2$ then $\Lambda_{\rm max}(x)=\Lambda_{\rm min}(x)$, % \label{cond:3} \item[(iv)] $u(x)$ is a constant multiple of $v(x)$. %\label{cond:4} \end{itemize} \end{theorem} \begin{proof} Direct computations show \begin{equation}\label{eq:proof1} \begin{aligned} \operatorname{div}\left(u a(x)\norm{\nabla u}^{p-2} \nabla u\right) &= ul(u)-c(x)|u|^p +\langle \nabla u, a(x)\norm{\nabla u}^{p-2}\nabla u \rangle \\ &\geq ul(u)-c(x)|u|^p +\lambda_{\rm min}(x)\norm{\nabla u}^{p}. \end{aligned} \end{equation} Evaluating the divergence and using definition of the operator $L$ we obtain \begin{equation} \begin{aligned} &-\operatorname{div}\Big(|u|^p\frac{A(x)\|\nabla v\|^{p-2} \nabla v}{|v|^{p-2}v}\Big) \\ &=-p|u|^{p-2}u\big\langle \nabla u, \frac{A(x)\|\nabla v\|^{p-2}\nabla v}{|v|^{p-2}v}\big\rangle -\frac{|u|^p}{|v|^{p-2}v}\operatorname{div} \left(A(x)\|\nabla v\|^{p-2}\nabla v\right)\\ &\quad - (1-p)\frac{|u|^p}{|v|^p}\ss{A(x) \|\nabla v\|^{p-2}\nabla v}{\nabla v}\\ &=-\frac{|u|^p}{|v|^{p-2}v}L(v) +C(x)|u|^p -p|u|^{p-2}u\big\langle \nabla u, \frac{A(x)\|\nabla v\|^{p-2} \nabla v}{|v|^{p-2}v}\big\rangle \\ &\quad +(p-1)\frac{|u|^p}{|v|^p}\ss{A(x)\|\nabla v\|^{p-2}\nabla v}{\nabla v}. \end{aligned} \label{eq:-div} \end{equation} To estimate the last two terms, in terms of the product $K(x)\norm{\nabla u}^p$, we use the Young inequality \begin{equation} \frac{p-1}pX^{\frac p{p-1}}- XY\geq -\frac 1p Y^p\label{eq:Young} \end{equation} and split the proof into two cases. \textbf{Case 1:} First we consider general case $p>1$. Schwarz inequality and the fact that norm of the matrix $A$ is $\Lambda_{\rm max}$ imply \begin{align*} |u|^{p-2}u\big\langle \nabla u, \frac{A(x)\|\nabla v\|^{p-2} \nabla v}{|v|^{p-2}v}\big\rangle &\leq |u|^{p-1}\norm{\nabla u} \big\|\frac{A(x)\|\nabla v\|^{p-2}\nabla v}{|v|^{p-2}v}\big\|\\ &\leq |u|^{p-1}\|\nabla u\|\Lambda_{\rm max}(x) \frac{\|\nabla v\|^{p-1}}{|v|^{p-1}}. \end{align*} Using this inequality and Young inequality we can find an apriori bound for last two terms at the right-hand side of \eqref{eq:-div} as follows \begin{align*} &\frac{p-1}{p}\frac{|u|^p}{|v|^p}\ss{A(x)\|\nabla v\|^{p-2}\nabla v}{\nabla v} -|u|^{p-2}u \big\langle\nabla u, \frac{A(x)\|\nabla v\|^{p-2}\nabla v}{|v|^{p-2}v} \big\rangle \\ &\geq\frac{p-1}p \big|\frac{u}{v}\big|^p \|\nabla v\|^p \Lambda_{\rm min}(x)-|u|^{p-1}\norm{\nabla u}\Lambda_{\rm max} (x)\frac{\|\nabla v\|^{p-1}}{|v|^{p-1}} \\ &=\frac{p-1}p \Big[\Big(\big|\frac{u}{v}\big| \|\nabla v\| \Lambda_{\rm min}^{1/p}(x)\Big)^{p-1}\Big]^{\frac {p}{p-1}} -|u|^{p-1}\norm{\nabla u}\Lambda_{\rm max} (x)\frac{\|\nabla v\|^{p-1}}{|v|^{p-1}} \\ &=\frac{p-1}p \Big[\Big(\big|\frac{u}{v}\big| \|\nabla v\| \Lambda_{\rm min}^{1/p}(x)\Big)^{p-1}\Big]^{\frac {p}{p-1}}\\ &\quad -\Big(\big|\frac{u}{v}\big| \|\nabla v\| \Lambda_{\rm min}^{1/p}(x)\Big)^{p-1} \norm{\nabla u}\Lambda_{\rm max}(x)\Lambda_{\rm min}^{\frac{1-p}p}(x) \\ &\geq -\frac 1p \Lambda_{\rm max}^p(x)\Lambda_{\rm min}^{1-p}(x) \norm{\nabla u}^p. \end{align*} \textbf{Case 2:} In this second case we consider $1

