When I read the paper "On the attractor for a semilinear wave equation with critical exponent and nonlinear boundary dissipation" by Igor Chueshov,Matthias Eller and Irena Lasiecka,I encounter a difficulty: Suppose that $\Omega$ is a simple connected domain with smooth boundary, the authors introduced a Robin Laplacian operator $$\Delta_{R}:L^{2}(\Omega)\to L^{2}(\Omega).$$This is an unbounded operator with the domain $$D(\Delta_{R})=\{u\in H^{2}(\Omega):\partial_{\nu}u+u=0\ on\ \partial\Omega\}.$$Moreover, the Robin Laplacian can be extended to a continuous operator $\Delta_{R}:H^{1}(\Omega)\to H^{1}(\Omega)'$ by $$(-\Delta_{R}u,v)_{L^{2}(\Omega)}=(\nabla u,\nabla v)_{L^{2}(\Omega)}+<u,v>_{L^{2}(\partial\Omega)}.$$ Then the authors say "this extension is the duality map $H^{1}(\Omega)$ into $(H^{1}(\Omega))'$" when we equip $H^{1}(\Omega)$ the norm $$\|u\|=\sqrt{(\nabla u,\nabla v)_{L^{2}(\Omega)}+<u,v>_{L^{2}(\partial\Omega)}}.$$ So, firstly, I want to know if this norm is equivalent to the usual Sobolev norm $$\|u\|=\sqrt{(\nabla u,\nabla v)_{L^{2}(\Omega)}+(u,v)_{L^{2}(\Omega)}}$$ when $u$ satisfies the robin boundary conditon? Next, the authors said $$D((-\Delta_{R}))^{\frac{1}{2}}\sim H^{1}(\Omega),$$ why? the authors offered an reference but it is french, but I don't understand french. the reference is "Grisvard,P. Characterisation de Quelques Esoaces d'interpolation. Archives Rational Mechanics and Analysis 1967,26,40-63"

Any comments and hints are welcome, thank you very much!!!


1 Answer 1


I] First the equivalence of norms.

We set, \begin{align*} \lVert u\rVert_{\mathrm{H}^1(\Omega)} := \left( \lVert u\rVert_{\mathrm{L}^2(\Omega)}^2+\lVert \nabla u\rVert_{\mathrm{L}^2(\Omega)}^2\right)^{\frac{1}{2}}\text{ and } \widetilde{\lVert u\rVert}_{\Omega}:=\left( \lVert u\rVert_{\mathrm{L}^2(\partial\Omega)}^2+\lVert \nabla u\rVert_{\mathrm{L}^2(\Omega)}^2\right)^{\frac{1}{2}}\text{ . } \end{align*}

  1. By the trace Theorem [2, Thm 1, p.273], there exists a constant $C>0$, such that for all $u\in\mathrm{H}^1(\Omega)$, \begin{align*} \widetilde{\lVert u\rVert}_{\Omega}=\left( \lVert u\rVert_{\mathrm{L}^2(\partial\Omega)}^2+\lVert \nabla u\rVert_{\mathrm{L}^2(\Omega)}^2\right)^{\frac{1}{2}} &\leqslant \left( C^2\lVert u\rVert_{\mathrm{H}^1(\Omega)}^2+\lVert \nabla u\rVert_{\mathrm{L}^2(\Omega)}^2\right)^{\frac{1}{2}}\\ &\leqslant \left(1+ C^2\right)^{\frac{1}{2}} \lVert u\rVert_{\mathrm{H}^1(\Omega)}\text{.} \end{align*}
  2. For the reverse inequality, we argue by contradiction, assuming that $\frac{\lVert \cdot\rVert_{\mathrm{H}^1(\Omega)}}{\widetilde{\lVert \cdot\rVert}_{\Omega}}$ is not bounded (i.e. there is no $c_0>0$ such that ${\lVert \cdot\rVert_{\mathrm{H}^1(\Omega)}} \leqslant c_0{\widetilde{\lVert \cdot\rVert}_{\Omega}}$). Therefore, up to a renormalization, it should exist a sequence $(u_k)_{k\in\mathbb{N}} \subset \mathrm{H}^1(\Omega)$ such that \begin{align*} \widetilde{\lVert u_k\rVert}_{\Omega}\underset{k\rightarrow +\infty}{\longrightarrow} 0\text{ and }\forall k\in\mathbb{N},\,\,\lVert u_k \rVert_{\mathrm{H}^1(\Omega)}=1\,\text{.} \end{align*}

