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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!

Senna
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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.

  • Ok, this was my attempt after a long time searching and thinking: As you said, I proved fristly the existence of solution using the energy method. For the uniqueness, I did the following: I proved that the functional $E$ is stricly convex using the definition ($E(tu+(1-t)v) < tE(u) + (1-t)E(v),\ u\neq v, 0<t<1$) and some algebra. From the strictly convexity of $E$, easily follows that if $E$ has a minimum, then is unique. After that, I proved that every weak solution of the problem is actually a minimizer of $E$ on $H_0^1$, therefore we end up with the uniqueness of solution. – Senna May 01 '19 at 17:33
  • This is what you did (more or less) since positive hessian implies strict. convexity, isn't it? – Senna May 01 '19 at 17:36
  • Another approach, following the idea of the comment in https://math.stackexchange.com/questions/2713540/discuss-uniqueness-of-solution-of-delta-u-u-int-omegau2ydy-f, is: define $w= u - v $ where $u$ and $v$ are two solutions, then $w$ satisfies $$ -\Delta w + |u|^{p-1}u - |v|^{p-1}v =0 $$, and so $$ \int |\nabla w |^2\ dx + \int w [u|u|^{p-1}-v|v|^{p-1}] \ dx = 0$$ from this follows $$ \int uv(|u|^{p-1} - |v|^{p-1})\ dx = \Vert \nabla w\Vert ^2_{L^2} + \Vert u \Vert^{p-1}{L^{p-1}} + \Vert v \Vert^{p-1}{L^{p-1}} $$ and now I got stucked. I would like to get a contradiction from that. – Senna May 01 '19 at 17:42
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    @R.N.Marley Yes, it seems like we had the same idea. Let me think about your second approach, I'll post it as an answer if I figure it out. . . –  May 01 '19 at 18:16
  • @R.N.Marley See my other answer –  May 01 '19 at 21:12
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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 $u = v$.

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    Excellent! Thanks you :) – Senna May 01 '19 at 21:42
  • Sorry, I dont understand the step where you use Hölder. Can you give me some details? As far as I am concerned, Hölder is $$ \int fg \leq \Vert f \Vert_p \Vert g \Vert_q $$ where $p^{-1} + q^{-1} = 1$, so what $f$ and $g$ did you pick? – Senna May 02 '19 at 10:47
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    I'll show you how I got the $\lVert u \rVert_{L^{p+1}} \lVert v \rVert_{L^{p+1}}^p$ term: the other one is analogous. By the usual Holder inequality, $$\int u|v|^{p-1} v;dx \leq \lVert u \rVert_{L^{p+1}} \lVert v|v|^{p-1} \rVert_{L^{\frac{p+1}{p}}}$$ But, the last term is $$\lVert v|v|^{p-1} \rVert_{L^{\frac{p+1}{p}}} = \left(\int |v|^{p \frac{p+1}{p}};dx\right)^{\frac{p}{p+1}} = \left(\left(\int |v|^{p+1};dx\right)^{1/p}\right)^p = \lVert v \rVert_{L^{p+1}}^p$$ –  May 02 '19 at 13:56
  • I don't really think of it that way, though. I see it more like a statement that we have an 'iterated Holder' inequality, namely that $$\int f_1 f_2 \cdots f_p f_{p+1} ;dx \leq \lVert f_1 \rVert_{L^{p+1}}\lVert f_2 \rVert_{L^{p+1}}\cdots \lVert f_p \rVert_{L^{p+1}}\lVert f_{p+1} \rVert_{L^{p+1}}$$ (You can prove this by induction using the ordinary Holder inequality). –  May 02 '19 at 14:01