Sharp Interior and Boundary Regularity of the SMGTJ-Equation with Dirichlet or Neumann Boundary Control

Abstract

We consider the third order (in time) linear equation known as SMGTJ-equation, as defined on a multidimensional bounded domain and subject to either Dirichlet or Neumann boundary control. We then establish corresponding sharp interior and boundary regularity results.

Appendices

Appendix A

• 1. Cosine Operators. While we refer to standard work [6, 10, 11, 17, 22, 36, 40] etc for the topic of cosine operator theory on Banach space, we include here only a few results which are used and invoked in the text with reference to a Hilbert space H ($H=L^2(\Omega )$ in Part A; $H=L^2(\Omega )/\mathbb {R}$ in Part B). In line with the text, we let ($-A$) be the (strictly positive) self-adjoint infinitesimal generator of a strongly continuous (self-adjoint) cosine operator family ${\mathcal {C}}(t)$ with sine operator ${\mathcal {S}}(t)x = \int \nolimits _0^t {\mathcal {C}}(\tau )x d\tau , x \in H$, with $A^{\frac{1}{2}}{\mathcal {S}}(t)$ strongly continuous:

  $$\begin{aligned} {\mathcal {S}}(t-\tau ) = {\mathcal {S}}(t) {\mathcal {C}}(\tau ) - {\mathcal {C}}(t){\mathcal {S}}(\tau ) \end{aligned}$$

  (A.1a)
  $$\begin{aligned} {\mathcal {C}}(t-\tau ) = {\mathcal {C}}(t) {\mathcal {C}}(\tau ) - A{\mathcal {S}}(t){\mathcal {S}}(\tau ), \quad \tau , t \in \mathbb {R} \end{aligned}$$

  (A.1b)

  We have

  $$\begin{aligned} \dfrac{d^2 {\mathcal {C}}(t) x}{dt^2} = -A{\mathcal {C}}(t)x, x \in {\mathcal {D}}(A); \qquad \dfrac{d {\mathcal {C}}(t) x}{dt} = -A{\mathcal {S}}(t)x, x \in {\mathcal {D}}(A^{\frac{1}{2}}), \end{aligned}$$

  (A.2)

  ${\mathcal {C}}(t)$ is even on H, ${\mathcal {C}}(0)=I$; ${\mathcal {S}}(t)$ is odd on H, ${\mathcal {S}}(0)=0$. The above formulae (A.2) on H with $H \supset {\mathcal {D}}(A) \rightarrow H$ can be extended to $[{\mathcal {D}}(A)]'$ with A now the extension $A_e: H \rightarrow [{\mathcal {D}}(A)]'$, which we still denote by A.

• 2. Representation formulae of non-homogeneous boundary control for wave (second order hyperbolic) equations [25,26,27, 31, 41], [23, Sect. 3]

  Dirichlet case We return to the Dirichlet non-homogeneous w-problem in (2.15a)–(2.15c). Let D be the Dirichlet map in (2.21a), (2.21b) and $(-A)$ be the Dirichlet Laplacian in (2.2). Then

  $$\begin{aligned} w(t) = A \int \limits _0^t {\mathcal {S}}(t-\tau )Dg(\tau )d\tau ; \quad w_t(t) = A \int \limits _0^t {\mathcal {C}}(t-\tau )Dg(\tau )d\tau . \end{aligned}$$

  (A.3)

  Neumann case We now return to the Neumann non-homogeneous w-problem in (7.5a)–(7.5c). Let N be the Neumann map in (7.17a), (7.17b) and $(-A)$ be the Neumann Laplacian in (7.2). Then

  $$\begin{aligned} w(t) = A \int \limits _0^t {\mathcal {S}}(t-\tau )Ng(\tau )d\tau ; \quad w_t(t) = A \int \limits _0^t {\mathcal {C}}(t-\tau )Ng(\tau )d\tau . \end{aligned}$$

  (A.4)

• 3. Operator formulae for traces

  Let $(-A)$ be the Dirichlet Laplacian in (2.2) and D the Dirichlet map in (2.21a), (2.21b). Then [41], [29, p. 181]

  $$\begin{aligned} D^* A^* \phi = -\dfrac{\partial \phi }{\partial \nu }, \quad \phi \in {\mathcal {D}}(A), \end{aligned}$$

  (A.5)

  which can be extended to all $\phi \in H^{\frac{3}{2}+\epsilon }(\Omega ) \cap H_0^1(\Omega ), \epsilon >0$. Let now $(-A)$ be the Neumann Laplacian in (7.2) and N the Neumann map in (7.17a), (7.17b). Then [41], [29, p. 196]

  $$\begin{aligned} N^* A^* \phi = \phi \big |_{\Gamma } \quad \phi \in {\mathcal {D}}(A), \end{aligned}$$

  (A.6)

  which can be extended to all $\phi \in H^{\frac{3}{2}+\epsilon }(\Omega ) \cap H_0^1(\Omega ), \epsilon >0$, with $\left. \frac{\partial \phi }{\partial \nu }\right| _{\Gamma } = 0$.

