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I am reposting a question from Math Overflow, because it seems it gets no attention.

Let $\Omega\subset \mathbb R^{d=3}$ is a bounded and Lipschitz domain. Let $u\in H_g^1(\Omega)$ satisfy the weak formulation $$a(u,v)+\int\limits_{\Omega}{f(u)vdx}=0,\,\forall v\in H_0^1(\Omega)\cap L^\infty(\Omega)\quad\quad (*)$$ where $a(u,v)=\int\limits_{\Omega}{\nabla u\cdot\nabla vdx}$, $f(u)=\sinh (u)=\frac{e^u-e^{-u}}{2}$, and $H_g^1(\Omega)=\{u\in H^1(\Omega):\,u=g \text{ on } \partial \Omega\}$.

Note that there is no polynomial growth condition on the nonlinearity $f$ and so $f(u)$ is not supposed to be in $L^2(\Omega),\,\forall u\in H_g^1(\Omega)$. In $d=3$, there are functions $u\in H^1(\Omega)$ s.t $e^u\notin L^1(\Omega)$ (for instance $u=\ln\left(\frac{1}{r^\alpha}\right)\in H_0^1(B_1),\,\alpha\ge 3$ over the unit ball $B_1$).

Question: Is it true that equation $(*)$ is satisfied for all $v\in H_0^1(\Omega)$ ? (If there is an a priori estimate which gives $u\in L^\infty(\Omega)$ then $(∗)$ is obviously satisfied for all $v\in H_0^1(\Omega)$)

If it would help, we may assume that $f(u)\in L^1(\Omega)$.

If I have a positive answer to the question, this means that equation $(*)$ has a unique solution: suppose $u_1,u_2$ are solutions to $(*)$, then $a(u_1-u_2,v)+(f(u_1)-f(u_2),v)=0,\,\forall v\in H_0^1(\Omega)$ and so we can test with $u_1-u_2\in H_0^1(\Omega)$ and use the monotonicity property $(f(u_1)-f(u_2),u_1-u_2)\ge 0$ to conclude that $u_1=u_2$.

My try: From equation $(*)$ we see that $\left|\int\limits_{\Omega}{f(u)vdx}\right|\leq \|u\|_1\|v\|_1,\,\forall v\in H_0^1(\Omega)\cap L^\infty(\Omega)$ and therefore $f(u)$ defines a bounded linear functional over the dense subspace $H_0^1(\Omega)\cap L^\infty(\Omega)$ by the formula $\langle F,v\rangle=\int\limits_{\Omega}{f(u)vdx}$ and so it is uniquely extendable over the whole $H_0^1(\Omega)$ by continuity to the functional $\bar F\in H^{-1}(\Omega)$. However, the extension is not necessarily reprezentable by the same formula. Now, let $v\in H_0^1(\Omega)$ is arbitrary and $\{v_n\}\subset H_0^1(\Omega)\cap L^\infty(\Omega)$ s.t $$\begin{array}{|l} &v_n(x)\to v(x),\,a.e\\ & \|v_n-v\|_1\to 0 \end{array}$$ Then, $\int\limits_{\Omega}{f(u)v_ndx}=-a(u,v_n),\,\forall n$ and $a(u,v_n)\to a(u,v)\Rightarrow \int\limits_{\Omega}{f(u)v_ndx}\to -a(u,v)=:A$. Now, I want to conclude that $A=\int\limits_{\Omega}{f(u)vdx}$. We also have $f(u(x))v_n(x)\to f(u(x))v(x),\,a.e$ in $\Omega$ and $\int\limits_{\Omega}{f(u)v_ndx}=\langle F,v_n\rangle\to \langle \bar F,v\rangle$. But from here I still cannot show that $\int\limits_{\Omega}{f(u)v_ndx}\to \int\limits_{\Omega}{f(u)vdx}$.

