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Let $\Omega$ be a 2D region. Now we have a partial differential equation system describing the characteristics of the region:

\begin{align*} \nabla \cdot (h_0^3 P_0 \nabla P_0) &= 0 \\ \nabla \cdot [h_0^3 \nabla (P_0 P_1)] &= 3 h_1 \nabla \cdot (h_0^2 P_0 \nabla P_0) \end{align*} where $ P_0 $ is determined by a boundary condition, $ P_0>0 $, also, $P_1=0$ on the boundary $\partial \Omega$. $h_0$ is an arbitrary distribution function, $ h_0>0 $, while $h_1$ is a constant value and $h_1 \ll h_0$.

How to prove the following integral inequality: $$ \frac{\iint P_1 \mathrm{d} \Omega }{h_1} < \iint \frac{ P_0 }{h_0} \mathrm{d} \Omega $$

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  • $\begingroup$ $h_0$ is just an arbitrary function, nothing to do with Probability theory. $\endgroup$
    – Yangong Wu
    Commented Sep 17, 2020 at 9:46
  • $\begingroup$ Can we prove this by constructing intermediate variables like this: \begin{align} \nabla \cdot (h_0^3 \nabla P_{0c}) &= 0 \\ \nabla \cdot [h_0^3 \nabla ( P_{1c})] &= 3 h_1 \nabla \cdot (h_0^2 \nabla P_{0c} ) \end{align} and we try to prove the following integral inequality: $$ \frac{\iint P_1 \mathrm{d} \Omega }{h_1} < \frac{\iint P_{1c} \mathrm{d} \Omega }{h_1} < \iint \frac{ P_{0c} }{h_0} \mathrm{d} \Omega < \iint \frac{ P_0 }{h_0} \mathrm{d} \Omega $$ $\endgroup$
    – Yangong Wu
    Commented Dec 2, 2020 at 17:08
  • $\begingroup$ The middle P1c is less than P0c can be explained by physical meaning. $\endgroup$
    – Yangong Wu
    Commented Dec 2, 2020 at 17:09

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