dirichlet trace operator $\gamma _0 :H^{1, \frac{1}{2}}_{;0,}(Q_0) \rightarrow H^{\frac{1}{2},\frac{1}{4}} (\Sigma _0)$ has a continuous right inverse operator $\mathcal {E}_0:H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _0) \rightarrow H^{1, \frac{1}{2}}_{;0,} (Q_0),$ satisfying $\gamma _0 \mathcal {E}_0 v = v$ for all $v \in H^{\frac{1}{2},\frac{1}{4}}(\Sigma _0)$.

3.2 Dirichlet Trace Operator on Non-Cylindrical Domains

In this section, we denote the (interior) Dirichlet trace operator with respect to a non-cylindrical domain by $\gamma _{0,t}$ to distinguish it from the Dirichlet trace operator with respect to a cylindrical domain introduced above. When no confusion can happen, we will drop the subscript t in the trace operator for a non-cylindrical domain.

For a smooth function $u\in C^1(\overline{Q}_T)$, defined on a non-cylindrical domain, we set

$$\begin{aligned} \gamma _{0,t} u (t, \mathbf {x}_t) \mathrel {\mathrel {\mathop :}=}\lim _{\Omega _t \ni \mathbf {y}_t \rightarrow \mathbf {x}_t \in \Gamma _t} u (t, \mathbf {y}_t ). \end{aligned}$$

It obviously holds

$$\begin{aligned} \gamma _{0,t} u (t, \mathbf {x}_t)&= \lim _{\begin{array}{c} \Omega _0 \ni \mathbf {y}\rightarrow \mathbf {x}\in \Gamma _0, \\ {\varvec{\kappa }}(t,\mathbf {x}) = \mathbf {x}_t \end{array}} u \big (t, {\varvec{\kappa }}(t,\mathbf {y}) \big )\\&= \gamma _{0}(u \circ {\varvec{\kappa }}) (t, \mathbf {x}) = \gamma _{0}(u \circ {\varvec{\kappa }}) \big (t, {\varvec{\kappa }}^{-1} (t,\mathbf {x}_t) \big ) \end{aligned}$$

for the diffeomorphism ${\varvec{\kappa }}$ from (2.1). By density of the smooth functions in the Sobolev spaces, we can also extend this notion to Sobolev spaces. Moreover, we have the same mapping properties for $\gamma _{0,t}$ as for $\gamma _0$, since

$$\begin{aligned} \Vert \gamma _{0,t} u \Vert _{H^{\frac{1}{2},\frac{1}{4}} (\Sigma _T)}&= \Vert \gamma _{0,t} u\circ {\varvec{\kappa }}\Vert _{H^{\frac{1}{2},\frac{1}{4}}(\Sigma _0)} \\&= \big \Vert \big (\gamma _{0}(u \circ {\varvec{\kappa }}) \circ {\varvec{\kappa }}^ {-1} \big ) \circ {\varvec{\kappa }}\big \Vert _{H^{\frac{1}{2},\frac{1}{4}}(\Sigma _0)} \\&\lesssim \Vert u \circ {\varvec{\kappa }}\Vert _{H^{1, \frac{1}{2}}(Q_0)} \\&= \Vert u \Vert _{H^{1, \frac{1}{2}}(Q_T)}. \end{aligned}$$

Note that the hidden constant changes from line to line and depends on the diffeomorphism ${\varvec{\kappa }}$, because we used the norm equivalence on the cylindrical and non-cylindrical domain as well as the mapping property of the Dirichlet trace operator on the cylindrical domain.

Due to this consideration, all the properties of Sect. 3.1 remain valid for the Dirichlet trace operator on non-cylindrical domains. The surjectivity follows for example from the following consideration: Let $v \in H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T)$. By the definition of the norm, we thus have $v \circ {\varvec{\kappa }}\in H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _0)$. By the surjectivity of the Dirichlet trace operator with respect to $Q_0$, there exists a $w\in H^{1, \frac{1}{2}}_{;0,} (Q_0)$ with $\gamma _{0}w = v \circ {\varvec{\kappa }}$. Due to the bijectivity of ${\varvec{\kappa }}$, we may define

$$\begin{aligned} \hat{w}\mathrel {\mathrel {\mathop :}=}w\circ {\varvec{\kappa }}^{-1}\in H^{1, \frac{1}{2}}_{;0,}(Q_T). \end{aligned}$$

We thus have

$$\begin{aligned} \gamma _{0,t} \hat{w}(t, \mathbf {x}_t) = \gamma _0 w \big (t, {\varvec{\kappa }}^{-1} (t,\mathbf {x}_t)\big ) = v (t,\mathbf {x}_t), \end{aligned}$$

from where the subjectivity and hence the existence of the right inverse operator $\mathcal {E}_0$ follows.

3.3 Existence and Uniqueness of Dirichlet Problem

We consider the following Dirichlet problem with homogeneous initial datum

$$\begin{aligned} \begin{aligned} (\partial _t - \Delta ) v&= f&\ \ \text {in } Q_T, \\ \gamma _0 v&= q&\ \ \text {on } \Sigma _T, \\ v(0, \cdot )&= 0&\ \ \text {in } \Omega _0. \end{aligned} \end{aligned}$$

(3.1)

We have the following existence and uniqueness theorem for its solution.

Theorem 3.4

Let $f \in H^{-1, -\frac{1}{2}}_{;0,}(Q_T)$ and $q \in H^{\frac{1}{2},\frac{1}{4}}(\Sigma _T)$. Then, there exists a unique solution $v \in H^{1,\frac{1}{2}}_{;0,} (Q_T)$, satisfying the boundary condition in (3.1) and

$$\begin{aligned} \begin{aligned} S(v, \varphi ) \mathrel {\mathrel {\mathop :}=}&\int _0^T\int _{\Omega _t} \{\nabla v \cdot \nabla \varphi + \partial _t v \varphi \}\, \mathrm {d}\mathbf {x}\mathrm {d}t\\ =&\int _0^T \int _{\Omega _t} f u \, \mathrm {d}\mathbf {x}\mathrm {d}t \ \text {for all}\ \varphi \in H^{1, \frac{1}{2}}_{0;,0}(Q_T). \end{aligned} \end{aligned}$$

(3.2)

Proof

A proof of this statement based on the proof of [3, Lemma 2.8] can be found in [1]. $\square$

4 Neumann Trace Operator

Similarly as we defined the Dirichlet trace operator, we can also introduce an (interior) Neumann trace operator. In the following, we will first introduce this concept on cylindrical domains. Then, we will introduce the notion of a Neumann trace on a non-cylindrical domain formally and rigorously.

4.1 Neumann Trace Operator on Cylindrical Domains

We first introduce the Neumann trace operator, also called the conormal derivative, on a cylindrical domain along the lines of [3]. Let us define the space

$$\begin{aligned} H^{1,\frac{1}{2}} (Q_0; \mathcal {L}) \mathrel {\mathrel {\mathop :}=}\big \lbrace u \in H^{1, \frac{1}{2}} (Q_0) :\mathcal {L}u \in L^2(Q_0) \big \rbrace , \end{aligned}$$

where $\mathcal {L}\mathrel {\mathrel {\mathop :}=}\partial _t - \Delta$ is the partial differential operator under consideration. The norm on this space is given by

$$\begin{aligned} \Vert u \Vert _{H^{1, \frac{1}{2}}(Q_0; \mathcal {L})}^2 \mathrel {\mathrel {\mathop :}=}\Vert u \Vert _{H^{1, \frac{1}{2}} (Q_0)}^2 + \big \Vert (\partial _t - \Delta ) u \big \Vert _{L^2 (Q_0)}^2. \end{aligned}$$

According to [3, Lemma 2.16], the bilinear form

$$\begin{aligned} b(u,v) \mathrel {\mathrel {\mathop :}=}\int _{Q_0} \big \{\langle \nabla u, \nabla v \rangle - (\partial _t - \Delta ) u v\big \}\, \mathrm {d}\mathbf {x}\mathrm {d}t + d(u,v) \end{aligned}$$

is continuous on $H^{1, \frac{1}{2}}(\mathbb {R}\times \Omega _0; \mathcal {L}) \times H^{1, \frac{1}{2}} (\mathbb {R}\times \Omega _0)$, where

$$\begin{aligned} d (u,v) \mathrel {\mathrel {\mathop :}=}\int _{\mathbb {R}\times \Omega _0} \partial _t u v \, \mathrm {d}\mathbf {x}\mathrm {d}t. \end{aligned}$$

The bilinear form d(u, v) has a continuous extension from $C_0^\infty (\mathbb {R}^{d+1})\times$ $C_0^\infty (\mathbb {R}^{d+1})$ to $H^\frac{1}{2} \big (\mathbb {R}; L^2 (\Omega _0) \big )\times H^\frac{1}{2}\big (\mathbb {R}; L^2 (\Omega _0) \big )$ and it holds $d(u,v) = - d(v,u)$ for all $u,v \in H^{\frac{1}{2}} \big (\mathbb {R}; L^2 (\Omega _0) \big )$, compare [3, Lemma 2.6].

The (interior) Neumann trace is defined for $u \in C^1 (\overline{Q}_0)$ by

$$\begin{aligned} \gamma ^{{\text {int}}}_1 u (t, \mathbf {x}) \mathrel {\mathrel {\mathop :}=}\lim _{\Omega _0 \ni \mathbf {y}\rightarrow \mathbf {x}\in \Gamma _0} \big \langle \nabla _{\mathbf {y}} u (t, \mathbf {y}), \mathbf {n}_{\mathbf {x}} \big \rangle \quad \text {for } (t, \mathbf {x}) \in \Sigma _0 \end{aligned}$$

and coincides with the normal derivative on $\Sigma _0$, thus $\gamma _1^{{\text {int}}} u = \partial u/\partial \mathbf {n}$ on $\Sigma _0$, see [4, Section 3.3] and also [20, Satz 8.7] for the elliptic case. Since it holds

$$\begin{aligned} b(u,v) = \int _{\Sigma _0} \frac{\partial u}{\partial \mathbf{n}} v \, \mathrm {d}\sigma \mathrm {d}t \end{aligned}$$

for $u,v \in C_0^2 (\mathbb {R}\times \overline{\Omega }_0)$, we can extend this definition as follows, corresponding to [3, Definition 2.17].

Definition 4.1

Let $u \in H^{1, \frac{1}{2}} (\mathbb {R}\times \Omega _0; \mathcal {L})$. Then, the Neumann trace operator $\gamma _1 u \in H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _0)$ is the continuous linear form on $H^{\frac{1}{2}, \frac{1}{2}} (\Sigma _0)$ defined by $\gamma _1^{{\text {int}}} u :\varphi \mapsto b (u, \mathcal {E}_0 \varphi ),$ where $\mathcal {E}_0$ is the extension operator given in Lemma 3.3.

