Zero Extension for the Dirichlet Problem of the Biharmonic Equation

Abstract

In this paper, we consider whether the zero extension of a solution to the Dirichlet problem for the biharmonic equation in a smaller domain remains a solution to the corresponding extended problem in a larger domain. We analyze classical and strong solutions, and present a necessary and sufficient condition under each framework, respectively.

1. Introduction

The Dirichlet problem for the Biharmonic equation is a classical problem arising from elasticity [1,2,3]. It can describe the equilibrium state of an elastic thin plate with fixed boundaries [4], as well as the motion of low-speed viscous fluids (Stokes flow) with fixed interfaces [5]. Numerous classical results exist concerning the biharmonic equation [6,7] and its Green’s function [8,9]. Here, we consider the zero extension Dirichlet problem for the Biharmonic Equation, which is a natural generalization of the extension problem for the Poisson equation [10].

Let $\Omega$ and $\tilde{\Omega }$ be two smooth bounded domains in ${\mathbb{R}}^n$, and $\Omega$ is a compact subset of $\tilde{\Omega }$, which means $\Omega$ is completely contained in $\tilde{\Omega }$. Assume that u is a solution to the following Dirichlet problem

(1)$\left\{\begin{array}{cc}{\Delta }^{2}u=f\hfill & \mathrm{in}\Omega ,\hfill \\ {\displaystyle u=0,\frac{\partial u}{\partial \mathbf{n}}=0}\hfill & \mathrm{on}\partial \Omega ,\hfill \end{array}\right.$

where $f(x)$ is a given function in the smaller domain $\Omega$.

Define the zero extensions of u and f from the smaller domain $\Omega$ to the larger domain $\tilde{\Omega }$

(2)$\tilde{u}(x)=\left\{\begin{array}{cc}u(x)\hfill & x\in \Omega ,\hfill \\ 0\hfill & x\in \overline{\tilde{\Omega }}\backslash \Omega ,\hfill \end{array}\right.\tilde{f}(x)=\left\{\begin{array}{cc}f(x)\hfill & x\in \Omega ,\hfill \\ 0\hfill & x\in \tilde{\Omega }\backslash \Omega .\hfill \end{array}\right.$

Consider the corresponding Dirichlet problem

(3)$\left\{\begin{array}{cc}{\Delta }^{2}v=\tilde{f}\hfill & \mathrm{in}\tilde{\Omega },\hfill \\ {\displaystyle v=0,\frac{\partial v}{\partial \mathbf{n}}=0}\hfill & \mathrm{on}\partial \tilde{\Omega }.\hfill \end{array}\right.$

General speaking, even if f is sufficiently smooth, the extended function $\tilde{u}$ of solution u may not be a solution for (3).

For example, let $\Omega =B(0,{\displaystyle \frac{1}{2}}),\tilde{\Omega }=B(0,1)f\in C_0^{\infty }(B(0,{\displaystyle \frac{1}{2}}))$ be a nonnegative and nonzero function. It is obvious that $\tilde{f}\in C_0^{\infty }(B(0,1))$. Then, a unique classical solution v exists for (3).

Let $\tilde{G}(x,y)$ be Green’s function of (3) in $\tilde{\Omega }$ (see Definition 2.26. of [11]). We know

$v(x)={\int }_{B(0,1)}\tilde{G}(x,y)\tilde{f}(y)dy={\int }_{B(0,{\displaystyle \frac{1}{2}})}\tilde{G}(x,y)f(y)dy.$

Note

$\tilde{G}(x,y)={\displaystyle \frac{1}{4n{e}_n}}{|x-y|}^{4-n}{\int }_1^{{\displaystyle \frac{\left|\left|x\right|y-\frac{x}{\left|x\right|}\right|}{|x-y|}}}(v^2-1)v^{1-n}dv>0,\forall x,y\in B(0,1),$

where $e_n={\displaystyle \frac{{\pi }^{\frac{n}{2}}}{\Gamma (1+\frac{n}{2})}}$ (see Lemma 2.27 of [11]). Therefore, we have

$v(x)>0,\forall x\in B(0,1),$

which implies that $\tilde{u}(x)$ cannot be a solution of (3), since $\tilde{u}(x)=0$, $x\in \tilde{\Omega }\backslash \Omega$.

It is interesting to consider the necessary and sufficient conditions on $f(x)$ that guarantee the zero extension $\tilde{u}(x)$ is still a solution of (3).

In this paper, we provide a complete answer to this question. We establish necessary and sufficient conditions ensuring that the zero extension of a solution to the biharmonic equation in a smaller domain remains a solution to the corresponding extended problem in a larger domain. We prove the following results under the frameworks of classical and strong solutions.

We will introduce this definition before stating our results.

Definition 1.

Let $f(x),g(x)$ be measurable in Ω. We say $f(x)$ is orthogonal to $g(x)$ if

${\int }_{\Omega }f(x)g(x)dx=0.$

To make sure that (1) and (3) admit classical solutions and ensure the extension, we first assume f is Hölder continuous in $\overline{\Omega }$ and f equals 0 on the boundary $\partial \Omega$.

Theorem 1.

