The fourth-order elliptic partial differential operator \(\Delta ^{2}-\kappa^{2}\Delta\), for \(\kappa>0\), corresponds to the fourth-order elliptic equation:
$$ \bigl(\Delta^{2}-\kappa^{2}\Delta\bigr)[u]=0. $$
(3.1)
Using the multiplication rule on the Clifford algebra \(\operatorname{Cl}(V_{3,3})\), equation (3.1) may also be written as
$$DDL_{\kappa}L_{-\kappa}[u]=DDL_{-\kappa}L_{\kappa}[u]=L_{-\kappa }L_{\kappa}DD[u]=L_{\kappa}L_{-\kappa}DD[u]=0, $$
where \(L_{\kappa}=D+\kappa\) and \(L_{-\kappa}=D-\kappa\).
Lemma 3.1
[7, 8]
Let
$$ E(\kappa, \mathbf{x})=\frac{1}{4\pi\kappa^{2}}\frac{e^{-\kappa\|\mathbf {x}\|}-1}{\|\mathbf{x}\|}. $$
(3.2)
Then the kernel function
\(E(\kappa, \mathbf{x})\)
is the fundamental solution to (3.1) in
\(\mathbb{R}^{3}\).
Let
$$ \left \{ \textstyle\begin{array}{l} H_{1}(\kappa, \mathbf{x})=\frac{1}{4\pi}\frac{\mathbf{x}}{\|\mathbf{x}\| ^{3}}, \\ H_{2}(\kappa, \mathbf{x})=-\frac{1}{4\pi}\frac{1}{\|\mathbf{x}\|}, \\ H_{3}(\kappa, \mathbf{x})=\frac{1}{4\pi\kappa^{2}} [(\frac{\mathbf {x}}{\|\mathbf{x}\|^{3}} +\kappa\frac{1}{\|\mathbf{x}\|})(e^{-\kappa\|\mathbf{x}\|}-1)+\kappa \frac{\mathbf{x}}{\|\mathbf{x}\|^{2}}e^{-\kappa\|\mathbf{x}\|} ], \\ H_{4}(\kappa, \mathbf{x})=-E(\kappa, \mathbf{x})=-\frac{1}{4\pi\kappa ^{2}}\frac{e^{-\kappa\|\mathbf{x}\|}-1}{\|\mathbf{x}\|}, \end{array}\displaystyle \right . $$
(3.3)
and
$$ \left \{ \textstyle\begin{array}{l} E_{1}(\kappa, \mathbf{x})=\frac{1}{4\pi} [\frac{\mathbf{x}}{\| \mathbf{x}\|^{3}} +\kappa\frac{\mathbf{x}}{\|\mathbf{x}\|^{2}}+\kappa\frac{1}{\|\mathbf {x}\|} ]e^{-\kappa\|\mathbf{x}\|}, \\ E_{2}(\kappa, \mathbf{x})=-\frac{1}{4\pi}\frac{e^{-\kappa\|\mathbf{x}\| }}{\|\mathbf{x}\|}, \\ E_{3}(\kappa, \mathbf{x})=\frac{1}{4\pi\kappa^{2}} [\frac{\mathbf {x}}{\|\mathbf{x}\|^{3}}(e^{-\kappa\|\mathbf{x}\|}-1) +\kappa\frac{\mathbf{x}}{\|\mathbf{x}\|^{2}}e^{-\kappa\|\mathbf{x}\| } ], \\ E_{4}(\kappa, \mathbf{x})=-E(\kappa, \mathbf{x})=-\frac{1}{4\pi\kappa ^{2}}\frac{e^{-\kappa\|\mathbf{x}\|}-1}{\|\mathbf{x}\|}, \end{array}\displaystyle \right . $$
(3.4)
where \(\kappa>0\), \(\mathbf{x}\in\mathbb{R}^{3}\setminus\{0\}\).
Lemma 3.2
Let
\(H_{i}(\kappa, \mathbf{x})\)
and
\(E_{i}(\kappa, \mathbf{x})\)
be as in (3.3) and (3.4), \(i=1,2,3,4\). Then
$$ \left \{ \textstyle\begin{array}{l} L_{-\kappa}[H_{4}(\kappa, \mathbf{x})]=[H_{4}(\kappa, \mathbf {x})]L_{-\kappa}=H_{3}(\kappa, \mathbf{x}), \\ L_{\kappa}[H_{3}(\kappa, \mathbf{x})]=[H_{3}(\kappa, \mathbf {x})]L_{\kappa}=H_{2}(\kappa, \mathbf{x}), \\ L_{-\kappa}[E_{2}(\kappa, \mathbf{x})]=[E_{2}(\kappa, \mathbf {x})]L_{-\kappa}=E_{1}(\kappa, \mathbf{x}), \\ L_{\kappa}[E_{1}(\kappa, \mathbf{x})]=[E_{1}(\kappa, \mathbf {x})]L_{\kappa}=0, \end{array}\displaystyle \right . $$
and
$$ \left \{ \textstyle\begin{array}{l} D[E_{4}(\kappa, \mathbf{x})]=[E_{4}(\kappa, \mathbf{x})]D=E_{3}(\kappa, \mathbf{x}), \\ D[E_{3}(\kappa, \mathbf{x})]=[E_{3}(\kappa, \mathbf{x})]D=E_{2}(\kappa, \mathbf{x}), \\ D[H_{2}(\kappa, \mathbf{x})]=[H_{2}(\kappa, \mathbf{x})]D=H_{1}(\kappa, \mathbf{x}), \\ D[H_{1}(\kappa, \mathbf{x})]=[H_{1}(\kappa, \mathbf{x})]D=0, \end{array}\displaystyle \right . $$
here
\(\kappa>0\), \(\mathbf{x}\in\mathbb{R}^{3}\setminus\{0\}\).
Remark 3.3
Let
$$ H_{3}^{*}(\kappa, \mathbf{x})= \frac{1}{4\pi\kappa^{2}} \biggl[\biggl(\frac {\mathbf{x}}{\|\mathbf{x}\|^{3}} -\kappa\frac{1}{\|\mathbf{x}\|}\biggr) \bigl(e^{-\kappa\|\mathbf{x}\|}-1\bigr)+\kappa \frac{\mathbf{x}}{\|\mathbf{x}\|^{2}}e^{-\kappa\|\mathbf{x}\|} \biggr] $$
(3.5)
and
$$ E_{1}^{*}(\kappa, \mathbf{x})= \frac{1}{4\pi} \biggl[\frac{\mathbf{x}}{\| \mathbf{x}\|^{3}} +\kappa\frac{\mathbf{x}}{\|\mathbf{x}\|^{2}}-\kappa \frac{1}{\|\mathbf {x}\|} \biggr]e^{-\kappa\|\mathbf{x}\|}. $$
(3.6)
When \(H_{3}^{*}(\kappa, \mathbf{x})\), \(E_{1}^{*}(\kappa, \mathbf{x})\) replace \(H_{3}(\kappa, \mathbf{x})\), \(E_{1}(\kappa, \mathbf{x})\) in (3.3), (3.4), respectively, we have the following results:
$$ \left \{ \textstyle\begin{array}{l} L_{\kappa}[H_{4}(\kappa, \mathbf{x})]=[H_{4}(\kappa, \mathbf {x})]L_{\kappa}=H^{*}_{3}(\kappa, \mathbf{x}), \\ L_{-\kappa}[H^{*}_{3}(\kappa, \mathbf{x})]=[H^{*}_{3}(\kappa, \mathbf {x})]L_{-\kappa}=H_{2}(\kappa, \mathbf{x}), \\ L_{\kappa}[E_{2}(\kappa, \mathbf{x})]=[E_{2}(\kappa, \mathbf {x})]L_{\kappa}=E^{*}_{1}(\kappa, \mathbf{x}), \\ L_{-\kappa}[E^{*}_{1}(\kappa, \mathbf{x})]=[E^{*}_{1}(\kappa, \mathbf {x})]L_{-\kappa}=0. \end{array}\displaystyle \right . $$
(3.7)
Let Ω be an open non-empty subset of \(\mathbb{R}^{3}\) with a Lyapunov boundary, \(u(\mathbf{x})=\sum_{A}e_{A}u_{A}(\mathbf{x})\), where \(u_{A}(\mathbf{x})\) are real functions. \(u(\mathbf{x})\) is called a Hölder continuous function on Ω̅ if the following condition is satisfied:
$$ \bigl\Vert u(\mathbf{x}_{1})-u(\mathbf{x}_{2})\bigr\Vert = \biggl[\sum_{A}\bigl\Vert u_{A}(\mathbf{x}_{1})-u_{A}( \mathbf{x}_{2})\bigr\Vert \biggr]^{\frac{1}{2}}\leq C\Vert \mathbf{x}_{1}-\mathbf{x}_{2}\Vert ^{\alpha}, $$
where, for any \(\mathbf{x}_{1}, \mathbf{x}_{2}\in\overline{\Omega}\), \(\mathbf{x}_{1}\neq\mathbf{x}_{2}\), \(0<\alpha\leq1\), C is a positive constant independent of \(\mathbf{x}_{1}\), \(\mathbf{x}_{2}\).
