Rate of convergence in the large diffusion limit for the heat equation with a dynamical boundary condition

Abstract

We study the heat equation on a half-space or on an exterior domain with a linear dynamical boundary condition. Our main aim is to establish the rate of convergence to solutions of the Laplace equation with the same dynamical boundary condition as the diffusion coefficient tends to infinity.

Keywords

Heat equation dynamical boundary condition large diffusion limit

1. Introduction

We consider the heat equation with a dynamical boundary condition, $\begin{array}{l} (1) & \begin{matrix} ε \partial_{t} u_{ε} - Δ u_{ε} = 0 in Ω \times (0, \infty), \\ \partial_{t} u_{ε} + \partial_{ν} u_{ε} = 0 on \partial Ω \times (0, \infty), \\ u_{ε} (\cdot, 0) = φ in Ω, \\ u_{ε} (\cdot, 0) = φ_{b} on \partial Ω \end{matrix} \end{array}$ and the connection with its elliptic counterpart, $\begin{array}{l} (2) & \begin{matrix} - Δ u = 0 in Ω \times (0, \infty), \\ \partial_{t} u + \partial_{ν} u = 0 on \partial Ω \times (0, \infty), \\ u (\cdot, 0) = φ_{b} on \partial Ω . \end{matrix} \end{array}$ Here, $φ \in L^{\infty} (Ω)$ , $φ_{b} \in L^{\infty} (\partial Ω)$ , $ε \in (0, 1)$ .

The expected relation is, of course, that solutions of (1) converge to those of (2) as $ε \to 0$ . Indeed, for the case of the half-space $Ω = R_{+}^{N} : = R^{N - 1} \times R_{+}$ , $N ⩾ 2$ , this has been proven recently in [12]. This means, in particular, that the influence of the initial function φ is lost in the limit. In the present paper we are interested in a detailed description of the loss of influence of φ. More precisely, we investigate the rate of convergence of solutions of (1) to solutions of (2) as $ε \to 0$ .

We will deal with the case $Ω = R_{+}^{N}$ , $N ⩾ 2$ , and the radially symmetric setting for $Ω = R^{3} ∖ B_{1} (0)$ , $B_{1} (0) : = {x \in R^{3} | | x | < 1}$ . It turns out that the rate of convergence is of order $\sqrt{ε}$ in both cases.

For bounded domains Ω, results on convergence as $ε \to 0$ were established in [18] by a method that is completely different from the one used in [12] and in this paper. The rate of convergence was not studied in [18].

Various aspects of analysis of parabolic equations with dynamical boundary conditions have been treated by many authors, for existence, uniqueness and regularity see for example [1,4,5,16,23,28–30], for blow-up of solutions [2,16,24,33–35], for asymptotic behaviour of solutions [15,17,18,20,32,36], and for the mean curvature flow with dynamical boundary conditions see [21,22]. Some of similar issues for elliptic equations with dynamical boundary conditions were considered in [3,6–11,13,14,19,24–27,31,37], for instance.

Given $ϕ \in L^{\infty} (Ω)$ , by $S_{1} (t) ϕ : = θ (t)$ we denote the bounded solution of $\begin{array}{l} \partial_{t} θ = Δ θ in Ω \times (0, \infty), \\ θ = 0 on \partial Ω \times (0, \infty), \\ θ (\cdot, 0) = ϕ in Ω . \end{array}$

For $ψ \in L^{\infty} (\partial Ω)$ , we want to have that $S_{2} (t) ψ : = p (t)$ , where p solves $\begin{array}{l} - Δ p = 0 in Ω \times [0, \infty), \\ \partial_{t} p + \partial_{ν} p = 0 on \partial Ω \times (0, \infty), \\ p (\cdot, 0) = ψ in \partial Ω, \end{array}$ and hence define $\begin{matrix} [S_{2} (t) ψ] (x) = \frac{2 Γ (\frac{N}{2})}{\sqrt{π} Γ (\frac{N - 1}{2})} \int_{R^{N - 1}} {(x_{N} + t)}^{1 - N} {(1 + \frac{| x^{'} - y^{'} |^{2}}{{(x_{N} + t)}^{2}})}^{- \frac{N}{2}} ψ (y^{'}) d y^{'}, \end{matrix}$ for $Ω = R_{+}^{N}$ , $x = (x^{'}, x_{N}) \in R^{N - 1} \times R_{+}$ , and $\begin{matrix} [S_{2} (t) ψ] (r) = \frac{ψ}{r e^{t}}, r > 1, t ⩾ 0 . \end{matrix}$ for radially symmetric $ψ \in L^{\infty} (\partial Ω)$ (i.e. $ψ \in R$ ) in case of $Ω = R^{3} ∖ B_{1} (0)$ .

We furthermore introduce $\begin{matrix} F_{1} [φ_{b}] (x, t) : = \partial_{t} [S_{2} (t) φ_{b}] (x), φ_{b} \in L^{\infty} (\partial Ω), x \in Ω, t > 0, \end{matrix}$ and $\begin{matrix} F_{2} [v] (x, t) : = - [S_{2} (0) \partial_{ν} v |_{\partial Ω} (\cdot, t)] (x) - \int_{0}^{t} \partial_{t} [S_{2} (t - s) \partial_{ν} v |_{\partial Ω} (\cdot, s)] (x) d s \end{matrix}$ for $x \in Ω$ , $t > 0$ and functions $v : \overline{Ω} \to R$ that have a normal derivative $\partial_{ν} v \in L^{\infty} (\partial Ω)$ on $\partial Ω$ .

If $\begin{matrix} (3) & w_{ε} (x, t) = [S_{2} (t) φ_{b}] (x) - \int_{0}^{t} [S_{2} (t - s) \partial_{ν} v_{ε} (s)] (x) d s, \end{matrix}$ then $\begin{matrix} \partial_{t} w_{ε} = F_{1} [φ_{b}] + F_{2} [v_{ε}] \end{matrix}$ and $\begin{matrix} (4) & Δ w_{ε} = 0, [\partial_{t} w_{ε} + \partial_{ν} w_{ε}] |_{\partial Ω} = [- \partial_{ν} v_{ε}] |_{\partial Ω}, w_{ε} (\cdot, 0) |_{\partial Ω} = φ_{b} . \end{matrix}$ If $\begin{matrix} (5) & v_{ε} (x, t) : = [S_{1} (\frac{t}{ε}) Φ] (x) - \int_{0}^{t} [S_{1} (\frac{t - s}{ε}) \partial_{t} w_{ε} (s)] (x) d s, \end{matrix}$ where $Φ = φ - S_{2} (0) φ_{b}$ , then $v_{ε}$ solves $\begin{matrix} (6) & ε \partial_{t} v_{ε} = Δ v_{ε} - ε \partial_{t} w_{ε} = Δ v_{ε} - ε F_{1} [φ_{b}] - ε F_{2} [v_{ε}], v_{ε} |_{\partial Ω} = 0, v_{ε} (\cdot, 0) = Φ, \end{matrix}$ so that $u_{ε} : = v_{ε} + w_{ε}$ is a classical solution of (1).

Definition 1.
Let $φ \in L^{\infty} (Ω)$ and $φ_{b} \in L^{\infty} (\partial Ω)$ . Let $T \in (0, \infty]$ and $\begin{matrix} v_{ε}, \partial_{ν} v_{ε}, w_{ε} \in C (\overline{Ω} \times (0, T)) . \end{matrix}$ We call $(v_{ε}, w_{ε})$ a solution of (6), (4) – and $u_{ε} : = v_{ε} + w_{ε}$ a solution of (1) – in $Ω \times (0, T)$ if (5) and (3) are satisfied. The solutions are called global-in-time solutions if $T = \infty$ .

Note: Here ν is to be understood as (smooth) vector field defined on all of $\overline{Ω}$ , which on $\partial Ω$ coincides with the outer unit normal. In the context of the cases treated in this article, of course, $\partial_{ν} = - \partial_{r}$ for $Ω = R^{3} ∖ B_{1} (0)$ and $\partial_{ν} = - \partial_{x_{N}}$ for $Ω = R_{+}^{N}$ .

The following was shown in [12, Theorem 1.1 and Corrollary 1.1]:
Theorem 2.
Let $N ⩾ 2$ and $Ω = R_{+}^{N}$ ; let $ε \in (0, 1)$ , $φ \in L^{\infty} (Ω)$ and $φ_{b} \in L^{\infty} (\partial Ω)$ . Then problem ( 6 ), ( 4 ) has a unique global-in-time solution $(v_{ε}, w_{ε})$ satisfying $\begin{matrix} (7) & sup_{0 < t < T} ({‖ v_{ε} (\cdot, t) ‖}_{L^{\infty} (Ω)} + \sqrt{\frac{t}{ε}} {‖ \partial_{ν} v_{ε} (\cdot, t) ‖}_{L^{\infty} (Ω)} + {‖ w_{ε} (\cdot, t) ‖}_{L^{\infty} (Ω)}) < \infty \end{matrix}$ for any $T > 0$ . Furthermore, $v_{ε}$ and $w_{ε}$ are bounded and smooth in $\overline{Ω} \times I$ for any bounded interval $I \subset (0, \infty)$ . Moreover, for every $T > 0$ there is $C (T) > 0$ such that for every $ε \in (0, 1)$ and φ, $φ_{b}$ as above we have $\begin{array}{l} sup_{0 < t < T} ({‖ v_{ε} (\cdot, t) ‖}_{L^{\infty} (Ω)} + \sqrt{\frac{t}{ε}} {‖ \partial_{ν} v_{ε} (\cdot, t) ‖}_{L^{\infty} (Ω)} + {‖ w_{ε} (\cdot, t) ‖}_{L^{\infty} (Ω)}) \\ (8) & ⩽ C (T) (‖ φ ‖_{L^{\infty} (Ω)} + ‖ φ_{b} ‖_{L^{\infty} (\partial Ω)}) . \end{array}$

