Linear PDE with integral term exercise

Question

An exercise from a course I have, I am having problems with it.

Let $\Omega$ be a bounded regular interval of $\mathbb{R}^2$. Let $f\in\mathcal{C}^0(\overline{\Omega})$ and $a\in\mathbb{R}$. We consider the problem, for $u\in\mathcal{C}^2(\overline{\Omega})$: $$ \begin{equation} \left\{ \begin{aligned} -\Delta u(x)+\int_{\Omega}u(y)dy=f(x),~\forall x\in\Omega\\ \frac{\partial u}{\partial n}(\sigma)=a,~\forall\sigma\in\partial\Omega \end{aligned} \right. \end{equation} $$

1. Is this a linear problem ?
2. Show that is $u$ is a solution, then we have $$\vert\Omega\vert\int_{\Omega}u(x)dx=\int_{\Omega}f(x)dx+\vert\partial\Omega\vert a$$ where $\vert\Omega\vert$ denotes the area of $\Omega$ and $\vert\partial\Omega\vert$ the length of $\partial\Omega$.
3. If $x_0\in\overline{\Omega}$ is such that $u(x_0)=\inf\Bigl\{u(x),x\in\overline{\Omega}\Bigr\}$, show that if $a>0$, then $x_0\notin\partial\Omega$.
4. Show that if a solution $u$ exists, it is unique.
5. Find a "relatively easy" solution when $f=c\in\mathbb{R}$ is constant. Conclude in that case.

For the next questions, we take $\Omega=\Bigl\{(x_1,x_2)\in\mathbb{R}^2,x_1^2+x_2^2<1\Bigr\}$.

6. Construct a solution in the case $f(x)=0$ and $a=1$.
7. Construct a solution in the case $f(x)=\vert x\vert^3$, and $a=0$.
8. Show that in general the system has a unique solution if $f$ is of the form $f(x)=\tilde{f}(\vert x\vert)$ where $\tilde{f}:[0,1]\to\mathbb{R}$ is continuous.

We now consider the system for $a$ and $f$ given above: $$ \begin{equation} \left\{ \begin{aligned} -\Delta u(x)+\Biggl(\int_{\Omega}u(y)dy\Biggr)^2u(x)=f(x),~\forall x\in\Omega\\ \frac{\partial u}{\partial n}(\sigma)=a,~\forall\sigma\in\partial\Omega \end{aligned} \right. \end{equation} $$ where $\Omega$ still denotes the open unit ball in $\mathbb{R}^2$.

9. Is this a linear problem ?
10. Find a relation for the integral term $\int_\Omega u(y)dy$ which depends only on $f$ and $a$.
11. Show that if a solution $u$ exists, it is unique.
12. Show that if $f=0$ and $a=0$, then necessarely $u=0$.
13. Show that if $f>0$ and $a>0$, then $u\geqslant0$.
14. Discuss the case $f\geqslant 0$ and $a\geqslant 0$.