Lemma 3.4
Let
\(f, g\in C^{1}(\Omega, \operatorname{Cl}(V_{3,3}))\cap C(\overline{\Omega}, \operatorname{Cl}(V_{3,3}))\). Then
$$\begin{aligned} \int _{\partial\Omega}f\,d\sigma_{\mathbf{y}}g =& \int _{\Omega }[f]L_{\kappa}g\,dV+ \int _{\Omega}fL_{-\kappa}[g]\,dV \\ =& \int _{\Omega}[f]L_{-\kappa}g\,dV+ \int _{\Omega}fL_{\kappa }[g]\,dV, \end{aligned}$$
where
dV
denotes Lebesgue volume measure, dσ
stands for
\(\operatorname{Cl}(V_{3,3})\)-valued 2-differential form.
Proof
From Stokes’ theorem in Clifford analysis in [29], the results can be directly proved. □
Theorem 3.5
Suppose that Ω is an open bounded non-empty subset of
\(\mathbb {R}^{3}\)
with a Lyapunov boundary
∂Ω, \(u\in C^{4}(\Omega, \operatorname{Cl}(V_{3,3}))\cap C^{3}(\overline{\Omega}, \operatorname{Cl}(V_{3,3}))\). Then
$$\begin{aligned}& \sum_{i=1}^{3}(-1)^{i-1} \int _{\partial\Omega}H_{i}(\kappa , \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf{y}}D^{i-1}[u](\mathbf{y}) \\& \qquad {}- \int _{\partial\Omega}H_{4}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma _{\mathbf{y}}L_{-\kappa}\Delta[u](\mathbf{y}) \\& \qquad {}+ \int _{\Omega}H_{4}(\kappa, \mathbf{y}-\mathbf{x}) \bigl(\Delta ^{2}-\kappa^{2}\Delta\bigr)[u](\mathbf{y})\,dV \\& \quad = \left \{ \textstyle\begin{array}{l@{\quad}l} u(\mathbf{x}), & \mathbf{x}\in\Omega, \\ 0, & \mathbf{x}\in\mathbb{R}^{3}\setminus\overline{\Omega}, \end{array}\displaystyle \right . \end{aligned}$$
(3.8)
where
\(H_{i}(\kappa, \mathbf{y}-\mathbf{x})\) (\(i=1,2,3,4\)) are as in (3.3).
Proof
Let \(\mathbf{x}\in\mathbb{R}^{3}\setminus\overline{\Omega}\). Using Lemma 3.4 and Lemma 3.2, we get
$$\begin{aligned}& \int _{\Omega}H_{4}(\kappa, \mathbf{y}-\mathbf{x}) \bigl(\Delta^{2}-\kappa^{2}\Delta\bigr)[u](\mathbf{y})\,dV \\& \quad = \int _{\Omega}H_{4}(\kappa, \mathbf{y}- \mathbf{x})L_{\kappa }[L_{-\kappa}\Delta u](\mathbf{y})\,dV \\& \quad = \int _{\partial\Omega}H_{4}(\kappa, \mathbf{y}-\mathbf{x}) \,d \sigma_{\mathbf{y}}L_{-\kappa}\Delta[u](\mathbf{y}) \\& \qquad {} - \int _{\Omega} \bigl[H_{4}(\kappa, \mathbf{y}- \mathbf{x})\bigr]L_{-\kappa}\cdot L_{-\kappa }\Delta[u](\mathbf{y})\,dV \\& \quad = \int _{\partial\Omega}H_{4}(\kappa, \mathbf{y}-\mathbf{x}) \,d \sigma_{\mathbf{y}}L_{-\kappa}\Delta[u](\mathbf{y}) \\& \qquad {} - \int _{\Omega} H_{3}(\kappa, \mathbf{y}-\mathbf{x}) L_{-\kappa}\Delta[u](\mathbf{y})\,dV. \end{aligned}$$
(3.9)
Assume
$$ I_{1}= \int _{\Omega} H_{3}(\kappa, \mathbf{y}-\mathbf{x}) L_{-\kappa}\Delta[u](\mathbf{y})\,dV. $$
Applying Lemma 3.4 and Lemma 3.2 once again, we continue to calculate the integral \(I_{1}\) and get
$$\begin{aligned} I_{1} =& \int _{\partial\Omega} H_{3}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf{y}}\Delta [u](\mathbf{y}) - \int _{\Omega} \bigl[H_{3}(\kappa, \mathbf{y}-\mathbf{x}) \bigr]L_{\kappa}\cdot\Delta[u](\mathbf {y})\,dV \\ =& \int _{\partial\Omega} H_{3}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf{y}}\Delta [u](\mathbf{y}) - \int _{\Omega} H_{2}(\kappa, \mathbf{y}-\mathbf{x}) \Delta[u](\mathbf{y})\,dV \\ =& \int _{\partial\Omega} H_{3}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf{y}}\Delta [u](\mathbf{y}) - \int _{\partial\Omega} H_{2}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf{y}}D[u](\mathbf {y}) \\ &{} + \int _{\Omega} \bigl[H_{2}(\kappa, \mathbf{y}-\mathbf{x}) \bigr]D\cdot D[u](\mathbf{y})\,dV \\ =& \int _{\partial\Omega} H_{3}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf{y}}\Delta [u](\mathbf{y}) - \int _{\partial\Omega} H_{2}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf{y}}D[u](\mathbf {y}) \\ &{}+ \int _{\partial\Omega} H_{1}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf{y}}u(\mathbf{y}), \end{aligned}$$
(3.10)
From (3.9) and (3.10), in this case, the result follows.