In particular, we obtain

  • $\nabla u_k \underset{k\rightarrow +\infty}{\longrightarrow} 0$ in $\mathrm{L}^2(\Omega)$;
  • $ {u_k}_{|_{\partial\Omega}} \underset{k\rightarrow +\infty}{\longrightarrow} 0$ in $\mathrm{L}^2(\partial\Omega)$;
  • The sequence $(u_k)_{k\in\mathbb{N}}$ is bounded in $\mathrm{H}^1(\Omega)$.

Hence, from the last point above, since $\Omega$ is bounded, we may apply Rellich-Kondrachov Theorem [1, Thm 9.16, p.285],so that by compactness of the embedding $\mathrm{H}^1(\Omega)\hookrightarrow \mathrm{L}^2(\Omega)$, there exists a function $u\in\mathrm{L}^2(\Omega)$, and an extracted subsequence such that \begin{align*} {u_{k_{j}}} \underset{j\rightarrow +\infty}{\longrightarrow} u\text{ in } \mathrm{L}^2(\Omega)\text{ . } \end{align*}

Up to extract subsequences again, we may apply a corollary of the Riesz-Fischer Theorem [1, Thm 4.9, p.94], to get the existence of $\phi\in\mathrm{L}^2(\Omega)$, $\varphi\in\mathrm{L}^2(\partial\Omega)$, such that

  • ${u_{k_{j}}} \underset{j\rightarrow +\infty}{\longrightarrow} u$, $\nabla {u_{k_{j}}} \underset{j\rightarrow +\infty}{\longrightarrow} 0$ a.e. on $\Omega$ with $\lvert{u_{k_{j}}}\rvert+\lvert \nabla {u_{k_{j}}}\rvert \leqslant \phi$.
  • ${u_{k_{j}}}_{|_{\partial\Omega}} \underset{j\rightarrow +\infty}{\longrightarrow} 0$ a.e. on $\partial\Omega$ with $\lvert{u_{k_{j}}}_{|_{\partial\Omega}}\rvert \leqslant \varphi$.

By the trace theorem and the green formula for all $1\leqslant \ell \leqslant n$, we may write for all $j\in\mathbb{N}$, and $v\in\mathrm{C}^1(\overline{\Omega})$, \begin{align*} \int_{\Omega}{u_{k_j}\,(\partial_{x_\ell}v)}=-\int_{\Omega}{(\partial_{x_\ell}u_{k_j})\,v}+\int_{\partial\Omega}{{u_{k_j}}_{|_{\partial\Omega}}v\,\nu_\ell}\text{ , } \end{align*} so that the Dominated Convergence Theorem yields \begin{align*} \int_{\Omega}{u\,(\partial_{x_\ell}v)}=0\text{ . } \end{align*}

From above equality, we deduce in particular that $\nabla u =0 \in\mathrm{L}^2(\Omega)$, so $u\in\mathrm{H}^1(\Omega)$. The continuity of the trace gives us necessarily $u\in\mathrm{H}^1_0(\Omega)$, but since $\nabla u =0$, the Poincaré inequality implies that $u=0$ almost everywhere. To conclude, we recall that \begin{align*} \lVert u_{k_j} \rVert_{\mathrm{H}^1(\Omega)}=1\,\text{ , } \end{align*} which is a contradiction since we just proved that ${u_{k_{j}}} \underset{j\rightarrow +\infty}{\longrightarrow} 0$ in $\mathrm{H}^1(\Omega)$.

Remark : The proof is still valid for the $\mathrm{L}^p$ Sobolev space $\mathrm{H}^{1,p}$, $1\leqslant p<+\infty$.

II] The Robin Laplacian and its squareroot

I'm going to introduce the Laplacian with Robin boundary conditions in a different but similar and equivalent way.

