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Let $\Omega$ be a bounded domain. Assume that $f:\mathbb{R}\to\mathbb{R}$ is a function with $f(0)>0, f^{\prime}(0)\geq0$ and $f^{\prime\prime}(u)>0$ for $u>0$. Let $u\in C^2(\Omega)\cap C(\overline{\Omega})$ be a solution of the following problem: \begin{equation*} \begin{split} -\Delta u&=f(u),\quad&\text{in }\Omega\\ u&=0,\quad&\text{on }\partial\Omega. \end{split} \end{equation*} Show that $u>0$ in $\Omega$. If $f$ is monotone on $\Omega$, I could use the maximum principle but since $f^{\prime\prime}>0$ only holds on $(0,\infty)$, I do not know how to start:

If I assume there is $x_0\in\Omega$ so that $u(x_0)\leq 0$, then $\min_{\overline{\Omega}}u\leq 0$, but I do not know anything about the sign of $f(u(x_0))$..

Maybe a variational approach is useful:

Consider the Energy functional \begin{equation} I(u)=\frac{1}{2}\int_{\Omega}|Du|^2-F(u)dx \end{equation} with $F(u)=\int_0^uf(s)ds.$ Suppose $u$ is a solution of the Poisson equation above. If I am able to show $I(|u|)\leq I(u)$, we get $u\geq 0$ on $\Omega$.

john_psl1298
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  • Is $f$ continuous on $\Bbb R$? Then it might help that $f(u) > 0$ for $u > -\epsilon$ with some positive $\epsilon$. – Martin R Jul 23 '25 at 13:34
  • @MartinR $f$ is supposed to be $C^2$. Can you give me a hint on how to use $f(u)>0$ near zero? – john_psl1298 Jul 23 '25 at 18:25
  • Do we know anything about $f$ on $(-\infty,0)$ ? – K.defaoite Jul 24 '25 at 01:03
  • @K.defaoite We just know that $f$ is twice differentiable on $(-\infty,0)$. – john_psl1298 Jul 24 '25 at 06:23
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    Do you know that this is true or is it a conjecture? I just wonder because the corresponding statement in one dimension seems to be false. $u(x) = -1-\cos(x)$ on $[-\pi, \pi]$ satisfies $-u''(x) = f(u(x))$ with $f(u) = 1+u$ for $u \le 0$. – Martin R Jul 24 '25 at 08:44
  • @MartinR That does not satisfy the suppositions of this problem, because $f''\equiv 0$. In the problem we require the strict inequality $f''>0$. – K.defaoite Jul 24 '25 at 14:55
  • @K.defaoite: $f(u) = 1+u$ is only needed for $u \le 0$, buf $f''(u) > 0$ is only required for $u > 0$. You can define $f(u)$ for $u > 0$ so that all conditions are satisfied. – Martin R Jul 24 '25 at 15:05
  • @MartinR I apologize, thanks for the correction. – K.defaoite Jul 24 '25 at 15:20
  • @MartinR I was thinking that this conjecture might hold if we also require that $f\in C^\infty (\mathbb R)$, but on second thought I think it should be possible to construct a function $f:\Bbb R\to \Bbb R$ satisfying all the requirements such that $f\in C^\infty (\Bbb R)$ and $f(z)=1+z$ for $z<0$ , using interpolation – K.defaoite Jul 24 '25 at 20:19

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Here is a counterexample. The idea is to find a radially symmetric function $u$ on some disk $\overline{B_R(0)}$ which is negative in the interior of the disk, zero on the boundary, and satisfies $$ -\Delta u(x, y) = u(x, y) + C $$ with some positive constant $C$. Then we can define $f$ as $f(u) = u+C$ for $u \le 0$ and extend that to a strictly convex function on $[0, \infty)$.

Let $\Omega = B_r(0)$ be the disk entered at the origin with radius $3$, and define $u$ on $\Omega$ as $$ u(x, y) = -J_0\left(\sqrt{x^2+y^2}\right) - C $$ where $J_0$ is the zero-th Bessel function of the first kind and $C = - J_0(3) > 0$. $u$ is strictly negative in $\Omega$ and zero on the boundary.

From the formula for the Laplace-operator in polar coordinates and the differential equation for the Bessel functions we get that $$ -\Delta u(r) = \Delta J_0(r) = J_0''(r) + \frac 1r J_0'(r) = -J_0(r) = u(r) + C $$ for $r > 0$. It can be verified that this holds also at the origin.

Finally define $f: \Bbb R \to \Bbb R$ as $f(z) = z+C + g(z)$ where $g: \Bbb R \to \Bbb R$ is a sufficiently smooth function with $g(z) = 0$ for $z \le 0$ and $g''(z) > 0$ for $z > 0$.

  • If $f \in C^2(\Bbb R)$ is sufficient then we can set $g(z) = z^4$ for $z > 0$.
  • If $f \in C^\infty(\Bbb R)$ is required then a possible candidate is $g(z) = z \cdot \exp(-1/z)$ for $z > 0$.

In any case: $-\Delta u = f(u)$ in $\Omega$ and $u = 0$ on $\partial \Omega$, but $u$ is strictly negative in $\Omega$.

Martin R
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