Now, let \(\mathbf{x}\in\Omega\) and take \(r>0\) such that \(B(\mathbf{x}, r)\subset\Omega\). Invoking the previous case, we may write
$$\begin{aligned}& \sum_{i=1}^{3}(-1)^{i-1} \int _{\partial(\Omega\setminus B(\mathbf{x}, r))}H_{i}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf {y}}D^{i-1}[u](\mathbf{y}) \\& \quad {} - \int _{\partial(\Omega\setminus B(\mathbf{x}, r))}H_{4}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf{y}}L_{-\kappa}\Delta[u](\mathbf {y}) \\& \quad {} + \int _{\Omega\setminus B(\mathbf{x}, r)}H_{4}(\kappa, \mathbf {y}-\mathbf{x}) \bigl(\Delta^{2}-\kappa^{2}\Delta\bigr)[u](\mathbf{y})\,dV=0. \end{aligned}$$
(3.11)
Here we take the limits for \(r\rightarrow0\). As regards the weak singularity of \(H_{4}(\kappa, \mathbf{y}-\mathbf{x})\), it follows that
$$\begin{aligned}& \lim_{r\rightarrow0} \int _{\Omega\setminus B(\mathbf{x}, r)}H_{4}(\kappa, \mathbf{y}-\mathbf{x}) \bigl(\Delta^{2}-\kappa^{2}\Delta \bigr)[u](\mathbf{y})\,dV \\& \quad = \int _{\Omega}H_{4}(\kappa, \mathbf{y}-\mathbf{x}) \bigl(\Delta ^{2}-\kappa^{2}\Delta\bigr)[u](\mathbf{y})\,dV. \end{aligned}$$
(3.12)
Furthermore we write
$$\begin{aligned}& \sum_{i=1}^{3}(-1)^{i-1} \int _{\partial(\Omega\setminus B(\mathbf{x}, r))}H_{i}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf {y}}D^{i-1}[u](\mathbf{y}) \\& \qquad {}- \int _{\partial(\Omega\setminus B(\mathbf{x}, r))}H_{4}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf{y}}L_{-\kappa }\Delta[u](\mathbf{y}) \\& \quad =\sum_{i=1}^{3}(-1)^{i-1} \int _{\partial\Omega }H_{i}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf {y}}D^{i-1}[u](\mathbf{y}) \\& \qquad {}- \int _{\partial\Omega}H_{4}(\kappa, \mathbf{y}-\mathbf {x})\,d \sigma_{\mathbf{y}}L_{-\kappa}\Delta[u](\mathbf{y}) \\& \qquad {}-\sum_{i=1}^{3}(-1)^{i-1} \int _{\partial B(\mathbf {x}, r)}H_{i}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf {y}}D^{i-1}[u](\mathbf{y}) \\& \qquad {}+ \int _{\partial B(\mathbf{x}, r)}H_{4}(\kappa, \mathbf {y}-\mathbf{x})\,d \sigma_{\mathbf{y}}L_{-\kappa}\Delta[u](\mathbf{y}). \end{aligned}$$
(3.13)
We denote
$$\begin{aligned} I_{2} \triangleq&-\sum_{i=1}^{3}(-1)^{i-1} \int _{\partial B(\mathbf{x}, r)}H_{i}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf {y}}D^{i-1}[u](\mathbf{y}) \\ &{}+ \int _{\partial B(\mathbf{x}, r)}H_{4}(\kappa, \mathbf {y}-\mathbf{x})\,d \sigma_{y}L_{-\kappa}\Delta[u](\mathbf{y}). \end{aligned}$$
Applying the Stokes formula and the Lebesgue differentiation theorem, we have
$$ \lim_{r\rightarrow0} I_{2}=-u(\mathbf{x}). $$
(3.14)
Combining (3.11) with (3.12)-(3.14), we get the desired result. □
Theorem 3.6
Suppose that Ω is an open bounded non-empty subset of
\(\mathbb {R}^{3}\)
with a Lyapunov boundary
∂Ω, \(u\in C^{4}(\Omega, \operatorname{Cl}(V_{3,3}))\cap C^{3}(\overline{\Omega}, \operatorname{Cl}(V_{3,3}))\). Then
$$\begin{aligned}& \int _{\partial\Omega}E_{1}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma _{\mathbf{y}}u(\mathbf{y}) - \int _{\partial\Omega}E_{2}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma _{\mathbf{y}}L_{-\kappa}[u](\mathbf{y}) \\& \qquad {}+ \int _{\partial\Omega}E_{3}(\kappa, \mathbf{y}-\mathbf {x})\,d \sigma_{\mathbf{y}}L_{\kappa}L_{-\kappa}[u](\mathbf{y}) \\& \qquad{} - \int _{\partial\Omega}E_{4}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma _{\mathbf{y}}DL_{\kappa}L_{-\kappa}[u](\mathbf{y}) \\& \qquad {}+ \int _{\Omega}E_{4}(\kappa, \mathbf{y}-\mathbf{x})\Delta \bigl(\Delta -\kappa^{2}\bigr)[u](\mathbf{y})\,dV \\& \quad = \left \{ \textstyle\begin{array}{l@{\quad}l} u(\mathbf{x}), &\mathbf{x}\in\Omega, \\ 0, & \mathbf{x}\in\mathbb{R}^{3}\setminus\overline{\Omega}, \end{array}\displaystyle \right . \end{aligned}$$
(3.15)
where
\(E_{i}(\kappa, \mathbf{y}-\mathbf{x})\) (\(i=1,2,3,4\)) are as in (3.4).
Proof
The result can be similarly proved to Theorem 3.5. □
Applying Theorems 3.5 and 3.6, we directly have the following results.
Theorem 3.7
Suppose that Ω is an open bounded non-empty subset of
\(\mathbb {R}^{3}\)
with a Lyapunov boundary
∂Ω, \(u\in C^{4}(\Omega, \operatorname{Cl}(V_{3,3}))\cap C^{3}(\overline{\Omega}, \operatorname{Cl}(V_{3,3}))\), and
\((\Delta^{2}-\kappa^{2}\Delta)[u]=0\)
in Ω. Then
$$\begin{aligned}& \sum_{i=1}^{3}(-1)^{i-1} \int _{\partial\Omega}H_{i}(\kappa ,\mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf{y}}D^{i-1}[u](\mathbf{y}) \\& \qquad {}- \int _{\partial\Omega}H_{4}(\kappa,\mathbf{y}-\mathbf{x})\,d \sigma _{\mathbf{y}}L_{-\kappa}\Delta[u](\mathbf{y}) \\& \quad = \left \{ \textstyle\begin{array}{l@{\quad}l} u(\mathbf{x}), & \mathbf{x}\in\Omega, \\ 0, & \mathbf{x}\in\mathbb{R}^{3}\setminus\overline{\Omega}, \end{array}\displaystyle \right . \end{aligned}$$
(3.16)
where
\(H_{i}(\kappa, \mathbf{y}-\mathbf{x})\) (\(i=1,2,3,4\)) are as in (3.3).
Theorem 3.8
Suppose that Ω is an open bounded non-empty subset of
\(\mathbb {R}^{3}\)
with a Lyapunov boundary
∂Ω, \(u\in C^{4}(\Omega, \operatorname{Cl}(V_{3,3}))\cap C^{3}(\overline{\Omega}, \operatorname{Cl}(V_{3,3}))\), and
\((\Delta^{2}-\kappa^{2}\Delta)[u]=0\)
in Ω. Then
$$\begin{aligned}& \int _{\partial\Omega}E_{1}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma _{\mathbf{y}}u(\mathbf{y}) - \int _{\partial\Omega}E_{2}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma _{\mathbf{y}}L_{-\kappa}[u](\mathbf{y}) \\& \qquad {}+ \int _{\partial\Omega}E_{3}(\kappa, \mathbf{y}-\mathbf {x})\,d \sigma_{\mathbf{y}}L_{\kappa}L_{-\kappa}[u](\mathbf{y}) \\& \qquad {}- \int _{\partial\Omega}E_{4}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma _{\mathbf{y}}DL_{\kappa}L_{-\kappa}[u](\mathbf{y}) \\& \quad = \left \{ \textstyle\begin{array}{l@{\quad}l} u(\mathbf{x}), & \mathbf{x}\in\Omega, \\ 0, & \mathbf{x}\in\mathbb{R}^{3}\setminus\overline{\Omega}, \end{array}\displaystyle \right . \end{aligned}$$
(3.17)
where
\(E_{i}(\kappa, \mathbf{y}-\mathbf{x})\) (\(i=1,2,3,4\)) are as in (3.4).
