# biharmonic equation with L^1 data and Navier Condition

I am reading an article that, a section of it is mentioned below . I have some question about this section. I will ask my question after the section below. I am thanksed if some one could help me , because really I need help.

$\Omega \subset \mathbb{R}^N$ is a smooth bounded domain. Let $f \in L^1(\Omega)$ , $f \geq 0$ , we know we can find sequence $\{f_n \} \in C_0^{\infty}(\Omega) \subset L^2(\Omega)$ such that $f_n(x) \leq f(x) \hspace{.2cm} \forall x\in \Omega$ and $f_n \to f \hspace{.2cm} L^1(\Omega)$. Now Let $u_n \in W^{2,2}(\Omega) \cap W_0^{1,2}(\Omega) \cap L^{\infty}(\Omega)$ be a weak solution of the problem $$\begin{cases} ‎\Delta^2u_n=‎‎f_n & in‎\hspace{.2cm}‎ \Omega \\ u_n>0 & in ‎\hspace{.2cm}‎ \Omega \\ u_n=\Delta u_n =0 & on‎\hspace{.2cm}‎ \partial \Omega ‎ \end{cases}$$

then $u_n$ verifies $$‎\int_{\Omega} ‎(-\Delta u_n)(-\Delta ‎\phi)‎‎‎‎\mathrm{d}x =‎ ‎\int_{\Omega} ‎f_n ‎\phi‎‎‎‎ ‎‎\forall ‎\phi ‎\in‎ ‎W^{2,2}(\Omega) ‎\cap ‎W_0^{1,2}(\Omega)$$

Noting by $-\Delta u_n = v_n$ such that $-\Delta v_n = f_n$ , and choosing appropriate test function we can proof that $\{ v_n \}$ is a bounded sequence in $W_0^{1,q}(\Omega) \hspace{.2cm} \forall \hspace{.1cm} 1\leq q < \dfrac{N}{N-1}$ . So there exists $v \in W_0^{1,q}(\Omega)$ and a subsequence(for simplicity we do not change the indeces) like $\{ v_n \}$ such that $v_n \rightharpoonup v$ weakly in $W_0^{1,q}(\Omega)$ for any $1 \leq q < \dfrac{N}{N-1}$ .

with choosing appropriate test function we could show that $\nabla v_n \to \nabla v$ strongly in $W_0^{1,q}(\Omega)$ for any $1 \leq q < \dfrac{N}{N-1}$ . So $v_n \to v$ strongly in $W_0^{1,q}(\Omega)$ for any $1 \leq q < \dfrac{N}{N-1}$

Now we could go to limit and conclude $v$ is a positive solution of $-\Delta v = f$ . Now thaking into account that $-\Delta u_n = v_n$ and applying Rellich-Kondrachov's theorem we conclude that there exists $u \in W^{2,q}(\Omega) \hspace{.2cm} \forall \hspace{.2cm} 1\leq q < \frac{N}{N-1}$ with $v = - \Delta u$ such that $$u_n \to u \hspace{.2cm} in \hspace{.2cm} W^{2,q}(\Omega) \hspace{.25cm} \forall \hspace{.4cm} 1\leq q < \frac{N}{N-1}$$ Now by strong Maximum Principle then $u_n >0$ and so $u >0$. this show that $u$ is a positive solution of the problem

$$\begin{cases} ‎\Delta^2u=‎‎f & in‎\hspace{.2cm}‎ \Omega \\ u>0 & in ‎\hspace{.2cm}‎ \Omega \\ u=\Delta u =0 & on‎\hspace{.2cm}‎ \partial \Omega ‎ \end{cases}$$

My Questions:

1 . Why $v$ is positive? ( I know $v_n >0$ in $\Omega$ , but $v$ as a P.W. limit may be zero )

2 . I do not know that how taking into account that $- \Delta u_n = v_n$ and applying Rellich-Kondrachov's theorem gives $u \in W^{2,q}$ such that $u_n \to u \hspace{.2cm} in \hspace{.2cm} W^{2,q}(\Omega) \hspace{.25cm} \forall \hspace{.4cm} 1\leq q < \frac{N}{N-1}$. ( I think if I could show $\{ u_n \}$ is bounded in $W^{3,q}$ then because $W^{3,q}$ is compactly embedded in $W^{2,q}$ then we reach to desired aim. but I do not know how to show boundedness of sequence in $W^{3,q}$ , Or maybe showing boundedness of sequence is not correct approach.)

3 . Why $u$ is positive ? ( I know $u_n > 0$ , but $u$ as a P.W. limit of $u_n$ maybe zero )

This is the link of article page 9th .

• Do you mean $\mathbb{R}^N$ instead of $\mathbb{R}^\mathbb{N}$? – user35593 Jul 27 '15 at 11:21
• @user35593 : yes – Finish Jul 27 '15 at 11:23
• here is a good book for fourth order problems: Filippo Gazzola, Hans-Christoph Grunau and Guido Sweers www1.mate.polimi.it/~gazzola/book_GGS.pdf – Math604 Jul 27 '15 at 16:00
• This isn't what you are asking but... Consider using a the cut-off $f_m(x)=f(x)$ if $f(x) \le m$ and $f_m(x)=m$ otherwise. So $f_m(x) \nearrow f(x)$. Let $-\Delta v_m=f_m$ in $\Omega$ with $v_m=0$ on $\partial \Omega$. Then note that $v_m$ is increasing and this makes showing $v$ is positive maybe easier. – Math604 Jul 28 '15 at 7:11
• @Math604 : if I assume cut-off function $f_m(x)$ as you mentioned , then exactly $f_m(x) \in L^2$ and lax-millgiram lemma ensure existence of solution $u_m \in W^{2,2} \cap W_0^{1,2}$ but I can not show $u_m \in L^{\infty}$. Can you say why $u_m \in L^{\infty}$ – Finish Jul 28 '15 at 9:22