# Weak formulation of a nonlinear problem with test functions in a dense subspace of $H_0^1$

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.

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.

• 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 3 '16 at 17:14
• At which part of the proof do you struggle? – gerw Jun 3 '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 4 '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 4 '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 4 '16 at 22:08