In this article, as usual dS denotes the Lebesgue surface measure. Using Theorem 3.7 or 3.8, we have the following result.
Corollary 3.9
Suppose that
\((\Delta^{2}-\kappa^{2}\Delta)[u](\mathbf{x})=0\)
in
\(\mathbb{R}^{3}\). Then
$$\begin{aligned} u(\mathbf{x}) =&\frac{1+\kappa R}{4\pi R^{2}(1+\kappa R+\frac{\kappa^{2} R^{2}}{3})} \int _{\partial B(\mathbf{x}, R)} u(\mathbf{y})\,dS \\ &{}+\frac{\kappa^{2}}{4\pi R(1+\kappa R+\frac{\kappa^{2}R^{2}}{3})} \int _{B(\mathbf{x}, R)}u(\mathbf{y})\,dV +\frac{\frac{1}{\kappa^{2}}+\frac{R}{\kappa}+\frac{R^{2}}{3}-\frac {e^{\kappa R}}{\kappa^{2}}}{1+\kappa R+\frac{\kappa^{2}R^{2}}{3}}\Delta [u]( \mathbf{x}). \end{aligned}$$
Proof
For arbitrary \(\mathbf{x}\in\mathbb{R}^{3}\), Theorem 3.7 and Stokes’ formula imply
$$\begin{aligned} u(\mathbf{x}) =&\frac{1}{4\pi R^{2}} \int _{\partial B(\mathbf{x}, R)}u(\mathbf{y})\,dS +\frac{1}{4\pi R} \int _{B(\mathbf{x}, R)}\Delta[u](\mathbf{y})\,dV \\ &{}+\biggl(\frac{e^{-\kappa R}-1}{4\pi\kappa^{2}R^{2}}+\frac{e^{-\kappa R}}{4\pi \kappa R}\biggr) \int _{\partial B(\mathbf{x}, R)}\Delta[u](\mathbf {y})\,dS \\ &{}+\frac{e^{-\kappa R}-1}{4\pi\kappa^{2}R} \int _{B(\mathbf{x}, R)}\Delta^{2}[u](\mathbf{y})\,dV. \end{aligned}$$
(3.18)
Using the mean value formula for harmonic functions and the condition \((\Delta^{2}-\kappa^{2}\Delta)[u]=0\), from (3.18), we have
$$\begin{aligned} u(\mathbf{x}) =&e^{-\kappa R}\frac{1+\kappa R}{4\pi R^{2}} \int _{\partial B(\mathbf{x}, R)}u(\mathbf{y})\,dS +e^{-\kappa R} \frac{\kappa^{2}}{4\pi R} \int _{B(\mathbf{x}, R)}u(\mathbf{y})\,dV \\ &{}+\biggl(\frac{1-e^{\kappa R}}{\kappa^{2}}+\frac{R}{\kappa}+\frac {R^{2}}{3} \biggr)e^{-\kappa R}\Delta[u](\mathbf{x}) \\ &{}-\biggl(1-e^{\kappa R}+\kappa R+\frac{R^{2}\kappa^{2}}{3}\biggr)e^{-\kappa R}u( \mathbf{x}). \end{aligned}$$
(3.19)
Thus the result follows. □
Theorem 3.10
Suppose that
\((\Delta^{2}-\kappa^{2}\Delta)[u]=0\)
in
\(\mathbb{R}^{3}\)
and
\(\lim_{r\rightarrow\infty}\frac{\Lambda(r,u)}{r^{m}}=l<\infty\), \(m\in\mathbf{N}^{*}\). Then
$$ \left \{ \textstyle\begin{array}{l} \liminf_{r\rightarrow\infty}\frac{M(r, D[u])}{r^{m-1}}< \infty, \\ \Delta[u](\infty)=0, \\ L_{\pm\kappa}\Delta[u](\infty)=0. \end{array}\displaystyle \right . $$
(3.20)
Proof
For arbitrary \(\mathbf{x}\in\mathbb{R}^{3}\), by Corollary 3.9, we have
$$\begin{aligned} \Delta[u](\mathbf{x}) =&\frac{1+\kappa R+\frac{\kappa^{2}R^{2}}{3}}{\frac {1}{\kappa^{2}}+\frac{R}{\kappa}+\frac{R^{2}}{3}-\frac{e^{\kappa R}}{\kappa^{2}}}u(\mathbf{x}) \\ &{}- \frac{(1+\kappa R)}{4\pi R^{2}(\frac{1}{\kappa^{2}}+\frac{R}{\kappa }+\frac{R^{2}}{3}-\frac{e^{\kappa R}}{\kappa^{2}})} \int _{\partial B(\mathbf{x}, R)} u(\mathbf{y})\,dS \\ &{}-\frac{\kappa^{2}}{4\pi R(\frac{1}{\kappa^{2}}+\frac{R}{\kappa}+\frac {R^{2}}{3}-\frac{e^{\kappa R}}{\kappa^{2}})} \int _{B(\mathbf{x}, R)}u(\mathbf{y})\,dV. \end{aligned}$$
(3.21)
Taking \(R=\|\mathbf{x}\|\) in (3.21), we obtain
$$\begin{aligned} \bigl\Vert \Delta[u](\mathbf{x})\bigr\Vert \leq&\biggl\vert \frac{1+\kappa\|\mathbf{x}\|+\frac {\kappa^{2}\|\mathbf{x}\|^{2}}{3}}{\frac{1}{\kappa^{2}} +\frac{\|\mathbf{x}\|}{\kappa}+\frac{\|\mathbf{x}\|^{2}}{3}-\frac {e^{\kappa\|\mathbf{x}\|}}{\kappa^{2}}}\biggr\vert \bigl\Vert u(\mathbf{x})\bigr\Vert \\ &{}+\biggl\vert \frac{1+\kappa\|\mathbf{x}\|}{\frac{1}{\kappa^{2}}+\frac{\| \mathbf{x}\|}{\kappa}+\frac{\|\mathbf{x}\|^{2}}{3}-\frac{e^{\kappa\| \mathbf{x}\|}}{\kappa^{2}}}\biggr\vert \max_{\|\mathbf{y}\|\leq2\|\mathbf{x}\|} \bigl\{ \bigl\Vert u(\mathbf{y})\bigr\Vert \bigr\} \\ &{}+\biggl\vert \frac{\kappa^{2}\|\mathbf{x}\|^{2}}{\frac{3}{\kappa^{2}}+\frac {3\|\mathbf{x}\|}{\kappa}+\|\mathbf{x}\|^{2}-3\frac{e^{\kappa\|\mathbf {x}\|}}{\kappa^{2}}}\biggr\vert \max_{\|\mathbf{y}\|\leq2\|\mathbf{x}\|} \bigl\{ \bigl\Vert u(\mathbf{y})\bigr\Vert \bigr\} . \end{aligned}$$
(3.22)
Denoting \(\|\mathbf{x}\|=r\), we get from (3.22)
$$\begin{aligned} \max_{\|\mathbf{x}\|=r}\bigl\{ \bigl\Vert \Delta[u]( \mathbf{x})\bigr\Vert \bigr\} \leq& \biggl\vert \frac{1+\kappa r+\frac{\kappa^{2}r^{2}}{3}}{\frac{1}{\kappa^{2}} +\frac{r}{\kappa}+\frac{r^{2}}{3}-\frac{e^{r\kappa}}{\kappa^{2}}} \biggr\vert \max_{\|\mathbf{x}\|=r}\bigl\{ \bigl\Vert u(\mathbf{x})\bigr\Vert \bigr\} \\ &{}+\biggl\vert \frac{1+r\kappa}{\frac{1}{\kappa^{2}}+\frac{r}{\kappa}+\frac {r^{2}}{3}-\frac{e^{r\kappa}}{\kappa^{2}}}\biggr\vert \max_{\|\mathbf{y}\|\leq2r} \bigl\{ \bigl\Vert u(\mathbf {y})\bigr\Vert \bigr\} \\ &{}+\biggl\vert \frac{\kappa^{2}r^{2}}{\frac{3}{\kappa^{2}}+\frac{3r}{\kappa }+r^{2}-3\frac{e^{r\kappa}}{\kappa^{2}}}\biggr\vert \max_{\|\mathbf{y}\|\leq2r} \bigl\{ \bigl\Vert u(\mathbf {y})\bigr\Vert \bigr\} . \end{aligned}$$
