Rate convergence of the heat equation as diffusion tends to zero

Question

Is there a good reference for the following problem? Consider any smooth bounded domain $\Omega$ and solve the heat equation

\begin{align} \partial_t u^\kappa &= \kappa \Delta u^\kappa,\\ u^\kappa|_{\partial\Omega}&=0,\\ u^\kappa|_{t=0}&=u_0. \end{align}

What is the rate of convergence as $\kappa\to 0$ of $u^\kappa(t)$ to $u_0$?

Answer 1

In a nutshell: The rate of convergence depends on the smoothness of $u_0$.

Here are the details.

Setting:

• Throughout I assume that $\kappa > 0$ is a real number, i.e. it does not depend on the spatial variable.

• I'm going to consider the equation in the space $L^2(\Omega)$ (but similar observations also hold on $L^p(\Omega)$ for $p \in [1,\infty)$).

• Nomenclature: For functions $f: (0,\infty) \to L^2(\Omega)$ and $r: (0,\infty) \to (0,\infty)$ I will write $f \in \mathcal{O}(r)$ if there exists a real number $c \ge 0$ such that $\|f(s)\| \le c r(s)$ for all sufficiently small $s \in (0,\infty)$ (where $\|\cdot\|$ denotes the norm in $L^2(\Omega)$).

• Let $A$ denote the realisation of the Dirichet Laplace operator on $L^2(\Omega)$, i.e. \begin{align*} D(A) & = \{u \in H^1_0(\Omega): \; \Delta u \in L^2(\Omega)\}, \\ Au & = \Delta u; \end{align*} here $\Delta u$ is to be understood in the sense of distributions. Note that $D(A)$ coincides with $H^2(\Omega) \cap H^1_0(\Omega)$ if $\Omega$ has $C^2$-boundary (by elliptic regularity theory).

• The operator $\kappa A$ (with domain $D(\kappa A) = D(A)$) generates a $C_0$-semigroup $(e^{t \kappa A})_{t \ge 0}$; each operator $e^{t \kappa A}$ has norm at most $1$. For every $u_0 \in L^2(\Omega)$ the (mild) solution of the heat equation in the question is given by the mapping \begin{align*} [0,\infty) \ni t \mapsto e^{t \kappa A}u_0 \in L^2(\Omega); \end{align*} this is even a classical solution if $u_0 \in D(A)$.

To answer the question we can use the following idea: If we fix a time $t \in (0,\infty)$, then the behaviour of $e^{t\kappa A}$ as $\kappa \to 0$ is the same as the behaviour of $e^{sA}$ as $s \to 0$ (set $s := t\kappa$); thus, the question is essentially "how continuous" the mapping $s \mapsto e^{sA}u_0$ is at $s = 0$.

Now, the answer consists of a negative and a positive result:

Theorem 1 (negative result). There is no rate which works for all $u_0 \in L^2(\Omega)$. More precisely: Fix $t \in (0,\infty)$. For every continuous function $r: (0,\infty) \to (0,\infty)$ which fulfils $r(s) \to 0$ as $s \to 0$ there exists a vector $u_0 \in L^2(\Omega)$ such that the mapping $\kappa \mapsto e^{t\kappa A}u_0 - u_0$ is not contained in $\mathcal{O}(r)$.

Proof. Assume to the contrary that we can find a continuous function $r: (0,\infty) \to (0,\infty)$ which fulfils $r(s) \to 0$ as $s \to 0$ as well as $\big(\kappa \mapsto e^{t\kappa A}u_0 - u_0\big) \in \mathcal{O}(r)$ for each $u_0 \in L^2(\Omega)$. Then it follows from the uniform boundedness principle that the operator family $\big((e^{t \kappa A} - \operatorname{id})/r(\kappa)\big)_{\kappa \in (0,1]}$ is bounded with respect to the operator norm (here, $\operatorname{id}$ denotes the identity operator on $L^2(\Omega)$). Hence, $e^{t\kappa A}$ converges to $\operatorname{id}$ with respect to the operator norm as $\kappa \to 0$. Consequently, $e^{sA}$ converges to $\operatorname{id}$ with respect to operator norm as $s \to 0$. By standard $C_0$-semigroup theory this can only be true if $A$ is a bounded operator, which is not fulfilled. This proves Theorem 1.

Theorem 2 (positive result). Let $u_0 \in D(A)$. Then \begin{align*} \|e^{t\kappa A} u_0 - u_0\| \le t\kappa \|\Delta u_0\| \end{align*} for all $t,\kappa \in (0,\infty)$. Hence, the convergence of $e^{t\kappa A}u_0$ to $u_0$ as $\kappa \to 0$ has a linear rate.

Proof. Fix $t,\kappa \in (0,\infty)$. Since $u_0 \in D(A)$, we have \begin{align*} e^{t\kappa A}u_0 - u_0 = \int_0^{\kappa t} e^{sA}A u_0 \, ds \end{align*} (see for instance [Engel/Nagel: One-Parameter Semigroups for Linear Evolution Equations (2000), Lemma II.1.3(iv)]). The assertion now readily follows by taking the norm on both sides of this equation (use that $Au_0 = \Delta u_0$ and that $\|e^{sA}\| \le 1$ for each $s$).