Notice that we can also introduce the conormal derivative $\gamma _1^{{\text {int}}} u \in H^{-\frac{1}{2}, -\frac{1}{4}}(\Sigma _0)$ as the unique solution of a variational problem, as it is done in [5, Section 3.4]. This variational problem can for example be obtained by applying $\gamma _1^{{\text {int}}} u$ to $\varphi$. According to [3, Proposition 2.18], the Neumann trace has the following properties.

Lemma 4.2

• (i) The map

  $$\begin{aligned} \gamma _1^{{\text {int}}}:H^{1, \frac{1}{2}} (\mathbb {R}\times \Omega _0; \mathcal {L}) \rightarrow H^{-\frac{1}{2}, -\frac{1}{4}} (\mathbb {R}\times \Gamma _0) \end{aligned}$$

  is continuous and by restriction also the map

  $$\begin{aligned} \gamma _1^{{\text {int}}}:H^{1, \frac{1}{2}} (Q_0; \mathcal {L}) \rightarrow H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _0) \end{aligned}$$

  is continuous.

• (ii) If $u \in C^2 (\overline{Q}_0)$, then $\gamma _1^{{\text {int}}} u = (\partial u/\partial \mathbf {n})|_{\Sigma _0}$ due to the Green formula.

4.2 Neumann Trace Operator on Non-Cylindrical Domains

On time-dependent boundaries, one could consider the usual Neumann trace, as it is done for example in [6, Section 6.1]. Instead, we follow here the idea of [18] and employ a velocity corrected Neumann trace. We first formally introduce this Neumann trace and afterwards characterize its properties rigorously.

4.2.1 Formal Derivation

For a time dependent spatial surface we define two Neumann trace operators

$$\begin{aligned} \gamma _1^\pm \varphi \mathrel {\mathrel {\mathop :}=}\frac{\partial \varphi }{\partial \mathbf {n}_t} \mp \frac{1}{2} \langle \mathbf {V}, \mathbf {n}_t \rangle \varphi . \end{aligned}$$

(4.1)

In order to motivate this definition, consider the boundary value problem

$$\begin{aligned} \begin{aligned} (\partial _t - \Delta ) u&= f&\ \ \text {in } Q_T, \\ \gamma _1 u&= g&\ \ \text {on } \Sigma _T, \\ u(0, \cdot )&= 0&\ \ \text {in } \Omega _0, \end{aligned} \end{aligned}$$

(4.2)

where we leave it a priori open what $\gamma _1$ means. Let us formally derive the weak formulation of the Neumann problem (4.2) by multiplying with a test function v with $v(T, \cdot ) = 0$ in $\Omega _T$ and using Reynolds’ transport theorem

$$\begin{aligned}&\int _0^T \int _{\Omega _t} f v \, \mathrm {d}\mathbf {x}\mathrm {d}t = \int _0^T \int _{\Omega _t} (\partial _t - \Delta ) u v \, \mathrm {d}\mathbf {x}\mathrm {d}t \\&\qquad = \int _0^T \int _{\Omega _t} \big \{\langle \nabla u, \nabla v \rangle + \partial _t (uv) - u \partial _t v\big \} \, \mathrm {d}\mathbf {x}\mathrm {d}t - \int _0^T \int _{\Gamma _t} \frac{\partial u}{\partial \mathbf {n}} v \, \mathrm {d}\sigma \mathrm {d}t \\&\qquad = \int _0^T \int _{\Omega _t} \big \{\langle \nabla u, \nabla v \rangle - u \partial _t v\big \} \, \mathrm {d}\mathbf {x}\mathrm {d}t + \int _0^T \frac{\mathrm {d}}{\mathrm {d}t} \int _{\Omega _t} uv \, \mathrm {d}\mathbf {x}\mathrm {d}t\\&\qquad \quad - \int _0^T \int _{\Gamma _t} \bigg \{\frac{\partial u}{\partial \mathbf {n}} v + u v \langle \mathbf {V}, \mathbf {n}\rangle \bigg \} \, \mathrm {d}\sigma \mathrm {d}t. \end{aligned}$$

Due to the fundamental theorem of calculus and the vanishing initial and end condition of u and v, respectively, we obtain the variational equation

$$\begin{aligned} a(u,v) = \int _0^T \int _{\Omega _t} f v \, \mathrm {d}\mathbf {x}\mathrm {d}t + \int _0^T \int _{\Gamma _t} \bigg \{\frac{\partial u }{\partial \mathbf {n}} v + \frac{1}{2} uv \langle \mathbf {V}, \mathbf {n}\rangle \bigg \} \, \mathrm {d}\sigma \mathrm {d}t \end{aligned}$$

with bilinear form

$$\begin{aligned} a(u,v) \mathrel {\mathrel {\mathop :}=}\int _0^T \int _{\Omega _t}\big \{\langle \nabla u, \nabla v \rangle - u \partial _t v\big \} \, \mathrm {d}\mathbf {x}\mathrm {d}t - \frac{1}{2}\int _0^T \int _{\Gamma _t} uv \langle \mathbf {V}, \mathbf {n}\rangle \, \mathrm {d}\sigma \mathrm {d}t.\end{aligned}$$

Thus, if we set the previously unspecified trace in (4.2) as $\gamma _1^{-}$, we arrive at

$$\begin{aligned} a(u,v) = \int _0^T \int _{\Omega _t} f v \, \mathrm {d}\mathbf {x}\mathrm {d}t + \int _0^T \int _{\Gamma _t} g v \, \mathrm {d}\sigma \mathrm {d}t. \end{aligned}$$

With the additonal boundary integral term it is easy to see that the bilinear form is bounded and coercive in the $H^{1, \frac{1}{2}}$-norm, so Lion’s projection theorem guarantees existence and uniqueness of this Neumann problem. However, for the cylindrical case, [3, Lemma 2.21] states that this strategy does not yield satisfactory results, since one has to make stronger assumptions on the regularity of the input data. Therefore, as in [3], we will proof the existence and uniqueness of solutions by using a boundary integral formulation (see Corollary 5.14).

4.2.2 Rigorous Derivation

We assume ${\varvec{\kappa }}$ to be defined on $\mathbb {R}\times \mathbb {R}^d$ and not only on $[0,T] \times \mathbb {R}^d$. Moreover, for sake of simplicity in representation, we always consider functions u and v throughout this section which satisfy

$$\begin{aligned} \int _{\mathbb {R}} \frac{\mathrm {d}}{\mathrm {d}t} \int _{\Omega _t} u v \, \mathrm {d}\mathbf {x}\mathrm {d}t = 0. \end{aligned}$$

(4.3)

This assumption stems from the fact that we would like to integrate by parts in time. Later on, we will consider a finite time interval (0, T) and equip u and v with the appropriate zero initial and end conditions. Extending u and v by zero for $t<0$ and $t>T$, respectively, leads then to the fulfillment of (4.3).

Let us define

$$\begin{aligned} d(u,v) \mathrel {\mathrel {\mathop :}=}\int _{\mathbb {R}} \int _{\Omega _t} \partial _t u v \, \mathrm {d}\mathbf {x}\mathrm {d}t + \frac{1}{2} \int _{\mathbb {R}} \int _{\Gamma _t} \langle \mathbf {V}, \mathbf {n}\rangle u v\,\mathrm {d}\sigma \mathrm {d}t. \end{aligned}$$

(4.4)

Notice that the additional boundary term is a speciality of the time-dependent boundary. We shall first state the analogue of [3, Lemma 2.6].

Lemma 4.3

The bilinear form d(u, v) has a continuous extension from $C_0^{\infty } (\mathbb {R}^{d+1})\times C_0^{\infty } (\mathbb {R}^{d+1})$ to $H^{1,\frac{1}{2}} \big (\bigcup _{t \in \mathbb {R}}(\lbrace t \rbrace \times \Omega _t)\big )\times H^{1,\frac{1}{2}}\big (\bigcup _{t \in \mathbb {R}}(\lbrace t \rbrace \times \Omega _t)\big )$, and it holds

$$\begin{aligned} d(u,v) = - d(v,u). \end{aligned}$$

(4.5)

Proof

The use of Reynolds’ transport theorem allows us to compute

$$\begin{aligned} d(u,v)&= \int _{\mathbb {R}} \int _{\Omega _t} \partial _t u v \, \mathrm {d}\mathbf {x}\mathrm {d}t + \frac{1}{2} \int _{\mathbb {R}} \int _{\Gamma _t} \langle \mathbf {V}, \mathbf {n}\rangle u v\,\mathrm {d}\sigma \mathrm {d}t \\&= \int _{\mathbb {R}} \int _{\Omega _t} \big \{\partial _t (uv) - u \partial _t v\big \} \, \mathrm {d}\mathbf {x}\mathrm {d}t + \frac{1}{2} \int _{\mathbb {R}} \int _{\Gamma _t} \langle \mathbf {V}, \mathbf {n}\rangle u v\,\mathrm {d}\sigma \mathrm {d}t \\&= \int _{\mathbb {R}} \frac{\mathrm {d}}{\mathrm {d}t} \int _{\Omega _t} u v \, \mathrm {d}\mathbf {x}\mathrm {d}t - \int _{\mathbb {R}} \int _{\Omega _t} u \partial _t v \, \mathrm {d}\mathbf {x}\mathrm {d}t - \frac{1}{2} \int _{\mathbb {R}} \int _{\Gamma _t} \langle \mathbf {V}, \mathbf {n}\rangle u v \, \mathrm {d}\sigma \mathrm {d}t. \end{aligned}$$

The assumption (4.3) hence implies

$$\begin{aligned} d(u,v) = - \int _{\mathbb {R}} \int _{\Omega _t} u \partial _t v \, \mathrm {d}\mathbf {x}\mathrm {d}t - \frac{1}{2} \int _{\mathbb {R}} \int _{\Gamma _t} \langle \mathbf {V}, \mathbf {n}\rangle u v \, \mathrm {d}\sigma \mathrm {d}t, \end{aligned}$$

from where (4.5) follows immediately. The rest is in complete analogy to [3, Lemma 2.6], but we need higher regularity in the spatial variable instead just $L^2(\Omega _0)$ like in [3], because the boundary term in the definition of d(u, v) has to be well-defined. $\square$

As in Sect. 4.1, we introduce the space

$$\begin{aligned} H^{1, \frac{1}{2}} (Q_T; \mathcal {L}) \mathrel {\mathrel {\mathop :}=}\big \lbrace u \in H^{1, \frac{1}{2}} (Q_T) :\mathcal {L}u \in L^2 (Q_T) \big \rbrace , \end{aligned}$$

where $\mathcal {L}\mathrel {\mathrel {\mathop :}=}\partial _t - \Delta$ is the differential operator on the non-cylindrical domain. We state the analogue of [3, Lemma 2.16] in the case of a non-cylindrical domain, the proof of which is obvious.