Assume $f\in C^{\alpha }(\overline{\Omega })(0<\alpha <1)$ with ${f|}_{\partial \Omega }=0$. Let $u\in C^{4,\alpha }(\overline{\Omega })$ be the unique solution of (1), and let the functions $\tilde{u},\tilde{f}$ be defined as in (2). Then, $\tilde{u}$ is a classical solution of (3) if and only if f is orthogonal to every biharmonic function g in Ω that can be continuously extended to $\partial \Omega$, i.e.,

(4)${\int }_{\Omega }f(x)g(x)dx=0$

for any $g\in C^4(\Omega )\cap C^1(\overline{\Omega })$ satisfying ${\Delta }^{2}g=0$ in Ω.

Remark 1.

If $f\in {\Delta }^2{C}_0^{\infty }(\Omega )$, which means that there exists a function $w\in C_0^{\infty }(\Omega )$, such that $f={\Delta }^{2}w$, then f is orthogonal to any function $g\in C^2(\Omega )\cap C(\overline{\Omega })$ satisfying ${\Delta }^{2}g=0$ in Ω.

Next, we assume f is in a Lebesgue space to guarantee that (1) and (3) admit strong solutions.

Theorem 2.

Assume $f\in L^p(\Omega )$$(1<p<+\infty )$. Let $u\in W^{4,p}(\Omega )\cap W_0^{2,p}(\Omega )$ be the unique solution of (1) and functions $\tilde{u},\tilde{f}$ be defined in (2). Then, $\tilde{u}$ is the strong solution of (3) if and only if f is orthogonal to any biharmonic function g in Ω.

Remark 2.

Let $f\in {\Delta }^2{W}_0^{4,p}(\Omega )$, which means that there exists a function $w\in W_0^{4,p}(\Omega )$, such that $f={\Delta }^{2}w$, then f is orthogonal to any biharmonic function $g\in C^4(\Omega )\cap C^1(\overline{\Omega })$.

2. Proof of Main Results

In this section, we first prove a lemma, which will be used later.

Lemma 1.

Let $g\in C^4(\Omega )\cap C^1(\overline{\Omega })$ be a biharmonic function and Ω be a bounded $C^{4,\alpha }$ domain in ${\mathbb{R}}^n$. Then, for any $\epsilon >0$, there exists a biharmonic function $g_{\epsilon }\in C^4(\overline{\Omega })$, such that

$\underset{\overline{\Omega }}{max}\left|g\left(x\right)-g_{\epsilon }\left(x\right)\right|\le \epsilon .$

Proof. 

As $g\in C^1(\overline{\Omega })$, then for any $\delta >0$, there exists a function $g_{\delta }\in C^{\infty }(\overline{\Omega })$, which satisfies

(5)$\begin{array}{c}\hfill \underset{\partial \Omega }{max}\left|g\left(x\right)-g_{\delta }\left(x\right)\right|\le {\displaystyle \frac{\delta }{2}}and\underset{\partial \Omega }{max}\left|{\displaystyle \frac{\partial g\left(x\right)}{\partial \mathbf{n}}}-{\displaystyle \frac{\partial g_{\delta }\left(x\right)}{\partial \mathbf{n}}}\right|\le {\displaystyle \frac{\delta }{2}}.\end{array}$

We solve the following problem

(6)$\left\{\begin{array}{cc}\hfill {\Delta }^2{g}_{\epsilon }=0& in\Omega ,\hfill \\ \hfill g_{\epsilon }=g_{\delta },{\displaystyle \frac{\partial g_{\epsilon }}{\partial \mathbf{n}}}={\displaystyle \frac{\partial g_{\delta }}{\partial \mathbf{n}}}& on\partial \Omega .\hfill \end{array}\right.$

The problem (6) is solvable and there exists a unique solution $g_{\epsilon }\in C^4(\overline{\Omega })$ [12].

Then, for any $x\in \Omega$, $g_{\epsilon }(x)$ and $g(x)$ can be expressed as

$g_{\epsilon }\left(x\right)={\int }_{\partial \Omega }{K}_0\left(x,y\right)g_{\delta }\left(y\right)d{\sigma }_y+{\int }_{\partial \Omega }{K}_1\left(x,y\right){\displaystyle \frac{\partial g_{\delta }\left(y\right)}{\partial \mathbf{n}}}d{\sigma }_y,$

$g\left(x\right)={\int }_{\partial \Omega }{K}_0\left(x,y\right)g\left(y\right)d{\sigma }_y+{\int }_{\partial \Omega }{K}_1\left(x,y\right){\displaystyle \frac{\partial g\left(y\right)}{\partial \mathbf{n}}}d{\sigma }_y,$

where $K_0$, $K_1$ are the Poisson kernels [9].

Therefore, for any $x\in \Omega$, we have

(7)$\begin{array}{cc}\hfill |g(x)-g_{\epsilon }(x)| \le & \underset{\partial \Omega }{max}\left|g-g_{\delta }\right|{\int }_{\partial \Omega }\left|K_0\left(x,y\right)\right|d{\sigma }_y\hfill \\ \hfill & +\underset{\partial \Omega }{max}\left|{\displaystyle \frac{\partial g}{\partial \mathbf{n}}}-{\displaystyle \frac{\partial g_{\delta }}{\partial \mathbf{n}}}\right|{\int }_{\partial \Omega }\left|K_1\left(x,y\right)\right|d{\sigma }_y.\hfill \end{array}$

Fix any $x\in \Omega$ and denote $d_x=dist(x,\partial \Omega )$, it follows from [9] that

(8)$\begin{array}{c}\hfill \left|K_j(x,y)\right|\le C{\displaystyle \frac{d_x^2}{{|x-y|}^{n-j+1}}},j=0,1,y\in \partial \Omega ,\end{array}$

where C is a positive constant.