Finally, $u_{ε} : = v_{ε} + w_{ε}$ is a classical solution of ( 1 ) and for every compact set K in $\overline{Ω}$ and $0 < τ_{1} < τ_{2} < \infty$ it holds that $\begin{matrix} (9) & lim_{ε \to 0} sup_{τ_{1} < t < τ_{2}} {‖ u_{ε} (\cdot, t) - S_{2} (t) φ_{b} ‖}_{L^{\infty} (K)} = 0 . \end{matrix}$

In this paper we establish an analogous result for radially symmetric solutions when the domain Ω is the exterior of the unit ball in $R^{3}$ .
Theorem 3.
Let $Ω = R^{3} ∖ B_{1} (0)$ , $ε \in (0, \frac{1}{\sqrt{π}})$ and let the functions $φ \in L^{\infty} (Ω)$ and $φ_{b} \in L^{\infty} (\partial Ω)$ be radially symmetric.
Then problem ( 6 ), ( 4 ) has a unique radially symmetric global-in-time solution $(v_{ε}, w_{ε})$ satisfying ( 7 ) and $u_{ε} : = v_{ε} + w_{ε}$ is a classical solution of ( 1 ).

Furthermore, if $\begin{matrix} (10) & sup_{ρ ⩾ 1} | ρ φ (ρ) | < \infty, \end{matrix}$ and $0 < τ_{1} < τ_{2} < \infty$ , then $\begin{matrix} (11) & lim_{ε \to 0} sup_{τ_{1} < t < τ_{2}} {‖ u_{ε} (\cdot, t) - S_{2} (t) φ_{b} ‖}_{L^{\infty} (Ω)} = 0 . \end{matrix}$ Moreover, for every $ε_{0} \in (0, \frac{1}{\sqrt{π}})$ and $T > 0$ there is $C (T) > 0$ such that for every $ε \in (0, ε_{0})$ and φ, $φ_{b}$ as above the following holds: $\begin{array}{l} sup_{0 < t < T} ({‖ v_{ε} (\cdot, t) ‖}_{L^{\infty} (Ω)} + \sqrt{\frac{t}{ε}} {‖ \partial_{ν} v_{ε} (\cdot, t) ‖}_{L^{\infty} (Ω)} + {‖ w_{ε} (\cdot, t) ‖}_{L^{\infty} (Ω)}) \\ (12) & ⩽ C (T) (‖ φ ‖_{L^{\infty} (Ω)} + ‖ φ_{b} ‖_{L^{\infty} (\partial Ω)} + sup_{ρ ⩾ 1} | ρ φ (ρ) |) . \end{array}$

The role of assumption (10) is explained in Section 7.

Our remaining results are concerned with the question what the optimal rate of convergence in (9) and (11) is.
Theorem 4.
Let $Ω = R_{+}^{N}$ , $N ⩾ 2$ . Let $φ \in L^{\infty} (Ω)$ , $φ_{b} \in L^{\infty} (\partial Ω)$ , $K \subset \overline{Ω}$ compact and $0 < τ_{1} < τ_{2} < \infty$ . Then there is $C > 0$ such that $\begin{matrix} sup_{τ_{1} < t < τ_{2}} {‖ u_{ε} (\cdot, t) - S_{2} (t) φ_{b} ‖}_{L^{\infty} (K)} ⩽ C \sqrt{ε}, for every ε \in (0, 1) . \end{matrix}$
Theorem 5.
Let $Ω = R^{3} ∖ B_{1} (0)$ . Let $φ \in L^{\infty} (Ω)$ be radially symmetric and such that ( 10 ) holds. Assume further that $φ_{b}$ is constant and $ε_{0} \in (0, \frac{1}{\sqrt{π}})$ , $0 < τ_{1} < τ_{2} < \infty$ . Then there is $C > 0$ such that $\begin{matrix} sup_{τ_{1} < t < τ_{2}} {‖ u_{ε} (\cdot, t) - S_{2} (t) φ_{b} ‖}_{L^{\infty} (Ω)} ⩽ C \sqrt{ε}, for every ε \in (0, ε_{0}) . \end{matrix}$

The upper bounds from Theorems 4 and 5 are sharp.
Theorem 6.
Let $Ω = R_{+}^{N}$ , $N ⩾ 2$ and $φ_{b} \equiv 0$ . There is $φ \in L^{\infty} (Ω)$ such that for every $x_{0} > 0$ there is $τ_{2} > 0$ such that for every compact set $K \subset R^{N - 1} \times (x_{0}, \infty) \times (0, τ_{2})$ there are $ε_{0} > 0$ and $C > 0$ such that $\begin{matrix} u_{ε} (x, t) - [S_{2} (t) φ_{b}] (x) = u_{ε} (x, t) ⩾ C \sqrt{ε} \end{matrix}$ for every $ε \in (0, ε_{0})$ and every $(x, t) \in K$ .
Theorem 7.
Let $Ω = R^{3} ∖ B_{1} (0)$ and $φ_{b} \equiv 0$ . Then there is $φ \in L^{\infty} (Ω)$ such that for every $r > 1$ there is $τ_{2} > 0$ such that for every compact set $K \subset (R^{3} ∖ B_{r} (0)) \times (0, τ_{2})$ there are $C > 0$ and $ε_{0} > 0$ such that $\begin{matrix} u_{ε} (x, t) - [S_{2} (t) φ_{b}] (x) = u_{ε} (x, t) ⩾ C \sqrt{ε} \end{matrix}$ for every $ε \in (0, ε_{0})$ and every $(x, t) \in K$ .

The paper is organized as follows. In Section 2 we introduce some notation, in Section 3 we recall some estimates in the case of the half-space, and in Section 4 we derive analogous estimates for the exterior domain. In Section 5 we prove Theorem 3 and in Section 6 Theorems 4 and 5. Section 7 is devoted to a remark on the long-time behaviour of solutions of the heat equation on the exterior domain and Sections 8 and 9 to the proofs of Theorems 6 and 7, respectively.
2. Preparation: Introducing further notation. The space $X_{T}$

With the abbreviations $\begin{matrix} D_{ε} [ψ] (x, t) : = \int_{0}^{t} [S_{1} (\frac{t - s}{ε}) F_{1} [ψ] (\cdot, s)] (x) d s, ψ \in L^{\infty} (\partial Ω), \end{matrix}$ and $\begin{matrix} (13) & \tilde{D_{ε}} [ψ] (x, t) = \int_{0}^{t} [S_{1} (\frac{t - s}{ε}) F_{2} [ψ] (\cdot, s)] (x) d s, ψ \in L^{\infty} (Ω), \end{matrix}$ we introduce $\begin{matrix} (14) & Q_{ε} [v] (x, t) : = [S_{1} (\frac{t}{ε}) Φ] (x) - D_{ε} [φ_{b}] (x, t) - \tilde{D_{ε}} [v] (x, t) . \end{matrix}$

The proofs will be based on a fixed point argument for $Q_{ε}$ in the space $X_{T}$ , which we define as $\begin{matrix} X_{T} : = {v \in C (\overline{Ω} \times (0, T)) | \partial_{x_{N}} v \in C (\overline{Ω} \times (0, T)), ‖ v ‖_{X_{T}} < \infty} \end{matrix}$ if $Ω = R_{+}^{N}$ , and as $\begin{matrix} X_{T} : = {v \in C (\overline{Ω} \times (0, T)) | v radially symmetric, \partial_{ν} v \in C (\overline{Ω} \times (0, T)), ‖ v ‖_{X_{T}} < \infty} \end{matrix}$ for $Ω = R^{3} ∖ B_{1} (0)$ . In both cases, we equip it with the norm $\begin{matrix} (15) & ‖ v ‖_{X_{T}} : = sup_{t \in (0, T)} E_{ε} [v] (t), \end{matrix}$ where $\begin{matrix} (16) & E_{ε} [v] (t) : = {‖ v (\cdot, t) ‖}_{L^{\infty} (Ω)} + \sqrt{\frac{t}{ε}} {‖ \partial_{ν} v (\cdot, t) ‖}_{L^{\infty} (\partial Ω)}, \end{matrix}$ and observe that it thereby becomes a Banach space.

3. Estimates: Recalling the case $Ω = R_{+}^{N}$

In this section we recall the necessary estimates from [12] for $Ω = R_{+}^{N}$ . They have been used in the proof of Theorem 2 in [12] and will be essential for our proof of Theorems 4 and 6. We will take care of the corresponding results for $Ω = R^{3} ∖ B_{1} (0)$ – and their proofs – in the next section.