Lemma 4.4

The bilinear form

$$\begin{aligned} b^-(u,v) \mathrel {\mathrel {\mathop :}=}\int _0^T \int _{\Omega _t} \big \{\langle \nabla u, \nabla v \rangle - (\partial _t - \Delta ) u v\big \} \, \mathrm {d}\mathbf {x}\mathrm {d}t + d (u,v) \end{aligned}$$

with d(u, v) being defined in (4.4) is continuous on $H^{1,\frac{1}{2}}\big ( \bigcup _{t \in \mathbb {R}}(\lbrace t \rbrace \times \Omega _t); \partial _t - \Delta \big ) \times H^{1, \frac{1}{2}} \big ( \bigcup _{t \in \mathbb {R}} (\lbrace t \rbrace \times \Omega _t) \big )$. If $u, v \in C_0^2 \big ( \bigcup _{t \in \mathbb {R}} (\lbrace t \rbrace \times \overline{\Omega }_t)\big )$, we have

$$\begin{aligned} b^-(u,v) = \int _0^T \int _{\Gamma _t} \bigg \{\frac{\partial u}{\partial \mathbf {n}} v + \frac{1}{2} \langle \mathbf {V}, \mathbf {n}\rangle u v\bigg \} \, \mathrm {d}\sigma \mathrm {d}t \end{aligned}$$

by means of Green’s formula.

In complete analogy to the Neumann trace operator in the cylindrical case, we will define $\gamma _1^- u$, which is one of the two required Neumann trace operators.

Definition 4.5

Given $u \in H^{1, \frac{1}{2}}\big ( \bigcup _{t \in \mathbb {R}}(\lbrace t \rbrace \times \Omega _t); \partial _t - \Delta \big )$, we denote by $\gamma _1^- u \in H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T)$ the continuous linear form on $H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T)$ defined by $\gamma _1^- u:\varphi \mapsto b^-(u, \mathcal {E}_0 \varphi )$, where $\mathcal {E}_0$ is the extension operator as mentioned in Sect. 3.2.

The following lemma is the non-cylindrical equivalent to [3, Proposition 2.18].

Lemma 4.6

The map

$$\begin{aligned} \gamma _1^- :H^{1, \frac{1}{2}} \bigg ( \bigcup _{t \in \mathbb {R}} \big (\lbrace t \rbrace \times \Omega _t \big ); \partial _t - \Delta \bigg ) \rightarrow H^{-\frac{1}{2}, -\frac{1}{4}} \bigg ( \bigcup _{t \in \mathbb {R}} \big (\lbrace t \rbrace \times \Gamma _t \big )\bigg ) \end{aligned}$$

is continuous and by restriction also the map

$$\begin{aligned} \gamma _1^- :H^{1, \frac{1}{2}} (Q_T; \partial _t - \Delta )\rightarrow H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T) \end{aligned}$$

is continuous. Moreover, if $u \in C^2 (\overline{Q}_T)$, then it holds

$$\begin{aligned} \gamma _1^- u = \frac{\partial u}{\partial \mathbf {n}} + \frac{1}{2} \langle \mathbf {V}, \mathbf {n}\rangle u. \end{aligned}$$

Proof

As in [3], the continuity is a consequence of the continuity of the bilinear form $b(\cdot , \cdot )$ (cf. Lemma 4.4). The second statement follows immediately from Green’s first formula. $\square$

Remark 4.7

In view of the reformulation of the heat equation in terms of boundary integral equations, we will moreover encounter a second Neumann trace operator, which we denote by $\gamma _1^+$. It can be achieved analogously to above by considering the differential operator $\partial _t + \Delta$ instead of $\partial _t - \Delta$. The former operator for example arises when considering a time reversal of the latter one. With

$$\begin{aligned} b^{+ }(u,v) \mathrel {\mathrel {\mathop :}=}\int _0^T \int _{\Omega _t} \big \{\langle \nabla u, \nabla v \rangle + (\partial _t + \Delta ) u v\big \} \, \mathrm {d}\mathbf {x}\mathrm {d}t - d(u,v), \end{aligned}$$

we can state the analogue of Lemma 4.4, namely the continuity of $b^{+}(\cdot , \cdot )$ in the appropriate space and for u and v smooth enough, we have

$$\begin{aligned} b^{+} (u,v) = \int _0^T \int _{\Gamma _t} \bigg \{\frac{\partial u}{\partial \mathbf {n}} v - \frac{1}{2} \langle \mathbf {V}, \mathbf {n}\rangle u v\bigg \} \, \mathrm {d}\sigma \mathrm {d}t. \end{aligned}$$

With this property at hand, we can define the trace operator $\gamma _1^+$ in analogy to Definition 4.5. For u smooth enough, it then holds

$$\begin{aligned} \gamma _1^+ u = \frac{\partial u}{\partial \mathbf {n}} - \frac{1}{2} \langle \mathbf {V}, \mathbf {n}\rangle u. \end{aligned}$$

The existence of two Neumann trace operators is a speciality of the time-dependent boundary.

Likewise to [3, Formula (2.35)], given $u\in H^{1, \frac{1}{2}}\big (\bigcup _{t \in \mathbb {R}}(\lbrace t \rbrace \times \Omega _t); \partial _t - \Delta \big )$ and $v \in H^{1, \frac{1}{2}}\big ( \bigcup _{t \in \mathbb {R}} (\lbrace t \rbrace \times \Omega _t)\big )$, we obtain Green’s first formula

$$\begin{aligned} \int _{\mathbb {R}} \int _{\Omega _t}\langle \nabla u, \nabla v \rangle \, \mathrm {d}\mathbf {x}\mathrm {d}t + d(u,v) = \int _{\mathbb {R}} \int _{\Omega _t} (\partial _t - \Delta ) u v \, \mathrm {d}\mathbf {x}\mathrm {d}t + \langle \gamma _1^- u, \gamma _0 v \rangle .\nonumber \\ \end{aligned}$$

(4.6)

By restriction, this formula also holds for $u \in H^{1, \frac{1}{2} }_{;0,} (Q_T; \partial _t - \Delta )$ and $v \in H^{1, \frac{1}{2}}_{;,0} (Q_T)$, but not, as was pointed out in [3], when u, v are both in $H^{1,\frac{1}{2}}_{;0,} (Q_T)$.

In complete analogy to (4.6), Green’s formula for $u \in H^{1, \frac{1}{2}}\big ( \bigcup _{t \in \mathbb {R}} (\lbrace t \rbrace \times \Omega _t); \partial _t +\Delta \big )$ and $v \in H^{1, \frac{1}{2}} \big (\bigcup _{t \in \mathbb {R}}(\lbrace t \rbrace \times \Omega _t)\big )$ reads

$$\begin{aligned} \int _{\mathbb {R}} \int _{\Omega _t}\langle \nabla u, \nabla v \rangle \, \mathrm {d}\mathbf {x}\mathrm {d}t - d(u,v) = \int _{\mathbb {R}} \int _{\Omega _t} (- \partial _t - \Delta ) u v \, \mathrm {d}\mathbf {x}\mathrm {d}t + \langle \gamma _1^+ u, \gamma _0 v \rangle . \end{aligned}$$

Again, by restriction, this formula also holds for $u \in H^{1, \frac{1}{2} }_{;,0} (Q_T; \partial _t + \Delta )$ and $v \in H^{1, \frac{1}{2}}_{;0,} (Q_T)$.

We are now in the position to state Green’s formulas for a finite time interval, the time-independent analogues of which are given in [3, Proposition 2.19]. Notice, however, that [3] introduces a time reversal map. For a time-dependent domain, this approach does not make sense, since the integration over a time forward tube of a time reversed entity is not always well defined. Therefore, we choose a slightly different approach to obtain a further Green formula.

Lemma 4.8

• (i) Let $u \in H^{1, \frac{1}{2}}_{;0,}\big ( \bigcup _{t \in \mathbb {R}_+}(\lbrace t \rbrace \times \Omega _t); \partial _t - \Delta \big )$ and $v \in H^{1, \frac{1}{2}}_{;,0}\big (\bigcup _{- \infty< t < t_0}(\lbrace t \rbrace \times \Omega _t)\big )$. Then, for $t_0>0$, there holds Green’s first formula

  $$\begin{aligned} \int _0^{t_0} \int _{\Omega _t}\langle \nabla u, \nabla v\rangle \, \mathrm {d}\mathbf {x}\mathrm {d}t + d (u, v) = \int _0^{t_0} \int _{\Omega _t} (\partial _t - \Delta ) u v \, \mathrm {d}\mathbf {x}\mathrm {d}t + \langle \gamma _1^- u, \gamma _0 v \rangle . \end{aligned}$$

• (ii) Let $u \in H^{1, \frac{1}{2}}_{;,0}\big ( \bigcup _{- \infty< t < t_0}(\lbrace t \rbrace \times \Omega _t); \partial _t + \Delta \big )$ and $v \in H^{1, \frac{1}{2}}_{;0,}\big (\bigcup _{t \in \mathbb {R}_+}(\lbrace t \rbrace \times \Omega _t)\big )$. Then, for $t_0>0$, there holds Green’s alternative first formula

  $$\begin{aligned} \int _0^{t_0} \int _{\Omega _t}\langle \nabla u, \nabla v \rangle \, \mathrm {d}\mathbf {x}\mathrm {d}t - d(u,v) = \int _0^{t_0} \int _{\Omega _t} (- \partial _t - \Delta ) u v \, \mathrm {d}\mathbf {x}\mathrm {d}t + \langle \gamma _1^+ u, \gamma _0 v \rangle . \end{aligned}$$

• (iii) Let $u \in H^{1, \frac{1}{2}}_{;0,}\big ( \bigcup _{t \in \mathbb {R}_+}(\lbrace t \rbrace \times \Omega _t); \partial _t - \Delta \big )$ and $v \in H^{1, \frac{1}{2}}_{;,0}\big ( \bigcup _{- \infty< t < t_0}(\lbrace t \rbrace \times \Omega _t); \partial _t + \Delta \big )$. Then, for $t_0>0$, there holds Green’s second formula

  $$\begin{aligned} \int _0^{t_0} \int _{\Omega _t} \big \{(\partial _t - \Delta ) u v + u (\partial _t + \Delta ) v\big \} \, \mathrm {d}\mathbf {x}\mathrm {d}t = \langle \gamma _0 u, \gamma _1^+ v \rangle - \langle \gamma _1^- u, \gamma _0 v \rangle . \end{aligned}$$

Proof

Statements (i) and (ii) are clear. Statement (iii) follows then immediately from these by interchanging v and u in (ii) and using (4.5). $\square$

We need the tube equivalent of [3, Lemma 2.22]. In there, the space $\widetilde{C}^{\infty } (\overline{Q}_0) \mathrel {\mathrel {\mathop :}=}C_0^\infty \big ((0,T] \times \overline{\Omega }_0 \big )$ is defined as the space of the restrictions of functions in $C_0^\infty (\mathbb {R}_+ \times \mathbb {R}^d)$ to $\overline{Q}_0$. This space $\widetilde{C}^\infty (\overline{Q}_0)$ is dense in $H^{1, \frac{1}{2}}_{;0,} (Q_0; \partial _t - \Delta )$ according to [3, Lemma 2.22]. As we only consider a $C^2$-mapping between the reference cylinder and the tube, we will prove the analogue result only for $C^2$-functions.