My work

1. Yes: if $u$ and $v$ are two solutions of the homogeneous problem, then $u+v$ is also a solution of the homogeneous problem.
2. Multiplying by a smooth enough function $v$, integrating, then integrating by parts the Laplacan term, we get $$\int_\Omega \nabla u\nabla v-\int_{\partial\Omega}\frac{\partial u}{\partial n}v+\int_\Omega u\int_\Omega v=\int_\Omega fv$$ If we take $v=1$, we have the relation.
3. If $x_0\in\partial\Omega$, then $\nabla u (x_0)\leqslant 0$ because it is an infimum. But because $x_0$ is at the boundary, $\nabla u(x_0)\cdot n=\frac{\partial u}{\partial n}(x_0)=a>0$, which is a contradition. Thus $x_0\notin\partial\Omega$.
4. Because it is a linear problem, we can show that if $u$ is a solution to the homogeneous equation, then $u=0$. This will imply that the solution, if it exists, is unique. From the equation above, taking $v=u$ gives: $$\int_\Omega \vert\nabla u\vert^2+\Biggl(\int_\Omega u\Biggr)^2=0$$ This implies that $u$ is a constant with a null integral, thus $u=0$.
5. Well, I could not find this "easy" solution. From the relation at 2), we can obtain $$-\Delta u=a\frac{\vert\partial\Omega\vert}{\vert\Omega\vert}$$ so we can choose $u(x_1,x_2)=-\frac{a\vert\partial\Omega\vert}{4\vert\Omega\vert}(x_1^2+y_1^2)$ to satisfy this relation. However, this does not give $\frac{\partial u}{\partial n}=0$ on the boundary, and $c$ has disappeared from the equation.
6. Well, same thing as question 5) really...
7. Same thing again. It does not seem I have the right method.
8. $f$ is invariant by rotation, and the Laplacian is also invariant by rotation. This means that the problem is invariant by rotation. If I note $u(x)=\tilde{u}(\vert x\vert)$, then $\tilde{u}$ satisfies $$-\frac{1}{r}\frac{d}{dr}\Bigl(r\frac{d}{dr}\tilde{u}\Bigr)+\int_\Omega\tilde{u}=\tilde{f}$$ which means that $$r\frac{d}{dr}\tilde{u}=\Bigl(\int_\Omega\tilde{u}(y)dy\Bigr)\Bigl(\int_{0}^{r}sds\Bigl)-\Bigl(\int_{0}^{r}\tilde{f(s)}sds\Bigr)+A$$ where $A$ is a constant. Next: $$\frac{d}{dr}\tilde{u}=\frac{1}{r}\Biggl[\Bigl(\int_\Omega\tilde{u}(y)dy\Bigr)\frac{r^2}{2}-\Bigl(\int_{0}^{r}\tilde{f(s)}sds\Bigr)\Biggr]+\frac{A}{r}$$ thus $$\tilde{u}(r)=\Bigl(\int_\Omega\tilde{u}(y)dy\Bigr)\Bigr(\int_{0}^{r}\frac{s}{2}ds\Bigl)-\Biggl(\int_{0}^r\frac{\Bigl(\int_{0}^{s}\tilde{f}(z)dz\Bigr)}{s}ds\Biggr)+\ln(A)+B$$ Because $u$ is continuous, $A=0$ and we can take $B$ such that $\tilde{u}(1)=0$. Then we have $$\tilde{u}(r)=\Bigl(\int_\Omega\tilde{u}(y)dy\Bigr)\frac{r^2}{4}-\int_0^{r}\frac{1}{s}\int_{s}^{1}s\tilde{f}(s)ds$$ This gives a unique solution $u$.
9. This is not a linear problem due to the term of the integral squared (and then the multiplication by $u$).
10. As in question 2), we multiply by a smooth enough function $v$, integrate, then integrate by parts to get $$\int_{\Omega}\nabla u\nabla v-\int_{\partial\Omega}\frac{\partial u}{\partial n}v+\Bigl(\int u)^2\int uv=\int fv$$ Taking $v=1$, we have $$-\vert\partial\Omega\vert a+\Bigl(\int u\Bigr)^3=\int f$$
11. Well, no idea about this one. I tried to look at $v_1-v_2$ if both are solutions but because of the integral squared I do not have anything relevant...
12. The relation in 10) gives $\Bigl(\int u\Bigr)^3=0$, hence $\int u=0$. We are faced against an homogeneous Laplace equation with Von Neumann conditions on the boundary that are $0$. By multiplying by u, integrating and integrating by parts, we obtain $\int\vert\nabla u\vert^2=0$ which means that $u$ is constant. It has a null integral, thus $u=0$.
13. This is the maximum principle. If $u(x_0)=\inf\Big\{u(x),x\in\overline{\Omega}\big\}$ then: if $x_0$ is on the boundary, then $\nabla u(x_0)\leqslant 0$ and $\frac{\partial u}{\partial n}(x_0) = a>0$ Thus $x_0\notin\partial\Omega$. So $x_0\in\mathring{\Omega}$, thus $\nabla u(x_0)=0$ and $\Delta u(x_0)\geqslant 0$. This means that $f+\Delta u=\Bigl(\int_\Omega u)\Bigr)^2u(x_0)\geqslant0$, which means that $u(x_0)\geqslant 0$. Thus $u\geqslant 0$.
14. I guess we can have some cause some perturbations to $u$ with a $v_\varepsilon$ so that $v_\varepsilon$ satisfies the system but with $f_\varepsilon>0$ and $a_\varepsilon>0$, and such that $\lim\limits_{\varepsilon\to 0}v_\varepsilon\to u$, but I could not find any way to do that properly.

Questions

Do you have any hints regarding questions 5 to 7 (which are almost the same question) and 11 ? Thank you !