Using (8), let $x_0\in \partial \Omega$ be the point such that $d_x=|x-x_0|$. Then,

${\int }_{\partial \Omega }\left|K_0\left(x,y\right)\right|d{\sigma }_y\le C{\int }_{\partial \Omega }{\displaystyle \frac{d_x^2}{{\left|x-y\right|}^{n+1}}}d{\sigma }_y.$

Take some ${\delta }_1>0$ to be determined later.

If $x\in {\Omega }_{{\displaystyle \frac{{\delta }_1}{2}}}$, where ${\Omega }_{{\displaystyle \frac{{\delta }_1}{2}}}=\{x\in \Omega |dist(x,\partial \Omega )\ge {\displaystyle \frac{{\delta }_1}{2}}\}$, then

(9)$\begin{array}{c}\hfill {\int }_{\partial \Omega }{\displaystyle \frac{d_x^2}{{\left|x-y\right|}^{n+1}}}d{\sigma }_y\le d_x^{1-n}\left|\partial \Omega \right|\le {\left({\displaystyle \frac{{\delta }_1}{2}}\right)}^{1-n}\left|\partial \Omega \right|\le C_0,\end{array}$

where $|\partial \Omega |$ is the Lebesgue measure of $\partial \Omega$ and $C_0=C_0({\delta }_1,n,\partial \Omega )>0.$

If $x\in \Omega \backslash {\Omega }_{{\displaystyle \frac{{\delta }_1}{2}}}$, then

$\begin{array}{cc}\hfill {\int }_{\partial \Omega }{\displaystyle \frac{d_x^2}{{\left|x-y\right|}^{n+1}}}d{\sigma }_y\le & {\int }_{\partial \Omega \backslash B_{{\delta }_1}\left(x_0\right)}{\displaystyle \frac{d_x^2}{{\left|x-y\right|}^{n+1}}}d{\sigma }_y\hfill \\ \hfill & +{\int }_{\partial \Omega \cap B_{{\delta }_1}\left(x_0\right)}{\displaystyle \frac{d_x^2}{{\left|x-y\right|}^{n+1}}}d{\sigma }_y\hfill \\ \hfill =& I_1+I_2.\hfill \end{array}$

Since $|y-x|\ge |y-x_0|-d_x\ge {\displaystyle \frac{{\delta }_1}{2}}$ for any $y\in \partial \Omega \backslash B_{{\delta }_1}\left(x_0\right)$, we estimate $I_1$ by

(10)$\begin{array}{c}\hfill I_1\le {\int }_{\partial \Omega \backslash B_{{\delta }_1}\left(x_0\right)}{\displaystyle \frac{d_x^2}{{\left({\delta }_1/\phantom{{\delta }_{1}2}2\right)}^{n+1}}}d{\sigma }_y\le {\left({\displaystyle \frac{{\delta }_1}{2}}\right)}^{1-n}\left|\partial \Omega \right|\le C_0.\end{array}$

Now, to estimate $I_2$, we use the method of “straighten out the boundary”. Without loss of generality, we assume $x_0=0$ and x are on the $x_n$-axis, that is, $x=(0,\dots ,0,d_x)$. Since $\Omega$ is a $C^{4,\alpha }$ domain, there exists a $C^{4,\alpha }$ mapping $\phi :{\mathbb{R}}^{n-1}\to \mathbb{R}$ such that

$\partial \Omega \cap B_{{\delta }_1}(0)\subset \{y=(y_1,\dots ,y_n)\in {\mathbb{R}}^n|y_n=\phi (y_1,\dots ,y_{n-1}),|y^{\prime }|\le \tilde{C}{\delta }_1\},$

where $y^{\prime }=(y_1,\dots ,y_{n-1})\in {\mathbb{R}}^{n-1}$ and $\tilde{C}>1$ is a constant.

For any $y\in \partial \Omega \cap B_{{\delta }_1}(0)$, we denote $\tilde{B}$ the ball in ${\mathbb{R}}^{n-1}$ and define

$\begin{array}{cc}\hfill \psi :\partial \Omega \cap B_{{\delta }_1}(0)& \to {\tilde{B}}_{\tilde{C}{\delta }_1}(0)\hfill \\ \hfill y& \mapsto z\hfill \end{array}$

in the following way

$\left\{\begin{array}{c}{z}_i=y_i,i=1,\dots ,n-1,\hfill \\ z_n=y_n-\phi (y_1,\dots ,y_{n-1}).\hfill \end{array}\right.$

It is obvious that $\psi$ is a $C^{4,\alpha }$ mapping and $det(\psi )=det({\psi }^{-1})=1$.