2$. In this case \eqref{eq:PICineq} becomes equality if and only if all the following equalities hold: \begin{gather} \ss{\nabla u}{a(x)\nabla u}=\lambda_{\rm min}\norm{\nabla u}^2,\label{eq:pod1}\\ uv\ss{\nabla u}{\nabla v}=|uv|\norm{\nabla u}\|\nabla v\|,\label{eq:pod2}\\ \norm{A(x)\nabla v}=\Lambda_{\rm max}(x)\|\nabla v\|,\label{eq:pod3}\\ \ss{\nabla v}{A(x)\nabla v}=\Lambda_{\rm min}\|\nabla v\|^2,\label{eq:pod4}\\ \big|\frac{u}{v}\big|^p \|\nabla v\|^p \Lambda_{\rm min}(x)=\Lambda_{\rm max}^p(x)\Lambda_{\rm min}^{1-p}(x)\norm{\nabla u}^p.\label{eq:pod5} \end{gather} Equations \eqref{eq:pod1} and \eqref{eq:pod4} imply that $\nabla u$ and $\nabla v$ are eigenvectors of matrices $a(x)$ and $A(x)$ belonging to $\lambda_{\rm min}(x)$ and $\Lambda_{\rm min}(x)$, respectively. Equation \eqref{eq:pod3} implies that $\nabla v$ is also an eigenvector of $A(x)$ belonging to $\Lambda_{\rm max}(x)$ and hence $\Lambda_{\rm max}(x)=\Lambda_{\rm min}(x)$. Using this fact, \eqref{eq:pod5} becomes \begin{equation} \label{eq:pod5b} \big|\frac{u}{v}\big| \|\nabla v\|=\norm{\nabla u}. \end{equation} Equation \eqref{eq:pod2} implies that $\nabla u$ is a scalar multiple of $\nabla v$, i.e., there exists a function $\rho(x)$ such that $\nabla u=\rho\nabla v$. Now \eqref{eq:pod5b} and \eqref{eq:pod2} imply that $u(x)=\rho(x)v(x)$. Evaluating the gradient we get \[ \nabla u=v\nabla \rho+\rho\nabla v \] and \[ \norm{\nabla u}\leq |v|\norm{\nabla \rho}+ |\rho|\|\nabla v\|. \] From here and from the fact that $v$ does not have zeros on $\Omega$ we conclude that $\norm{\nabla \rho}=0$ and $\rho$ is a constant function. Hence all conditions (i)--(iv) hold. On the other hand, it is easy to see, that if (i)--(iv) hold, then \eqref{eq:pod1}--\eqref{eq:pod5} are satisfied and hence equality holds in \eqref{eq:PICineq}. \textbf{Case 2:} $p\leq 2$. Similarly to the previous case we find that \eqref{eq:PICineq} becomes equality if and only if \eqref{eq:pod1}, \eqref{eq:pod2}, \eqref{eq:pod3} and the following equations hold: \begin{gather} \ss{A(x)\nabla v}{A^{-1}(x)A(x)\nabla v} =\Lambda_{\rm max}^{-1}(x)\norm{A(x)\nabla v}^2,\label{eq:pod6}\\ \big|\frac{u}{v}\big|^p\|\nabla v\|^{p-2} \frac{\norm{A(x)\nabla v}^2}{\Lambda_{\rm max}(x)} ={\Lambda_{\rm max}^{p-1}(x)} \norm{\nabla u}^p \|\nabla v\|^{p-2} \norm{A(x)\nabla v}^{2-p} .\label{eq:pod7} \end{gather} As in the previous case \eqref{eq:pod1} and \eqref{eq:pod3} imply that $\nabla u$ and $\nabla v$ are eigenvectors of matrices $a(x)$ and $A(x)$ belonging to $\lambda_{\rm min}(x)$ and $\Lambda_{\rm max}(x)$, respectively. This also implies that \eqref{eq:pod6} holds and \eqref{eq:pod7} reduces into \eqref{eq:pod5b}. The remaining part is identical to the previous case and the proof that (i), (ii), and (iv) imply \eqref{eq:pod1}--\eqref{eq:pod3}, \eqref{eq:pod6} and \eqref{eq:pod7} is easy. Theorem is proved. \end{proof} \begin{remark}[Riccati inequality] \label{rmk2.1} \rm If $L(v)=0$ and $u\equiv 1$, then \eqref{eq:-div} becomes the generealized Riccati equation for the vector variable $\vec w(x):=A(x)\frac{\|\nabla v\|^{p-2}\nabla v}{|v|^{p-2}v}$: \[ \operatorname{div} \vec w +C(x)+(p-1)\norm{A^{-1}(x)\vec w}^{q-2}\ss{\vec w}{A^{-1}(x)\vec w}=0. \] Recall that the eigenvalues of the matrix $A^{-1}(x)$ are reciprocal