I cannot use Lebesgue DCT, because I do not have a dominating function for $|f(u(x))v_n(x)|$. I thought also about the generalized Lebesgue DCT and maybe Vitali convergence theorem, but couldn't do it.

If we can use some type of regularity theory or maximum principle (which I do not know for this equation) and show that $u\in L^\infty(\Omega)$ or just that $f(u)\in L^{6/5}(\Omega)$, then $$\left|\int\limits_{\Omega}{f(u)(v_n-v)dx}\right|\leq \|f(u)\|_{L^{6/5}(\Omega)}\|v_n-v\|_{L^6(\Omega)}\leq C\|f(u)\|_{L^{6/5}(\Omega)}\|v_n-v\|_1\to 0$$ and I would be done.

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Svetoslav
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1 Answers1

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You can show your regulary if $g \in W^{1,p}(\Omega)$, $p > d \ge 2$, by using a trick by Stampacchia.

First, we note $$a(u - g,v) + \int_\Omega f(u) v \, \mathrm{d} x = a(-g,v).$$

Now, let $K > 0$ and set $$(u - g)_K = \begin{cases} u - g & \text{if } |u - g| \le K, \\ \operatorname{sign}(u-g) \, K & \text{else}.\end{cases}$$ Then, $(u-g)_K \in H_0^1(\Omega) \cap L^\infty(\Omega)$.

Now, we choose $v = (u-g)_K$ and use $$a(u-g, (u-g)_K) = a((u-g)_K,(u-g)_K)$$ and $$\int_\Omega f(u) \, (u-g)_K \, \mathrm{d}x = \int_\Omega (f(u) - f(g)) \, (u-g)_K \, \mathrm{d}x + \int_\Omega f(g) \, (u-g)_K \, \mathrm{d}x \\\ge \int_\Omega f(g) \, (u-g)_K \, \mathrm{d}x$$ by monotonicity of $f$.

This shows $$ | (u-g)_K |_{H_0^1}^2 \le -\int_\Omega \nabla g \nabla (u-g)_K \, \mathrm{d}x - \int_\Omega f(g) \, (u - g)_K \, \mathrm{d}x$$ for all $K > 0$ and $\nabla g \in L^p(\Omega)$, $f(g) \in L^\infty(\Omega)$.

This estimate is enough to conclude $u - g \in L^\infty(\Omega)$. The detailed arguments can be found in many books, e.g., "An introduction to variational inequalities and their applications" by Kinderlehrer and Stampacchia, proof of Theorem B.2.

It is a remarkable feature of this trick that only the monotonicity of $f$ is used.

