$-\Delta u - \alpha u^{1/3} = 0$ implies $u \equiv 0$ if $\alpha$ is small Let $\Omega$ be a domain in $\mathbb{R}^{d}$ with smooth boundary. Let $u(x)$ be a $H^{1}(\Omega)$ solution of the equation $-\Delta u - \alpha u^{1/3} = 0$, $u|_{\partial \Omega} = 0$. The problem I am working on is to show that $u \equiv 0$ if $\alpha > 0$ is small.
The hint in this problem is to use Poincare's Inequality. Multiplying the given PDE by $u$, we have $-u\Delta u - \alpha u^{4/3} = 0$. Then $$0 = \int_{\Omega}-u\Delta u - \alpha u^{4/3}\, dx = \int_{\Omega}|\nabla u|^{2} - \alpha u^{4/3}\, dx$$ and hence $\alpha\int_{\Omega}u^{4/3}\, dx = \int_{\Omega}|\nabla u|^{2}\, dx \geq \frac{1}{C}\int_{\Omega}u^{2}\, dx$ where $C$ is the constant from Poincare's Inequality. Rewriting, this implies $$\int_{\Omega}u^{2/3}u^{4/3}\, dx \leq C\alpha\int_{\Omega}u^{4/3}\, dx$$ and then
$$\int_{\Omega}u^{4/3}(C\alpha - u^{2/3})\, dx \geq 0.$$ My question is: Are there suggestions on how to finish? Am I approaching this problem correctly?
 A: This is not true. Let's prove it for a more general setting: assume that $\beta>0$ is such that $\beta+1 < 2$, then for any $\alpha>0$ there exists nontrivial variational solution $u$ of the problem
$$ \left\{ \begin{array}{ccc} -\Delta u-\alpha |u|^{\beta-1}u=0 &\mbox{ if $x\in\Omega$}, \\ u=0 &\mbox{if $x\in\partial\Omega$}. \end{array} \right. $$
To prove it, define the corresponding energy functional: $$ F_\beta(u)=\frac{1}{2}\int_\Omega|\nabla u|^2-\frac{\alpha}{\beta+1}\int_\Omega |u|^{\beta+1}, \quad \forall u\in H_0^1(\Omega).  $$
Note that the condition $\beta+1 < 2$ implies that $F_\beta$ is well defined. Now use Poincare's inequality and the continuous inclusion of $L^2(\Omega)$ into $L^{\beta+1}(\Omega)$ to conclude that $$ \int_\Omega |u|^{\beta+1}\le c\left(\int_\Omega |u|^2\right)^{\frac{\beta+1}{2}}\le C\left(\int_\Omega |\nabla u|^2\right)^{\frac{\beta+1}{2}}, \quad \forall u\in H_0^1(\Omega).\tag{1}  $$
Hence, we conclude from (1) that $F_\beta$ is coercive for any $\alpha$.
On the other hand, as long as we have that $\beta+1 < 2$,
the compact embedding of $H_0^1(\Omega)$ in $L^2(\Omega)$ implies that $F_\beta$ is sequentially lower continuous with respect to the weak topology, i.e., if $u_n\to u$ weak in $H_0^1(\Omega)$ then, $F_\beta(u)\le \liminf F_\beta (u_n)$. We combine these two facts to conclude that $F_\beta$ has a unique minimizer $u \in H_0^1(\Omega)$, which is the unique variational solution of our problem.
Let us show now that such minimizer $u \in H_0^1(\Omega)$ is nontrivial. For this end, take arbitrary function $w \in H_0^1(\Omega)$ and observe that $$ F_\beta(tw) = \frac{t^2}{2}\int_\Omega|\nabla w|^2- \frac{\alpha \ t^{\beta+1}}{\beta+1}\int_\Omega |w|^{\beta+1} = C_1 t^2 - C_2 t^{\beta + 1} < 0, $$ for sufficiently small $t$, due to the fact that $\beta+1<2$.
From here it follows that $$ F_\beta(u) \leq F_\beta(tw) < F(0) = 0, $$ which implies the desired nontriviality of $u$.