We recall the following fact from [3, Chapter 1] (combination of many results) :

Definition 0: Let $\mathrm{H}$ be a complex Hilbert space. Let $\mathrm{D}(\mathfrak{a})$ be a subspace of $\mathrm{H}$. Let $\mathfrak{a}: \mathrm{D}(\mathfrak{a}) \times \mathrm{D}(\mathfrak{a}) \rightarrow \mathbb{K}$ be a sesquilinear form we say that:

a) $\mathfrak{a}$ is densely defined if $\mathrm{D}(\mathfrak{a})$ is dense in $\mathrm{H}$.

b) $\mathfrak{a}$ is accretive if \begin{align*} \Re \mathfrak{a}(u, u) \geq 0 \text { for all } u \in \mathrm{D}(\mathfrak{a}) . \end{align*}

c) $\mathfrak{a}$ is continuous if there exists a non-negative constant $M$ such that \begin{align*} |\mathfrak{a}(u, v)| \leq M\|u\|_{\mathfrak{a}}\|v\|_{\mathfrak{a}} \text { for all } u, v \in \mathrm{D}(\mathfrak{a}) \end{align*} where $\|u\|_{\mathfrak{a}}:=\sqrt{\Re \mathfrak{a}(u, u)+\|u\|^2}$.

d) $\mathfrak{a}$ is closed if $\left(\mathrm{D}(\mathfrak{a}),\|\cdot\|_{\mathfrak{a}}\right)$ is a complete space.

Theorem A: Let $\mathrm{H}$ be a complex Hilbert space. Let $\mathrm{D}(\mathfrak{a})$ be a subspace of $\mathrm{H}$, and consider $\mathfrak{a}\,:\,\mathrm{D}(\mathfrak{a})\times \mathrm{D}(\mathfrak{a})\longrightarrow \mathbb{C}$ a sesquilinear form. If $(\mathrm{D}(\mathfrak{a}),\mathfrak{a})$ is a densely defined, accretive, continuous, closed and symmetric sesquilinear form then there exists a unique densely defined, self-adjoint, closed operator $(\mathrm{D}(A),A)$ such that \begin{align*} \mathfrak{a}(u,v)=\langle Au,v\rangle \text{ , } \forall(u,v)\in\mathrm{D}(A)\times\mathrm{D}(\mathfrak{a}) \text{.} \end{align*} Moreover, $\mathrm{D}(A)$ is dense in $\mathrm{D}(\mathfrak{a})$.

Now, we want to apply above theorem to \begin{alignat*}{2} \mathfrak{a}_\mathcal{R} \,\colon\, \mathrm{H}^1(\Omega)\times\mathrm{H}^1(\Omega) &\longrightarrow \mathbb{C}\\ (u,v)\qquad &\longmapsto \int_{\Omega} \nabla u \cdot\overline{\nabla v} + \int_{\partial\Omega} \,u\,\overline{v} \end{alignat*} setting $\mathrm{D}(\mathfrak{a}_\mathcal{R})=\mathrm{H}^1(\Omega)$ which is obvisouly dense in $\mathrm{H}=\mathrm{L}^2(\Omega)$.

  1. Let us check assumptions of the Theorem A. The symmetry, and the density assumptions are already checked.
  • Accretivity (Coercivity) and closedness : By what have been achieved in the first part above for all $u\in\mathrm{D}(\mathfrak{a}_\mathcal{R})=\mathrm{H}^1(\Omega)$, we have \begin{align*} \mathfrak{a}_\mathcal{R}(u,u) + \lVert u\rVert_{\mathrm{L}^2(\Omega)}^2 = \lVert u\rVert_{\mathrm{L}^2(\partial\Omega)}^2 +\lVert \nabla u\rVert_{\mathrm{L}^2(\Omega)}^2 + \lVert u\rVert_{\mathrm{L}^2(\Omega)}^2 \simeq_{\Omega} \lVert u\rVert_{\mathrm{H}^1(\Omega)}^2 \end{align*} which yields directly the coercivity and the closedness.