gerw
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  • By the way, I managed to prove it by cutting off the test function in Theorem B.2. In this way it got quite complicated, but finally it worked. I couldn't prove it with the simpler cut-off test function that you suggested in your answer. I would like to ask you ( only in case you have done it before with the test function you suggested, otherwise do not waste time) how it is done. Thanks. – Svetoslav Jun 03 '16 at 17:14
  • At which part of the proof do you struggle? – gerw Jun 03 '16 at 18:07
  • Let $B(k):={x\in\Omega: |u-g|\ge k}$ and $A(k):={x\in\Omega: |u-g|<k}$. Then the RHS is estimated by $C(|\Omega|^{1/2}+|A(k)|^{1/3})|\nabla (u-g)k|{L^2(\Omega)}\quad (1)$ if we use the estimate $\int_\Omega f(u) , (u-g)K , \mathrm{d}x \ge \int\Omega f(g) , (u-g)K , \mathrm{d}x$ that you wrote above. Or we can even improve this estimate if we take $k\ge |g|{L^\infty(\Omega)}$ to estimate the RHS by $C(|A(k)|^{1/2}+|A(k)|^{1/3})|\nabla (u-g)k|{L^2(\Omega)}\quad (2)$. Now, we get $|\nabla (u-g)k|{L^2(\Omega)}\leq (1)$ or $\leq (2)$ – Svetoslav Jun 04 '16 at 21:44
  • Now, the LHS is estimated from below like: $|\nabla (u-g)k|{L^2(\Omega)}\ge C|(u-g)k|{H^1(\Omega)}\ge C|(u-g)k|{L^6(\Omega)}=\left(\int\limits_{\Omega}{|(u-g)k|^6dx}\right)^{1/6}=\left(\int\limits{A(k)}{(u-g)^6dx} + \int\limits_{B(k)}{k^6dx}\right)^{1/6}$. If $h>k$ we have $\ge \left(\int\limits_{A(h)\setminus A(k)\cup B(h)}{k^6}\right)^{1/6}=k(|A(h)|-|A(k)|+|B(h)|)^{1/6}\ge $ here we can continue in several ways, like $\ge (h-k)(|A(h)|-|A(k)|+|B(h)|)^{1/6},,k<h<2k,\quad ()$ or just leave it $\ge k|B(k)|^{1/6}\quad (*)$. – Svetoslav Jun 04 '16 at 22:02
  • On the other hand, if $h<k$ we can estimate the LHS from below like: $|\nabla (u-g)k|{L^2(\Omega)}\ge C|(u-g)k|{H^1(\Omega)}\ge C|(u-g)k|{L^6(\Omega)}=\left(\int\limits_{\Omega}{|(u-g)k|^6dx}\right)^{1/6‌​}=\left(\int\limits{A(k)}{(u-g)^6dx} + \int\limits_{B(k)}{k^6dx}\right)^{1/6}\ge \left(\int\limits_{A(k)\setminus A(h)}{(u-g)^6dx}+\int\limits_{B(k)}{k^6dx}\right)^{1/6}\ge h|B(h)|^{1/6}\quad (*)$. Or if in addition $h<k<2h$ we can get $\ge (k-h)|B(h)|^{1/6} (**)$. – Svetoslav Jun 04 '16 at 22:08
  • Finally, combining the estimates (1), (2) for the RHS with the estimates $(),(),(),(**)$ for the LHS we get like 8 similar inequalities. And I couldn't conclude from any of them that either $|B(h)|=0$ or $|A(h)|=|\Omega|$ for some big $h$. BTW, in my first comment, in the last line, it is $|\nabla(u-g)k|{L^2(\Omega)}^2\leq (1)$ or $\leq (2)$ – Svetoslav Jun 04 '16 at 22:15
  • For the LHS, you use the same estimate as in Kinderlehrer/Stampacchia, to arrive at $(h-k)^2 [meas(B(h))]^{2/2^*} \le C , |(u-g)k|{H^1}^2$. For the RHS, you use $RHS \le C , ( |g|{W^{1,p}} , |(u-g)_k|{W^{1,p'}} + |f(g)|{L^\infty} , |(u-g)_k|{L^1}$. – gerw Jun 06 '16 at 10:58
  • Now, you use the Sobolev embedding theorem on the RHS to extract some powers of $meas(B(k))$. Then, you should be able to use Lemma B.1 from Kinderlehrer, Stampacchia to conclude $meas(B(k)) = 0$ for some $k$ large enough. – gerw Jun 06 '16 at 11:00
  • For the RHS, you can also look at the last lines on p. 64 of Kinderlehrer/Stampacchia. – gerw Jun 06 '16 at 11:06