Note that the Cauchy inequality yields

${\displaystyle \frac{1}{2}}{|x-y|}^2\le {\left|x\right|}^2+{\left|y\right|}^2\le 5{|y-x|}^{2}foranyy\in \partial \Omega \cap B_{{\delta }_1}(0).$

Therefore, after changing the variables, we have

(11)$\begin{array}{cc}\hfill I_2& \le C{\int }_{\partial \Omega \cap B_{{\delta }_1}(0)}{\displaystyle \frac{d_x^2}{{\left(d_x^2+{\left|y\right|}^2\right)}^{\frac{n+1}{2}}}}d{\sigma }_y\hfill \\ \hfill & \le C{\int }_{{\tilde{B}}_{\tilde{C}{\delta }_1}(0)}{\displaystyle \frac{d_x^2}{{\left(d_x^2+{\left|{\psi }^{-1}(z)\right|}^2\right)}^{\frac{n+1}{2}}}}d{\tilde{\sigma }}_z\hfill \\ \hfill & \le C{\int }_0^{\tilde{C}{\delta }_1}{\displaystyle \frac{d_x^2{r}^{n-2}}{{\left(d_x^2+r^2\right)}^{\frac{n+1}{2}}}}dr\hfill \\ \hfill & \le C{d}_x^2{\int }_0^{\infty }{\displaystyle \frac{1}{{(d_x+r)}^3}}dr\hfill \\ \hfill & \le C_1.\hfill \end{array}$

where $C_1=C_1(n)>0$.

By the same argument, using (8) for $j=1$, we can compute

(12)$\begin{array}{c}\hfill {\int }_{\partial \Omega }\left|K_1\left(x,y\right)\right|d{\sigma }_y\le C_2,\end{array}$

where $C_2=C_2({\delta }_1,n)>0.$

Taking $C=max(1,C_0,C_1,C_2)$, it follows from (9)–(12) that

${\int }_{\partial \Omega }\left|K_0\left(x,y\right)\right|d{\sigma }_y\le Cand{\int }_{\partial \Omega }\left|K_1\left(x,y\right)\right|d{\sigma }_y\le C.$

Hence, by virtue of (5) and (7), choosing $\delta ={\displaystyle \frac{\epsilon }{C}}$, we obtain

$\underset{\overline{\Omega }}{max}\left|g\left(x\right)-g_{\epsilon }\left(x\right)\right|\le \epsilon .$

□

Classical Solutions

Now, we are ready to prove Theorem 1.

Proof. 

(1) Necessity.

As $f\in C^{\alpha }(\overline{\Omega })$ and ${f|}_{\partial \Omega }=0$, we know $\tilde{f}\in C^{\alpha }(\overline{\tilde{\Omega }})$. Then, a unique solution $v\in C^{4,\alpha }\left(\overline{\tilde{\Omega }}\right)\cap C^1\left(\overline{\tilde{\Omega }}\right)$ exists for (3) [12].

Let $u\in C^{4,\alpha }(\overline{\Omega })$ be the classical solution of (1). If $\tilde{u}$ is the classical solution of (3), then $\tilde{u}=v\in C^4(\overline{\tilde{\Omega }})$, which implies $D^3{u|}_{\partial \Omega }=0$, $D^2{u|}_{\partial \Omega }=0$, ${Du|}_{\partial \Omega }=0$.

First, we assume that $g\in C^4(\overline{\Omega })$ satisfies ${\Delta }^{2}g=0$. Integration by parts yields that

$\begin{array}{ccc}\hfill {\int }_{\Omega }f(x)g(x)dx& =& {\int }_{\Omega }{\Delta }^{2}u(x)g(x)dx\hfill \\ & =& -{\int }_{\Omega }\nabla (\Delta u(x))·\nabla g(x)dx\hfill \\ & =& {\int }_{\Omega }\Delta u(x)\Delta g(x)dx\hfill \\ & =& -{\int }_{\Omega }\nabla u(x)·\nabla (\Delta g(x))dx\hfill \\ & =& {\int }_{\Omega }u(x){\Delta }^{2}g(x)dx\hfill \\ & =& 0.\hfill \end{array}$

Next, for $g\in C^4(\Omega )\cap C^1(\overline{\Omega })$ satisfying ${\Delta }^{2}g=0$ in $\Omega$, by Lemma 1 we find $g_{\epsilon }\in C^4(\overline{\Omega })$ satisfying ${\Delta }^2{g}_{\epsilon }=0$, such that

$\underset{\overline{\Omega }}{max}|g(x)-g_{\epsilon }(x)|<\epsilon .$

Recalling that

${\int }_{\Omega }f(x)g_{\epsilon }(x)dx=0,$

and sending $\epsilon \to 0$, we have

${\int }_{\Omega }f(x)g(x)dx=0.$

Remark 3.

If $g(x)\in C^4(\Omega )\cap C^1(\overline{\Omega })$ satisfying ${\Delta }^{2}g=0$ in Ω, the following equation

${\int }_{\Omega }f(x)g(x)dx={\int }_{\Omega }u(x){\Delta }^{2}g(x)dx$

does not hold.

(2) 

Sufficiency

Now, f is orthogonal to any biharmonic function in $\Omega$, which can be continuously extended to $\partial \Omega$. Then, f is orthogonal to any harmonic function in $\Omega$, which is continuous on $\overline{\Omega }$.

