Mathematics Asked by Danilo Gregorin Afonso on November 26, 2021
The following is in Appendix B of Struwe’s Variational Methods
Let $u$ be a solution of $-Delta u = g(x, u(x))$ in a domain $Omega subset mathbb R^N$, $N geq 3$, where $g$ is a Carathéodory function with subcritical superlinear growth.
Theorem:
Let $Omega subset mathbb R^N$ be a smooth open set and let $g: Omega times mathbb R to mathbb R$ be a Carathéodory function such that
$$
|g(x, u(x))| leq a(x)(1 + |u(x)|) quad text{ a.e. in } Omega
$$
for some $0 leq a in L_{loc}^{N/2}(Omega)$. Let $u in H^1_{loc}(Omega)$ be a weak solution to $-Delta u = g(x, u)$. Then $u in L^q_{loc}(Omega)$ for all $1 < q < infty$. If $u in H_0^1(Omega)$ and $a in L^{N/2}(Omega)$, then $u in L^q(Omega)$ for all $1 < q < infty$.
The proof goes as follows:
Take $eta in C_c^infty(Omega)$, $s geq 0$ and $L geq 0$ and let
$$
varphi = u min {|u|^{2s}, L^2} eta^2 in H_0^1(Omega)
$$
Testing the equation against $varphi$ yields
$$
int_Omega |nabla u|^2 min{|u|^{2s}, L^2} eta^2 dx + frac s2 int_{{|u|^sleq L }} |nabla(|u|^2)|^2 |u|^{2s – 2} eta ^2 dx \
leq -2 int_Omega nabla u u min {|u|^{2s}, L^2} nabla eta eta dx + int_Omega a(1 + 2|u|^2)min {|u|^{2s}, L^2}eta^2 dx \
(*) quad {leq} frac 12 int_Omega |nabla u|^2 min{|u|^{2s}, L^2}eta^2 dx + c int_Omega |u|^2 min{|u|^{2s}, L^2} |nabla eta|^2 dx \
quad + 3 int_Omega a|u|^2 min{|u|^{2s}, L^2} eta^2 dx + int_Omega |a|eta^2 dx
$$
Why does $(*)$ hold?
Thanks in advance and kind regards.
I got this with the hint by MaoWao and the help of David Stolnicki.
Recall Young's Inequality with an epsilon: $$ ab leq frac 1{2varepsilon} a^2 + frac varepsilon2 b^2. $$ We apply this inequality to the first term, which gives begin{align*} - 2 int_Omega nabla u u min{|u|^{2s}, L^2} nabla eta eta dx leq & frac 12 int_Omega |nabla u|^2 min {|u|^{2s}, L^2} eta^2 dx \ & + c int_Omega u^2 min{|u|^{2s}, L^2} |nabla eta|^2 dx. end{align*} As for the second term, we can estimate it as begin{align*} int_Omega a(1+|u|^2 min{|u|^{2s}, L^2} eta^2 dx = & int_Omega a min{|u|^{2s}, L^2} eta^2 dx \ & + 2 int_Omega a |u|^2 min{|u|^{2s}, L^2} eta^2 dx \ = & 3 int_Omega a|u|^2 min{|u|^{2s}, L^2} eta^2 dx \ & + int_Omega amin{|u|^{2s}, L^2} eta^2 (1 - |u|^2) dx \ leq & 3 int_Omega a|u|^2 min{|u|^{2s}, L^2} eta^2 dx \ & + int_Omega a eta^2 dx. end{align*} Therefore, begin{align} begin{split} & int_Omega |nabla u|^2 min{|u|^{2s}, L^2} eta^2 dx + frac s2 int_{{|u|^sleq L }} |nabla(|u|^2)|^2 |u|^{2s - 2} eta ^2 dx \ & qquad leq frac 12 int_Omega |nabla u|^2 min {|u|^{2s}, L^2} eta^2 dx + c int_Omega u^2 min{|u|^{2s}, L^2} |nabla eta|^2 dx \ & qquad quad + 3 int_Omega a|u|^2 min{|u|^{2s}, L^2} eta^2 dx + int_Omega a eta^2 dx. end{split} end{align}
Answered by Danilo Gregorin Afonso on November 26, 2021
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