Appendix B The Dual Problem of the Boundary Non-homogeneous Problem (2.1a)–(2.1c). A PDE-Approach

In this Appendix we consider the following two problems:

Problem #1 (2.1a)–(2.1c)

Problem #2 With $T>0$ arbitrary,

The v-problem (B.2a)–(B.2d) is dual to the y-problem (B.1a)–(B.1d) for zero I.C.: $y_0=y_1=y_2=0$, in the sense specified below

Theorem B.1

(i) Under the appropriate regularity assumptions on the data: $\{y_0,y_1,y_2\}, g$, and $\{v_0,v_1,v_2\}$—to be made explicit below—the following identity holds true, where $\langle \ , \ \rangle _{\Omega }$ denotes the duality pairing with respect to $H=L^2(\Omega )$ and $\langle \ , \ \rangle _{\Gamma }$ denotes the duality pairing with respect to $L^2(\Gamma )$:

$$\begin{aligned} \left\langle y_{tt}(T) + \alpha y_t(T), v_0 \right\rangle _{\Omega } - \left\langle y_{t}(T) + \alpha y(T), v_1 \right\rangle _{\Omega } + \left\langle y(T), v_2 \right\rangle _{\Omega } -b\left\langle y(T), \Delta v_0 \right\rangle _{\Omega } \end{aligned}$$

$$\begin{aligned} + \left\langle y_0, -v_{tt}(0) + \alpha v_t(0) + b\Delta v(0) \right\rangle _{\Omega } + \left\langle y_1, v_{t}(0) - \alpha v(0) \right\rangle _{\Omega } - \left\langle y_2, v(0) \right\rangle _{\Omega } \end{aligned}$$

$$\begin{aligned} - \left\langle c^2 \dfrac{\partial y}{\partial \nu } + b \dfrac{\partial y_t}{\partial \nu }, v \right\rangle _{L^2(\Sigma )} + \left\langle c^2 y + b y_t, \dfrac{\partial v}{\partial \nu } \right\rangle _{L^2(\Sigma )} = 0. \end{aligned}$$

(B.3)

(ii) Consider the Dirichlet non-homogeneous condition (B.1c) with zero I.C. $y_0=y_1=y_2=0$, coupled with the corresponding homogeneous Dirichlet condition (B.2c). Then identity (B.3) specializes to

$$\begin{aligned} \left\langle y_{tt}(T) + \alpha y_t(T), v_0 \right\rangle _{\Omega } - \left\langle y_{t}(T) + \alpha y(T), v_1 \right\rangle _{\Omega } + \left\langle y(T), v_2 \right\rangle _{\Omega } -b\left\langle y(T), \Delta v_0 \right\rangle _{\Omega } \nonumber \\ = - \left\langle c^2 g + b g_t, \dfrac{\partial v}{\partial \nu } \right\rangle _{L^2(0,T;L^2(\Gamma ))}. \end{aligned}$$

(B.4)

(iii) Consider the Neumann non-homogeneous condition (B.1d) with zero I.C. $y_0=y_1=y_2=0$, coupled with the corresponding homogeneous Neumann condition (B.2d). Then identity (B.3) specializes to

$$\begin{aligned} \begin{aligned} \left\langle y_{tt}(T) + \alpha y_t(T), v_0 \right\rangle _{\Omega } - \left\langle y_{t}(T) + \alpha y(T), v_1 \right\rangle _{\Omega } + \left\langle y(T), v_2 \right\rangle _{\Omega }&\\ -b\left\langle y(T), \Delta v_0 \right\rangle _{\Omega }&\\ =&\left\langle c^2 g + b g_t, v \right\rangle _{L^2(0,T;L^2(\Gamma ))}. \end{aligned} \end{aligned}$$

(B.5)

Proof

Step 1 Multiply (B.1a) by v and integrate by parts. We obtain:

(B.6)

(B.7)

(B.8)

(B.9)