values of the eigenvalues of the matrix $A(x)$ and thus \begin{gather*} \frac{1}{\Lambda_{\rm max}(x)}\norm{\vec w}\leq \norm{A^{-1}(x)\vec w}\leq \frac{1}{\Lambda_{\rm min}(x)}\norm{\vec w},\\ \frac{1}{\Lambda_{\rm max}(x)}\norm{\vec w}^2\leq \ss{\vec w}{A^{-1}(x)\vec w}\leq \frac{1}{\Lambda_{\rm min}(x)}\norm{\vec w}^2. \end{gather*} Combinations of these estimates allow to derive various types of Riccati inequalities. \end{remark} \begin{remark} \label{rmk2.2} \rm Remark that if the matrices $a(x)$, $A(x)$ are scalar multiples of identity matrix (say $a(x)=\tilde a(x)I$ and $A(x)=\tilde A(x)I$ where $\tilde a$ and $\tilde A$ are scalar functions) as in \cite{JKY2000}, then $\lambda_{\rm max} (x)=\lambda_{\rm min}(x)=\tilde a(x)$ and $K(x)=\Lambda_{\rm max}(x)=\Lambda_{\rm min}(x)=\tilde A(x)$. In this case we have the following identity for the first term from the right hand side of \eqref{eq:PICineq}: $[\lambda_{\rm min} (x)-K(x)]\norm{\nabla u}^p=[\tilde a(x)-\tilde A(x)]\norm{\nabla u}^p$. \end{remark} Immediately from the proof of Theorem \ref{th:main} we obtain the following statement, where only the ``second part'' \eqref{eq:proof2} of the Picone inequality \eqref{eq:PICineq} is considered. A closer examination of the proof reveals that condition (i) in Theorem~\ref{th:main} is needed only for the equality in the ``first part'' \eqref{eq:proof1} of \eqref{eq:PICineq}, while the other three conditions (ii)--(iv) mean the equality in \eqref{eq:proof2}. \begin{corollary} \label{cor:main} Let $u\in C^1(\overline\Omega)$, $v\in D_L(\Omega)$, $v\neq 0$ on $\Omega$ and let $K$ be the function defined in \eqref{eq:K}. Then the following inequality \begin{equation} \label{eq:PICineq2} \operatorname{div}\Big(|u|^p\frac{A(x)\|\nabla v\|^{p-2} \nabla v}{|v|^{p-2}v}\Big) \leq \frac{|u|^p}{|v|^{p-2}v}L(v) -C(x)|u|^p +K(x)\norm{\nabla u}^p \end{equation} holds for every $x\in\Omega$. The inequality in \eqref{eq:PICineq2} can be replaced by equality if and only if conditions (ii)--(iv) of Theorem \ref{th:main} hold. \end{corollary} \begin{remark} \label{rmk2.3} \rm Note that \begin{equation} \Big(\frac{\Lambda_{\rm max}(x)}{\Lambda_{\rm min}(x)}\Big)^{p-1}\Lambda_{\rm max}(x)\geq \Lambda_{\rm max}(x) \end{equation} The quotient $\frac{\Lambda_{\rm max}(x)}{\Lambda_{\rm min}(x)}$ is conditioned number of the matrix $A(x)$ and this number shows, that the inequality for the case $p\leq 2$ is sharper than inequality for $p>2$ (which holds in fact for every $p>1$). In addition, if $\Lambda_{\rm max}(x)=\Lambda_{\rm min}(x)$, then there is no difference between cases $p>2$ and $p\leq 2$ in Theorem \ref{th:main} and Corollary \ref{cor:main}. \end{remark} \begin{remark} \label{rmk2.4} \rm If $p>2$ and $A$ is not a scalar multiple of identity matrix, then condition (iii) of Theorem \ref{th:main} fails and \eqref{eq:PICineq} never becomes equality. \end{remark} \section{Applications of Picone inequality} As a consequence of the Picone inequality derived in the previous section we have the following version of Leighton-type comparison theorem. \begin{theorem} \label{th:compar_leighton} Let $u$ be a nontrivial solution of $l(u)=0$ such