  • Continuity : By what have been achieved in the first part above for all $u,v\in\mathrm{D}(\mathfrak{a}_\mathcal{R})=\mathrm{H}^1(\Omega)$, we apply Cauchy-Schwarz to obtain \begin{align*} \lvert\mathfrak{a}_\mathcal{R}(u,v)\lvert &\leqslant \lVert u\rVert_{\mathrm{L}^2(\partial\Omega)} \lVert v\rVert_{\mathrm{L}^2(\partial\Omega)}+\lVert \nabla u\rVert_{\mathrm{L}^2(\Omega)}\lVert \nabla v\rVert_{\mathrm{L}^2(\Omega)}\\ &\lesssim_{\Omega} \lVert u\rVert_{\mathrm{H}^1(\Omega)}\lVert v\rVert_{\mathrm{H}^1(\Omega)} \end{align*}

Therefore by above Theorem A, there exists a unique densely defined, self-adjoint, closed operator operator $(\mathrm{D}(A_\mathcal{R}),A_\mathcal{R})$ such that \begin{align*} \langle \nabla u, \nabla v\rangle_{\Omega} + \langle u, v\rangle_{\partial\Omega} = \langle A_{\mathcal{R}}u,v\rangle_{\Omega} \text{ , } \forall(u,v)\in\mathrm{D}(A_\mathcal{R})\times\mathrm{H}^1(\Omega) \text{.} \end{align*} Moreover, $\mathrm{D}(A_\mathcal{R})$ is dense in $\mathrm{H}^1(\Omega)$.

For now until the end, we set \begin{align*} \mathrm{D}(\Delta_{\mathcal{R}}) &:= \{ u\in\mathrm{H}^1(\Omega)\,|\, \Delta u\in\mathrm{L}^2(\Omega) \text{ and } {\partial_{\nu}u}_{|_{\partial\Omega}} + u_{|_{\partial\Omega}}=0\}\\ -\Delta_{\mathcal{R}}u&:=-\Delta u, \qquad u\in\mathrm{D}(\Delta_{\mathcal{R}}) \text{.} \end{align*}

  1. We want to prove that $(\mathrm{D}(A_\mathcal{R}),A_\mathcal{R})=(\mathrm{D}(\Delta_{\mathcal{R}}),-\Delta_{\mathcal{R}})$. We just prove $(\mathrm{D}(A_\mathcal{R}),A_\mathcal{R})\subset(\mathrm{D}(\Delta_{\mathcal{R}}),-\Delta_{\mathcal{R}})$.

Let $u\in \mathrm{D}(A_\mathcal{R})$. First we have \begin{align} \langle \nabla u, \nabla v\rangle_{\Omega} + \langle u, v\rangle_{\partial\Omega} = \langle A_{\mathcal{R}}u,v\rangle_{\Omega} \text{ , } \forall v\in\mathrm{H}^1(\Omega) \text{.} \end{align} In particular, \begin{align*} \langle \nabla u, \nabla v\rangle_{\Omega} = \langle A_{\mathcal{R}}u,v\rangle_{\Omega} \text{ , } \forall v\in\mathrm{C}_c^\infty(\Omega) \text{,} \end{align*} so that $A_\mathcal{R}u= - \Delta u$ in $\mathcal{D}'(\Omega)$, hence in $\mathrm{L}^2(\Omega)$, so that if we get back on the first identity, we deduce \begin{align*} \langle \nabla u, \nabla v\rangle_{\Omega} + \langle u, v\rangle_{\partial\Omega} = \langle -\Delta u,v\rangle_{\Omega} \text{ , } \forall v\in\mathrm{H}^1(\Omega) \text{.} \end{align*} We can specialise it into \begin{align*} \langle \nabla u, \nabla v\rangle_{\Omega} + \langle u, v\rangle_{\partial\Omega} &= \langle -\Delta u,v\rangle_{\Omega} \text{ , } \forall v\in\mathrm{C}^\infty(\overline{\Omega})\\ \langle \nabla u, \nabla v\rangle_{\Omega} - \langle -\Delta u,v\rangle_{\Omega} + \langle u, v\rangle_{\partial\Omega} &= 0\\ \langle \partial_{\nu} u,v\rangle_{\partial\Omega} + \langle u, v\rangle_{\partial\Omega} &= 0 \end{align*} therefore ${\partial_{\nu}u}_{|_{\partial\Omega}} + u_{|_{\partial\Omega}}=0$, and we have $(\mathrm{D}(A_\mathcal{R}),A_\mathcal{R})\subset(\mathrm{D}(\Delta_{\mathcal{R}}),-\Delta_{\mathcal{R}})$.

