In this paper, we consider positive supersolutions of the semilinear fourth-order problem$\{\begin{array}{cc}\hfill {(-\mathrm{\Delta })}^{2}u=\rho (x)f(u)\hfill & \hfill \text{in}\mathrm{\Omega },\hfill \\ \hfill -\mathrm{\Delta }u>0\hfill & \hfill \text{in}\mathrm{\Omega },\hfill \end{array}$ where Ω is a domain in ${\mathbb{R}}^N$ (bounded or not), $f:D_f=[0,a_f)\to [0,\infty )$ $(0<a_f⩽+\infty )$ is a non-decreasing continuous function with $f(u)>0$ for $u>0$ and $\rho :\mathrm{\Omega }\to \mathbb{R}$ is a positive function. Using a maximum principle-based argument, we give explicit estimates on positive supersolutions that can easily be applied to obtain Liouville-type results for positive supersolutions either in exterior domains, or in unbounded domains Ω with the property that ${\sup }_{x\in \mathrm{\Omega }}\mathrm{dist}(x,\partial \mathrm{\Omega })=\infty$. In particular, we consider the above problem with $f(u)=u^p$ ($p>0$) and with different weights $\rho (x)=|x{|}^a$, $e^{a{x}_1}$ or $x_1^m$ (m is an even integer). Also, when f is convex and $\rho :\mathrm{\Omega }\to (0,\infty )$ is smooth with $\mathrm{\Delta }(\sqrt{\rho })>0$, then under an extra condition between f and ρ we show that every positive supersolution u of this problem with $u=0$ on ∂Ω (Ω bounded) satisfies the inequality $-\mathrm{\Delta }u\ge \sqrt{2\rho (x)F(u)}$ for all $x\in \mathrm{\Omega }$, where $F(t):={\int }_0^t(f(s)-f(0))ds$.

在本文中，我们考虑半线性四阶问题的正超解$\{\begin{array}{cc}\hfill {(-\mathrm{\Delta })}^2你=\rho (X)F(你)\hfill & \hfill \text{在}\mathrm{\Omega },\hfill \\ \hfill -\mathrm{\Delta }你>0\hfill & \hfill \text{在}\mathrm{\Omega },\hfill \end{array}$ 其中 Ω 是一个域 ${\mathbb{电阻}}^N$ （有界或无界）， $F：D_F=[0,{\mathrm{一种}}_F)\to [0,\infty )$ $(0<{\mathrm{一种}}_F⩽+\infty )$ 是一个非递减的连续函数 $F(你)>0$ 为了 $你>0$ 和 $\rho ：\mathrm{\Omega }\to \mathbb{电阻}$是正函数。使用基于最大原理的论证，我们给出了对正超解的明确估计，这些估计可以很容易地应用于在外部域或无界域 Ω 中获得正超解的 Liouville 型结果，其性质为${\mathrm{超}}_{X\in \mathrm{\Omega }}\mathrm{区}(X,\partial \mathrm{\Omega })=\infty$. 特别地，我们考虑上述问题$F(你)={你}^{磷}$ ($磷>0$) 和不同的权重 $\rho (X)=|X{|}^{\mathrm{一种}}$, ${\mathrm{电子}}^{\mathrm{一种}{X}_1}$ 或者 $X_1^{米}$（m是偶数）。此外，当f是凸的并且$\rho ：\mathrm{\Omega }\to (0,\infty )$ 是光滑的 $\mathrm{\Delta }(\sqrt{\rho })>0$，然后在f和ρ之间的一个额外条件下，我们证明了这个问题的每个正超解u$你=0$ on ∂Ω (Ω bounded) 满足不等式 $-\mathrm{\Delta }你\ge \sqrt{2\rho (X)F(你)}$ 对所有人 $X\in \mathrm{\Omega }$， 在哪里 $F(吨)：={\int }_0^{吨}(F(秒)-F(0))d秒$.