Lemma 8.
Let $Ω = R_{+}^{N}$ . Then for every $ϕ \in L^{\infty} (Ω)$ , $\begin{matrix} (17) & sup_{t > 0} {‖ S_{1} (t) ϕ ‖}_{L^{\infty} (Ω)} ⩽ ‖ ϕ ‖_{L^{\infty} (Ω)} \end{matrix}$ and $\begin{matrix} (18) & sup_{t > 0} t^{\frac{1}{2}} {‖ \partial_{x_{N}} [S_{1} (t) ϕ] ‖}_{L^{\infty} (Ω)} ⩽ ‖ ϕ ‖_{L^{\infty} (Ω)} . \end{matrix}$ Moreover, for every $L > 0$ there is $C > 0$ such that for every $ϕ \in L^{\infty} (Ω)$ , $x \in R^{N - 1} \times (0, L)$ and $t > 0$ , $\begin{matrix} (19) & | [S_{1} (t) ϕ] (x) | ⩽ C t^{- \frac{1}{2}} ‖ ϕ ‖_{L^{\infty} (Ω)} \end{matrix}$
Proof.
[12, (2.1) and (2.2) and proof of (2.3), respectively]. □
Lemma 9.
Let $Ω = R_{+}^{N}$ and $ψ \in L^{\infty} (\partial Ω)$ . Then for every $t ⩾ 0$ $\begin{matrix} {‖ S_{2} (t) ψ ‖}_{L^{\infty} (Ω)} ⩽ ‖ ψ ‖_{L^{\infty} (\partial Ω)} . \end{matrix}$
Proof.
[12, (2.8)]. □
Lemma 10.
Let $Ω = R_{+}^{N}$ . There is $C > 0$ such that $\begin{matrix} (20) & {‖ D_{ε} [ψ] (\cdot, t) ‖}_{L^{\infty} (Ω)} ⩽ C t^{\frac{1}{4}} (ε^{\frac{1}{2}} + t^{\frac{3}{4}}) ‖ ψ ‖_{L^{\infty} (\partial Ω)} \end{matrix}$ for all $ψ \in L^{\infty} (\partial Ω)$ , $ε > 0$ and $t > 0$ .

There is $C > 0$ such that $\begin{matrix} (21) & {‖ \partial_{x_{N}} D_{ε} [ψ] (\cdot, t) ‖}_{L^{\infty} (Ω)} ⩽ C ε^{\frac{3}{4}} t^{- \frac{1}{4}} ‖ ψ ‖_{L^{\infty} (\partial Ω)} \end{matrix}$ for all $ψ \in L^{\infty} (\partial Ω)$ , $ε > 0$ , $t > 0$ .

For every $L > 0$ there is $C > 0$ such that $\begin{matrix} (22) & D_{ε} [ψ] (x, t) ⩽ C ‖ ψ ‖_{L^{\infty} (\partial Ω)} \sqrt{ε} (t^{\frac{1}{4}} + t^{\frac{1}{2}}) \end{matrix}$ for every $ψ \in L^{\infty} (\partial Ω)$ , $ε > 0$ , $t > 0$ , $x \in R^{N - 1} \times (0, L)$ .
Proof.
[12, (2.10), Lemma 2.4 and proof of (2.11), respectively] □
Lemma 11.
Let $Ω = R_{+}^{N}$ . Then there is $C > 0$ such that for every $T > 0$ , $v \in X_{T}$ , $x \in Ω$ , $t \in (0, T)$ $\begin{matrix} (23) & | \tilde{D_{ε}} [v] (x, t) | ⩽ C ε^{\frac{1}{2}} ‖ v ‖_{X_{T}} (t^{\frac{1}{2}} + ε^{\frac{1}{2}} t^{\frac{3}{4}} + t) \end{matrix}$ and $\begin{matrix} | \partial_{x_{N}} \tilde{D_{ε}} [v] (x, t) | ⩽ C \sqrt{\frac{t}{ε}} ‖ v ‖_{X_{T}} ({(ε t)}^{\frac{1}{2}} + {(ε t)}^{\frac{7}{8}} + {(ε t)}^{\frac{3}{4}}) . \end{matrix}$
Proof.
[12, proof of (3.9) and of (3.10), respectively] □

4. Estimates: $R^{3} ∖ B_{1} (0)$

The assertions of the lemmata in this section parallel those in the previous section. Before we begin dealing with their statements and proofs, let us first bring some of the quantities that have been defined in the introduction in the explicit form in which the radially symmetric 3-dimensional setting allows us to express them.

If ψ is a radially symmetric function on $\partial B_{1} (0)$ , we can interpret it as a real number and write $\begin{matrix} [S_{2} (t) ψ] (r) = \frac{ψ}{r e^{t}}, r > 1, t ⩾ 0 . \end{matrix}$ This means that also $\begin{array}{l} F_{1} [φ_{b}] (r, t) = - \frac{φ_{b}}{r e^{t}}, \\ F_{2} [v] (r, t) = \frac{v_{r} (1, t)}{r} + \int_{0}^{t} \frac{v_{r} (1, s)}{r e^{t - s}} d s, \end{array}$ and $\begin{matrix} Φ (r) = φ (r) - \frac{φ_{b}}{r} . \end{matrix}$

Also $S_{1}$ can be written explicitly: $\begin{matrix} [S_{1} (t) φ] (r) = \frac{1}{r \sqrt{4 π t}} \int_{1}^{\infty} (e^{- \frac{{(r - ρ)}^{2}}{4 t}} - e^{- \frac{{(r + ρ - 2)}^{2}}{4 t}}) ρ φ (ρ) d ρ, r > 1, t > 0 . \end{matrix}$ This representation is, of course, based on the fact that for every radially symmetric solution u of the heat equation in $R^{3}$ , the function $(r, t) \mapsto r u (r, t)$ solves the one-dimensional heat equation.

Lemma 12.
Let $Ω = R^{3} ∖ B_{1} (0)$ . Then for every radially symmetric $ϕ \in L^{\infty} (Ω)$ $\begin{matrix} (24) & sup_{t > 0} {‖ S_{1} (t) ϕ ‖}_{L^{\infty} (Ω)} ⩽ ‖ ϕ ‖_{L^{\infty} (Ω)} . \end{matrix}$ If, moreover, ${sup}_{ρ ⩾ 1} | ρ φ (ρ) | ⩽ K$ for $K > 0$ , then $\begin{matrix} (25) & | [S_{1} (t) φ] (r) | ⩽ \frac{(r - 1) K}{r \sqrt{π t}} for every r > 1, t > 0, \end{matrix}$ and $\begin{matrix} (26) & | \partial_{r} [S_{1} (t) φ] (r) | ⩽ \frac{2 K}{\sqrt{π t}} for every r > 1, t > 0 . \end{matrix}$
Proof.
We obtain (24) from explicit computations as follows: $\begin{array}{l} | [S_{1} (t) φ] (r) | & = | \frac{1}{r \sqrt{4 π t}} \int_{1}^{\infty} (e^{- \frac{{(r - ρ)}^{2}}{4 t}} - e^{- \frac{{(r + ρ - 2)}^{2}}{4 t}}) ρ φ (ρ) d ρ | \\ ⩽ \frac{‖ φ ‖_{L^{\infty} (Ω)}}{r \sqrt{4 π t}} \int_{1}^{\infty} (e^{- \frac{{(r - ρ)}^{2}}{4 t}} - e^{- \frac{{(r + ρ - 2)}^{2}}{4 t}}) ρ d ρ \\ = \frac{‖ φ ‖_{L^{\infty} (Ω)}}{r \sqrt{π}} (\int_{\frac{1 - r}{2 \sqrt{t}}}^{\infty} (2 \sqrt{t} z + r) e^{- z^{2}} d z - \int_{\frac{r - 1}{2 \sqrt{t}}}^{\infty} (2 \sqrt{t} z + 2 - r) e^{- z^{2}} d z) \\ = \frac{2 \sqrt{t}}{r \sqrt{π}} ‖ φ ‖_{L^{\infty} (Ω)} \int_{\frac{1 - r}{2 \sqrt{t}}}^{\frac{r - 1}{2 \sqrt{t}}} z e^{- z^{2}} d z + \frac{r ‖ φ ‖_{L^{\infty} (Ω)}}{r \sqrt{π}} \int_{\frac{1 - r}{2 \sqrt{t}}}^{\infty} e^{- z^{2}} d z \\ + \frac{(r - 2) ‖ φ ‖_{L^{\infty} (Ω)}}{r \sqrt{π}} \int_{\frac{r - 1}{2 \sqrt{t}}}^{\infty} e^{- z^{2}} d z \\ = (1 - \frac{2}{r \sqrt{π}} \int_{\frac{r - 1}{2 \sqrt{t}}}^{\infty} e^{- z^{2}} d z) ‖ φ ‖_{L^{\infty} (Ω)}, r ⩾ 1, t > 0 . \end{array}$ If we not only control $‖ φ ‖_{L^{\infty} (Ω)}$ , but even ${sup}_{ρ ⩾ 1} | ρ φ (ρ) |$ , we can proceed slightly differently: $\begin{array}{l} | [S_{1} (t) φ] (r) | & = | \frac{1}{r \sqrt{4 π t}} \int_{1}^{\infty} (e^{- \frac{{(r - ρ)}^{2}}{4 t}} - e^{- \frac{{(r + ρ - 2)}^{2}}{4 t}}) ρ φ (ρ) d ρ | \\ ⩽ \frac{{sup}_{ρ ⩾ 1} | ρ φ (ρ) |}{r \sqrt{4 π t}} \int_{1}^{\infty} (e^{- \frac{{(r - ρ)}^{2}}{4 t}} - e^{- \frac{{(r + ρ - 2)}^{2}}{4 t}}) d ρ \\ = \frac{{sup}_{ρ ⩾ 1} | ρ φ (ρ) |}{r \sqrt{π}} (\int_{\frac{1 - r}{2 \sqrt{t}}}^{\infty} e^{- z^{2}} d z - \int_{\frac{r - 1}{2 \sqrt{t}}}^{\infty} e^{- z^{2}} d z) \\ (27) & = \frac{2 {sup}_{ρ ⩾ 1} | ρ φ (ρ) |}{r \sqrt{π}} \int_{0}^{\frac{r - 1}{2 \sqrt{t}}} e^{- z^{2}} d z ⩽ \frac{(r - 1) {sup}_{ρ ⩾ 1} | ρ φ (ρ) |}{r \sqrt{π t}} \end{array}$ holds for every $r > 1$ , $t > 0$ and (25) follows.

