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

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

(1)

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 $\tilde{\Omega }$

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

(2)

Consider the corresponding Dirichlet problem

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

(3)

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.,

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

(4)

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

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

(5)

We solve the following problem

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

(6)

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

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

(7)

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

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

(8)

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

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

(9)

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

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

(10)

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

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

(11)

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

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

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

(12)

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

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

(13)

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

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

(14)

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

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

(15)

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

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

(16)

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

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

(17)

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

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

(18)

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

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

(19)

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.