TITLE: Convex solutions of the Poisson equation
QUESTION [6 upvotes]: Let $D$ be a planar, bounded, convex open domain. Given a positive function $f:D\to(0,+\infty)$, let us consider the Poisson equation
$$\Delta u=f\quad\hbox{in }D.$$
Not specifying any boundary condition, it admits a bunch of solutions $u=u_0+v$, where $u_0$ is a particular one (for instance the solution of the Dirichlet BVP with $u_0=0$ on $\partial D$), and $v$ is an arbitrary harmonic function.

Does there exist a convex solution (actually strongly convex in the sense that ${\rm D}^2u(x)$ is positive definite for every $x\in D$) ?

Let me refine the question as follows. For this, I denote $\nu$ and $\tau$ the unit normal and tangential vector fields along $\partial D$.

Does there exist a solution of the Laplace equation, satisfying the 2nd-order boundary condition ${\rm D}^2u(\nu,\tau)=0$ along $\partial D$ ? If so, is it a convex function ?

REPLY [7 votes]: I apologize for having posted this question too early. I realize that the answer to the first question is negative.
Actually suppose that $D=D(0;R)$ is a disk and $f=f(r)$ is a radial function. If a convex solution $u$ existed, then certainly $u_\theta(x):=u(R_\theta x)$ would be another one, where $R_\theta$ is a rotation. Then
$$U:=\frac1{2\pi}\int_0^{2\pi}u_\theta\,d\theta$$
is again a convex solution, now a radial one $U=v(r)$. It satisfies $v''+\frac1r v'=f$, and the spectrum of ${\rm D}^2U$ is $\{v'',\frac1rv'\}$. Thus the existence of a convex solution for every smooth positive radial $f$ amounts to saying that $rv''+v'\ge0$ over $(0,R)$ for $v'(0)=0$, implies $v',v''\ge0$. This is obviously false.
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