For the estimate of the radial derivative let us first observe that for every $t > 0$ and $r ⩾ 1$ $\begin{array}{l} \partial_{r} [S_{1} (t) φ] (r) = & - \frac{1}{r^{2} \sqrt{4 π t}} \int_{1}^{\infty} (e^{- \frac{{(r - ρ)}^{2}}{4 t}} - e^{- \frac{{(r + ρ - 2)}^{2}}{4 t}}) ρ φ (ρ) d ρ \\ + \frac{1}{r \sqrt{4 π t}} \int_{1}^{\infty} (- \frac{r - ρ}{2 t} e^{- \frac{{(r - ρ)}^{2}}{4 t}} + \frac{r + ρ - 2}{2 t} e^{- \frac{{(r + ρ - 2)}^{2}}{4 t}}) ρ φ (ρ) d ρ \\ = & - \frac{1}{r} [S_{1} (t) φ] (r) \\ (28) & + \frac{1}{r \sqrt{4 π t}} \int_{1}^{\infty} (- \frac{r - ρ}{2 t} e^{- \frac{{(r - ρ)}^{2}}{4 t}} + \frac{r + ρ - 2}{2 t} e^{- \frac{{(r + ρ - 2)}^{2}}{4 t}}) ρ φ (ρ) d ρ . \end{array}$ We see that here for every $t > 0$ and $r ⩾ 1$ $\begin{array}{l} | \int_{1}^{\infty} (- \frac{r - ρ}{2 t} e^{- \frac{{(r - ρ)}^{2}}{4 t}} + \frac{r + ρ - 2}{2 t} e^{- \frac{{(r + ρ - 2)}^{2}}{4 t}}) ρ φ (ρ) d ρ | \\ ⩽ (2 \int_{\frac{1 - r}{2 \sqrt{t}}}^{\infty} | z | e^{- z^{2}} d z + + 2 \int_{\frac{r - 1}{2 \sqrt{t}}}^{\infty} z e^{- z^{2}} d z) sup_{ρ ⩾ 1} | ρ φ (ρ) | \\ (29) & = 2 \int_{R} | z | e^{- z^{2}} d z sup_{ρ ⩾ 1} | ρ φ (ρ) | = 2 sup_{ρ ⩾ 1} | ρ φ (ρ) | \end{array}$

Hence, inserting (27) and (29) into (28), we arrive at $\begin{matrix} | \partial_{r} [S_{1} (t) φ] (r) | ⩽ \frac{(r - 1) {sup}_{ρ ⩾ 1} | ρ φ (ρ) |}{r^{2} \sqrt{π t}} + \frac{2 {sup}_{ρ ⩾ 1} | ρ φ (ρ) |}{r \sqrt{4 π t}} ⩽ \frac{2}{\sqrt{π t}} sup_{ρ ⩾ 1} | ρ φ (ρ) | \end{matrix}$ for every $r ⩾ 1$ , $t > 0$ . □
Lemma 13.
Let $Ω = R^{3} ∖ B_{1} (0)$ and let $ψ \in L^{\infty} (\partial Ω)$ be radially symmetric (constant). Then for every $t ⩾ 0$ , $\begin{matrix} {‖ S_{2} (t) ψ ‖}_{L^{\infty} (Ω)} ⩽ ‖ ψ ‖_{L^{\infty} (\partial Ω)} . \end{matrix}$
Proof.
This is obvious from the explicit form of $S_{2} (t) ψ = \frac{ψ}{r e^{t}}$ , $r ⩾ 1$ , $t ⩾ 0$ . □
Lemma 14.
Let $Ω = R^{3} ∖ B_{1} (0)$ . Then there is $C > 0$ such that for every $ψ \in R$ and every $t > 0$ , $ε > 0$ , $\begin{matrix} (30) & {‖ D_{ε} [ψ] (\cdot, t) ‖}_{L^{\infty} (Ω)} ⩽ C \sqrt{ε t} | ψ | \end{matrix}$ and $\begin{matrix} (31) & {‖ \partial_{r} D_{ε} [ψ] (\cdot, t) ‖}_{L^{\infty} (Ω)} ⩽ C \sqrt{ε t} | ψ | . \end{matrix}$
Proof.
We use the explicit representation for $D_{ε} [ψ]$ to see that for every $r > 1$ , $t > 0$ , $ε > 0$ , $\begin{array}{l} D_{ε} [ψ] (r, t) & = - \int_{0}^{t} S_{1} (\frac{t - s}{ε}) \frac{ψ}{r e^{s}} d s \\ = - \int_{0}^{t} \frac{1}{r \sqrt{4 π \frac{t - s}{ε}}} \int_{1}^{\infty} (e^{- \frac{{(r - ρ)}^{2}}{4 \frac{t - s}{ε}}} - e^{- \frac{{(r + ρ - 2)}^{2}}{4 \frac{t - s}{ε}}}) ρ \frac{ψ}{ρ e^{s}} d ρ d s \\ = - \frac{ψ}{r} \sqrt{\frac{ε}{4 π}} \int_{0}^{t} \frac{e^{- s}}{\sqrt{t - s}} \int_{1}^{\infty} (e^{- \frac{{(r - ρ)}^{2}}{4 \frac{t - s}{ε}}} - e^{- \frac{{(r + ρ - 2)}^{2}}{4 \frac{t - s}{ε}}}) d ρ d s \\ = - \frac{ψ}{r} \sqrt{\frac{ε}{4 π}} \int_{0}^{t} \frac{e^{- s}}{\sqrt{t - s}} \sqrt{4 \frac{t - s}{ε}} (\int_{\frac{1 - r}{\sqrt{4 \frac{t - s}{ε}}}}^{\infty} e^{- z^{2}} d z - \int_{\frac{1 + r - 2}{\sqrt{4 \frac{t - s}{ε}}}}^{\infty} e^{- z^{2}} d z) \\ = - \frac{2 ψ}{r \sqrt{π}} \int_{0}^{t} e^{- s} \int_{0}^{\frac{(r - 1) \sqrt{ε}}{2 \sqrt{t - s}}} e^{- z^{2}} d z d s . \end{array}$ Due to the elementary estimates $\int_{0}^{L} e^{- z^{2}} d z ⩽ L$ for every $L > 0$ and $\int_{0}^{t} e^{- s} {(t - s)}^{- \frac{1}{2}} d s ⩽ 2 \sqrt{t}$ for every $t > 0$ , this implies (30).

