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Studying the Fisher-KPP evolution equation I came across the steady state elliptic problem which can be written in the following form:

$$ \begin{cases} -d\Delta Y(x)=r(x)Y(x)\left (1-\dfrac{Y(x)}{K(x)}\right ),\ x\in\Omega \\ \dfrac{\partial Y}{\partial \nu}=0, \ x\in\partial\Omega\end{cases} $$

Hereabove $\Omega\subset\mathbb{R}^N$ is a bounded smooth domain and $K\in L^{\infty}(\Omega)$ is the carrying capacity. $r=r(x)$ is a function from $L^{\infty}(\Omega)$ and $d>0$ is a positive constant. We assume that there are two constants $k_0, r_0>0$ such that $K(x)>k_0$ and $r(x)>r_0$ for almost all $x\in\Omega$.

Since for the logistic equation from ODE $y'(t)=ry(t)\left (1-\dfrac{y(t)}{k}\right )$ the stationary solution $y(t)=k>0$ for all $t\geq 0$ is the unique strict positive and globally asimptotically stable I think the for the spatial model this will hold too. So presumably there will be a unique strict positive solution $Y$ for the elliptic problem which is globally asimptotically stable. I cannot prove this. For any advice I will be thankful.

I know that some monotone techniques will be useful here but I cannot see how. I tried to use upper and lower solutions for this problem and some eigenvalue problems but I did not succeed in proving anything relevant.

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  • $\begingroup$ What are $d,r$? $\endgroup$ Commented Oct 1, 2022 at 10:49
  • $\begingroup$ $d$ is a positive constant and $r=r(x)$ is a bounded strict positive function. I have edited my post. $\endgroup$
    – Bogdan
    Commented Oct 1, 2022 at 13:34
  • $\begingroup$ Thanks, but then what is the point of $d$? Isn’t it possible to standardize the PDE by dividing “d” on both sides and embedding “d” into “r”? $\endgroup$ Commented Oct 1, 2022 at 15:53
  • $\begingroup$ Of course, you're right! $d$ is the diffusion coefficient and $r$ is the growth rate of a population. Because it came out of an parabolic problem , where one cannot simplify the problem by dividing by $d$, I decided to write it here too. $\endgroup$
    – Bogdan
    Commented Oct 1, 2022 at 16:16
  • $\begingroup$ Thanks for the clarification. $\endgroup$ Commented Oct 1, 2022 at 16:20

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