Lemma 4.9

The space

$$\begin{aligned} \widetilde{C}^2 (\overline{Q}_T) \mathrel {\mathrel {\mathop :}=}\Big \lbrace u :u \circ {\varvec{\kappa }}\in C^2_0 \big ( (0,T] \times \overline{\Omega }_0 \big ) \Big \rbrace \end{aligned}$$

is dense in $H^{1, \frac{1}{2}}_{;0,} (Q_T; \partial _t - \Delta )$.

Proof

We modify the proof of [3, Lemma 2.22] appropriately. According to this proof, $C_0^\infty \big ( (0,T] \times \overline{\Omega }_0 \big )$ is dense in $H^{1,\frac{1}{2}}_{;0,} (Q_0)$. Therefore, also $C_0^2 \big ((0,T] \times \overline{\Omega }_0 \big )$ is dense in $H^{1,\frac{1}{2}}_{;0,} (Q_0)$. Due to the definition of the spaces on the tube via the mapping ${\varvec{\kappa }}$ and the resulting equivalence of norms, we also obtain that $\widetilde{C}^2 (\overline{Q}_T)$ is dense in $H^{1,\frac{1}{2}}_{;0,} (Q_T)$. Similarly, we obtain that

$$\begin{aligned} \widetilde{C}^2 (Q_T) \mathrel {\mathrel {\mathop :}=}\Big \lbrace u :u \circ {\varvec{\kappa }}\in C_0^2 \big ( (0,T] \times \Omega _0 \big ) \Big \rbrace \end{aligned}$$

is dense in $H^{1,\frac{1}{2}}_{0;0,} (Q_T)$.

Let $\mathcal {R}$ be an extension operator from $H^{1, \frac{1}{2}}_{;0,} (Q_T)$ to $H^{1, \frac{1}{2}} (\mathbb {R}^{d+1})$. It thus holds $(\mathcal {R}u)|_{Q_T} = u$. As in [3], let us choose $\mathcal {R}$ such that ${\text {supp}}\mathcal {R}\subset [0,\infty ) \times \mathbb {R}^d$.¹ In that way, we can identify $H^{1, \frac{1}{2}}_{;0,} (Q_T)$ with a closed subspace of $H^{1, \frac{1}{2}}_{;0,} (\mathbb {R}_+ \times \mathbb {R}^d)$. The map $u \mapsto \big ( \mathcal {R}u, (\partial _t - \Delta ) u \big )$ identifies $H^{1, \frac{1}{2}}_{;0,} (Q_T; \partial _t - \Delta )$ with a closed subspace of $H^{1, \frac{1}{2}}_{;0,} (\mathbb {R}_+ \times \mathbb {R}^d) \times L^2 (\mathbb {R}_+ \times \mathbb {R}^d)$. Due to this identification, we find for every bounded linear functional $\ell :H^{1, \frac{1}{2}}_{;0,} (Q_T; \partial _t - \Delta )\rightarrow \mathbb {R}$ some $f \in \big ( H^{1, \frac{1}{2}}_{;0,} (\mathbb {R}_+ \times \mathbb {R}^d) \big )' = H^{-1,-\frac{1}{2}} (\mathbb {R}_+ \times \mathbb {R}^d)$ and $g \in L^2 (\mathbb {R}_+ \times \mathbb {R}^d)$ such that it holds

$$\begin{aligned} \langle \ell , u \rangle = \langle f, \mathcal {R}u \rangle + \int _{\mathbb {R}_+ \times \mathbb {R}^d} g (\partial _t - \Delta ) u \, \mathrm {d}(t,\mathbf {x}) \end{aligned}$$

for all $u \in H^{1, \frac{1}{2}}_{;0,} (Q_T; \partial _t - \Delta )$. Since $\ell$ acts only on u, which is supported on $\overline{Q}_T$, we may assume that ${\text {supp}}f \subset \overline{Q}_T$ and ${\text {supp}}g \subset \overline{Q}_T$.

We shall suppose next that it holds $\langle \ell , \varphi \rangle = 0$ for all $\varphi \in \widetilde{C}^2 (\overline{Q}_T)$. If we can show $\ell = 0$, we obtain the desired density result in accordance with [19, Korollar III.1.9]. For all $\varphi \in C_0^2(\mathbb {R}_+ \times \mathbb {R}^d)$, we conclude

$$\begin{aligned} 0 = \langle \ell , \varphi \rangle = \langle f, \varphi \rangle + \int _{Q_T} g (\partial _t - \Delta ) \varphi \,\mathrm {d}(t,\mathbf {x}) = \langle f, \varphi \rangle + \int _{\mathbb {R}_+ \times \mathbb {R}^d} g(\partial _t - \Delta ) \varphi \,\mathrm {d}(t,\mathbf {x}). \end{aligned}$$

This equation states that $f = (\partial _t + \Delta ) g$ holds on $\mathbb {R}_+ \times \mathbb {R}^d$ in complete analogy to [3]. Due to $f \in H^{-1,-\frac{1}{2}} (\mathbb {R}_+ \times \mathbb {R}^d)$ and the differential operator, we find $g \in H^{1, \frac{1}{2}} (\mathbb {R}_+ \times \mathbb {R}^d)$ and, thus, $g|_{Q_T} \in H^{1, \frac{1}{2}}_{0;,0} (Q_T)$.

On a cylindrical domain, any function $h\in H^{1, \frac{1}{2}}_{0;,0} (Q_0)$ can be approximated by a series $h_n \in C_0^\infty \big ( (-\infty ,T) \times \Omega _0\big )$ (see [3, Proof of Lemma 2.22]). Hence, by choosing $h\mathrel {\mathrel {\mathop :}=}g\circ {\varvec{\kappa }}$ and setting $g_n \mathrel {\mathrel {\mathop :}=}h_n\circ {\varvec{\kappa }}^{-1}$, we can approximate $g|_{Q_T}\in H^{1, \frac{1}{2}}_{0;,0}(Q_T)$ by a series $g_n\in C_0^2 \big (\bigcup _{-\infty< t< T}(\lbrace t \rbrace \times \Omega _t)\big )$ in the norm of $H^{1, \frac{1}{2}}\big (\bigcup _{0< t< \infty }(\lbrace t \rbrace \times \Omega _t)\big )$. Thus, denoting by $\hat{g}_n$ the extension of $g_n$ by zero outside of $Q_T$, we find $(\partial _t + \Delta ) \hat{g}_n \rightarrow f$ in $H^{-1, -\frac{1}{2}} (\mathbb {R}_+ \times \mathbb {R}^d)$. We then conclude for any $u \in H^{1, \frac{1}{2}}_{;0,} (Q_T; \partial _t - \Delta )$ that

$$\begin{aligned} \langle \ell , u \rangle&= \lim _{n \rightarrow \infty } \left[ \big \langle (\partial _t + \Delta ) \hat{g}_n, \mathcal {R}u \big \rangle + \int _{Q_T} g_n (\partial _t - \Delta ) u \, \mathrm {d}(t,\mathbf {x}) \right] \\&= \lim _{n \rightarrow \infty } \left[ \int _{Q_T} (\partial _t + \Delta ) g_n u \, \mathrm {d}(t,\mathbf {x}) + \int _{Q_T} g_n (\partial _t - \Delta ) u \, \mathrm {d}(t,\mathbf {x}) \right] = 0. \end{aligned}$$

The expression above is equal to zero, because $u = 0$ for $t =0$, $g_n = 0$ for $t=T$, and $g_n$ has a zero boundary condition. $\square$

Remark 4.10

If we consider $t \mapsto T-t$ in Lemma 4.9, we obtain that $\widehat{C}^2 (\overline{Q}^T)$ is dense in $H^{1, \frac{1}{2}}_{;,0} (Q^T; - \partial _t - \Delta )$, where

$$\begin{aligned} \widehat{C}^2 (\overline{Q}_T) \mathrel {\mathrel {\mathop :}=}\Big \lbrace u :u \circ {\varvec{\kappa }}\in C^2_0 \big ( [0,T) \times \overline{\Omega }_0 \big ) \Big \rbrace \end{aligned}$$

and $Q^T$ is the time flipped $Q_T$. Hence, $\widehat{C}^2 (\overline{Q}_T)$ is dense in $H^{1, \frac{1}{2}}_{;,0} (Q_T; \partial _t + \Delta )$ as $Q_T$ was arbitrary.

Next, we will introduce a lemma concerning the trace maps, which will be later used in the proof of the jump relations. It is the analogue of [3, Lemma 2.23].

Lemma 4.11

The combined trace map $(\gamma _0, \gamma _1^+) :u \mapsto (\gamma _0 u, \gamma _1^+ u)$ maps $\widehat{C}^2 (\overline{Q}_T)$ onto a dense subspace of $H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T) \times H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T)$.