(3.23)
The inequality (3.23) can be rewritten as
$$\begin{aligned} M\bigl(r,\Delta[u]\bigr) \leq&\biggl\vert \frac{r^{m}+\kappa r^{m+1}+\frac{\kappa ^{2}r^{m+2}}{3}}{\frac{1}{\kappa^{2}} +\frac{r}{\kappa}+\frac{r^{2}}{3}-\frac{e^{r\kappa}}{\kappa^{2}}} \biggr\vert \frac{M(r,u)}{r^{m}} \\ &{}+\biggl\vert \frac{r^{m}+r^{m+1}\kappa}{\frac{1}{\kappa^{2}}+\frac{r}{\kappa }+\frac{r^{2}}{3}-\frac{e^{r\kappa}}{\kappa^{2}}}\biggr\vert \frac{\Lambda(2r,u)}{r^{m}} \\ &{}+\biggl\vert \frac{\kappa^{2}r^{m+2}}{\frac{3}{\kappa^{2}}+\frac{3r}{\kappa }+r^{2}-3\frac{e^{r\kappa}}{\kappa^{2}}}\biggr\vert \frac{\Lambda(2r,u)}{r^{m}}. \end{aligned}$$
(3.24)
In view of (3.23) and \(\lim_{r\rightarrow\infty}\frac {\Lambda(r,u)}{r^{m}}=l<\infty\), when \(r\rightarrow\infty\), we get \(\Delta[u](\infty)=0\).
Using Lemma 3.13 in [19] and \((\Delta-\kappa^{2})\Delta[u](\mathbf{x})=0\), we obtain
$$ L_{-\kappa}\Delta[u](\infty)=0 $$
(3.25)
and
$$ L_{\kappa}\Delta[u](\infty)=0. $$
(3.26)
From (3.25) and (3.26), we further obtain \(D^{3}[u](\infty)=0\).
We finally verify the remaining of the result of the theorem. For all \(\mathbf{x}\in\mathbb{R}^{3}\), from Theorem 3.7, Lemma 3.2, Remark 3.3, and Stokes’ formula it follows that
$$\begin{aligned} D[u](\mathbf{x}) =&\frac{1}{4\pi} \int_{\partial B(\mathbf{x}, R)}\frac {\mathbf{y}-\mathbf{x}}{\|\mathbf{y}-\mathbf{x}\|^{3}}\,d\sigma_{\mathbf {y}}D[u]( \mathbf{y}) \\ &{}+\frac{1}{4\pi} \int_{\partial B(\mathbf{x},R)}\frac{1}{\|\mathbf {y}-\mathbf{x}\|}e^{-\kappa\|\mathbf{y}-\mathbf{x}\|}\,d \sigma_{\mathbf {y}}\Delta[u](\mathbf{y}) \\ &{}+\frac{1}{4\pi\kappa^{2}} \int_{\partial B(\mathbf{x}, R)}\frac{\mathbf {y}-\mathbf{x}}{\|\mathbf{y}-\mathbf{x}\|^{3}} \bigl(e^{-\kappa\|\mathbf{y}-\mathbf{x}\|}-1\bigr)\,d \sigma_{\mathbf {y}}D^{3}[u](\mathbf{y}) \\ &{}+\frac{1}{4\pi\kappa} \int_{\partial B(\mathbf{x}, R)}\frac{\mathbf {y}-\mathbf{x}}{\|\mathbf{y}-\mathbf{x}\|^{2}}e^{-\kappa\|\mathbf {y}-\mathbf{x}\|}\,d \sigma_{\mathbf{y}} D^{3}[u](\mathbf{y}) \\ =&\frac{3}{4\pi R^{3}} \int_{\partial B(\mathbf{x},R)}\,d\sigma_{\mathbf {y}}u(\mathbf{y}) + \frac{1}{4\pi R^{3}} \int_{B(\mathbf{x},R)}(\mathbf{y}-\mathbf{x})\Delta [u](\mathbf{y})\,dV \\ &{}+\frac{e^{-\kappa R}}{4\pi R} \int_{\partial B(\mathbf{x},R)}\,d\sigma _{\mathbf{y}}\Delta[u](\mathbf{y}) \\ &{}+\frac{3(e^{-\kappa R}-1)}{4\pi\kappa^{2}R^{3}} \int_{\partial B(\mathbf {x},R)}\,d\sigma_{\mathbf{y}}\Delta[u](\mathbf{y}) \\ &{}+\frac{e^{-\kappa R}-1}{4\pi R^{3}} \int_{B(\mathbf{x},R)}(\mathbf {y}-\mathbf{x})\Delta[u](\mathbf{y})\,dV \\ &{}+\frac{3e^{-\kappa R}}{4\pi\kappa R^{2}} \int_{\partial B(\mathbf {x},R)}\,d\sigma_{\mathbf{y}}\Delta[u](\mathbf{y}) \\ &{}+\frac{\kappa e^{-\kappa R}}{4\pi R^{2}} \int_{B(\mathbf{x},R)}(\mathbf {y}-\mathbf{x})\Delta[u](\mathbf{y})\,dV \\ =&\frac{3}{4\pi R^{3}} \int _{\partial B(\mathbf{x},R)}\,d\sigma _{\mathbf{y}}u(\mathbf{y}) +\biggl( \frac{1+R\kappa}{4\pi R^{3}}\biggr)e^{-\kappa R} \int _{B(\mathbf {x},R)}(\mathbf{y}-\mathbf{x})\Delta[u](\mathbf{y})\,dV \\ &{}+\biggl[\frac{e^{-\kappa R}}{4\pi R}+\frac{3(e^{-\kappa R}-1)}{4\pi\kappa ^{2}R^{3}}+\frac{3e^{-\kappa R}}{4\pi\kappa R^{2}}\biggr] \int_{\partial B(\mathbf{x},R)}\,d\sigma_{\mathbf{y}}\Delta[u](\mathbf{y}). \end{aligned}$$
(3.27)
Taking \(R=\|\mathbf{x}\|\) in (3.27), in view of the maximum principle of the modified Helmholtz equation in [7, 18], it immediately follows that
$$\begin{aligned} \bigl\Vert D[u](\mathbf{x})\bigr\Vert \leq&24 \frac{\Lambda(2\Vert \mathbf{x}\Vert ,u)}{\Vert \mathbf {x}\Vert } +\frac{e^{-\kappa \Vert \mathbf{x}\Vert }}{3}\Vert \mathbf{x}\Vert M\bigl(2\Vert \mathbf{x}\Vert ,\Delta[u]\bigr) \\ &{}+\kappa\frac{e^{-\kappa \Vert \mathbf{x}\Vert }}{3}\Vert \mathbf{x}\Vert ^{2}M\bigl(2 \Vert \mathbf{x}\Vert ,\Delta[u]\bigr) +e^{-\kappa \Vert \mathbf{x}\Vert }\Vert \mathbf{x} \Vert M\bigl(2\Vert \mathbf{x}\Vert ,\Delta[u]\bigr) \\ &{}+\frac{3(1-e^{-\kappa \Vert \mathbf{x}\Vert })}{\kappa^{2}\Vert \mathbf{x}\Vert } M\bigl(2\Vert \mathbf{x}\Vert ,\Delta[u]\bigr)+ \frac{3e^{-\kappa \Vert \mathbf{x}\Vert }}{\kappa }M\bigl(2\Vert \mathbf{x}\Vert ,\Delta[u]\bigr). \end{aligned}$$
(3.28)
