Uniqueness of weak solution of a pde I have proved the existence of weak solution in $H_0^1(\Omega)$ of the problem $-\Delta u + |u|^{p-1}u=f$ on $\Omega$ and $u=0$  on $\partial\Omega$, where $\Omega$ is a bounded and smooth domain of $R^n$, $f\in L^2(\Omega)$ and $p>1$. I wonder which is the easiest technique to prove the unicity here. Any help is welcome, thanks in advance!
 A: I'm going to assume you used a variational method to find a weak solution (if not, look at Using Energy method in finding weak solutions of PDE).  In this case, the $u$ you found was the minimum of the energy
$$E[u] = \int_\Omega \frac{1}{2} |\nabla u |^2 + \frac{1}{p+1} |u|^{p+1} - fu \;dx$$
Moreover, $v$ solves
$$-\Delta v + |v|^{p-1} v = fv$$
if and only if the first variation of the energy at $v$, $\delta E[v]$, is zero.  We will show this can only happen at the minimizer $u$ by showing, roughly, that $E$ has positive Hessian.  In particular, we will show that the second variation of $E$, 
$$\delta^2E[u + t(u-v)](u-v) = \left.\left(\frac{d}{d\theta}\right)^2\right|_{\theta = 0} E[u + (t+\theta)(u-v)] > 0\tag{*}$$
for any $t$ where $(1-t)u \neq tv$.  By the Fundamental Theorem of Calculus, it will follow that
$$\delta E[v](u-v) = \delta E[u](u-v) + \int_0^1 \delta^2 E[u + t(u-v)](u-v)\;dt > 0$$
so $v$ cannot be a critical point of the energy.
To prove (*), we compute that
$$\delta^2 E[u + t(u-v)](u-v) = \int_\Omega |\nabla ((1-t)u + tv)|^2 + p|v|^{-1}((1-t)u + tv)^2\;dx$$
Since the first term is positive for $(1-t)u \neq tv$ and the second term in nonnegative, the result follows.
A: Here's an alternative answer based on your comment on my other answer.
Suppose $u$ and $v$ are two solutions, and let $w = u-v$.  Then, $w$ solves
$$-\Delta w + |u|^{p-1}u - |v|^{p-1}v = 0$$
Multiplying by $w$ and integrating gives
$$\int |\nabla w|^2 + |u|^{p+1} + |v|^{p+1} - vu|u|^{p-1} - uv|v|^{p-1}\;dx = 0$$
so, rearranging
$$\begin{align*}
 \lVert \nabla w \rVert_{L^2}^2 + \lVert u \rVert_{L^{p+1}}^{p+1} + \lVert v \rVert_{L^{p+1}}^{p+1}  &= \int uv(|u|^{p-1} + |v|^{p-1})\;dx\\
   &\leq \lVert u \rVert_{L^{p+1}}\lVert v \rVert_{L^{p+1}}^{p} + \lVert u \rVert_{L^{p+1}}^p\lVert u \rVert_{L^{p+1}}
\end{align*}$$
by Holder's inequality.  On the other hand, we have the elementary inequality
$$a^{p+1} + b^{p+1} \geq ab^p + a^p b \qquad a,b \geq 0$$
where equality holds only if $a = 0$, $b = 0$, or $a = b$.  Hence,
$$\lVert \nabla w \rVert_{L^2}^2 + \lVert u \rVert_{L^{p+1}}^{p+1} + \lVert v \rVert_{L^{p+1}}^{p+1} \leq \lVert u \rVert_{L^{p+1}}^{p+1} + \lVert v \rVert_{L^{p+1}}^{p+1}$$
from which it follows that $\nabla w = 0.$  By the Poincare inequality, $w = 0$ in $H^1$ and $u = v$.
