# A priori estimate of an inhomogeneous p-Laplace equation with Dirichlet boundary condition

I'm currently working on this Dirichlet problem:

\begin{cases} div(\sigma |\nabla u|^{p-2} \nabla u) = f &\quad {in }~ \Omega\\ u = g &\quad in~\partial\Omega \end{cases}

with $\sigma \in L^{\infty}_{+}(\Omega)$, boundary $g \in W^{1,p}(\Omega)$ and $f \in L^2(\Omega)$.

I think I've already proved that this problem has a unique solution, but I don't know how to prove the following a priori estimate

\begin{align} ||{u}||_{W^{1,p}} \leq C ({||g||}_{W^{1,p}} + {||f||}_{L^2}^{\frac{1}{p-1}}) \end{align}

This estimate is mentioned in this paper https://pdfs.semanticscholar.org/7399/da07c625d51aa7ee72840789916b036019d2.pdf (second page; equation (1.4)) but without giving proof. I guess it's seen "easy", but I just don't know how to get there. I know how to conclude an a priori estimate in the case $f=0$ or $g=0$, but since p-Laplace operator isn't linear, this doesn't help and I'm really clueless know. I hope someone can help me with with.

As far as i thought i can use the Poincaré inequality since $u - g \in W^{1,p}_0$:

\begin{align} ||u||_{W^{1,p}} &= ||u||_{L^p} + ||\nabla u||_{L^p}\\ & \leq ||g||_{L^p} + ||u - g||_{L^p} + ||\nabla u||_{L^p}\\ & \leq ||g||_{L^p} + C||\nabla(u - g)||_{L^p} + ||\nabla u||_{L^p} \end{align}

The problem is that i can't find an upper bound for $||\nabla u||_{L^p}$ using $f$...

The inequality in the paper that i found implies that i should found something like

\begin{align} ||\nabla u||^p_{L^p} \leq ||f||_{L^2}~||\nabla u||_{L^p} \end{align}

which only be valid, when the solution u is in $W^{1,p}_0$...

I hope someone can help me with with this...

Edit: Using fedjas advice i got

\begin{align} \int_{\Omega} f~(u - g) &= \int_{\Omega} div(\sigma |\nabla u|^{p-2} \nabla u) (u - g)\\ &= \int_{\Omega} \sigma |\nabla u|^{p-2} \nabla u \nabla (u - g)\\ &\leq ||f||_{L^2} ||u - g||_{L^p} \\ &\leq ||f||_{L^2} ~C ||\nabla (u - g)||_{L^p} \end{align}

Now if i would have $||\nabla (u - g)||_{L^p}^p \leq \int_{\Omega} f~(u - g)$ it would be good... BUT I dont know how to get there...

$%\int_{\Omega} f~(u - g) \leq ||\nabla (u - g)||_{L^p}^p$

because the mapping $x \mapsto |x|^p$ is convex we have always

$|x|^p \geq |y|^p + p |y|^{p-2}y (x-y)$ (characterization with first derivation)

and with $x = \nabla (u - g)$ and $y = \nabla u$ it follows

$|\nabla (u - g)|^p \geq |\nabla u|^p - p|\nabla u|^{p-2} \nabla u \nabla g$

don't know if it's useful here, but its valid for $p \geq 1$.

Furthermore i know that

$\frac{1}{p} (\int_{\Omega} |\nabla u|^p - |\nabla g|^p) \leq \int_{\Omega} f~(u - g)$

Because the solution u is the minimizer of the functional: $J(v) = \frac{1}{p} \int_{\Omega} |\nabla v|^2 - \int_{\Omega} fv$ for all $v \in W^{1,p}_g$ (space with needed boundary values) and therefore $J(u)\leq J(g)$

Nevertheless i still don't see how to put all of this together to achieve the desired result... :/

• Play $u-g\in W_0^{1,p}$ against $f$, use the equation, and integrate by parts. I hope $p$ is good enough for the Sobolev embedding $W^{1,p}\subset L^2$. Commented Dec 11, 2017 at 5:49
• Thanks for comment! :) I tried to follow your advice, but i am not sure if i 100% understood it (espacally the sobolev embedding part?). I edited my post and tried to do what you meant, but as you can see I still dont know how to get there Commented Dec 11, 2017 at 7:50