Proof

We mimic the respective proof from [3], but will not use a time reversal map. Let us assume a linear functional $(\chi , \psi ) \in H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T) \times H^{-\frac{1}{2}, -\frac{1}{4}}(\Sigma _T)$ that vanishes on the range of $(\gamma _1^+, \gamma _0)$. We need to show that $(\chi , \psi ) = (0,0)$, since then the density follows by [19, Korollar III.1.9]. To this end, assume

$$\begin{aligned} \langle \chi , \gamma _1^+ \varphi \rangle = \langle \psi , \gamma _0 \varphi \rangle \quad \text {for all } \varphi \in \widehat{C}^2 (\overline{Q}_T). \end{aligned}$$

(4.7)

Let

$$\begin{aligned} \mathcal {T}= (g \mapsto \mathcal {T}g) :H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T) \rightarrow H^{1, \frac{1}{2}}_{;0,} (Q_T) \end{aligned}$$

be the solution operator (see Theorem 3.4) of the Dirichlet problem

$$\begin{aligned} \begin{aligned} (\partial _t - \Delta ) (\mathcal {T}g)&= 0&\ \ \text {in } Q_T, \\ \gamma _0 (\mathcal {T}g)&= g&\ \ \text {on } \Sigma _T. \end{aligned} \end{aligned}$$

(4.8)

Moreover, let

$$\begin{aligned} \mathcal {S}= (f \mapsto \mathcal {S}f) :L^2 (Q_T) \rightarrow H^{1, \frac{1}{2}}_{0;,0} (Q_T) \end{aligned}$$

be the solution operator (see Theorem 3.4 used for the substitution $t \mapsto T-t$) of the Dirichlet problem

$$\begin{aligned} \begin{aligned} (\partial _t + \Delta ) (\mathcal {S}f)&= f&\ \ \text {in } Q_T, \\ \gamma _0 (\mathcal {S}f)&= 0&\ \ \text {on } \Sigma _T. \end{aligned} \end{aligned}$$

We can apply Green’s second formula from Lemma 4.8 to $u \mathrel {\mathrel {\mathop :}=}\mathcal {T}\chi$ and $v \mathrel {\mathrel {\mathop :}=}\mathcal {S}f$ for any $f \in L^2 (Q_T)$, since $u \in H^{1, \frac{1}{2}}_{;0,} (Q_T; \partial _t - \Delta )$ and $(\partial _t + \Delta ) v \in L^2 (Q_T)$. We obtain

$$\begin{aligned} \int _{Q_T} \big \{(\partial _t - \Delta ) u v + u (\partial _t + \Delta ) v \big \}\, \mathrm {d}(\mathbf {x}, t) = \langle \gamma _0 u, \gamma _1^+ v \rangle - \langle \gamma _1^- u, \gamma _0 v \rangle . \end{aligned}$$

Since $\gamma _0 v = 0$ and $\gamma _0 u = \chi$, as well as $(\partial _t - \Delta ) u = 0$ and $(\partial _t + \Delta ) v = f$, we obtain

$$\begin{aligned} \int _{Q_T} u f \, \mathrm {d}(t,\mathbf {x}) = \langle \chi , \gamma _1^+ v \rangle . \end{aligned}$$

Due to continuity and Remark 4.10, (4.7) holds also for all $\varphi \in H^{1, \frac{1}{2}}_{;,0} (Q_T;$ $\partial _t + \Delta )$ and, thus, also for $\varphi = \mathcal {S}f$. This implies

$$\begin{aligned} \langle \chi , \gamma _1^+ \mathcal {S}f \rangle = \langle \psi , \gamma _0 \mathcal {S}f \rangle . \end{aligned}$$

Since $\gamma _0 \mathcal {S}f = 0$, we thus obtain $\int _{Q_T} u f \, \mathrm {d}(t,\mathbf {x}) = 0$ for all $f \in L^2 (Q_T)$. Therefore, $0 = u = \mathcal {T}( \chi )$ and thus $\chi = \gamma _0 u = 0$. Looking again at (4.7) gives

$$\begin{aligned} \langle \psi , \gamma _0 \varphi \rangle = 0 \quad \text {for all } \varphi \in H^{1, \frac{1}{2}}_{;,0} (Q_T). \end{aligned}$$

The trace map $\gamma _0$ is not only surjective for $\varphi \in H_{;0,}^{1, \frac{1}{2}}(Q_T)$ as shown in Sect. 3.2, but also for $\varphi \in H_{;,0}^{1, \frac{1}{2}} (Q_T)$ if one considers the backward problem. We may hence conclude that $\psi = 0$. $\square$

Subsequently, we state the analogue of [3, Proposition 2.24].

Lemma 4.12

Green’s first formula given in (4.6) holds for all $u \in H^{1, \frac{1}{2}}_{;0,} (Q_T;$ $\partial _t - \Delta )$ and $v \in H^{1, \frac{1}{2}}_{;0,} (Q_T)$. If also $v \in H^{1, \frac{1}{2}}_{;0,} (Q_T; \partial _t - \Delta )$, we can write the Green’s formula as

$$\begin{aligned} \begin{aligned}&\int _{Q_T} \langle \nabla u, \nabla v \rangle \, \mathrm {d}(t,\mathbf {x}) - d(v,u) + \int _{\Omega _T} u(T, \mathbf {x}) v(T, \mathbf {x}) \, \mathrm {d}\mathbf {x}\\& = \langle \gamma _1^- u, \gamma _0 v \rangle + \int _{Q_T} (\partial _t - \Delta ) u v \, \mathrm {d}(t,\mathbf {x}). \end{aligned} \end{aligned}$$

(4.9)

Proof

We adopt the proof of [3, Proposition 2.24]. Given $u \in \widetilde{C}^2 (\overline{Q}_T)$ and $v \in \widetilde{C}^1(\overline{Q}_T)$², we find

$$\begin{aligned} \begin{aligned}&\int _{Q_T} \langle \nabla u, \nabla v \rangle \, \mathrm {d}(t,\mathbf {x}) + \int _{Q_T} \partial _t u v \, \mathrm {d}(t,\mathbf {x}) + \frac{1}{2} \int _0^T \int _{\Gamma _t} \langle \mathbf {V}, \mathbf {n}\rangle u v \mathrm {d}\sigma \mathrm {d}t \\& = \langle \gamma _1^- u, \gamma _0 v \rangle + \int _{Q_T} (\partial _t - \Delta ) u v \, \mathrm {d}(t,\mathbf {x}). \end{aligned} \end{aligned}$$

(4.10)

All terms are continuous with respect to v in the $H^{1,\frac{1}{2}}_{;0,} (Q_T)$-norm. Thus, by continuity, we can extend (4.10) to all $v \in H^{1, \frac{1}{2}}_{;0,} (Q_T)$. Let $v \in H^{1, \frac{1}{2}}_{;0,} (Q_T)$ be fixed. Then, all terms in (4.10) except the term containing the $\partial _t u$ are obviously continuous with respect to u in the norm of $H^{1, \frac{1}{2}}_{;0,} (Q; \partial _t- \Delta )$. Therefore, also the term containing $\partial _t u$ is continuous. Lemma 4.9 allows to extend (4.10) to all $u \in H^{1, \frac{1}{2}}_{;0,} (Q_T; \partial _t - \Delta )$. Thus, Green’s first formula holds as given in the claim.

For $u, v \in \widetilde{C}^2 (\overline{Q}_T)$, (4.9) holds. As in [3], the term $\int _{\Omega _T} u(T, \mathbf {x}) v(T, \mathbf {x})$ $\mathrm {d}(t,\mathbf {x})$ is continuous for u and v in the norm of $H^{1,\frac{1}{2}}_{;0,} (Q_T; \partial _t - \Delta ) \subset \mathcal {V}(Q_T)$ and $\mathcal {V}(Q_T)$ consists of functions $\varphi$, which always satisfy $\varphi \circ {\varvec{\kappa }}\in C \big ( [0,T]; L^2 (\Omega _0) \big )$. From here, the second claim follows. $\square$

5 The Calderón Operator

Let us first introduce the fundamental solution for the heat equation, which, in accordance with e.g. [18], reads

$$\begin{aligned} G (t, \tau , \mathbf {x}, \mathbf {y}) \mathrel {\mathrel {\mathop :}=}{\left\{ \begin{array}{ll} \frac{1}{(4\pi (t-\tau ))^{d/2}} \exp \left( - \frac{ \Vert \mathbf {x}- \mathbf {y}\Vert ^2}{4 (t-\tau )}\right) ,\ \ &{} \text { if } \tau < t, \\ 0, \ \ &{} \text { if } \tau \ge t.\end{array}\right. } \end{aligned}$$

Notice that this is equivalent to consider $\overline{G} (t-\tau , \mathbf {x}, \mathbf {y})$, where $\overline{G}$ is given by

$$\begin{aligned} \overline{G} (\tau , \mathbf {x}, \mathbf {y})\mathrel {\mathrel {\mathop :}=}\frac{1}{(4 \pi \tau )^{d/2}} \exp \left( -\frac{ \Vert \mathbf {x}- \mathbf {y}\Vert ^2}{4 \tau } \right) \frac{1}{2} (1 + {\text {sign}} \tau ), \end{aligned}$$

as introduced in [3, Formula (2.39)]. Moreover, let us denote

$$\begin{aligned} \widetilde{G}(t, \mathbf {x}) = {\left\{ \begin{array}{ll} \frac{1}{(4\pi t)^{d/2}} \exp \left( -\frac{\Vert \mathbf {x}\Vert ^2}{4 t}\right) ,\ \ &{} \text { if } t > 0, \\ 0, \ \ &{} \text { if } t \le 0. \end{array}\right. } \end{aligned}$$

(5.1)

For $u \in C^2\big (\bigcup _{0< t< \infty }(\lbrace t \rbrace \times \overline{\Omega }_t)\big )$ with $u(0, \mathbf {x}) = 0$ on $\Omega _0$, we have for $(t_0,\mathbf {x}_0) \in \bigcup _{0< t< \infty }( \lbrace t \rbrace \times \Omega _t)$ that

$$\begin{aligned} u (t_0, \mathbf {x}_0)&= \int _{Q_T} \int _{\Omega _t} (\partial _t - \Delta ) u (t, \mathbf {x}) \widetilde{G}(t_0 - t, \mathbf {x}_0 - \mathbf {x}) \, \mathrm {d}(t,\mathbf {x})\\&\quad + \int _{\Sigma _T} \widetilde{G}( t_0 - t, \mathbf {x}_0 - \mathbf {x}) \frac{\partial u}{\partial \mathbf {n}} (t, \mathbf {x}) \,\mathrm {d}\sigma _{(t,\mathbf {x})} \\&\quad - \int _{\Sigma _T} \frac{\partial \widetilde{G}}{\partial \mathbf {n}} (t_0 - t, \mathbf {x}_0 - \mathbf {x}) u (t, \mathbf {x}) \, \mathrm {d}\sigma _{(t,\mathbf {x})} \\&\quad + \int _{\Sigma _T} \widetilde{G}( t_0 - t, \mathbf {x}_0 - \mathbf {x}) \langle \mathbf {V}, \mathbf {n}\rangle (t, \mathbf {x}) u (\mathbf {y}, \tau ) \mathrm {d}\sigma _{(t,\mathbf {x})}, \end{aligned}$$

as it can be seen from Lemma 4.8 and the property of the fundamental solution. Moreover, we will only look at the case, for which $(\partial _t - \Delta ) u = 0$ holds.

We introduce the single and double layer potentials as

$$\begin{aligned} \widetilde{\mathcal {V}} \varphi (t_0, \mathbf {x}_0)&\mathrel {\mathrel {\mathop :}=}\langle \varphi , \gamma _0 G \rangle = \int _{\Sigma _T} G (t_0, t, \mathbf {x}_0, \mathbf {y}) \varphi (t, \mathbf {y}) \, \mathrm {d}\sigma _{(t,\mathbf {y})}, \\ \widetilde{\mathcal {K}} w (t_0, \mathbf {x}_0)&\mathrel {\mathrel {\mathop :}=}\langle \gamma _1^+ G, w \rangle = \int _{\Sigma _T} \gamma _{1, (t, \mathbf {y})}^+ G (t_0, t, \mathbf {x}_0, \mathbf {y}) w(t, \mathbf {y}) \, \mathrm {d}\sigma _{(t,\mathbf {y})}. \end{aligned}$$

Then, similarly to [3, Theorem 2.20], we obtain the representation formula from [18, Equation (6)] given in the following lemma.