Step 2 We sum up: and obtain

$$\begin{aligned} \begin{aligned}&\left\langle y_{tt}(T) + \alpha y_{t}(T), v(T)\right\rangle _{\Omega } - \left\langle y_{t}(T) + \alpha y(T), v_t(T)\right\rangle _{\Omega } + \left\langle y(T), v_{tt}(T)\right\rangle _{\Omega } - b\left\langle y(T), \Delta v(T)\right\rangle _{\Omega } \\ +&\left\langle y(0), -v_{tt}(0) + \alpha v_{t}(0) + b\Delta v(0)\right\rangle _{\Omega } + \left\langle y_1, v_{t}(0) - \alpha v(0)\right\rangle _{\Omega } - \left\langle y_2, v(0)\right\rangle _{\Omega } \\ -&\int \limits _Q y \left[ v_{ttt} - \alpha v_{tt} + c^2 \Delta v - b\Delta v_t \right] dQ - \left\langle c^2 \dfrac{\partial y}{\partial \nu } + b \dfrac{\partial y_t}{\partial \nu }, \nu \right\rangle _{L^2(\Sigma )} + \left\langle c^2 y + b y_t, \dfrac{\partial v}{\partial \nu } \right\rangle _{L^2(\Sigma )} = 0. \end{aligned} \end{aligned}$$

(B.10)

Step 3 The $\displaystyle \int \nolimits _Q$-term in (B.10) vanishes because of (B.2a). Next, we use the I.C. in (B.2b) for the v-problem at $t=T$, and identity (B.10) reduces to (B.3). Part (i) is proved.

Step 4 In the Dirichlet case, use $y\big |_{\Sigma } = 0$ in (B.1c) and $v\big |_{\Sigma } \equiv g$ in (B.2c). Then, identity (B.3) reduces to identity (B.4).

Step 5 In the Neumann case, use $\left. \dfrac{\partial y}{\partial \nu }\right| _{\Sigma } = g$ in (B.1d) and $\left. \dfrac{\partial v}{\partial \nu }\right| _{\Sigma } \equiv 0$ in (B.2d). Then identity (B.3) reduces to identity (B.5).

The next is a preliminary result.

Corollary B.2

With reference to the Dirichlet-Problem # 1 in (B.1a)–(B.1d) with I.C. $y_0=y_1=y_2=0$ and corresponding Dirichlet Problem # 2 in (B.2a)–(B.2d), assume

$$\begin{aligned} g \in H^1(0,T_1; L^2(\Gamma )), \quad \{v_0, v_1, v_2\} \in U_3 = {\mathcal {D}}(A) \times {\mathcal {D}}(A^{\frac{1}{2}}) \times H. \end{aligned}$$

(B.11)

Then, for any $t, 0 < t \le T_1$:

$$\begin{aligned} y(t), y_{t}(t), y_{tt}(t) \in H \times [{\mathcal {D}}(A^{\frac{1}{2}})]' \times [{\mathcal {D}}(A)]'. \end{aligned}$$

(B.12)

Proof

We have Theorem 6.1 for any $0< T_1 < \infty$:

$$\begin{aligned} \{v_0, v_1, v_2\} \in U_3 \Longrightarrow \{v, v_t, v_{tt}\} \in C\left( [0,T_1];U_3 = {\mathcal {D}}(A) \times {\mathcal {D}}(A^{\frac{1}{2}}) \times H\right) \end{aligned}$$

(B.13)

so that just by trace theory

$$\begin{aligned} \dfrac{\partial v}{\partial \nu } \in C\left( [0,T];H^{\frac{1}{2}}(\Gamma )) \right) \end{aligned}$$

(B.14)

Thus, the RHS of identity (B.4) is well defined by (B.11), (B.14), finite on any finite time interval. Here T is an arbitrary point $0< T \le T_1$. We then focus on the LHS of identity (B.4) to make sure that each term is well defined as a duality pairing w.r.t. $H=L^2(\Omega )$. We obtain

$$\begin{aligned} \left. \begin{aligned} \left. \begin{aligned} v_2 = v_{tt}(T) \in H&\Longrightarrow y(T) \in H \\[2mm] v_1 = v_{t}(T) \in {\mathcal {D}}(A^{\frac{1}{2}})&\Longrightarrow y_t(T) + \alpha y(T) \in [{\mathcal {D}}(A^{\frac{1}{2}})]' \end{aligned} \right\} \Longrightarrow y_t(T) \in [{\mathcal {D}}(A^{\frac{1}{2}})]' \\[2mm] v_0 = v(T) \in {\mathcal {D}}(A) \Longrightarrow y_{tt}(T) + \alpha y_t(T) \in [{\mathcal {D}}(A)]' \end{aligned}\right\} \Longrightarrow y_{tt}(T) \in [{\mathcal {D}}(A)]'. \end{aligned}$$

(B.15)

This takes care of the first three therms on the LHS of (B.4). Notice then that the fourth term $\left\langle y(T), \Delta v_0 \right\rangle _{\Omega }$ is likewise automatically well-posed with $v_0 \in {\mathcal {D}}(A), \Delta v_0 \in H = L^2(\Omega ), y(T) \in H$.$\square$