Let $G(x,y)$ be Green’s function of (1) in $\Omega$. We know

$G(x,y)=\Gamma (y-x)-\varphi (x,y),$

where $\Gamma (y-x)$ is the fundamental solution (see chapter 1 [13]) and

$\left\{\begin{array}{cc}{\Delta }_y^2\varphi (x,y)=0\hfill & y\in \Omega ,\hfill \\ {\displaystyle \varphi (x,y)=\Gamma (y-x),\frac{\partial \varphi (x,y)}{\partial n}=\frac{\partial \Gamma (y-x)}{\partial n}}\hfill & y\in \partial \Omega .\hfill \end{array}\right.$

Therefore, it follows from (4) that

(13)$u(x)={\int }_{\Omega }G(x,y)f(y)dy={\int }_{\Omega }\Gamma (x-y)f(y)dy,\forall x\in \Omega .$

Let $\tilde{G}(x,y)$ be Green’s function of (3) in $\tilde{\Omega }$, which is

$\tilde{G}(x,y)=\Gamma (x-y)-\tilde{\varphi }(x,y),$

where $\tilde{\varphi }(x,y)$ is the solution for the boundary value problem ${\Delta }_y^2\tilde{\varphi }(x,y)=0$$\mathtt{in} \tilde{\Omega }$; $\tilde{\varphi }(x,y)=\Gamma (x-y),{\displaystyle \frac{\partial \tilde{\varphi }(x,y)}{\partial n}}={\displaystyle \frac{\partial \Gamma (x-y)}{\partial n}}$, $\mathtt{on} \partial \tilde{\Omega }.$ Then,

$\begin{array}{ccc}\hfill v(x)& =& {\int }_{\tilde{\Omega }}\tilde{G}(x,y)\tilde{f}(y)dy\hfill \\ & =& {\int }_{\Omega }(\Gamma (x-y)-\tilde{\varphi }(x,y)f(y)dy.\hfill \end{array}$

Therefore, it follows from (4) that

(14)$\begin{array}{c}\hfill v(x)={\int }_{\Omega }\Gamma (x-y)f(y)dy,\forall x\in \tilde{\Omega }.\end{array}$

Case 1. $x\in \Omega$.

It follows from (13) and (14) that $v(x)=u(x)$.

Case 2. $x\in \tilde{\Omega }\backslash \Omega$.

When $y\in \Omega$, $\Gamma (x-y)$ is a biharmonic function in $\Omega$.

It follows from (14), (4) that

$v(x)=0,\forall x\in \tilde{\Omega }\backslash \overline{\Omega }.$

In view of the continuity of $v(x)$, we have

$v(x)=0,\forall x\in \tilde{\Omega }\backslash \Omega .$

Combining the two cases above, we find that

$\tilde{u}(x)=v(x),x\in \tilde{\Omega },$

which implies that $\tilde{u}(x)$ is the unique classical solution of (3). □

3. Strong Solutions

Classical solutions, which satisfy the governing equation pointwise with sufficient regularity ($C^2$ solutions), are appropriate for physical models relying on continuum assumptions, such as in classical fluid dynamics. In contrast, strong solutions (typically residing in Sobolev spaces $W^{k,p}$) become necessary when addressing problems with material discontinuities, singular coefficients, or irregular domains.

Next, we use an approximation argument to prove Theorem 2.

Proof. 

Now, $f\in L^p(\Omega )$ and then $\tilde{f}\in L^p(\tilde{\Omega })$. The unique solution $u\in W^{4,p}(\Omega )\cap W_0^{2,p}(\Omega )$ exists for (1) (see Chapter 3 [8] or see Chapter 9 [12]).

(1) 

Necessity

Now, $v(x)\in W^{4,p}(\tilde{\Omega })\cap W_0^{2,p}(\tilde{\Omega })$ is the strong solution for (3).

Let $\eta (x):{\mathbb{R}}^n\to [0,\infty )$ be a mollifier satisfying

(i). $\eta (x)\in C_0^{\infty }({\mathbb{R}}^n)$,

(ii). ${\displaystyle {\int }_{B_1}\eta (x)dx=1}$, where $B_1$ is the unit ball centered at the origin.

For $\epsilon >0$, denote

${\eta }_{\epsilon }(x)={\displaystyle \frac{1}{{\epsilon }^n}}\eta ({\displaystyle \frac{x}{\epsilon }}).$

Then, $\mathrm{Supp}{\eta }_{\epsilon }(x)\subset B_{\epsilon }$, and ${\displaystyle {\int }_{B_{\epsilon }}{\eta }_{\epsilon }(x)dx=1}$, where $B_{\epsilon }$ is the ball with radius $\epsilon$, centered at the origin.

We extend $\tilde{u}$, $\tilde{f}$ from $\tilde{\Omega }$ to ${\mathbb{R}}^n$ by setting $\tilde{u}(x)=0$, $\tilde{f}(x)=0$, $x\in {\mathbb{R}}^n\backslash \tilde{\Omega }$. We define

$\begin{array}{ccc}\hfill u_{\epsilon }(x)& =& {\int }_{{\mathbb{R}}^n}{\eta }_{\epsilon }(x-y)\tilde{u}(y)dy={\int }_{\Omega }{\eta }_{\epsilon }(x-y)u(y)dy;\hfill \\ \hfill f_{\epsilon }(x)& =& {\int }_{{\mathbb{R}}^n}{\eta }_{\epsilon }(x-y)\tilde{f}(y)dy={\int }_{\Omega }{\eta }_{\epsilon }(x-y)f(y)dy.\hfill \end{array}$

Choose ${\displaystyle \epsilon <\frac{1}{2}\mathrm{dist}(\partial \tilde{\Omega },\partial \Omega )}$, and denote ${\Omega }_{\epsilon }=\{x\in {\mathbb{R}}^n|\mathtt{dist}(x,\Omega )<\epsilon \}$.

