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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 .

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  • $\begingroup$ Do you mean $\mathbb{R}^N$ instead of $\mathbb{R}^\mathbb{N}$? $\endgroup$ – user35593 Jul 27 '15 at 11:21
  • $\begingroup$ @user35593 : yes $\endgroup$ – Finish Jul 27 '15 at 11:23
  • $\begingroup$ 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 $\endgroup$ – Math604 Jul 27 '15 at 16:00
  • $\begingroup$ 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. $\endgroup$ – Math604 Jul 28 '15 at 7:11
  • $\begingroup$ @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} $ $\endgroup$ – Finish Jul 28 '15 at 9:22

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