Denoting \(\|\mathbf{x}\|=r\), we conclude from (3.28)
$$\begin{aligned} M\bigl(r,D[u]\bigr) \leq&24\frac{\Lambda(2r,u)}{r}+e^{-\kappa r} \biggl(\frac{4r}{3}+\frac {\kappa r^{2}}{3}+\frac{3}{\kappa}\biggr)M\bigl(2r, \Delta[u]\bigr) \\ &{} +\frac{3(1-e^{-\kappa r})}{\kappa^{2} r}M\bigl(2r,\Delta[u]\bigr). \end{aligned}$$
(3.29)
Then by (3.29), we have
$$\begin{aligned} \frac{M(r,D[u])}{r^{m-1}} \leq&24\cdot\frac{\Lambda(2r,u)}{r^{m}} + \frac{e^{-\kappa r}}{r^{m-1}}\biggl(\frac{4r}{3}+\frac{\kappa r^{2}}{3}+\frac {3}{\kappa} \biggr)M\bigl(2r,\Delta[u]\bigr) \\ &{}+\frac{3(1-e^{-\kappa r})}{\kappa^{2} r^{m}}M\bigl(2r,\Delta[u]\bigr). \end{aligned}$$
(3.30)
Applying the maximum principle of the modified Helmholtz equation and the condition \(\lim_{r\rightarrow\infty}\frac{\Lambda(r,u)}{r^{m}}=l<\infty\), we have \(\liminf_{r\rightarrow\infty}\frac {M(r,D[u])}{r^{m-1}}<\infty\). The proof is completed. □
Next, denote some integral operators as follows:
$$\begin{aligned}& (\mathcal{F}u) (\mathbf{x})\triangleq\sum_{i=1}^{3}(-1)^{i-1} \int _{\partial\Omega} H_{i}(\kappa,\mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf {y}}D^{i-1}[u](\mathbf{y}) \\& \hphantom{(\mathcal{F}u)(\mathbf{x})\triangleq{}}{}- \int_{\partial\Omega}H_{4}(\kappa,\mathbf{y}-\mathbf{x})\,d \sigma _{\mathbf{y}}L_{-\kappa}\Delta[u](\mathbf{y}),\quad \mathbf{x}\in \mathbb{R}^{3}\setminus\partial\Omega, \end{aligned}$$
(3.31)
$$\begin{aligned}& (\mathcal{S}u) (\mathbf{x})\triangleq\sum_{i=1}^{3}(-1)^{i-1} \int _{\partial\Omega} H_{i}(\kappa,\mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf {y}}D^{i-1}[u](\mathbf{y}) \\& \hphantom{(\mathcal{S}u)(\mathbf{x})\triangleq{}}{}- \int_{\partial\Omega}H_{4}(\kappa,\mathbf{y}-\mathbf{x})\,d \sigma _{\mathbf{y}}L_{-\kappa}\Delta[u](\mathbf{y}), \quad \mathbf{x}\in \partial\Omega, \end{aligned}$$
(3.32)
where \(H_{i}(\kappa, \mathbf{y}-\mathbf{x})\) (\(i=1,2,3,4\)) are as in (3.3) and the above singular integral is taken in the principal sense.
Lemma 3.11
Let Ω be an open, bounded non-empty subset of
\(\mathbb{R}^{3}\)
with Lyapunov boundary
∂Ω, \(u(\mathbf{x})\in H^{\alpha}(\partial\Omega, \operatorname{Cl}(V_{3,3}))\), \(0<\alpha \leq1\). Then, for
\(\mathbf{x}\in\partial\Omega\),
$$ \left \{ \textstyle\begin{array}{l} \lim_{\substack{\mathbf{x}\rightarrow\mathbf{x}_{0}\in\partial\Omega\\ \mathbf{x}\in\Omega}} (\mathcal{F}u)(\mathbf{x})=\frac{u(\mathbf{x}_{0})}{2}+(\mathcal {S}u)(\mathbf{x}_{0}), \\ \lim_{\substack{\mathbf{x}\rightarrow\mathbf{x}_{0}\in\partial\Omega\\ \mathbf{x}\in\mathbb{R}^{3}\setminus\overline{\Omega}}} (\mathcal{F}u)(\mathbf{x})=-\frac{u(\mathbf{x}_{0})}{2}+(\mathcal {S}u)(\mathbf{x}_{0}). \end{array}\displaystyle \right . $$
(3.33)
Proof
Applying the Plemelj formulas with parameter (Theorem 2.3 in [10]), the result follows. □
In the following, we denote
$$ u^{\pm}(\mathbf{x})=\lim_{\substack{\mathbf{y}\rightarrow\mathbf{x}\in \partial\Omega\\ \mathbf{x}\in\Omega^{\pm}}}u( \mathbf{y}), $$
(3.34)
where \(\Omega=\Omega^{+}\) and \(\Omega^{-}=\mathbb{R}^{3}\setminus \overline{\Omega}\).
Theorem 3.12
Assume that Ω is an open, bounded non-empty subset of
\(\mathbb {R}^{3}\)
with a Lyapunov boundary
∂Ω, \(u\in C^{4}(\Omega, \operatorname{Cl}(V_{3,3}))\cap C^{3}(\overline{\Omega}, \operatorname{Cl}(V_{3,3}))\), \(u\in C^{4}(\Omega^{-}, \operatorname{Cl}(V_{3,3}))\cap C^{3}(\overline{\Omega} ^{-}, \operatorname{Cl}(V_{3,3}))\)
and
\(u(\mathbf{x})\)
satisfies the following conditions:
$$ \left \{ \textstyle\begin{array}{l} (\Delta^{2}-\kappa^{2}\Delta)[u](\mathbf{x})=0, \quad \textit{in }\mathbb {R}^{3}\setminus\partial\Omega, \\ u^{+}({\mathbf{x}})=u^{-}(\mathbf{x})\in H^{\alpha_{1}}(\partial\Omega, \operatorname{Cl}(V_{3,3})), \quad \forall\mathbf{x}\in\partial\Omega, \\ D[u]^{+}({\mathbf{x}})=D[u]^{-}(\mathbf{x})\in H^{\alpha_{2}}(\partial \Omega, \operatorname{Cl}(V_{3,3})), \quad \forall\mathbf{x}\in\partial\Omega, \\ \Delta[u]^{+}({\mathbf{x}})=\Delta[u]^{-}(\mathbf{x})\in H^{\alpha _{3}}(\partial\Omega, \operatorname{Cl}(V_{3,3})), \quad \forall\mathbf{x}\in \partial\Omega, \\ L_{-\kappa}\Delta[u]^{+}(\mathbf{x})=L_{-\kappa}\Delta[u]^{-}(\mathbf {x})\in H^{\alpha_{4}}(\partial\Omega, \operatorname{Cl}(V_{3,3})), \quad \forall \mathbf{x}\in\partial\Omega, \end{array}\displaystyle \right . $$
(3.35)
where
\(0<\alpha_{i}\leq1\), \(i=1,2,3,4\), then
\((\Delta^{2}-\kappa ^{2}\Delta)[u]=0\)
in
\(\mathbb{R}^{3}\).