Remark : Notice that $\mathrm{D}(\Delta_{\mathcal{R}})$ is well defined. Since $u\in\mathrm{H}^1(\Omega)$, it ensures that $u_{|_{\partial\Omega}}$ makes sense in $\mathrm{H}^{\frac{1}{2}}(\partial\Omega)\subset\mathrm{L}^2(\partial\Omega)\subset\mathrm{H}^{-\frac{1}{2}}(\partial\Omega)$. The conditions $\nabla u, \Delta u\in \mathrm{L}^2(\Omega)$ ensure that $\partial_{\nu}u_{|_{\partial\Omega}}$ makes sense in $\mathrm{H}^{-\frac{1}{2}}(\partial\Omega)$, so that the equality ${\partial_{\nu}u}_{|_{\partial\Omega}} + u_{|_{\partial\Omega}}=0$ makes sense in $\mathrm{H}^{-\frac{1}{2}}(\partial\Omega)$ at least, but since $u_{|_{\partial\Omega}}\in \mathrm{L}^2(\partial\Omega)$, one also obtains $\partial_{\nu}u_{|_{\partial\Omega}} \in \mathrm{L}^2(\partial\Omega)$.

  1. The squareroot property : $\mathrm{D}((-\Delta_{\mathcal{R}})^\frac{1}{2})=\mathrm{H}^1(\Omega)$.

We recall that the squareroot of a densely defined, closed, non-negative self-adjoint operator $(\mathrm{D}(A),A)$ on a complex Hilbert space $\mathrm{H}$ is defined through the identity \begin{align*} \langle Au,u\rangle = \lVert A^{\frac{1}{2}}u\rVert^2 \text{ . } \end{align*}

In our specific case, we have for all $u\in\mathrm{D}(\Delta_{\mathcal{R}})$, \begin{align*} \lVert (-\Delta_{\mathcal{R}})^\frac{1}{2}u\rVert^2_{\mathrm{L}^2(\Omega)} = \langle -\Delta_{\mathcal{R}}u,u\rangle =\mathfrak{a}_{\mathcal{R}}(u,u) \text{ . } \end{align*}

But if you still had in mind, what we have proved in the part I], we have \begin{align*} \lVert (-\Delta_{\mathcal{R}})^\frac{1}{2}u\rVert^2_{\mathrm{L}^2(\Omega)} \simeq_{\Omega} \lVert u\rVert_{\mathrm{H}^1(\Omega)}^2 \text{ . } \end{align*}

But Theorem A also told us that $\mathrm{D}(\Delta_{\mathcal{R}})$ is dense in $\mathrm{D}(\mathfrak{a}_{\mathcal{R}})=\mathrm{H}^1(\Omega)$, from which we can conclude that the domain of the squareroot is exactly $\mathrm{H}^1(\Omega)$. More generally, the domain of the squareroot of a densely defined, closed, non-negative self-adjoint operator is always its form domain.

[1] Brezis, H., Functional analysis, Sobolev spaces and partial differential equations, Universitext. New York, NY: Springer (ISBN 978-0-387-70913-0/pbk; 978-0-387-70914-7/ebook). xiii, 599 p. (2011). ZBL1220.46002.

[2] Evans, L. C., Partial differential equations, Graduate Studies in Mathematics 19. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-4974-3/hbk). xxi, 749 p. (2010). ZBL1194.35001.

[3] Ouhabaz, E. M., Analysis of heat equations on domains, London Mathematical Society Monographs Series 31. Princeton, NJ: Princeton University Press (ISBN 0-691-12016-1/hbk). xi, 284 p. (2005). ZBL1082.35003.

  • $\begingroup$ Excellent!!!! Thank you for your fruitful description!!! I would like to take a day to digest your answer. $\endgroup$ Sep 27, 2022 at 14:47

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