  • Thanks for the response. Actually, I have read in detail Lemma B.1 and Theorem B.2 and everything is clear there. But in my case, the problem is that in the estimation of the RHS we can only get powers of the set $A(k)=\Omega\setminus B(k)$, as in my first comment. – Svetoslav Jun 06 '16 at 13:13
  • The best estimate, I can do is $RHS\leq |f(g)|{L^\infty(\Omega)}|\Omega|^{1/2}|(u-g)_k|{L^2(\Omega)}+|\nabla g|{L^2(A(k))}|\nabla(u-g)_k|{L^2(A(k))}$ $\leq (C_F|f(g)|{L^\infty(\Omega)}|\Omega|^{1/2}+|\nabla g|{L^2(A(k))})|\nabla(u-g)k|{L^2(\Omega)}$ which using that $\nabla g\in L^6(\Omega)$ (if we assume that $g\in H^2(\Omega)$ becomes $\leq (C_F|f(g)|{L^\infty(\Omega)}|\Omega|^{1/2}+|\nabla g|{L^6(A(k))} |A(k)|^{1/3})|\nabla(u-g)k|{L^2(\Omega)}\leq C(|\Omega|^{1/2}+|A(k)|^{1/3})|\nabla (u-g)k|{L^2(\Omega)}$ – Svetoslav Jun 06 '16 at 13:13
  • And then we divide both sides by $|\nabla (u-g)k|{L^2(\Omega)}$. Also, as I said in my second comment, if we take $k>|g|_{L^\infty}$, the estimate of the RHS can be improved, and instead $|\Omega|^{1/2}$ to become $|A(k)|^{1/2}$. But still the problem is that we only have $|A(k)|$ on the RHS, and not $|B(k)|$. – Svetoslav Jun 06 '16 at 13:16
  • You can use $\int_\Omega f(g) , (u-g)k \le |f(g)|{L^\infty} , |(u-g)k|{L^1(B(k))} \le C , |(u-g)k|{L^q} , meas(B(k))^s$ for some $q < d$ and some appropriate $s$. The $L^q$-norm of $(u-g)_k$ can be estimated by the $H^1$ norm due to Poincare. – gerw Jun 06 '16 at 13:27
  • The problem is that the inequality you wrote in your last comment should be: $\int_\Omega f(g) , (u-g)k \le |f(g)|{L^\infty} , |(u-g)k|{L^1(B(\Omega))} \le C , |(u-g)k|{L^q} , meas(\Omega))^s$, i.e it should be $\Omega$ and not $B(k)$. Or maybe there is something that I still cannot see. – Svetoslav Jun 06 '16 at 14:33
  • I just want to see if it can be proven with this test function. As I said, I proved it with a bit more complicated test function. I am still very grateful for showing me this trick and telling me the reference. – Svetoslav Jun 06 '16 at 14:37
  • By Hölder, you have $|(u-g)k|{L^1(B(k))} = \int_{B(k)} (u-g)k \cdot 1 , \mathrm{d}x \le |(u-g)_k|{L^q} , |1|{L^{q'}(B(k))} = |(u-g)_k|{L^q} , meas(B(k))^{1/q'}$. Or do I miss something? – gerw Jun 06 '16 at 20:24
  • The norm of $|(u-g)k|{L^1(\Omega)}$ is not equal to $|(u-g)k|{L^1(B(k))}$, because the functions $(u-g)_k$ are not zero in the set $A(k)={x\in\Omega: |u-g|< k}$. I think this is the problem. – Svetoslav Jun 06 '16 at 20:44
  • In my 3-rd comment from the end upwards, I have a typo- it is $L^1(\Omega)$, not $L^1(B(\Omega))$, i.e it should be like this: $\int_\Omega f(g) , (u-g)k \le |f(g)|{L^\infty} , |(u-g)k|{L^1(\Omega)} \le C , |(u-g)k|{L^q} , meas(\Omega)^s$ – Svetoslav Jun 06 '16 at 20:47
  • Oh, I see. Actually, Kinderlehrer and Stampacchia use a different test function, see the beginning of the proof of Theorem B.2. Maybe this is the problem with my answer! – gerw Jun 07 '16 at 06:18
  • Yes, but as I said, thanks a lot for the answer. I used truncation version of the function in the beginning of the proof of Theorem B.2 and it worked, although it was a bit more complicated. – Svetoslav Jun 07 '16 at 10:00