Concerning the radial derivative, for every $r > 1$ , $t > 0$ , $ε > 0$ , we have $\begin{array}{l} \partial_{r} D_{ε} [ψ] (r, t) & = \frac{2 ψ}{r^{2} \sqrt{π}} \int_{0}^{t} e^{- s} \int_{0}^{\frac{(r - 1) \sqrt{ε}}{2 \sqrt{t - s}}} e^{- z^{2}} d z d s - \frac{2 ψ}{r \sqrt{π}} \int_{0}^{t} e^{- s} \frac{\sqrt{ε}}{2 \sqrt{t - s}} e^{- \frac{{(r - 1)}^{2} ε}{4 (t - s)}} d s . \end{array}$ Here, due to $\frac{1}{r} ⩽ 1$ , the first term can be estimated by (30); the second obeys $\begin{matrix} | - \frac{2 ψ}{r \sqrt{π}} \int_{0}^{t} e^{- s} \frac{\sqrt{ε}}{2 \sqrt{t - s}} e^{- \frac{{(r - 1)}^{2} ε}{4 (t - s)}} d s | ⩽ \frac{| ψ | \sqrt{ε}}{r \sqrt{π}} \int_{0}^{t} e^{- s} {(t - s)}^{- \frac{1}{2}} d s ⩽ \frac{2 | ψ | \sqrt{ε} \sqrt{t}}{r \sqrt{π}} \end{matrix}$ for every $t > 0$ , $r > 1$ , $ε > 0$ . □
Lemma 15.
Let $Ω = R^{3} ∖ B_{1} (0)$ . Then there is $C > 0$ such that for every $T > 0$ $t \in (0, T)$ , $ε > 0$ and every radial function $v : \overline{Ω} \times (0, T) \to R$ with $\partial_{ν} v |_{\partial Ω} \in L^{\infty} (\partial Ω)$ , $\begin{matrix} (32) & {‖ \tilde{D_{ε}} [v] (\cdot, t) ‖}_{L^{\infty} (Ω)} ⩽ \sqrt{ε π} (1 + t) sup_{τ \in (0, T)} | \sqrt{τ} \partial_{r} v (1, τ) | \end{matrix}$ and $\begin{matrix} (33) & {‖ \partial_{r} \tilde{D_{ε}} [v] (\cdot, t) ‖}_{L^{\infty} (Ω)} ⩽ (1 + 2 \sqrt{t} + 2 t) \sqrt{ε π} sup_{τ \in (0, T)} | \sqrt{τ} \partial_{r} v (1, τ) | \end{matrix}$
Proof.
If we insert the explicit definitions of $S_{1}$ and $S_{2}$ into (13), we see that for every $r ⩾ 1$ , $t > 0$ , $ε > 0$ we have $\begin{array}{l} \tilde{D_{ε}} [v] (r, t) = & \int_{0}^{t} \int_{1}^{\infty} \frac{1}{r \sqrt{4 π \frac{t - s}{ε}}} (e^{- \frac{{(r - ρ)}^{2}}{4 \frac{t - s}{ε}}} - e^{- \frac{{(r + ρ - 2)}^{2}}{4 \frac{t - s}{ε}}}) ρ \frac{\partial_{r} v (1, s)}{ρ} d ρ d s \\ - \int_{0}^{t} \int_{1}^{\infty} \int_{0}^{s} \frac{1}{r \sqrt{4 π \frac{t - s}{ε}}} (e^{- \frac{{(r - ρ)}^{2}}{4 \frac{t - s}{ε}}} - e^{- \frac{{(r + ρ - 2)}^{2}}{4 \frac{t - s}{ε}}}) ρ \frac{\partial_{r} v (1, σ)}{ρ e^{s - σ}} d σ d ρ d s \\ = & \frac{2}{r \sqrt{π}} \int_{0}^{t} \partial_{r} v (1, s) \int_{0}^{\frac{r - 1}{\sqrt{4 \frac{t - s}{ε}}}} e^{- z^{2}} d z d s \\ - \frac{2}{r \sqrt{π}} \int_{0}^{t} \int_{0}^{s} e^{σ - s} \partial_{r} v (1, σ) d σ \int_{0}^{\frac{r - 1}{\sqrt{4 \frac{t - s}{ε}}}} e^{- z^{2}} d z d s, \end{array}$ so that $\begin{array}{l} | \tilde{D_{ε}} [v] (r, t) | ⩽ & \frac{2}{r \sqrt{π}} \int_{0}^{t} \frac{r - 1}{\sqrt{s} \sqrt{4 \frac{t - s}{ε}}} d s sup_{s \in (0, t)} | \sqrt{s} \partial_{r} v (1, s) | \\ + \frac{2}{r \sqrt{π}} \int_{0}^{t} \int_{0}^{s} \frac{e^{σ - s}}{\sqrt{σ}} d σ \frac{r - 1}{\sqrt{4 \frac{t - s}{ε}}} d s sup_{σ \in (0, t)} | \sqrt{σ} \partial_{r} v (1, σ) | \\ = & \frac{(r - 1) \sqrt{ε}}{r \sqrt{π}} (\int_{0}^{t} \frac{1}{\sqrt{s} \sqrt{t - s}} d s + \int_{0}^{t} \int_{0}^{s} \frac{e^{σ - s}}{\sqrt{σ}} d σ \frac{1}{\sqrt{t - s}} d s) \\ (34) & \times sup_{s \in (0, t)} | \sqrt{s} \partial_{r} v (1, s) | \end{array}$ Taking into account that $\begin{matrix} (35) & \int_{0}^{t} s^{- \frac{1}{2}} {(t - s)}^{- \frac{1}{2}} d s = \int_{0}^{1} σ^{- \frac{1}{2}} {(1 - σ)}^{- \frac{1}{2}} d σ = π for all t > 0 \end{matrix}$ and $\begin{matrix} (36) & \int_{0}^{t} \int_{0}^{s} \frac{e^{σ - s}}{\sqrt{σ}} d σ {(t - s)}^{- \frac{1}{2}} d s ⩽ \int_{0}^{t} \frac{2 \sqrt{s}}{\sqrt{t - s}} d s = 2 t \int_{0}^{1} \sqrt{\frac{σ}{1 - σ}} d σ = π t for all t > 0, \end{matrix}$ we see that (34) proves (32).

As to the radial derivative, we compute $\begin{array}{l} \partial_{r} \tilde{D_{ε}} [v] (r, t) \\ = - \frac{2}{r^{2} \sqrt{π}} \int_{0}^{t} \partial_{r} v (1, s) \int_{0}^{\frac{r - 1}{\sqrt{4 \frac{t - s}{ε}}}} e^{- z^{2}} d z d s + \frac{1}{r \sqrt{π}} \int_{0}^{t} \partial_{r} v (1, s) \frac{\sqrt{ε}}{\sqrt{t - s}} e^{- \frac{{(r - 1)}^{2} ε}{4 (t - s)}} d s \\ + \frac{2}{r^{2} \sqrt{π}} \int_{0}^{t} \int_{0}^{s} e^{σ - s} \partial_{r} v (1, σ) d σ \int_{0}^{\frac{r - 1}{\sqrt{4 \frac{t - s}{ε}}}} e^{- z^{2}} d z d s \\ - \frac{1}{r \sqrt{π}} \int_{0}^{t} \int_{0}^{s} e^{σ - s} \partial_{r} v (1, σ) d σ \frac{\sqrt{ε}}{\sqrt{t - s}} e^{- \frac{{(r - 1)}^{2} ε}{4 (t - s)}} d s for r ⩾ 1, t > 0 . \end{array}$ Here, we can simplify two integrals according to $\begin{array}{l} - \frac{2}{r^{2} \sqrt{π}} \int_{0}^{t} \partial_{r} v (1, s) \int_{0}^{\frac{r - 1}{\sqrt{4 \frac{t - s}{ε}}}} e^{- z^{2}} d z d s \\ + \frac{2}{r^{2} \sqrt{π}} \int_{0}^{t} \int_{0}^{s} e^{σ - s} \partial_{r} v (1, σ) d σ \int_{0}^{\frac{r - 1}{\sqrt{4 \frac{t - s}{ε}}}} e^{- z^{2}} d z d s = - \frac{1}{r} \tilde{D_{ε}} [v] (r, t), r ⩾ 1, t > 0 . \end{array}$ The next one can be estimated as $\begin{array}{l} | \frac{1}{r \sqrt{π}} \int_{0}^{t} \partial_{r} v (1, s) \frac{\sqrt{ε}}{\sqrt{t - s}} e^{- \frac{{(r - 1)}^{2} ε}{4 (t - s)}} d s | & ⩽ \frac{\sqrt{ε}}{r \sqrt{π}} sup_{s \in (0, t)} | \sqrt{s} \partial_{r} v (1, s) | \int_{0}^{t} s^{- \frac{1}{2}} {(t - s)}^{- \frac{1}{2}} d s \\ (37) & ⩽ \frac{2 \sqrt{ε π t}}{r} sup_{s \in (0, t)} | \sqrt{s} \partial_{r} v (1, s) |, r ⩾ 1, t > 0, \end{array}$ according to (35), and for the last we again rely on (36) to see that $\begin{array}{l} | - \frac{1}{r \sqrt{π}} \int_{0}^{t} \int_{0}^{s} e^{σ - s} \partial_{r} v (1, σ) d σ \frac{\sqrt{ε}}{\sqrt{t - s}} e^{- \frac{{(r - 1)}^{2} ε}{4 (t - s)}} d s | \\ ⩽ \frac{\sqrt{ε}}{r \sqrt{π}} \int_{0}^{t} sup_{σ \in (0, t)} | \sqrt{σ} \partial_{r} v (1, σ) | \int_{0}^{s} \frac{e^{σ - s}}{\sqrt{σ}} d σ {(t - s)}^{- \frac{1}{2}} d s \\ (38) & ⩽ \frac{t \sqrt{ε π}}{r} sup_{s \in (0, t)} | \sqrt{s} \partial_{r} v (1, s) | for all r ⩾ 1, t > 0 . \end{array}$ If we finally combine (32), (37) and (38), we can readily conclude (33). □
Lemma 16.
Let $Ω = R^{3} ∖ B_{1} (0)$ and $T > 0$ . Then for every $v \in X_{T}$ , $\begin{matrix} (39) & {‖ \tilde{D_{ε}} [v] ‖}_{X_{T}} ⩽ (ε (1 + T) + \sqrt{ε} (\sqrt{T} + 2 T + 2 T \sqrt{T})) \sqrt{π} ‖ v ‖_{X_{T}} . \end{matrix}$
Proof.
If we insert (32) and (33) into (15), we immediately obtain $\begin{array}{l} {‖ \tilde{D_{ε}} [v] ‖}_{X_{T}} & = sup_{t \in (0, T)} ({‖ \tilde{D_{ε}} [v] (\cdot, t) ‖}_{L^{\infty} (Ω)} + \sqrt{\frac{t}{ε}} {‖ \partial_{r} \tilde{D_{ε}} [v] (\cdot, t) ‖}_{L^{\infty} (Ω)}) \\ ⩽ sup_{t \in (0, T)} ((\sqrt{ε π} (1 + t) + \sqrt{\frac{t}{ε}} (1 + 2 \sqrt{t} + 2 t) \sqrt{ε π}) sup_{s \in (0, t)} | \sqrt{s} \partial_{r} v (1, s) |) \\ ⩽ sup_{t \in (0, T)} (\sqrt{ε π} (1 + t) + (\sqrt{t} + 2 t + 2 t \sqrt{t}) \sqrt{π}) \sqrt{ε} ‖ v ‖_{X_{T}} \\ ⩽ (ε (1 + T) + \sqrt{ε} (\sqrt{T} + 2 T + 2 T \sqrt{T})) \sqrt{π} ‖ v ‖_{X_{T}} \end{array}$ □

5. Existence. Proof of Theorem 3

Proof of Theorem 3.
(a) The explicit representation of $S_{1} (\frac{t}{ε}) Φ$ , $D_{ε} [φ_{b}]$ and $\tilde{D_{ε}} [v]$ and their derivatives show that (for each ε, T, radially symmetric $φ \in L^{\infty} (Ω)$ and $φ_{b} \in L^{\infty} (\partial Ω)$ ) $Q_{ε}$ maps $X_{T}$ into $X_{T}$ .