It is a simple fact that $u_{\epsilon }(x),f_{\epsilon }(x)\in C_0^{\infty }({\Omega }_{2\epsilon })$. Recalling

${\Delta }^2\tilde{u}=\tilde{f}\mathtt{in}\tilde{\Omega },$

we have

${\Delta }^2{u}_{\epsilon }=f_{\epsilon }\mathtt{in}{\mathbb{R}}^n.$

First, let $g\in C^4(\overline{\Omega })$ be a biharmonic function in $\Omega$. By Whitney’s extension theorem, we extend g to be $\tilde{g}$ from $\overline{\Omega }$ to $\overline{\tilde{\Omega }}$, such that $\tilde{g}\in C^4(\overline{\tilde{\Omega }})$ (see [14,15]). By using the fact that $u_{\epsilon }(x)\in C_0^{\infty }({\Omega }_{2\epsilon })$, we find that

${\int }_{{\Omega }_{2\epsilon }}\tilde{g}{\Delta }^2{u}_{\epsilon }dx={\int }_{{\Omega }_{2\epsilon }}{u}_{\epsilon }{\Delta }^2\tilde{g}dx,$

which implies

${\int }_{{\Omega }_{2\epsilon }}{f}_{\epsilon }\tilde{g}dx={\int }_{{\Omega }_{2\epsilon }\backslash \Omega }{u}_{\epsilon }{\Delta }^2\tilde{g}dx.$

On the other hand, we have

$\begin{array}{ccc}\hfill {\int }_{\Omega }fgdx& =& {\int }_{\Omega }(f-f_{\epsilon })gdx+{\int }_{{\Omega }_{2\epsilon }}{f}_{\epsilon }\tilde{g}dx-{\int }_{{\Omega }_{2\epsilon }\backslash \Omega }{f}_{\epsilon }\tilde{g}dx\hfill \\ & =& {\int }_{\Omega }(f-f_{\epsilon })gdx+{\int }_{{\Omega }_{2\epsilon }\backslash \Omega }{u}_{\epsilon }{\Delta }^2\tilde{g}dx-{\int }_{{\Omega }_{2\epsilon }\backslash \Omega }{f}_{\epsilon }\tilde{g}dx\hfill \\ & =& I_1+I_2+I_3.\hfill \end{array}$

By the Hölder inequality, we estimate the three terms $I_1$, $I_2$, and $I_3$ as below:

$\begin{array}{ccc}\hfill I_1& \le & \underset{\overline{\Omega }}{max}|g|{\left({\int }_{\Omega }{|f-f_{\epsilon }|}^{p}dx\right)}^{{\displaystyle \frac{1}{p}}}{|\Omega |}^{1-{\displaystyle \frac{1}{p}}};\hfill \\ \hfill I_2& \le & \underset{\overline{\tilde{\Omega }}}{max}|{\Delta }^2\tilde{g}|{\left({\int }_{{\Omega }_{2\epsilon }\backslash \Omega }{|u_{\epsilon }|}^{p}dx\right)}^{{\displaystyle \frac{1}{p}}}{|{\Omega }_{2\epsilon }\backslash \Omega |}^{1-{\displaystyle \frac{1}{p}}}\hfill \\ & \le & \underset{\overline{\tilde{\Omega }}}{max}|{\Delta }^2\tilde{g}|{\left({\int }_{\Omega }{|u|}^{p}dx\right)}^{{\displaystyle \frac{1}{p}}}{|{\Omega }_{2\epsilon }\backslash \Omega |}^{1-{\displaystyle \frac{1}{p}}};\hfill \\ \hfill I_3& \le & \underset{\overline{\tilde{\Omega }}}{max}|\tilde{g}|{\left({\int }_{{\Omega }_{2\epsilon }\backslash \Omega }{|f_{\epsilon }|}^{p}dx\right)}^{{\displaystyle \frac{1}{p}}}{|{\Omega }_{2\epsilon }\backslash \Omega |}^{1-{\displaystyle \frac{1}{p}}}\hfill \\ & \le & \underset{\overline{\tilde{\Omega }}}{max}|\tilde{g}|{\left({\int }_{\Omega }{|f|}^{p}dx\right)}^{{\displaystyle \frac{1}{p}}}{|{\Omega }_{2\epsilon }\backslash \Omega |}^{1-{\displaystyle \frac{1}{p}}}.\hfill \end{array}$

Sending $\epsilon \to 0$, we conclude that

${\int }_{\Omega }fgdx=0,$

which is (4).