that $u=0$ on $\partial\Omega$ and let \[ \label{eq:int-ineq} \int_{\Omega}\left[(\lambda_{\rm min}(x)-K(x))\norm{\nabla u}^p+(C(x)-c(x))|u|^p\right]\d x \geq 0. \] Then every solution of $L(v)=0$ has a zero in $\overline\Omega$. \end{theorem} \begin{proof} Suppose, by contradiction, that $v$ is a solution of $L(v)=0$ such that $v\neq 0$ in $\overline\Omega$. The functions $u$, $v$ satisfy assumptions of Theorem \ref{th:main} and since $v$ is not a constant multiple of $u$, the Picone inequality \eqref{eq:PICineq} holds strict. Integrating this inequality with using the Gauss-Ostrogradskii theorem we obtain \begin{align*} & \int_{\partial\Omega} \big\langle \frac{u}{|v|^{p-2}v} \left[ |v|^{p-2}v a(x)\norm{\nabla u}^{p-2}\nabla u- |u|^{p-2}uA(x)\|\nabla v\|^{p-2}\nabla v \right], \nu\big\rangle \d S \\ &> \int_{\Omega}\Bigl[ \lambda_{\rm min}(x)- K(x) \Bigr]{\|\nabla u\|}^p+\Bigl[C(x)-c(x)\Bigr]|u|^p \d x, \end{align*} where $\nu$ denotes the outside unit normal. The fact that $u=0$ on $\partial\Omega$ gives \[ \int_{\Omega}\left[(\lambda_{\rm min}(x)-K(x)) \norm{\nabla u}^p+(C(x)-c(x))|u|^p\right]\d x < 0, \] a contradiction. \end{proof} The next two statements follow directly from Theorem \ref{th:compar_leighton}. \begin{corollary} \label{th:CORcompar} Let $u$ be a nontrivial solution of $l(u)=0$ such that $u=0$ on $\partial\Omega$. \begin{enumerate} \item[(i)] If $\lambda_{\rm min}(x)\geq K(x)$ and $C(x)\geq c(x)$ in $\Omega$, then every solution of $L(v)=0$ has a zero in $\overline\Omega$. \medskip \item[(ii)] If $\lambda_{\rm min}(x)=\lambda_{\rm max}(x)$, then every solution of $l(u)=0$ has a zero in $\overline\Omega$. \end{enumerate} \end{corollary} Similarly as Theorem \ref{th:compar_leighton} follows from Theorem \ref{th:main}, the next theorem can be obtained from Corollary \ref{cor:main}. \begin{theorem} \label{th:functional} Suppose that there exists a nontrivial function $u\in C^1(\overline\Omega)$ such that $u=0$ on $\partial\Omega$ and \[ \label{eq:int-ineq2} \int_{\Omega}\left[K(x)\norm{\nabla u}^p-C(x)|u|^p\right]\d x \leq 0. \] Then every solution of $L(v)=0$ has a zero in $\overline\Omega$. \end{theorem} \begin{proof} Let $v$ be a nontrivial solution of $L(v)=0$ and suppose, by contradiction, that $v\neq 0$ in $\overline\Omega$. The functions $u$, $v$ satisfy the (strict) inequality \eqref{eq:PICineq2}. Integrating this inequality and using the Gauss-Ostrogradskii theorem similarly as in the proof of Theorem \ref{th:compar_leighton}, we have \[ \int_{\Omega}\left[K(x)\norm{\nabla u}^p-C(x)|u|^p\right]\d x > 0, \] a contradiction. \end{proof} As a direct consequence of the above theorem we obtain the following Wirtinger-type inequality. \begin{corollary} \label{th:wirtinger} If there exists a solution $v$ of $L(v)=0$ such that $v\neq 0$ in $\overline\Omega$, then \[ \int_{\Omega}\left[K(x)\norm{\nabla u}^p-C(x)|u|^p\right]\d x > 0, \] for any nontrivial function $u\in C^1(\overline\Omega,\mathbb{R})$ such that $u=0$ on $\partial\Omega$. \end{corollary} We finish this section with two statements which show that if the integral inequality in Theorem \ref{th:compar_leighton} or Theorem \ref{th:functional} is