Uniqueness of weak solution of a pde

Question

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!

Answer 1

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.

Answer 2

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 $u = v$.