Lemma 5.1

Let $u \in H^{1, \frac{1}{2}} (Q_T)$ with $(\partial _t - \Delta ) u = 0$ in $Q_T$. Then, we have the representation formula

$$\begin{aligned} u (t, \widetilde{\mathbf {x}}) = \widetilde{\mathcal {V}} \gamma _1^- u (t, \widetilde{\mathbf {x}}) - \widetilde{\mathcal {K}} u (t, \widetilde{\mathbf {x}}) \quad \text {for all } (t, \widetilde{\mathbf {x}}) \in Q_T. \end{aligned}$$

(5.2)

As in [3, pg. 514], we can rewrite the definition of the single layer potential by

$$\begin{aligned} \begin{aligned} \widetilde{\mathcal {V}} \varphi (t_0, \mathbf {x}_0)&= \langle \varphi , \gamma _0 G(t, t_0, \mathbf {x}, \mathbf {x}_0) \rangle \\&= \langle \gamma _0 ' \varphi , G(t, t_0, \mathbf {x}, \mathbf {x}_0) \rangle \\&= \widetilde{G} \star (\gamma _0 ' \varphi ) (t_0, \mathbf {x}_0), \end{aligned} \end{aligned}$$

(5.3)

where $\widetilde{G}$ is given in (5.1) and

$$\begin{aligned} \langle \gamma _0 ' \varphi , \chi \rangle = \langle \varphi , \gamma _0 \chi \rangle = \int _{\Sigma _T} \varphi \chi \, \mathrm {d}\sigma _{(t, \mathbf {x})} \end{aligned}$$

for all $\chi \in C_0^\infty (\mathbb {R}^{d+1})$. We will use this also for $\chi \in C^2_0 (\mathbb {R}^{d+1})$.

We would like to find the mapping properties of the single and double layer potentials, which are the equivalent of the results given in [3, Proposition 3.1, Remark 3.2, and Proposition 3.3].

Lemma 5.2

The mappings

$$\begin{aligned}&\widetilde{\mathcal {V}} :H^{-\frac{1}{2},-\frac{1}{4}} (\Sigma _T) \rightarrow H^{1, \frac{1}{2}}_{;0,} (Q_T; \partial _t - \Delta ),\\&\widetilde{\mathcal {K}} :H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T) \rightarrow H^{1, \frac{1}{2}}_{;0,} (Q_T; \partial _t - \Delta ) \end{aligned}$$

are continuous.

Proof

The statement for the single layer potential follows as in [3, pg. 514–515] in the case of a cylindrical domain. In there, the claim is proven by considering the problem on $\mathbb {R}^{d+1}$ using Fourier techniques and then restricting it appropriately, which can also be done in the case of a non-cylindrical domain. The statement for the double layer potential is derived in complete analogy to [3, pg. 515]. $\square$

We can take the traces $\gamma _0$ of the single and double layer potential. Let the radius R be large enough such that the boundary $\Gamma _t$ is contained in the ball $B_R \mathrel {\mathrel {\mathop :}=}\big \lbrace \mathbf {x}\in \mathbb {R}^d :\Vert \mathbf {x}\Vert < R \big \rbrace$ and set $\Omega _t^c \mathrel {\mathrel {\mathop :}=}B_R \backslash \overline{\Omega }_t$ and $Q_T^c \mathrel {\mathrel {\mathop :}=}\bigcup _{-\infty< t< T}(\lbrace t \rbrace \times \Omega _t^c)$. Lemma 5.2 provides also the continuity of the mappings

$$\begin{aligned}&\widetilde{\mathcal {V}} :H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T) \rightarrow H^{1, \frac{1}{2}}_{;0,} (Q_T^c; \partial _t - \Delta ),\\&\widetilde{\mathcal {K}} :H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T) \rightarrow H^{1, \frac{1}{2}}_{;0,} (Q_T^c; \partial _t - \Delta ). \end{aligned}$$

In order to state the tube analogue of [3, Theorem 3.4], we define the jumps as in [3, Formula (3.16)] in accordance with

$$\begin{aligned}{}[\gamma _0 u] \mathrel {\mathrel {\mathop :}=}\gamma _0(u|_{Q_T^c}) - \gamma _0 (u|_{Q_T}), \quad [\gamma _1^\pm u ] \mathrel {\mathrel {\mathop :}=}\gamma _1^\pm (u|_{Q_T^c}) - \gamma _1^\pm (u|_{Q_T}). \end{aligned}$$

We then have:

Lemma 5.3

For all $\psi \in H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T)$ and all $w \in H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T)$, there hold the jump relations

$$\begin{aligned} \begin{aligned}&[\gamma _0 \widetilde{\mathcal {V}} \psi ] = 0, \ \&[\gamma _1^- \widetilde{\mathcal {V}} \psi ] = - \psi , \\&[\gamma _0 \widetilde{\mathcal {K}} w ] = w, \ \&[\gamma _1^- \widetilde{\mathcal {K}} w] = 0. \end{aligned} \end{aligned}$$

Proof

We modify the proof of [3] by not using the time reversal map. Let $\psi \in H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T)$. We set $u \mathrel {\mathrel {\mathop :}=}\widetilde{\mathcal {V}}\psi$. Due to the mapping property of the single layer potential, we then have $u \in H^{1, \frac{1}{2}}_{;0,} \big ( (0,T) \times B_R (0) \big )$ and thus, by the trace lemma, we have $\gamma _0 (u|_{Q_T}) = \gamma _0 (u|_{Q_T^c})$.

Let us next consider the normal jump of $\widetilde{\mathcal {V}}$. From (5.3), we obtain by considering $u = \widetilde{\mathcal {V}} \psi$

$$\begin{aligned} (\partial _t - \Delta ) u = \gamma _0 ' \psi \end{aligned}$$

in $\mathbb {R}_+ \times \mathbb {R}^d$. We consider any test function $\varphi \in C^2_0 \big ( (0,T) \times B_R \big )$ and we obtain

$$\begin{aligned} \langle \psi , \gamma _0 \varphi \rangle = \langle \gamma _0' \psi , \varphi \rangle = \big \langle (\partial _t - \Delta ) u , \varphi \big \rangle = - \big \langle u, (\partial _t + \Delta ) \varphi \big \rangle , \end{aligned}$$

where the last equality holds due to the integration by parts on a cylindrical domain. We thus have

$$\begin{aligned} \langle \psi , \gamma _0 \varphi \rangle = - \int _{(0,T) \times B_R} (\partial _t + \Delta ) \varphi u \, \mathrm {d}(t,\mathbf {x}). \end{aligned}$$

(5.4)

On the other hand, we can use Green’s second formula, given in Lemma 4.8 in $Q_T$ and $Q_T^c$, where we use that $(\partial _t - \Delta ) u = 0$ in $Q_T \cup Q_T^c$. This yields

$$\begin{aligned} \int _{Q_T} (\partial _t + \Delta ) \varphi u \, \mathrm {d}(t,\mathbf {x}) = \langle \gamma _0 u, \gamma _1^+ \varphi \rangle - \langle \gamma _1^- u , \gamma _0 \varphi \rangle \end{aligned}$$

and

$$\begin{aligned} \int _{Q_T^c} (\partial _t + \Delta ) \varphi \ u \, \mathrm {d}(t,\mathbf {x}) = - \langle \gamma _0 u, \gamma _1^+ \varphi \rangle + \langle \gamma _1^- u, \gamma _0 \varphi \rangle . \end{aligned}$$

Adding these two expressions yields

$$\begin{aligned} \int _{(0,T) \times B_R} (\partial _t + \Delta ) \varphi u \, \mathrm {d}(t,\mathbf {x}) = \big \langle [ \gamma _1^- u ], \gamma _0 \varphi \rangle , \end{aligned}$$

(5.5)

where we used $[\gamma _0 u] = 0 = [\gamma _0 \varphi ] = [ \gamma _1^+ \varphi ]$. Comparing (5.4) with (5.5) results in $[\gamma _1^- u] = - \psi$.

We are left with proving the jump relations for the double layer potential. To that end, we choose $w \in H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T)$ and define $u \mathrel {\mathrel {\mathop :}=}\widetilde{\mathcal {K}} w$. Let $\varphi \in C_0^\infty (\mathbb {R}_+ \times B_R)$ be a test function. As above, we obtain

$$\begin{aligned} \int _{(0,T) \times B_R} (\partial _t + \Delta ) \varphi u \, \mathrm {d}(t,\mathbf {x}) = \big \langle [\gamma _1^- u], \gamma _0 \varphi \big \rangle - \big \langle [\gamma _0 u], \gamma _1^+ \varphi \big \rangle . \end{aligned}$$

(5.6)

For $\widetilde{\mathcal {K}}$, we obtain $\widetilde{\mathcal {K}} w = \widetilde{G} \star \big ( (\gamma _1^+)' w \big )$ similar to (5.3). Therefore, we have $(\partial _t - \Delta ) \widetilde{\mathcal {K}} w = (\gamma _1^+)' w$ in $\mathbb {R}_+ \times B_R$. From here, it follows that

$$\begin{aligned} - \int _{(0,T) \times B_R} (\partial _t + \Delta ) \varphi u \, \mathrm {d}(t,\mathbf {x})&= \big \langle (\partial _t - \Delta ) u, \varphi \big \rangle \nonumber \\&= \big \langle (\gamma _1^+)' w, \varphi \big \rangle = \langle w, \gamma _1^+ \varphi \rangle .\nonumber \\ \end{aligned}$$

(5.7)

Comparing (5.6) with (5.7) yields

$$\begin{aligned} \big \langle [\gamma _1^- u], \gamma _0 \varphi \big \rangle = \big \langle [\gamma _0 u] - w, \gamma _1^+ \varphi \big \rangle \end{aligned}$$

(5.8)

for all $\varphi \in C_0^2(\mathbb {R}_+ \times B_R)$. Applying Lemma 4.11 says that both sides of (5.8) have to vanish identically, from where $[\gamma _1^- u] =0$ and $[\gamma _0 u] = w$ follows. $\square$

Now, as in [3, Definition 3.5], we are in the position to define the boundary integral operators.