Proof
We only need to prove that for \(\forall\mathbf{x}_{0}\in\partial\Omega \), \((\Delta^{2}-\kappa^{2}\Delta)[u]=0\). Taking a constant \(r>0\), \(B(\mathbf{x}_{0}, r)\) is an open ball with the center at \(\mathbf {x}_{0}\) and radius r such that \(\Omega\subset B(\mathbf{x}_{0}, r)\). It is clear that \(\partial\Omega \cup\partial B(\mathbf{x}_{0}, r)\) is a Lyapunov boundary.
Let
$$ \left \{ \textstyle\begin{array}{l} u(\mathbf{x})=u^{+}({\mathbf{x}})=u^{-}(\mathbf{x}), \\ D[u](\mathbf{x})=D[u]^{+}({\mathbf{x}})=D[u]^{-}(\mathbf{x}), \\ \Delta[u](\mathbf{x})=\Delta[u]^{+}({\mathbf{x}})=\Delta[u]^{-}(\mathbf {x}), \\ L_{-\kappa}\Delta[u](\mathbf{x})=L_{-\kappa}\Delta[u]^{+}(\mathbf {x})=L_{-\kappa}\Delta[u]^{-}(\mathbf{x}), \end{array}\displaystyle \right . $$
here \(\mathbf{x}\in\partial\Omega\). In view of Theorem 3.7, it follows that
$$\begin{aligned}& u(\mathbf{x}_{1})=\sum_{i=1}^{3}(-1)^{i-1} \int _{\partial \Omega}H_{i}(\kappa,\mathbf{y}- \mathbf{x}_{1})\,d\sigma_{\mathbf {y}}D^{i-1}[u](\mathbf{y}) \\& \hphantom{u(\mathbf{x}_{1})={}}{}- \int_{\partial\Omega}H_{4}(\kappa,\mathbf{y}- \mathbf{x}_{1})\,d\sigma _{\mathbf{y}}L_{-\kappa}\Delta[u]( \mathbf{y}), \quad \mathbf {x}_{1}\in\Omega, \end{aligned}$$
(3.36)
$$\begin{aligned}& u(\mathbf{x}_{2})=\sum_{i=1}^{3}(-1)^{i-1} \int_{\partial\Omega \cup\partial B(\mathbf{x}_{0}, r)} H_{i}(\kappa,\mathbf{y}- \mathbf{x}_{2})\,d\sigma_{\mathbf {y}}D^{i-1}[u](\mathbf{y}) \\& \hphantom{u(\mathbf{x}_{2})={}}{}- \int_{\partial\Omega\cup\partial B(\mathbf{x}_{0}, r)}H_{4}(\kappa ,\mathbf{y}- \mathbf{x}_{2})\,d\sigma_{\mathbf{y}}L_{-\kappa}\Delta [u]( \mathbf{y}), \\& \hphantom{u(\mathbf{x}_{2})={}}\mathbf{x}_{2}\in B(\mathbf{x}_{0}, r) \setminus\overline{\Omega}. \end{aligned}$$
(3.37)
Combining (3.36) with (3.37) and using Lemma 3.11, we obtain
$$\begin{aligned}& u^{+}(\mathbf{x}_{0})=\lim_{\mathbf{x}_{1}\rightarrow\mathbf {x}_{0}}u( \mathbf{x}_{1}) \\& \hphantom{u^{+}(\mathbf{x}_{0})}=\frac{u(\mathbf{x}_{0})}{2} +\sum_{i=1}^{3}(-1)^{i-1} \int _{\partial\Omega}H_{i}(\kappa ,\mathbf{y}- \mathbf{x}_{0})\,d\sigma_{\mathbf{y}}D^{i-1}[u](\mathbf{y}) \\& \hphantom{u^{+}(\mathbf{x}_{0})={}}{}- \int_{\partial\Omega}H_{4}(\kappa,\mathbf{y}- \mathbf{x}_{0})\,d\sigma _{\mathbf{y}}L_{-\kappa}\Delta[u]( \mathbf{y}), \end{aligned}$$
(3.38)
$$\begin{aligned}& u^{-}(\mathbf{x}_{0})=\lim_{\mathbf{x}_{2}\rightarrow\mathbf{x}_{0}} u( \mathbf{x}_{0}) \\& \hphantom{u^{-}(\mathbf{x}_{0})}=\frac{u(\mathbf{x}_{0})}{2} +\sum_{i=1}^{3}(-1)^{i-1} \int _{\partial\Omega\cup \partial B(\mathbf{x}_{0}, r)} H_{i}(\kappa,\mathbf{y}- \mathbf{x}_{0})\,d\sigma_{\mathbf {y}}D^{i-1}[u](\mathbf{y}) \\& \hphantom{u^{-}(\mathbf{x}_{0})={}}{}- \int _{\partial\Omega\cup\partial B(\mathbf{x}_{0}, r)}H_{4}(\kappa,\mathbf{y}- \mathbf{x}_{0})\,d\sigma_{\mathbf{y}}L_{-\kappa }\Delta[u]( \mathbf{y}). \end{aligned}$$
(3.39)
From (3.38) and (3.39), we derive
$$\begin{aligned} u(\mathbf{x}_{0}) =&\sum_{i=1}^{3}(-1)^{i-1} \int_{\partial B(\mathbf{x}_{0}, r)} H_{i}(\kappa,\mathbf{y}- \mathbf{x}_{0})\,d\sigma_{\mathbf {y}}D^{i-1}[u](\mathbf{y}) \\ &{}- \int_{\partial B(\mathbf{x}_{0}, r)}H_{4}(\kappa,\mathbf{y}-\mathbf {x}_{0})\,d\sigma_{\mathbf{y}}L_{-\kappa}\Delta[u](\mathbf{y}). \end{aligned}$$
(3.40)
Therefore \(\Delta(\Delta-\kappa^{2})[u](\mathbf{x}_{0})=0\), the result follows. □
Theorem 3.13
Let Ω be an open bounded non-empty subset of
\(\mathbb{R}^{3}\)
with Lyapunov boundary
∂Ω, \(u\in C^{4}(\Omega^{-}, \operatorname{Cl}(V_{3,3}))\cap C^{3}(\overline{\Omega}^{-}, \operatorname{Cl}(V_{3,3}))\), \((\Delta^{2}-\kappa^{2}\Delta)[u]=0\)
in
\(\Omega^{-}\), and
$$ \left \{ \textstyle\begin{array}{l} u({\mathbf{x}})\in H^{\alpha_{1}}(\partial\Omega, \operatorname{Cl}(V_{3,3})), \\ D[u](\mathbf{x})\in H^{\alpha_{2}}(\partial\Omega, \operatorname{Cl}(V_{3,3})), \\ \Delta[u](\mathbf{x})\in H^{\alpha_{3}}(\partial\Omega, \operatorname{Cl}(V_{3,3})), \\ L_{-\kappa}\Delta[u](\mathbf{x})\in H^{\alpha_{4}}(\partial\Omega, \operatorname{Cl}(V_{3,3})), \\ \lim_{r\rightarrow\infty}\frac{\Lambda(r,u)}{r^{m}}=l< \infty , \quad m\in\mathbf{N}^{*}, \end{array}\displaystyle \right . $$
where
\(0<\alpha_{i}\leq1\), \(i=1, 2, 3, 4\). Then
$$ \left \{ \textstyle\begin{array}{l} \liminf_{r\rightarrow\infty}\frac{M(r, D[u])}{r^{m-1}}< \infty, \\ \Delta[u](\infty)=0, \\ L_{-\kappa}[u]\Delta(\infty)=0,\qquad L_{\kappa}\Delta[u](\infty)=0. \end{array}\displaystyle \right . $$