Moreover, $Q_{ε}$ is a contraction: Let $ε_{0} \in (0, \frac{1}{\sqrt{π}})$ , let $T_{} > 0$ be so small that $(ε_{0} (1 + T_{}) + \sqrt{ε_{0}} ({\sqrt{T}}_{} + 2 T_{} + 2 T_{} \sqrt{T_{}}) \sqrt{π} = : q < 1$ . Then, according to Lemma 16, for every $ε \in (0, ε_{0})$ , every $T \in (0, T_{})$ , and every $v_{1}, v_{2} \in X_{T}$ , $\begin{matrix} {‖ Q_{ε} [v_{1}] - Q_{ε} [v_{2}] ‖}_{X_{T}} = {‖ \tilde{D_{ε}} [v_{1} - v_{2}] ‖}_{X_{T}} ⩽ q ‖ v_{1} - v_{2} ‖_{X_{T}} . \end{matrix}$ Banach’s fixed point theorem hence yields a unique solution $v_{ε}$ in $X_{T}$ (for $T \in (0, T_{})$ ), and $w_{ε}$ is defined according to (3). Since $T_{}$ was independent of the initial data, successive application of the same reasoning finally provides a global-in-time solution.

(b) For the second part of the theorem – and, in particular, for more quantitative and ε-independent information, which by way of Lemma 16 will rely on a uniform bound for $‖ v_{ε} ‖_{X_{T}}$ – the mere existence result obtained above is insufficient.

We hence restrict the class of admissible initial data and will attempt to apply the fixed point theorem not on $X_{T}$ , but on a bounded subset thereof.

Let $ε_{0}$ , $T_{}$ and q be as before. Let $\begin{matrix} M > \frac{1}{1 - q} (‖ φ ‖_{L^{\infty} (Ω)} + (1 + \frac{2}{\sqrt{π}} + 2 C) | φ_{b} | + \frac{2}{\sqrt{π}} sup_{ρ ⩾ 1} | ρ φ (ρ) |), \end{matrix}$ where C is the constant in Lemma 14.

If we can show that for every $T \in (0, min {1, T_{}})$ , $ε \in (0, ε_{0})$ , $Q_{ε}$ maps the set $\begin{matrix} B_{M} : = {v \in X_{T} | ‖ v ‖_{X_{T}} ⩽ M} \end{matrix}$ into itself, Banach’s fixed point theorem does not only prove existence of a solution $v_{ε}$ , but also shows $\begin{matrix} ‖ v_{ε} ‖_{X_{T}} ⩽ M, \end{matrix}$ which finally proves (12).

We therefore turn our attention to the proof of $Q_{ε} (B_{M}) \subset B_{M}$ : Due to (16), (24) and (26), we have $\begin{array}{l} E_{ε} [S_{1} (\frac{\cdot}{ε}) Φ] (t) & = {‖ S_{1} (\frac{t}{ε}) Φ ‖}_{L^{\infty} (Ω)} + \sqrt{\frac{t}{ε}} {‖ \partial_{r} S_{1} (\frac{t}{ε}) Φ ‖}_{L^{\infty} (Ω)} \\ ⩽ ‖ Φ ‖_{L^{\infty} (Ω)} + \sqrt{\frac{t}{ε}} \frac{2}{\sqrt{π}} \sqrt{\frac{ε}{t}} sup_{ρ ⩾ 1} | ρ Φ (ρ) | for every t > 0, ε > 0, \end{array}$ where $\begin{matrix} sup_{ρ ⩾ 1} | ρ Φ (ρ) | = sup_{ρ ⩾ 1} | ρ φ (ρ) - ρ \frac{φ_{b} (ρ)}{ρ} | ⩽ sup_{ρ ⩾ 1} | ρ φ (ρ) | + | φ_{b} | \end{matrix}$ and $\begin{matrix} ‖ Φ ‖_{L^{\infty} (Ω)} ⩽ ‖ φ ‖_{L^{\infty} (Ω)} + sup_{ρ ⩾ 1} | \frac{φ_{b}}{ρ} | ⩽ ‖ φ ‖_{L^{\infty} (Ω)} + | φ_{b} | . \end{matrix}$

If we furthermore insert (30) and (31) into (15), we see that with C from Lemma 14 for any $ε \in (0, 1)$ , $t \in (0, 1)$ we have $\begin{array}{l} E_{ε} [D_{ε} [φ_{b}]] (t) & = {‖ D_{ε} [φ_{b}] (\cdot, t) ‖}_{L^{\infty} (Ω)} + \sqrt{\frac{t}{ε}} {‖ \partial_{r} D_{ε} [φ_{b}] (\cdot, t) ‖}_{L^{\infty} (Ω)} \\ ⩽ C \sqrt{ε t} | φ_{b} | + \sqrt{\frac{t}{ε}} C \sqrt{ε t} | φ_{b} | ⩽ C (\sqrt{ε t} + t) | φ_{b} | ⩽ 2 C | φ_{b} | . \end{array}$ Finally, the choice of q and Lemma 16 ensure $\begin{array}{l} {‖ \tilde{D_{ε}} [v] ‖}_{X_{T}} ⩽ q ‖ v ‖_{X_{T}} . \end{array}$

In conclusion, for $ε \in (0, ε_{0})$ , $T \in (0, min {1, T_{}})$ and every $v \in X_{T}$ , $\begin{array}{l} ‖ Q_{ε} v ‖_{X_{T}} & ⩽ ‖ φ ‖_{L^{\infty} (Ω)} + | φ_{b} | + \frac{2}{\sqrt{π}} sup_{ρ ⩾ 1} | ρ φ (ρ) | + | φ_{b} | + 2 C | φ_{b} | + q ‖ v ‖_{X_{T}} \\ ⩽ (1 - q) M + q ‖ v ‖_{X_{T}}, \end{array}$ so that $‖ Q_{ε} v ‖_{X_{T}} ⩽ M$ whenever $‖ v ‖_{X_{T}} ⩽ M$ .

We postpone the proof of (11), seeing that it is a corollary of Theorem 5. □
Remark 17.
The proof of the existence result in Theorem 2 in [12] proceeds analogously. In place of (24) and (26), (30), (31) and Lemma 16, one has to use (17), (18), (20), (21) and Lemma 11.
6. Upper bounds. Proofs of Theorems 4 and 5

The estimates we have given in Sections 3 and 4 mainly deal with $v_{ε}$ . In preparation of the proofs of the estimates of $u_{ε} = v_{ε} + w_{ε}$ let us state the following lemma, which for $Ω = R_{+}^{N}$ also was part of [12, proof of Theorem 1.1 (c)]:

Lemma 18.
Let $Ω = R_{+}^{N}$ , $N ⩾ 2$ , or $Ω = R^{3} ∖ B_{1} (0)$ , and let $t > 0$ , $ε > 0$ . Then for every $v_{ε} \in X_{t}$ and $w_{ε}$ as in ( 3 ) we have $\begin{matrix} (40) & {‖ w_{ε} (\cdot, t) - S_{2} (t) φ_{b} ‖}_{L^{\infty} (Ω)} ⩽ 2 \sqrt{ε t} ‖ v_{ε} ‖_{X_{t}} . \end{matrix}$
Proof.
According to (3), Lemma 9 (for $Ω = R_{+}^{N}$ ) or Lemma 13 (for $Ω = R^{3} ∖ B_{1} (0)$ ) and (15), we can estimate $\begin{array}{l} {‖ w_{ε} (\cdot, t) - S_{2} (t) φ_{b} ‖}_{L^{\infty} (Ω)} & = {‖ \int_{0}^{t} S_{2} (t - s) \partial_{ν} v_{ε} (\cdot, s) |_{\partial Ω} d s ‖}_{L^{\infty} (Ω)} \\ ⩽ \int_{0}^{t} {‖ \partial_{ν} v_{ε} (\cdot, s) ‖}_{L^{\infty} (\partial Ω)} d s ⩽ ‖ v_{ε} ‖_{X_{t}} \sqrt{ε} \int_{0}^{t} \frac{1}{\sqrt{s}} d s = 2 \sqrt{ε t} ‖ v_{ε} ‖_{X_{t}} . \end{array}$ □
Proof of Theorem 4.
We let $K \subset \overline{Ω}$ be compact. Then by the definition of $u_{ε}$ (cf. (14) and Definition 1, or proof of Theorem 3 and Remark 17) we have $\begin{array}{l} {‖ u_{ε} (\cdot, t) - S_{2} (t) φ_{b} ‖}_{L^{\infty} (K)} & = {‖ v_{ε} (\cdot, t) + w_{ε} (\cdot, t) - S_{2} (t) φ_{b} ‖}_{L^{\infty} (K)} \\ ⩽ {‖ v_{ε} (\cdot, t) ‖}_{L^{\infty} (K)} + {‖ w_{ε} (\cdot, t) - S_{2} (t) φ_{b} ‖}_{L^{\infty} (K)} \\ ⩽ {‖ S_{1} (\frac{t}{ε}) Φ ‖}_{L^{\infty} (K)} + {‖ D_{ε} [φ_{b}] (\cdot, t) ‖}_{L^{\infty} (K)} \\ (41) & + {‖ \tilde{D_{ε}} [v_{ε}] (\cdot, t) ‖}_{L^{\infty} (Ω)} + {‖ w_{ε} (\cdot, t) - S_{2} (t) φ_{b} ‖}_{L^{\infty} (Ω)} \end{array}$ for every $t > 0$ . Here we can apply (19), (22), (23) and (40) so as to obtain $C_{1} = C_{1} (K) > 0$ , $C_{2} = C_{2} (K) > 0$ and $C_{3} > 0$ such that $\begin{array}{l} {‖ S_{1} (\frac{t}{ε}) Φ ‖}_{L^{\infty} (K)} + {‖ D_{ε} [φ_{b}] (\cdot, t) ‖}_{L^{\infty} (K)} + {‖ \tilde{D_{ε}} [v_{ε}] (\cdot, t) ‖}_{L^{\infty} (Ω)} \\ + {‖ w_{ε} (\cdot, t) - S_{2} (t) φ_{b} ‖}_{L^{\infty} (Ω)} ⩽ C_{1} \sqrt{\frac{ε}{t}} ‖ Φ ‖_{L^{\infty} (Ω)} + C_{2} ‖ φ_{b} ‖_{L^{\infty} (\partial Ω)} \sqrt{ε} (t^{\frac{1}{4}} + t^{\frac{1}{2}}) \\ + C_{3} ε^{\frac{1}{2}} ‖ v ‖_{X_{t}} (t^{\frac{1}{2}} + ε^{\frac{1}{2}} t^{\frac{3}{4}} + t) + 2 \sqrt{ε t} ‖ v_{ε} ‖_{X_{t}} \end{array}$ for every $t > 0$ and $ε \in (0, 1)$ . If we take the supremum over $t \in (τ_{1}, τ_{2})$ and use boundedness of $‖ v_{ε} ‖_{X_{τ_{2}}}$ according to (7), we thereby have proven Theorem 4. □
Proof of Theorem 5.
Similarly to (41), the construction of $u_{ε}$ in the proof of Theorem 3 (or directly Definition 1) allows us to estimate $\begin{array}{l} {‖ u_{ε} (\cdot, t) - S_{2} (t) φ_{b} ‖}_{L^{\infty} (Ω)} ⩽ & {‖ S_{1} (\frac{t}{ε}) Φ ‖}_{L^{\infty} (Ω)} + {‖ D_{ε} [φ_{b}] (\cdot, t) ‖}_{L^{\infty} (Ω)} \\ + {‖ \tilde{D_{ε}} [v] (\cdot, t) ‖}_{L^{\infty} (Ω)} + {‖ w_{ε} (\cdot, t) - S_{2} (t) φ_{b} ‖}_{L^{\infty} (Ω)} \end{array}$ for $ε \in (0, ε_{0})$ and $t > 0$ . If we abbreviate $K : = {sup}_{ρ ⩾ 1} | ρ Φ (ρ) |$ and employ (25), (30), (39), (40), we obtain $C > 0$ such that the last inequality turns into $\begin{array}{l} {‖ u_{ε} (\cdot, t) - S_{2} (t) φ_{b} ‖}_{L^{\infty} (Ω)} \\ ⩽ \frac{K}{\sqrt{π}} \sqrt{\frac{ε}{t}} + C \sqrt{ε t} | φ_{b} | + \sqrt{ε} (\sqrt{ε} (1 + t) + \sqrt{t} + 2 t + 2 t \sqrt{t}) \sqrt{π} ‖ v_{ε} ‖_{X_{t}} + 2 \sqrt{ε t} ‖ v_{ε} ‖_{X_{t}} \end{array}$ for every $ε \in (0, ε_{0})$ and $t > 0$ . Again, taking the supremum over $t \in (τ_{1}, τ_{2})$ and using the boundedness of $‖ v_{ε} ‖_{X_{τ_{2}}}$ according to (12), we can conclude Theorem 5. □

7. A remark on condition (10). Long-time behaviour of the heat equation on the exterior of the ball

Remark 19.
The main effect of the difference in geometry between $Ω = R^{3} ∖ B_{1} (0)$ and $Ω = R_{+}^{N}$ seems to lie in the additional condition ${sup}_{ρ ⩾ 1} | ρ φ (ρ) | < \infty$ to be posed on the initial data for the convergence result in the case of $Ω = R^{3} ∖ B_{1} (0)$ . In order to understand why this is not too unnatural, let us consider the solution of the heat equation emanating from initial data $φ = χ_{R^{3} ∖ B_{b} (0)}$ . $\begin{array}{l} S_{1} (t) χ_{R^{3} ∖ B_{b} (0)} (r) & = \frac{1}{r \sqrt{4 π t}} (\int_{b}^{\infty} ρ e^{- \frac{{(r - ρ)}^{2}}{4 t}} d ρ - \int_{b}^{\infty} ρ e^{- \frac{{(r + ρ - 2)}^{2}}{4 t}} d ρ) \\ = \frac{1}{r \sqrt{4 π t}} (\int_{\frac{b - r}{2 \sqrt{t}}}^{\infty} (2 \sqrt{t} z + r) e^{- z^{2}} d z - \int_{\frac{b + r - 2}{2 \sqrt{t}}}^{\infty} (2 \sqrt{t} z + 2 - r) e^{- z^{2}} 2 \sqrt{t} d z) \\ = \frac{1}{r \sqrt{π}} (2 \sqrt{t} \int_{\frac{b - r}{2 \sqrt{t}}}^{\frac{b + r - 2}{2 \sqrt{t}}} z e^{- z^{2}} d z + r \int_{\frac{b - r}{2 \sqrt{t}}}^{\infty} e^{- z^{2}} d z + (r - 2) \int_{\frac{b + r - 2}{2 \sqrt{t}}}^{\infty} e^{- z^{2}} d z) \\ = \frac{\sqrt{t}}{r \sqrt{π}} (e^{- \frac{{(b - r)}^{2}}{4 t}} - e^{- \frac{{(b + r - 2)}^{2}}{4 t}}) + \frac{1}{\sqrt{π}} \int_{\frac{b - r}{2 \sqrt{t}}}^{\infty} e^{- z^{2}} d z + \frac{(r - 2)}{r \sqrt{π}} \int_{\frac{b + r - 2}{2 \sqrt{t}}}^{\infty} e^{- z^{2}} d z . \end{array}$ This means that $\begin{matrix} S_{1} (t) φ \to 0 + \frac{1}{2} + \frac{r - 2}{2 r} = 1 - \frac{1}{r} as t \to \infty . \end{matrix}$ In particular, $S_{1} (\frac{t}{ε}) φ ↛ 0$ as $\frac{t}{ε} \to \infty$ .
8. Lower estimates in the halfspace: Proof of Theorem 6

Proof of Theorem 6.
In order to show optimality of the rate $\sqrt{ε}$ , we strive to find a lower estimate with the same rate, for one concrete example of initial data: We set $φ_{b} : = 0$ and $φ : = χ_{{x_{N} > b}}$ for some $b > 0$ . Then, apparently, $Φ = φ - S_{2} (0) φ_{b} = φ$ .

Due to $v_{ε}$ being a fixed point of $Q_{ε}$ (cf. Definition 1 and (14) or the construction in the proof of Theorem 3 and Remark 17), we have $\begin{matrix} v_{ε} (x, t) = Q_{ε} [v_{ε}] (x, t) = S_{1} (\frac{t}{ε}) φ (x) - 0 - \tilde{D_{ε}} [v_{ε}] (x, t) ⩾ S_{1} (\frac{t}{ε}) φ (x) - | \tilde{D_{ε}} [v_{ε}] (x, t) | . \end{matrix}$ Letting $y_{} : = (y^{'}, - y_{N})$ for $y = (y^{'}, y_{N}) \in R^{N - 1} \times R_{+}$ , we can write the first of these terms explicitly, cf. [12, (1.3) and (1.4)]: $\begin{array}{l} S_{1} (\frac{t}{ε}) Φ (x) & = \int_{R_{+}^{N}} {(\frac{4 π t}{ε})}^{- \frac{N}{2}} (e^{- \frac{| x - y |^{2}}{\frac{4 t}{ε}}} - e^{- \frac{| x - y |^{2}}{\frac{4 t}{ε}}}) φ (y^{'}, y_{N}) d y \\ = \int_{R_{+}^{N}} {(\frac{4 π t}{ε})}^{- \frac{N}{2}} (e^{- \frac{| x^{'} - y^{'} |^{2}}{\frac{4 t}{ε}}} e^{- \frac{| x_{N} - y_{N} |^{2}}{\frac{4 t}{ε}}} - e^{- \frac{| x^{'} - y^{'} |^{2}}{\frac{4 t}{ε}}} e^{- \frac{| x_{N} + y_{N} |^{2}}{\frac{4 t}{ε}}}) φ (y) d y \\ = \int_{R^{N - 1}} {(\frac{4 π t}{ε})}^{- \frac{N - 1}{2}} e^{- \frac{| x^{'} - y^{'} |^{2}}{\frac{4 t}{ε}}} d y^{'} \int_{b}^{\infty} {(\frac{4 π t}{ε})}^{- \frac{1}{2}} (e^{- \frac{{(x_{N} - y_{N})}^{2}}{\frac{4 t}{ε}}} - e^{- \frac{{(x_{N} + y_{N})}^{2}}{\frac{4 t}{ε}}}) d y_{N} \\ = \int_{\sqrt{\frac{ε}{4 t}} (b - x_{N})}^{\infty} {(\frac{4 π t}{ε})}^{- \frac{1}{2}} e^{- z^{2}} {(\frac{4 t}{ε})}^{\frac{1}{2}} d z - \int_{\sqrt{\frac{ε}{4 t} (b + x_{N})}}^{\infty} π^{- \frac{1}{2}} e^{- z^{2}} d z \\ = \frac{1}{\sqrt{π}} \int_{\sqrt{\frac{ε}{4 t}} (b - x_{N})}^{\sqrt{\frac{ε}{4 t}} (b + x_{N})} e^{- z^{2}} d z = \frac{x_{N} \sqrt{ε}}{\sqrt{π t}} I (ε, t, x_{N}), \end{array}$ if we abbreviate $\begin{matrix} I (ε, t, x_{N}) : = \frac{\sqrt{t}}{x_{N} \sqrt{ε}} \int_{\sqrt{\frac{ε}{4 t}} (b - x_{N})}^{\sqrt{\frac{ε}{4 t}} (b + x_{N})} e^{- z^{2}} d z . \end{matrix}$