Definition 5.4

Let $\psi \in H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T)$ and $w\in H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T)$. We can then define the single layer operator as

$$\begin{aligned} \mathcal {V}\psi \mathrel {\mathrel {\mathop :}=}\gamma _0 \widetilde{\mathcal {V}} \psi , \end{aligned}$$

the adjoint double layer operator as

$$\begin{aligned} \mathcal {K}' \psi \mathrel {\mathrel {\mathop :}=}\frac{1}{2} \left( \gamma _1^- (\widetilde{\mathcal {V}} \psi )|_{Q_T} + \gamma _1^- (\widetilde{\mathcal {V}} \psi )|_{Q_T^c} \right) , \end{aligned}$$

the double layer operator as

$$\begin{aligned} \mathcal {K}w \mathrel {\mathrel {\mathop :}=}\frac{1}{2} \left( \gamma _0 (\widetilde{\mathcal {K}} w)|_{Q_T} + \gamma _0(\widetilde{\mathcal {K}} w)|_{Q_T^c} \right) \end{aligned}$$

and the hypersingular operator as

$$\begin{aligned} \mathcal {D}w \mathrel {\mathrel {\mathop :}=}- \gamma _1^- \widetilde{\mathcal {K}} w. \end{aligned}$$

Likewise to [3, Theorem 3.7], we may conclude the mapping properties of these operators.

Theorem 5.5

The boundary integral operators from Definition 5.4 are continuous mappings as follows

$$\begin{aligned} \mathcal {V}:H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T)&\rightarrow H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T), \\ \mathcal {K}' :H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T)&\rightarrow H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T), \\ \mathcal {K}:H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T)&\rightarrow H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T), \\ \mathcal {D}:H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T)&\rightarrow H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T). \end{aligned}$$

Proof

The assertion follows immediately by using the mapping properties of the layer potentials from Lemma 5.2 as well as of the trace operators introduced in Sect. 3.2 and from Lemma 4.6. $\square$

The analogue of [3, Formulae (3.24)–(3.27)] reads:

Lemma 5.6

It holds

$$\begin{aligned} \gamma _0 (\widetilde{\mathcal {V}} \psi ) |_{Q_T}&= \gamma _0 (\widetilde{\mathcal {V}} \psi )|_{Q_T^c} = \mathcal {V}\psi , \\ \gamma _1^- (\widetilde{\mathcal {V}} \psi )|_{Q_T}&= \frac{1}{2} \psi + \mathcal {K}' \psi , \\ \gamma _1^- (\widetilde{\mathcal {V}} \psi )|_{Q_T^c}&= - \frac{1}{2} \psi + \mathcal {K}' \psi , \\ \gamma _0 (\widetilde{\mathcal {K}} w)|_{Q_T}&= - \frac{1}{2} w + \mathcal {K}w, \\ \gamma _0 (\widetilde{\mathcal {K}} w)|_{Q_T^c}&= \frac{1}{2} w + \mathcal {K}w, \\ \gamma _1^- (\widetilde{\mathcal {K}} w)|_{Q_T}&= \gamma _1^- (\mathcal {K}w)|_{Q_T^c} = - \mathcal {D}w. \end{aligned}$$

Proof

We just prove the second statement, as the other statements follow similarly. According to Lemma 5.3, we have

$$\begin{aligned}{}[\gamma _1^- \widetilde{\mathcal {V}} \psi ] = \gamma _1^- \widetilde{\mathcal {V}} \psi |_{Q_T^c} - \gamma _1^- \widetilde{\mathcal {V}} \psi |_{Q_T} = - \psi . \end{aligned}$$

Therefore,

$$\begin{aligned} \gamma _1^- \widetilde{\mathcal {V}} \psi |_{Q_T} = \psi + \gamma _1^- \widetilde{\mathcal {V}} \psi |_{Q_T^c}. \end{aligned}$$

By Definition 5.4, we have

$$\begin{aligned} \mathcal {K}' \psi \mathrel {\mathrel {\mathop :}=}\frac{1}{2} \left( \gamma _1^- \widetilde{\mathcal {V}} \psi |_{Q_T} + \gamma _1^- \widetilde{\mathcal {V}} \psi |_{Q_T^c} \right) . \end{aligned}$$

Substituting this into the expression above yields

$$\begin{aligned} \gamma _1^- \widetilde{\mathcal {V}} \psi |_{Q_T} = \psi + 2 \mathcal {K}' \psi - \gamma _1^- \widetilde{\mathcal {V}} \psi |_{Q_T}, \end{aligned}$$

from where the claim follows immediately. $\square$

Remark 5.7

Following [18, Formulae (7)–(10)], the relations in the interior given in Lemma 5.6 can also be written as

$$\begin{aligned} \gamma _0 \widetilde{\mathcal {V}} \psi (t, \mathbf {x})&= \int _{\Sigma _T} G(t, \tau , \mathbf {x}, \mathbf {y}) \psi (\tau , \mathbf {y}) \, \mathrm {d}\sigma _{(\tau ,\mathbf {y})},\\ \gamma _1^- \widetilde{\mathcal {V}} \psi (t, \mathbf {x})&= \frac{1}{2} \psi (t, \mathbf {x}) + \int _{\Sigma _T} \gamma _{1, (t,\mathbf {x})}^- G(t, \tau , \mathbf {x}, \mathbf {y}) \psi (\tau , \mathbf {y}) \, \mathrm {d}\sigma _{(\tau ,\mathbf {y})},\\ \gamma _0 \widetilde{\mathcal {K}} w (t, \mathbf {x})&= -\frac{1}{2} w(t, \mathbf {x}) + \int _{\Sigma _T} \gamma _{1, (\tau , \mathbf {y})}^+ G(t, \tau , \mathbf {x}, \mathbf {y}) w (\tau , \mathbf {y}) \, \mathrm {d}\sigma _{(\tau ,\mathbf {y})}, \\ \gamma _1^- \widetilde{\mathcal {K}} w (t, \mathbf {x})&= - \int _{\Sigma _T} \gamma _{1, (t, \mathbf {x})}^- \gamma _{1, (\tau , \mathbf {y})}^+ G(t, \tau , \mathbf {x}, \mathbf {y}) w (\tau , \mathbf {y}) \, \mathrm {d}\sigma _{(\tau ,\mathbf {y})}. \end{aligned}$$

We can take the traces in the representation formula (5.2) to obtain the Dirichlet data and the Neumann data of the solution u of the homogeneous heat equation. This yields

$$\begin{aligned} \gamma _0 u= & {} \frac{1}{2} \gamma _0 u - \mathcal {K}\gamma _0 u + \mathcal {V}\gamma _1^- u, \end{aligned}$$

(5.9)

$$\begin{aligned} \gamma _1^- u= & {} \mathcal {D}\gamma _0 u + \frac{1}{2} \gamma _1^- u + \mathcal {K}' \gamma _1^- u, \end{aligned}$$

(5.10)

compare also [18, Formulae (11) and (12)].

We define the Calderón projector and the associated involution $\mathcal {A}$ as

$$\begin{aligned} \mathcal {C}_{Q_T} \mathrel {\mathrel {\mathop :}=}\frac{1}{2} {\text {id}} + \mathcal {A}\mathrel {\mathrel {\mathop :}=}\frac{1}{2} {\text {id}} + \begin{bmatrix} - \mathcal {K}&{} \mathcal {V}\\ \mathcal {D}&{} \mathcal {K}' \end{bmatrix} \end{aligned}$$

and state the analogue of [3, Theorem 3.9] and [3, Corollary 3.10], the proof of which stays the same.

Theorem 5.8

The operator $\mathcal {C}_{Q_T}$ is a projection operator in the space

$$\begin{aligned} \mathcal {H}\mathrel {\mathrel {\mathop :}=}H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T) \times H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T) \end{aligned}$$

while the operator $\mathcal {A}:\mathcal {H}\rightarrow \mathcal {H}$ is an isomorphism, satisfying

$$\begin{aligned} \mathcal {A}^2 = \frac{1}{4} {\text {id}}. \end{aligned}$$

(5.11)

Moreover, the following statements are equivalent for $(w, \psi ) \in \mathcal {H}$:

• (i) There is a $u \in H^{1, \frac{1}{2}}_{;0,} (Q_T)$ with $(\partial _t - \Delta ) u = 0$ in $Q_T$ and $w = \gamma _0 u$, $\psi = \gamma _1^- u$ on $\Sigma _T$.

• (ii) It holds

  $$\begin{aligned} \begin{bmatrix} w \\ \psi \end{bmatrix} = \mathcal {C}_{Q_T} \begin{bmatrix}w \\ \psi \end{bmatrix}. \end{aligned}$$

As in [3], we can interchange the columns of the operator $\mathcal {A}$ to define the operator

$$\begin{aligned} A \mathrel {\mathrel {\mathop :}=}\begin{bmatrix} \mathcal {V}&{} - \mathcal {K}\\ \mathcal {K}' &{} \mathcal {D}\end{bmatrix}, \end{aligned}$$

which is an isomorphism of the space

$$\begin{aligned} \mathcal {H}' \mathrel {\mathrel {\mathop :}=}H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T) \times H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T) \end{aligned}$$

onto its dual space $\mathcal {H}$. Furthermore, we define the duality product between $\mathcal {H}'$ and $\mathcal {H}$ according to

$$\begin{aligned} \left\langle \begin{bmatrix} \psi \\ w \end{bmatrix}, \begin{bmatrix} v \\ \varphi \end{bmatrix} \right\rangle \mathrel {\mathrel {\mathop :}=}\langle \psi , v \rangle + \langle \varphi , w \rangle \end{aligned}$$

for all v, $w \in H^{\frac{1}{2}, \frac{1}{4}}(\Sigma _T)$ and $\varphi$, $\psi \in H^{-\frac{1}{2}, - \frac{1}{4}} (\Sigma _T)$. We are now in the position to state the analogue of [3, Theorem 3.11], which is the positive definiteness of the operator A.

Theorem 5.9

There exists a constant $\alpha > 0$ such that

$$\begin{aligned} \left\langle \begin{bmatrix} \psi \\ w \end{bmatrix}, A \begin{bmatrix} \psi \\ w \end{bmatrix} \right\rangle \ge \alpha \left( \Vert \psi \Vert _{H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T)}^2 + \Vert w \Vert _{H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T)}^2 \right) \end{aligned}$$

for all $[\psi , w]^{\intercal } \in \mathcal {H}'$.