Next, for $g\in C^4(\Omega )\cap C(\overline{\Omega })$ satisfying ${\Delta }^{2}g=0$ in $\Omega$. We use the same approximation argument as in the proof of Theorem 1 to obtain

${\int }_{\Omega }f(x)g(x)dx=0.$

(2) 

Sufficiency

Let $\left\{f_n\right\}\subset C_0^{\infty }(\Omega )$ be a sequence satisfying

$\underset{n\to \infty }{lim}{\parallel f_n-f\parallel }_{L^p(\Omega )}=0.$

Define

${\tilde{f}}_n(x)=\left\{\begin{array}{cc}{f}_n(x)& x\in \Omega ,\hfill \\ 0& x\in \tilde{\Omega }\backslash \Omega .\hfill \end{array}\right.$

We know that ${\tilde{f}}_n(x)\in C_0^{\infty }(\tilde{\Omega })$ and

$\underset{n\to \infty }{lim}{\parallel {\tilde{f}}_n-\tilde{f}\parallel }_{L^p(\tilde{\Omega })}=0.$

Let $\left\{u_n\right\}$ be the classical solutions for Dirichlet problems

(15)$\left\{\begin{array}{cc}{\Delta }^2{u}_n=f_n(x)& \mathrm{in}\Omega ,\hfill \\ {\displaystyle u_n=0,\frac{\partial u_n}{\partial n}=0}& \mathrm{on}\partial \Omega .\hfill \end{array}\right.$

It is obvious that $u_n\in W^{2,p}(\Omega )\cap W_0^{2,p}(\Omega )$. Let $u\in W^{4,p}(\Omega )\cap W_0^{2,p}(\Omega )$ be the unique solution of (1). This implies that (see [8])

$\begin{array}{c}\hfill \parallel u_n{-u\parallel }_{W^{4,p}(\Omega )}\le C{\parallel f_n-f\parallel }_{L^p(\Omega )},n=1,2,\dots .\end{array}$

Let $v_n$ be the classical solutions of the Dirichlet problem

(16)$\left\{\begin{array}{cc}-{\Delta }^2{v}_n={\tilde{f}}_n\hfill & \mathtt{in}\tilde{\Omega },\hfill \\ v_n=0,{\displaystyle \frac{\partial v_n}{\partial n}}=0\hfill & \mathtt{on}\partial \tilde{\Omega }.\hfill \end{array}\right.$

It is obvious that $v_n\in W^{4,p}(\tilde{\Omega })\cap W_0^{2,p}(\tilde{\Omega })$. Let $v\in W^{4,p}(\tilde{\Omega })\cap W_0^{2,p}(\tilde{\Omega })$, be the unique solution of system (3).

This implies that

$\begin{array}{ccc}\hfill \parallel v_n{-v\parallel }_{W^{4,p}(\tilde{\Omega })}& \le & C\parallel {\tilde{f}}_n-\tilde{f}{\parallel }_{L^p(\tilde{\Omega })}\hfill \\ & \le & C\parallel f_n{-f\parallel }_{L^p(\Omega )},n=1,2,\dots .\hfill \end{array}$

Let $G(x,y)$ and $\tilde{G}(x,y)$ be Green’s functions of ${\Delta }^2$ in $\Omega$ and $\tilde{\Omega }$. We know

(17)$G(x,y)=\Gamma (y-x)-\varphi (x,y),\tilde{G}(x,y)=\Gamma (y-x)-\tilde{\varphi }(x,y),$

where $\Gamma (y-x)$ is the fundamental solution, and $\varphi (x,y)$ is the solution for the boundary value problem ${\Delta }_y^2\varphi (x,y)=0$$\mathtt{in} \Omega$; $\varphi (x,y)=\Gamma (x-y),{\displaystyle \frac{\partial \varphi (x,y)}{\partial n}}={\displaystyle \frac{\partial \Gamma (x-y)}{\partial n}}$, $\mathtt{on} \partial \Omega .$ and $\tilde{\varphi }(x,y)$ is the solution for the boundary value problem ${\Delta }_y^2\tilde{\varphi }(x,y)=0$ $\mathtt{in} \tilde{\Omega }$; $\tilde{\varphi }(x,y)=\Gamma (x-y),{\displaystyle \frac{\partial \tilde{\varphi }(x,y)}{\partial n}}={\displaystyle \frac{\partial \Gamma (x-y)}{\partial n}}$, $\mathtt{on} \partial \tilde{\Omega }.$ Thus, we have $\varphi (x,y)\in C^4(\overline{\Omega })$ and $\tilde{\varphi }(x,y)\in C^4(\overline{\tilde{\Omega }})$.

Let $\phi (x)\in C_0^{\infty }(\Omega )$ be arbitrary. Since $u_k$ and $v_k$ are the classical solutions of (15) and (16). By virtue of Fubini’s theorem, we have

$\begin{array}{ccc}\hfill {\int }_{\Omega }{u}_k(x)\phi (x)dx& =& {\int }_{\Omega }\left[{\int }_{\Omega }G(x-y)f_k(y)dy\right]\phi (x)dx\hfill \\ & =& {\int }_{\Omega }\left[{\int }_{\Omega }G(x-y)\phi (x)dx\right]f_k(y)dy.\hfill \end{array}$

