The classical Diaz-Saa inequality found in [1] can be derived from the standard Picone inequality [2]. Let us try to use the generalized Picone inequality found in [3] and [4] to get a corresponding version of the Diaz-Saa inequality.

The generalized Picone inequality has the form

\[|\nabla u|^{p-2} \nabla u \nabla \left(\frac{v^{q}}{u^{q-1}}\right) \leq |\nabla v|^{p-2} \nabla v \nabla \left(\frac{v^{q-p+1}}{u^{q-p}}\right),\]

where $q$ and $p$ satisfy certain algebraic relations, see Theorem 1.8 in [4].

If we suppose that $u, v \in W_0^{1,p}(\Omega)$ are reasonably “good” functions so that one can integrate the inequality over $\Omega$ by parts, we arrive at

\[\int_\Omega (-\Delta_p u) \frac{v^{q}}{u^{q-1}} \,dx \leq \int_\Omega (-\Delta_p v) \frac{v^{q-p+1}}{u^{q-p}} \,dx,\]

and, analogously,

\[\int_\Omega (-\Delta_p v) \frac{u^{q}}{v^{q-1}} \,dx \leq \int_\Omega (-\Delta_p u) \frac{u^{q-p+1}}{v^{q-p}} \,dx.\]

Summing these inequalities and rearranging, we arrive at

\[\int_\Omega \left(\frac{-\Delta_p v}{v^{p-1}} - \frac{-\Delta_p u}{u^{p-1}}\right) (v^{2q-p} - u^{2q-p}) v^{p-q} u^{p-q} \,dx \geq 0.\]

If $q=p$, then we end up with the classical Diaz-Saa inequality.

Last modified: 09-Aug-25

Bibliography

  1. Díaz J.I., Saa J.E., Existence et unicité de solutions positives pour certaines équations elliptiques quasilinéaires, C. R. Acad. Sci. Paris Sér. I Math. 305 (1987), no. 12, 521–524 ↩

  2. Allegretto, W., & Huang, X. Y. (1998). A Picone’s identity for the $p$-Laplacian and applications. Nonlinear Analysis: Theory, Methods & Applications, 32(7), 819-830. ↩

  3. Il’yasov Y., On positive solutions of indefinite elliptic equations, C. R. Acad. Sci. Paris Ser. I Math. 333 (2001), no. 6, 533–538 ↩

  4. Bobkov, V., Tanaka, M. Generalized Picone inequalities and their applications to $(p,q)$-Laplace equations. Open Mathematics, 18(1), (2020) 1030-1044. ↩ ↩2