Proof

The proof of this theorem follows in complete analogy to the proof of [3, Theorem 3.11], when one uses Lemma 5.10 below which constitutes the tube analogue of [3, Lemma 2.15].³$\square$

Lemma 5.10

Let $u \in \mathcal {V}(Q_T)$ such that $(\partial _t - \Delta ) u = 0$ in $Q_T$. Then, there exist constants $m_1$, $m_2$, and $m_3$ such that

$$\begin{aligned} \Vert u \Vert _{H^{1, 0} (Q_T)} \le m_1 \Vert u\Vert _{H^{1, \frac{1}{2}} (Q_T)} \le m_2 \Vert u \Vert _{\mathcal {V}(Q_T)} \le m_3 \Vert u \Vert _{H^{1, 0}(Q_T)}. \end{aligned}$$

In other words, for functions $u \in \mathcal {V}(Q_T)$ satisfying the homogeneous heat equation, we have the equivalence of the norms in $\mathcal {V}(Q_T)$, $H^{1, 0} (Q_T)$, and $H^{1,\frac{1}{2}} (Q_T)$.

Proof

The first and second inequality follow directly from the equivalence of norms on $Q_T$ and $Q_0$, since the proof of [3] is based on the definition of the norm for the first inequality and the interpolation result (2.5) for the second inequality. Nonetheless, we cannot apply the equivalence of norms on $Q_T$ and norms on $Q_0$ directly for the third inequality, because we have $(\partial _t - \Delta )u = 0$ as an assumption, which is needed to show the third inequality. Mapping this differential operator from the tube onto the cylinder or vice versa will alter it. Therefore, we use the ideas of the proof of [3, Lemma 2.15], but adapt them to our context.

Transforming the partial differential equation $(\partial _t - \Delta ) u = 0$ from $Q_T$ back to $Q_0$ via the weak formulation (see [1]) yields

$$\begin{aligned} \partial _t (u \circ {\varvec{\kappa }}) -\mathcal {M}(u \circ {\varvec{\kappa }}) = 0\ \ \text {in } Q_0, \end{aligned}$$

where $\mathcal {M}$ is defined as

$$\begin{aligned} \mathcal {M}(u \circ {\varvec{\kappa }}) \mathrel {\mathrel {\mathop :}=}&{\text {div}} \Big ( ({\text {D}}\!{\varvec{\kappa }})^{-1} ({\text {D}}\!{\varvec{\kappa }})^{-\intercal } \nabla (u \circ {\varvec{\kappa }}) \Big ) + ({\text {D}}\!{\varvec{\kappa }})^{-\intercal } \nabla (u \circ {\varvec{\kappa }}) \cdot \partial _t {\varvec{\kappa }}\\&\quad + \frac{1}{\det ({\text {D}}\!{\varvec{\kappa }})} \nabla \big (\det ({\text {D}}\!{\varvec{\kappa }}) \big ) \cdot ({\text {D}}\!{\varvec{\kappa }})^{-1} ({\text {D}}\!{\varvec{\kappa }})^{- \intercal } \nabla (u \circ {\varvec{\kappa }}). \end{aligned}$$

By the standard theory, for fixed $t \in (0,T)$, we have that $\mathcal {M}:H^1 (\Omega _0) \rightarrow H^{-1}(\Omega _0)$ is bounded. Thus, for $u \in H^{1,0} (Q_0)$, we obtain $\mathcal {M}u\in H^{-1, 0}(Q_0)\mathrel {\mathrel {\mathop :}=}[H^{1,0} (Q_0)]'$ and we conclude

$$\begin{aligned} \Vert u \Vert _{\mathcal {V}(Q_T)}^2 = \Vert u \circ {\varvec{\kappa }}\Vert _{\mathcal {V}(Q_0)}^2&= \Vert u\circ {\varvec{\kappa }}\Vert _{H^{1,0}(Q_0)}^2 + \big \Vert \partial _t (u\circ {\varvec{\kappa }}) \big \Vert _{H^{-1,0}(Q_0)}^2 \\&= \Vert u \circ {\varvec{\kappa }}\Vert _{H^{1,0}(Q_0)}^2 + \big \Vert \mathcal {M}(u\circ {\varvec{\kappa }}) \big \Vert _{H^{-1,0}(Q_0)}^2\\&\lesssim \Vert u \circ {\varvec{\kappa }}\Vert _{H^{1,0}(Q_0)}^2 \\&= \Vert u \Vert _{H^{1,0}(Q_T)}^2. \end{aligned}$$

$\square$

Having the main result Theorem 5.9 at hand, we can state a few corollaries along the lines of [3, Corollary 3.13, Corollary 3.14, Remark 3.15, Corollary 3.16, Corollary 3.17].

Corollary 5.11

The single layer operator

$$\begin{aligned} \mathcal {V}:H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T) \rightarrow H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T) \end{aligned}$$

is an isomorphism and there exists $\alpha > 0$ such that

$$\begin{aligned} \langle \mathcal {V}\psi , \psi \rangle \ge \alpha \Vert \psi \Vert ^2_{H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T)} \quad \text {for all } \psi \in H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T). \end{aligned}$$

(5.12)

The hypersingular operator

$$\begin{aligned} \mathcal {D}:H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T) \rightarrow H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T) \end{aligned}$$

is an isomorphism and there exists $\alpha > 0$ such that

$$\begin{aligned} \langle \mathcal {D}w, w \rangle \ge \alpha \Vert w \Vert ^2_{H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T)} \quad \text {for all } w \in H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T). \end{aligned}$$

(5.13)

Proof

As in the proof of [3, Corollary 3.13], the coercivity estimates (5.12) and (5.13) result from Theorem 5.9 by using the special cases $w = 0$ and $\psi = 0$, respectively. In view of the continuity of $\mathcal {V}$ and $\mathcal {D}$, this leads to the invertibility of the operators. $\square$

Corollary 5.12

The operators

$$\begin{aligned} \frac{1}{2} {\text {id}} + \mathcal {K}, \ \ \frac{1}{2} {\text {id}} - \mathcal {K}&:H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T) \rightarrow H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T), \\ \frac{1}{2} {\text {id}} + \mathcal {K}', \ \ \frac{1}{2} {\text {id}} - \mathcal {K}'&:H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T) \rightarrow H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T) \end{aligned}$$

are isomorphisms. In addition, there holds the identity $\mathcal {V}^{-1} \mathcal {K}\mathcal {V}= \mathcal {K}' = \mathcal {D}\mathcal {K}\mathcal {D}^{-1}$.

Proof

The claim is implied by (5.11), compare the proof of [3, Corollary 3.14]. $\square$

Corollary 5.13

The unique solution $u \in H^{1, \frac{1}{2}}_{;0,} (Q_T)$ of the Dirichlet problem

$$\begin{aligned} \begin{aligned} (\partial _t - \Delta ) u&= 0 \ \&\text {in } Q_T, \\ \gamma _0 u&= g \ \&\text {on } \Sigma _T, \end{aligned} \end{aligned}$$

with $g \in H^{\frac{1}{2}, \frac{1}{4}}(\Sigma _T)$ can be represented

• (i) as $u = \widetilde{\mathcal {V}} \psi - \widetilde{\mathcal {K}} g$, where $\psi \in H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T)$ is the unique solution of the first kind integral equation

  $$\begin{aligned} \mathcal {V}\psi = \left( \frac{1}{2} {\text {id}} + \mathcal {K}\right) g. \end{aligned}$$

• (ii) as $u = \widetilde{\mathcal {V}} \psi - \widetilde{\mathcal {K}} g$, where $\psi \in H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T)$ is the unique solution of the second kind integral equation

  $$\begin{aligned} \left( \frac{1}{2} {\text {id}} - \mathcal {K}' \right) \psi = \mathcal {D}g. \end{aligned}$$

• (iii) as $u = \widetilde{\mathcal {V}} \psi$, where $\psi \in H^{-\frac{1}{2}, -\frac{1}{4}}(\Sigma _T)$ is the unique solution of the first kind integral equation

  $$\begin{aligned} \mathcal {V}\psi = g. \end{aligned}$$

• (iv) as $u = \widetilde{\mathcal {K}} w$, where $w \in H^{\frac{1}{2}, \frac{1}{4}}(\Sigma _T)$ is the unique solution of the second kind integral equation

  $$\begin{aligned} \left( \frac{1}{2} {\text {id}} - \mathcal {K}\right) w = - g. \end{aligned}$$

In (i) and (ii), it particularly holds $\psi = \gamma _1^- u$ on $\Sigma _T$.

Proof

We can again use directly the idea of the proof of [3, Corollary 3.16], which means the uniqueness results from above and the jump relations given in Lemma 5.6. $\square$

Corollary 5.14

The unique solution $u \in H^{1, \frac{1}{2}}_{;0,}(Q_T)$ of the Neumann problem

$$\begin{aligned} \begin{aligned} (\partial _t - \Delta ) u&= 0 \ \&\text {in } Q_T, \\ \gamma _1^- u&= h \ \&\text {on } \Sigma _T, \end{aligned} \end{aligned}$$

with $h \in H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T)$ can be represented

• (i) as $u = \widetilde{\mathcal {V}} h - \widetilde{\mathcal {K}} w$, where $w \in H^{\frac{1}{2}, \frac{1}{4}}(\Sigma _T)$ is the unique solution of the second kind integral equation

  $$\begin{aligned} \left( \frac{1}{2} {\text {id}} + \mathcal {K}\right) w = \mathcal {V}h. \end{aligned}$$

• (ii) as $u = \widetilde{\mathcal {V}} h - \widetilde{\mathcal {K}} w$, where $w \in H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T)$ is the unique solution of the first kind integral equation

  $$\begin{aligned} \mathcal {D}w = \left( \frac{1}{2} {\text {id}} - \mathcal {K}' \right) h. \end{aligned}$$

• (iii) as $u = \widetilde{\mathcal {V}} \psi$, where $\psi \in H^{-\frac{1}{2}, -\frac{1}{4}} (\Sigma _T)$ is the unique solution of the second kind integral equation

  $$\begin{aligned} \left( \frac{1}{2} {\text {id}} + \mathcal {K}' \right) \psi = h. \end{aligned}$$

• (iv) as $u = \widetilde{\mathcal {K}} w$, where $w \in H^{\frac{1}{2}, \frac{1}{4}} (\Sigma _T)$ is the unique solution of the first kind integral equation

  $$\begin{aligned} \mathcal {D}w = - h. \end{aligned}$$

In (i) and (ii), we have that $w = \gamma _0 u$ on $\Sigma _T$.

Proof

The proof follows by the same arguments as in the proof of [3, Corollary 3.17], which is similar to the respective proof for the Dirichlet problem. $\square$

6 Conclusion

In this article, we considered the heat equation on a time-varying (so-called non-cylindrical) domain. In contrast to the problem on a cylindrical domain, we used a modified Neumann trace operator containing a term which is dependent on the velocity of the moving surface. We were able to show the mapping properties of the layer operators by following the proofs of Costabel [3]. To this end, we heavily used the fact that the non-cylindrical domain is a mapped cylindrical domain. Then, using mapped anisotropic Sobolev spaces, we obtain analogous mapping properties and are also able to prove existence and uniqueness of solutions of the Dirichlet and of the Neumann problem.