Sending $k\to \infty$, we obtain that

$\begin{array}{ccc}\hfill {\int }_{\Omega }u(x)\phi (x)dx& =& {\int }_{\Omega }\left[{\int }_{\Omega }G(x-y)\phi (x)dx\right]f(y)dy\hfill \\ & =& {\int }_{\Omega }\left[{\int }_{\Omega }(\Gamma (x-y)-\varphi (x,y))\phi (x)dx\right]f(y)dy\hfill \\ & =& {\int }_{\Omega }\left[{\int }_{\Omega }\Gamma (x-y)\phi (x)dx\right]f(y)dy\hfill \\ & & -{\int }_{\Omega }\left[{\int }_{\Omega }\varphi (x,y)f(y)dy\right]\phi (x)dx.\hfill \end{array}$

It follows from (4) that

(18)$\begin{array}{ccc}\hfill {\int }_{\Omega }u(x)\phi (x)dx& =& {\int }_{\Omega }\left[{\int }_{\Omega }\Gamma (x-y)\phi (x)dx\right]f(y)dy.\hfill \end{array}$

Let $\psi (x)\in C_0^{\infty }(\tilde{\Omega })$ be arbitrary. Using the same argument as above, we conclude that

(19)$\begin{array}{ccc}\hfill {\int }_{\tilde{\Omega }}v(x)\psi (x)dx& =& {\int }_{\Omega }\left[{\int }_{\tilde{\Omega }}\Gamma (x-y)\psi (x)dx\right]f(y)dy.\hfill \end{array}$

Case 1. $x\in \Omega$.

Let $\psi (x)=\phi (x),x\in \Omega$ and $\psi (x)=0,x\in \tilde{\Omega }\backslash \overline{\Omega }$. By virtue of (18) and (19) we obtain that

$u(x)=v(x), \mathtt{a}.\mathtt{e}.x\in \Omega .$

Case 2. $x\in \tilde{\Omega }\backslash \overline{\Omega }$.

Choose ${\displaystyle \epsilon <\frac{1}{2}\mathrm{dist}(\partial \tilde{\Omega },\partial \Omega )}$, and denote ${\Omega }_{\epsilon }=\{x\in {\mathbb{R}}^n|\mathrm{dist}(x,\Omega )<\epsilon \}$. Let $\phi (x)\in C_0^{\infty }(\tilde{\Omega }\backslash {\Omega }_{2\epsilon })$ be arbitrary. Using the same argument as above, we conclude that

$\begin{array}{ccc}\hfill {\int }_{\tilde{\Omega }\backslash {\Omega }_{2\epsilon }}v(x)\phi (x)dx& =& {\int }_{\Omega }\left[{\int }_{\tilde{\Omega }\backslash {\Omega }_{2\epsilon }}\tilde{\Gamma }(x,y)\phi (x)dx\right]f(y)dy\hfill \\ & =& {\int }_{\tilde{\Omega }\backslash {\Omega }_{2\epsilon }}\left[{\int }_{\Omega }\tilde{\Gamma }(x,y)f(y)dy\right]\phi (x)dx.\hfill \end{array}$

By virtue of (4) we have

$\begin{array}{c}\hfill {\int }_{\tilde{\Omega }\backslash {\Omega }_{2\epsilon }}v(x)\phi (x)dx=0.\end{array}$

Sending $\epsilon \to 0$, we obtain that

$\begin{array}{c}\hfill {\int }_{\tilde{\Omega }\backslash \Omega }v(x)\phi (x)dx=0,\end{array}$

which implies that

$v(x)=0, \mathtt{a}.\mathtt{e}.x\in \tilde{\Omega }\backslash \overline{\Omega }.$

Combining the two cases above, we find that

$\tilde{u}(x)=v(x), \mathtt{a}.\mathtt{e}.x\in \tilde{\Omega },$

which implies that $\tilde{u}(x)$ is the unique strong solution for (3). □

4. Generalization

While this study focused on a biharmonic equation with the Dirichlet boundary, several questions remain unanswered for further investigation:

The proposed method could be extended to the following boundary problem of the ${\Delta }^m$ equation

$\left\{\begin{array}{cc}{\Delta }^{m}u=f\hfill & \mathrm{in} \Omega ,\hfill \\ {\displaystyle u=\frac{\partial u}{\partial n}=\cdots =\frac{{\partial }^{m-1}u}{\partial n^{m-1}}=0}\hfill & \mathrm{on} \partial \Omega ,\hfill \end{array}\right.$

where $m\ge 3$ is an integer and we can consider the zero extension problem.

The biharmonic equation is a linear differential equation, and it is convenient to use Green’s function. But, if it is a mixed boundary value problem or nonlinear equation, variational methods and functional analytic methods need to be used.

When $\Omega$ is not a compact subset of $\tilde{\Omega }$, what condition on the function $f(x)$ ensures that the zero function $\tilde{u}(x)$ is still a solution of (3).

5. Conclusions

This paper focuses on zero extension for the biharmonic equation of the Dirichlet problem. We established Theorems 1 and 2, which are the necessary and sufficient conditiosn under the frameworks of classical and strong solutions.

Author Contributions

S.X. carried out the mathematical studies and C.Y. drafted the manuscript. All of the authors contributed equally to the preparation of this paper. All authors have read and agreed to the published version of the manuscript.

Data Availability Statement

The original contributions presented in the study are included in the article; further inquiries can be directed to the corresponding author.

Conflicts of Interest

The authors declare no conflicts of interest.

Footnotes

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