It looks like you are totally new to such estimates, so let me show you the old $\varepsilon$ vs $C_\varepsilon$ trick. If you need to estimate some product $xy$, you can write $|xy|\le |x|^2+|y|^2$, but sometimes you would prefer a smaller constant on $x$ (that you would later subtract from the other side) and do not really care how big one you get on $y$. In this case you can write $$|xy|=|(\varepsilon^{1/2} x)(\varepsilon^{-1/2}y)|\le \varepsilon |x|^2+\varepsilon^{-1}|y|^2=\varepsilon |x|^2+C_\varepsilon |y|^2$$ You can do it with any positive powers, so you can equally well write $$|x|^a|y|^b\le\varepsilon|x|^{a+b}+C_\varepsilon |y|^{a+b}$$ (of course, $C_\varepsilon$ here depends on $a,b$ as well).
Now, you have your identity $$\int_\Omega f(u-g)=\int_\Omega\sigma|\nabla u|^{p-2}\langle\nabla u,\nabla u-\nabla g\rangle$$ Rewrite it as $$\int_\Omega\sigma|\nabla u|^{p}=\int_\Omega\sigma|\nabla u|^{p-2}\langle\nabla u,\nabla g\rangle+\int_\Omega f(u-g)\le \int_\Omega\sigma|\nabla u|^{p-1}|\nabla g|+\int_\Omega |f||u-g|$$ We certainly need $\sigma\ge \sigma_0>0$ in $\Omega$ (otherwise you can easily blow up everything). So, we have $$\sigma_0\int_\Omega|\nabla u|^{p}\le \int_\Omega\sigma|\nabla u|^{p-1}|\nabla g|+\int_\Omega |f||u-g|$$ Now play $\varepsilon$ vs $C\varepsilon$ and recall that $\sigma\le\sigma_1<+\infty$: $$\sigma|\nabla u|^{p-1}|\nabla g|\le \sigma_1(\varepsilon|\nabla u|^{p}+C_\varepsilon|\nabla g|^p)\,.$$ Assuming that $W^{1,p}\subset L^2$, we also have $$\int_\Omega |f||u-g|\le \|f\|_2\|u-g\|_2\le \|f\|_2\|u-g\|_{1,p}=(\|f\|^{1/(p-1)}_2)^{p-1}\|u-g\|_{1,p} \\ \le \varepsilon \|u-g\|_{1,p}^p+C_\varepsilon \|f\|_2^{p/(p-1)}.$$ Recalling the bound $$\|u-g\|^p_{1,p}\le C[\|\nabla u|^p_p+\|g\|_{1,p}^p]\,,$$ we get $$\sigma_0\int_\Omega|\nabla u|^{p}\le \sigma_1[\varepsilon\int_{\Omega}|\nabla u|^{p}+C_\varepsilon\int_{\Omega}|\nabla g|^p]+\varepsilon C[\int_{\Omega}|\nabla u|^{p}+\|g\|_{1,p}^p]+ CC_\varepsilon \|f\|_2^{p/(p-1)}$$ Now choose $\varepsilon$ so small that after moving all the terms with $\int_\Omega|\nabla u|^p$ from the right to the left, you still have some positive coefficient on the LHS.
• Wow! Thank you very very much! Yes, I am new too such estimates and this functional analysis stuff, so I am very gratefull for your detailed answer. But I still have one question regarding your mentioned bound $||u - g||^p_{1,p} \leq C [|| \nabla u||^p + ||g||^p_{1,p}]$... Can you explain why it is true? Because in my eyes it's not - at least not for arbitrary u and g? :/ I only have $||u - g||^p_{1,p} \leq C ||\nabla u - \nabla g||^p_p \leq C ( ||\nabla u||_p + || \nabla g ||_p)^p$ Commented Dec 12, 2017 at 13:29
• @ANZM91 What do you mean? $\|g\|_{1.p}^p\approx \|g\|_p^p+\|\nabla g\|_p^p$, so my RHS is larger than yours (and $\|g\|_p$ matters, by the way: think of large constants). Commented Dec 12, 2017 at 14:25
• Why is it larger? I have $(||\nabla u||_p + ||g||_p + ||\nabla g||_p ) ^p$ - the p-th power of the sum. Or is $| x - y| ^p \leq |x|^p + |y|^p$ always true? Sorry, but I am a bit confused at the moment >.< Commented Dec 12, 2017 at 14:49
• @ANZM91 Analysis is not algebra. If $a,b,p>0$, then we have $(a+b)^p=a^p+b^p$ for all practical purposes. My RHS is larger because it also includes $\|g\|_p^p$, which you lost somewhere (and it matters!). Commented Dec 12, 2017 at 16:50
• Ahhhh, maybe you did mean this way? \begin{align*} ||u - g||^p_{1,p} &\leq C_p || \nabla u - \nabla g ||^p_{p} \\ &\leq C_p(||\nabla u||_p + ||\nabla g||_p)^p \\ &= C_p \sum_{i=0}^{p} \binom{p}{i} ||\nabla u||_p^{p - i}||\nabla g||_p^i \\ &\leq C_p \binom{p}{p/2} (||\nabla u||^p_p + ||\nabla g||^p_p) \\ &\leq C (||\nabla u||^p_p + ||\nabla g||^p_p + ||\nabla g||) \\ &= \leq C (||\nabla u||^p_p + ||\nabla g||^p_{1,p}) \end{align*} By the way: In my case I have $p\geq 1$ so $W^{1,p} \subset L^2$ isnt working, is it? What do i have to change then? Commented Dec 12, 2017 at 16:53