Proof
For \(\mathbf{y}\in\partial\Omega\), let
$$ \left \{ \textstyle\begin{array}{l} u(\mathbf{y})=-f_{1}(\mathbf{y}), \\ D[u](\mathbf{y})=-f_{2}(\mathbf{y}), \\ \Delta[u](\mathbf{y})=-f_{3}(\mathbf{y}), \\ L_{-\kappa}\Delta[u](\mathbf{y})=-f_{4}(\mathbf{y}). \end{array}\displaystyle \right . $$
For \(\mathbf{x}\in\mathbb{R}^{3}\setminus\partial\Omega\), we get
$$ F(\mathbf{x})=\sum_{i=1}^{4}(-1)^{i-1} \int _{\partial\Omega} H_{i}(\kappa, \mathbf{y}-\mathbf{x})\,d \sigma_{\mathbf{y}}f_{i}(\mathbf{y}) $$
and
$$ \widetilde{F}(\mathbf{x})=\left \{ \textstyle\begin{array}{l@{\quad}l} -F(\mathbf{x}) & \mathbf{x}\in\Omega^{+}, \\ u(\mathbf{x})-F(\mathbf{x}), & \mathbf{x}\in\Omega^{-}, \end{array}\displaystyle \right . $$
in view of Lemma 3.2, it is easy to check that \((\Delta ^{2}-\kappa^{2}\Delta)[\widetilde{F}]=0\) in \(\mathbb{R}^{3}\setminus\partial\Omega\), combining Lemma 3.11, we get
$$ \left \{ \textstyle\begin{array}{l} [\widetilde{F}]^{+}(\mathbf{x})=[\widetilde{F}]^{-}(\mathbf{x})\in H^{\tilde{\alpha}_{1}}(\partial\Omega, \operatorname{Cl}(V_{3,3})), \\ D[\widetilde{F}]^{+}(\mathbf{x})=D[\widetilde{F}]^{-}(\mathbf{x})\in H^{\tilde{\alpha}_{2}}(\partial\Omega, \operatorname{Cl}(V_{3,3})), \\ \Delta[\widetilde{F}]^{+}(\mathbf{x})=\Delta[\widetilde{F}]^{-}(\mathbf {x})\in H^{\tilde{\alpha}_{3}}(\partial\Omega, \operatorname{Cl}(V_{3,3})), \\ L_{-\kappa}\Delta[\widetilde{F}]^{+}(\mathbf{x})=L_{-\kappa}\Delta [\widetilde{F}]^{-}(\mathbf{x})\in H^{\tilde{\alpha}_{4}}(\partial \Omega, \operatorname{Cl}(V_{3,3})). \end{array}\displaystyle \right . $$
Thus \((\Delta^{2}-\kappa^{2}\Delta)[\widetilde{F}]=0\) in \(\mathbb{R}^{3}\) where we use Theorem 3.12. Obviously, \(\lim_{r\rightarrow\infty}\frac{\Lambda(r,\widetilde{F}(\mathbf {x}))}{r^{m}}=l<\infty\), using Theorem 3.10, we arrive at
$$ \left \{ \textstyle\begin{array}{l} \liminf_{r\rightarrow\infty}\frac{M(r, D[\widetilde {F}])}{r^{m-1}}< \infty, \\ \Delta[\widetilde{F}](\infty)=0, \\ L_{-\kappa}\Delta[\widetilde{F}](\infty)=0,\qquad L_{\kappa}\Delta[\widetilde {F}](\infty)=0. \end{array}\displaystyle \right . $$
(3.41)
For \(\mathbf{x}\in\Omega^{-}\),
$$ \left \{ \textstyle\begin{array}{l} D[u](\mathbf{x})=D[\widetilde{F}](\mathbf{x}) \\ \hphantom{D[u](\mathbf{x})={}}{}+\frac{1}{4\pi} \int _{\partial\Omega}\frac{\mathbf{y}-\mathbf{x}}{\| \mathbf{y}-\mathbf{x}\|^{3}}\,d\sigma_{\mathbf{y}}f_{2}(\mathbf{y}) \\ \hphantom{D[u](\mathbf{x})={}}{}+\frac{1}{4\pi\kappa} \int _{\partial\Omega}[(\frac{\mathbf{y}-\mathbf {x}}{\|\mathbf{y}-\mathbf{x}\|^{3}} +\kappa\frac{\mathbf{y}-\mathbf{x}}{\|\mathbf{y}-\mathbf{x}\| ^{2}}+\kappa\frac{1}{\|\mathbf{y}-\mathbf{x}\|})e^{-\kappa\|\mathbf {y}-\mathbf{x}\|} -\frac{\mathbf{y}-\mathbf{x}}{\|\mathbf{y}-\mathbf{x}\|^{3}}] \,d\sigma_{\mathbf{y}}f_{3}(\mathbf{y}) \\ \hphantom{D[u](\mathbf{x})={}}{}+\frac{1}{4\pi\kappa^{2}}\int _{\partial\Omega }[\frac{\mathbf{y}-\mathbf{x}}{\|\mathbf{y}-\mathbf{x}\|^{3}} (e^{-\kappa\|\mathbf{y}-\mathbf{x}\|}-1)+\kappa\frac{\mathbf{y}-\mathbf {x}}{\|\mathbf{y}-\mathbf{x}\|^{2}} e^{-\kappa\|\mathbf{y}-\mathbf{x}\|}]\,d\sigma_{\mathbf{y}}f_{4}(\mathbf {y}), \\ \Delta[u](\mathbf{x})=\Delta[\widetilde{F}](\mathbf{x}) \\ \hphantom{\Delta[u](\mathbf{x})={}}{}+\frac{1}{4\pi}\int _{\partial\Omega} [\frac{\mathbf{y}-\mathbf{x}}{\|\mathbf{y}-\mathbf{x}\|^{3}}+\kappa\frac {\mathbf{y}-\mathbf{x}}{\|\mathbf{y}-\mathbf{x}\|^{2}} +\kappa\frac{1}{\|\mathbf{y}-\mathbf{x}\|}] e^{-\kappa\|\mathbf{y}-\mathbf{x}\|}\,d\sigma_{\mathbf{y}}f_{3}(\mathbf {y}) \\ \hphantom{\Delta[u](\mathbf{x})={}}{}+\frac{1}{4\pi}\int _{\partial\Omega} \frac{e^{-\kappa\|\mathbf{y}-\mathbf{x}\|}}{\|\mathbf{y}-\mathbf{x}\| }\,d\sigma_{\mathbf{y}}f_{4}(\mathbf{y}), \\ L_{-\kappa}\Delta[u](\mathbf{x})=L_{-\kappa}\Delta[\widetilde {F}](\mathbf{x}) \\ \hphantom{L_{-\kappa}\Delta[u](\mathbf{x})={}}{}+\frac{1}{4\pi}\int _{\partial\Omega} [\frac{\mathbf{y}-\mathbf{x}}{\|\mathbf{y}-\mathbf{x}\|^{3}}+\kappa\frac {\mathbf{y}-\mathbf{x}}{\|\mathbf{y}-\mathbf{x}\|^{2}} +\kappa\frac{1}{\|\mathbf{y}-\mathbf{x}\|}] e^{-\kappa\|\mathbf{y}-\mathbf{x}\|}\,d\sigma_{\mathbf{y}}f_{4}(\mathbf{y}). \end{array}\displaystyle \right . $$
(3.42)
Equations (3.41) and (3.42) imply that the result holds. □