From (23) and Lemma 18 we obtain $C > 0$ such that $\begin{array}{l} u_{ε} (x, t) & = v_{ε} (x, t) + w_{ε} (x, t) \\ ⩾ x_{N} \sqrt{\frac{ε}{π t}} I (ε, t, x_{N}) - C ε^{\frac{1}{2}} (t^{\frac{1}{2}} + ε^{\frac{1}{2}} t^{\frac{3}{4}} + t) ‖ v ‖_{X_{t}} - 2 \sqrt{ε t} ‖ v_{ε} ‖_{X_{t}} \end{array}$ for all $(x, t) \in R_{+}^{N} \times (0, \infty)$ . According to (8), for every $T > 0$ we can hence find $\tilde{C} > 0$ such that $\begin{array}{l} u_{ε} (x, t) & ⩾ x_{N} \sqrt{\frac{ε}{π t}} I (ε, t, x_{N}) - \tilde{C} ε^{\frac{1}{2}} (t^{\frac{1}{2}} + ε^{\frac{1}{2}} t^{\frac{3}{4}} + t) \\ = \sqrt{\frac{ε}{t}} (\frac{x_{N}}{\sqrt{π}} I (ε, t, x_{N}) - \tilde{C} (t) t^{\frac{1}{2}} (t^{\frac{1}{2}} + ε^{\frac{1}{2}} t^{\frac{3}{4}} + t)) \end{array}$ holds for every $x \in R_{+}^{N}$ and every $t \in (0, T)$ .

After these preparations, we can begin the actual proof of the statement of Theorem 6: Given $x_{0}$ we pick $τ_{2}$ such that $\tilde{C} (τ_{2}) τ_{2}^{\frac{1}{2}} (τ_{2}^{\frac{1}{2}} + τ_{2}^{\frac{3}{4}} + τ_{2}) < \frac{x_{0}}{2 \sqrt{π}}$ and for any compact $K \subset R^{N - 1} \times (x_{0}, \infty) \times (0, τ_{2})$ we let $\begin{matrix} C : = \frac{1}{2 inf {t | \exists x with (x, t) \in K}} (\frac{x_{0}}{2 \sqrt{π}} - C (τ_{2}) τ_{2}^{\frac{1}{2}} (τ_{2}^{\frac{1}{2}} + τ_{2}^{\frac{3}{4}} + τ_{2})) . \end{matrix}$ Noting that $I (ε, t, x_{N}) \to 1$ (uniformly with respect to $(x, t) \in K$ ) as $ε \to 0$ , we choose $ε_{0} \in (0, 1)$ so small that $I (ε, t, x_{N}) > \frac{1}{2}$ for all $(x, t) \in K$ and thus ensure that $\begin{matrix} u_{ε} (x, t) ⩾ C \sqrt{ε} for every (x, t) \in K and ε \in (0, ε_{0}), \end{matrix}$ as desired. □
9. Lower estimates in the exterior of the ball. Proof of Theorem 7

Proof of Theorem 7.

We introduce $φ (r) : = \frac{1}{r} χ_{{r > b}}$ for some $b > 1$ and $φ_{b} : = 0$ . Obviously, ${sup}_{r ⩾ 1} | r φ (r) |$ is finite, $Φ = φ$ and $D_{ε} [φ_{b}] = 0$ .

According to Definition 1 and (14), $v_{ε}$ is a fixed point of $Q_{ε}$ and thus $\begin{matrix} v_{ε} (x, t) = Q_{ε} [v_{ε}] (x, t) = S_{1} (\frac{t}{ε}) φ (x) - 0 - \tilde{D_{ε}} [v_{ε}] (x, t) ⩾ S_{1} (\frac{t}{ε}) φ (x) - | \tilde{D_{ε}} [v_{ε}] (x, t) | . \end{matrix}$

Again we compute the first term explicitly: $\begin{array}{l} S_{1} (\frac{t}{ε}) φ & = \frac{1}{r \sqrt{4 π \frac{t}{ε}}} (\int_{b}^{\infty} e^{- \frac{{(r - ρ)}^{2}}{\frac{4 t}{ε}}} \frac{ρ}{ρ} d ρ - \int_{b}^{\infty} e^{- \frac{{(r + ρ - 2)}^{2}}{\frac{4 t}{ε}}} d ρ) \\ = \frac{1}{r \sqrt{π}} \int_{\frac{b - r}{2 \sqrt{\frac{t}{ε}}}}^{\frac{b + r - 2}{2 \sqrt{\frac{t}{ε}}}} e^{- z^{2}} d z = \frac{r - 1}{r \sqrt{π}} \sqrt{\frac{ε}{t}} I (ε, t, r), \end{array}$ where $\begin{matrix} I (ε, t, r) : = \frac{\sqrt{t}}{(r - 1) \sqrt{ε}} \int_{\frac{b - r}{2 \sqrt{\frac{t}{ε}}}}^{\frac{b + r - 2}{2 \sqrt{\frac{t}{ε}}}} e^{- z^{2}} d z . \end{matrix}$

By (32) and Lemma 18 $\begin{array}{l} u_{ε} (r, t) = v_{ε} (r, t) + w_{ε} (r, t) & ⩾ \frac{r - 1}{r \sqrt{π}} \sqrt{\frac{ε}{t}} I (ε, t, r) - ε \sqrt{π} (1 + t) ‖ v_{ε} ‖_{X_{T}} - 2 \sqrt{ε t} ‖ v_{ε} ‖_{X_{t}} \end{array}$ for all $r ⩾ 1$ , $t > 0$ ; and according to (12) for every $T > 0$ we can find $\tilde{C} (T)$ such that $\begin{array}{l} u_{ε} (r, t) & ⩾ \frac{r - 1}{r \sqrt{π}} \sqrt{\frac{ε}{t}} I (ε, t, r) - \tilde{C} \sqrt{ε} (\sqrt{ε} (1 + t) + \sqrt{t}) \end{array}$ holds for every $r ⩾ 1$ , $t \in (0, T)$ and $ε \in (0, ε_{0})$ . With these, we can easily prove the statement of Theorem 7: Given $r > 1$ we let $τ_{2}$ be such that $\frac{r - 1}{r} \sqrt{\frac{1}{π}} \frac{1}{2} - \tilde{C} (τ_{2}) \sqrt{τ_{2}} (1 + τ_{2} + \sqrt{τ_{2}}) = : c_{1} > 0$ . For any compact $K \subset (R^{3} ∖ B_{r} (0)) \times (0, τ_{2})$ , we then use the (uniform in K) convergence $I (ε, t, r) \to 1$ as $ε \to 0$ to pick $ε_{0} \in (0, \frac{1}{\sqrt{π}})$ such that $I (ε, t, x) > \frac{1}{2}$ for all $(x, t) \in K$ . Then for $ε \in (0, ε_{0})$ and $(x, t) \in K$ , $\begin{array}{l} u_{ε} (x, t) & ⩾ \frac{r - 1}{r} \sqrt{\frac{ε}{π t}} (\frac{\sqrt{t}}{(r - 1) \sqrt{ε}} \int_{\frac{b - r}{2 \sqrt{\frac{t}{ε}}}}^{\frac{b + r - 2}{2 \sqrt{\frac{t}{ε}}}} e^{- z^{2}} d z) - \tilde{C} \sqrt{ε} (\sqrt{ε} (1 + t) + \sqrt{t}) \\ ⩾ (\frac{r - 1}{2 r} \frac{1}{\sqrt{π}} - \tilde{C} \sqrt{t} \sqrt{ε} (\sqrt{ε} (1 + t) + \sqrt{t})) \sqrt{\frac{ε}{t}} ⩾ \frac{c_{1}}{\sqrt{t}} \sqrt{ε} ⩾ \frac{c_{1}}{\sqrt{τ_{1}}} \sqrt{ε} \end{array}$ if we define $τ_{1} : = inf {t | \exists x : (x, t) \in K}$ . Finally setting $C : = \frac{c_{1}}{\sqrt{τ_{1}}}$ , we obtain Theorem 7. □

Footnotes

Acknowledgements

The first author was supported in part by the Slovak Research and Development Agency under the contract No. APVV-14-0378 and by the VEGA grant 1/0347/18. The second author was supported in part by the Grant-in-Aid for Scientific Research (A) (No. 15H02058) from Japan Society for the Promotion of Science. The third author was supported by the Grant-in-Aid for Young Scientists (B) (No. 16K17629) from Japan Society for the Promotion of Science.

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Rate of convergence in the large diffusion limit for the heat equation with a dynamical boundary condition

Abstract

Keywords

1. Introduction

3. Estimates: Recalling the case Ω = R + N

Footnotes

Acknowledgements

References

3. Estimates: Recalling the case $Ω = R_{+}^{N}$