strict, then the zero of the solution of $L(v)=0$ occurs in $\Omega$. The method of the proof is similar to that used in \cite{JKY2000} or \cite{Yoshida}. \begin{theorem} \label{th:functional-lim} Let $\partial\Omega \in C^1$ and suppose that there exists a nontrivial function $u\in C^1(\overline\Omega)$ such that $u=0$ on $\partial\Omega$ and \begin{equation} \label{eq:int-ineq2-strict} \int_{\Omega}\left[K(x)\norm{\nabla u}^p-C(x)|u|^p\right]\d x < 0. \end{equation} Then every solution of $L(v)=0$ has a zero in $\Omega$. \end{theorem} \begin{proof} Conditions imposed on $u$ and $\Omega$ imply that $u\in W_0^{1,p}(\Omega)$. Hence (see e.g. \cite{A-F}) there exists a sequence of functions $u_k\in C_0^{\infty}(\Omega)$ converging to $u$ in the norm \[ \norm{w}_p:=\Big(\int_{\Omega}[\norm{w}^p+\norm{\nabla w}^p] \d x \Big)^{1/p}. \] Suppose, by contradiction, that there exists a solution $v$ of $L(v)=0$ such that $v\neq 0$ in $\Omega$. From Corollary \ref{cor:main} it follows that for $x\in\Omega$, \[ \operatorname{div}\Big(|u_k|^p\frac{A(x)\|\nabla v\|^{p-2} \nabla v}{|v|^{p-2}v}\Big) \leq K(x)\norm{\nabla u_k}^p-C(x)|u_k|^p. \] Integrating this inequality over the set $\Omega_k\subset\Omega$ containing the (compact) support of $u_k$ and using the Gauss-Ostrogradskii theorem we have \begin{equation} \label{eq:ineq-seq} 0\leq \int_{\Omega_k} \left[K(x)\norm{\nabla u_k}^p-C(x) |u_k|^p\right]\d x = \int_{\Omega} \left[K(x)\norm{\nabla u_k}^p-C(x)|u_k|^p\right]\d x. \end{equation} Next, it can be shown in the same way as in \cite[Theorem 8.1.3]{Yoshida} that \begin{align*} &\big|\int_{\Omega} \left[K(x)\norm{\nabla u_k}^p-C(x)|u_k|^p\right]\d x -\int_{\Omega}\left[K(x)\norm{\nabla u}^p-C(x)|u|^p\right]\d x\big|\\ &\leq M \left(\norm{u_k}_p+\norm{u}_p\right)^{p-1}\norm{u_k-u}_p, \end{align*} where $M$ is a positive constant. Since $\norm{u_k-u}_p\to 0$ as $k\to\infty$, we have \[ \lim_{k\to\infty}\int_{\Omega} \left[K(x)\norm{\nabla u_k}^p-C(x)|u_k|^p\right]\d x =\int_{\Omega}\left[K(x)\norm{\nabla u}^p-C(x)|u|^p\right]\d x, \] which, together with \eqref{eq:ineq-seq}, contradicts assumption \eqref{eq:int-ineq2-strict}. \end{proof} \begin{corollary} \label{th:compar_leighton-lim} Let $\partial\Omega \in C^1$ and suppose that there exists a nontrivial solution $u$ of $l(u)=0$ such that $u=0$ on $\partial\Omega$ and \begin{equation} \label{eq:int-ineq-strict} \int_{\Omega}\left[(\lambda_{\rm min}(x)-K(x))\norm{\nabla u}^p+(C(x)-c(x))|u|^p\right]\d x > 0. \end{equation} Then every solution of $L(v)=0$ has a zero in $\Omega$. \end{corollary} \begin{proof} Inequality \eqref{eq:int-ineq-strict}, computation used in \eqref{eq:proof1} and Gauss-Ostrogradskii theorem imply \begin{align*} \int_{\Omega}\left[K(x)\norm{\nabla u}^p-C(x)|u|^p\right]\d x &< \int_{\Omega}\left[\lambda_{\rm min}(x)\norm{\nabla u}^p-c(x)|u|^p\right]\d x\\ &\leq \int_{\Omega}\left[\operatorname{div}\left(ua(x)\norm{\nabla u}^{p-2}\nabla u\right)-ul(u)\right] \d x\\ &=0. \end{align*} The statement now follows from Theorem \ref{th:functional-lim}. \end{proof} \section{An oscillation result} Recall that the half-linear differential equation \[ \left(s(t)|y'|^{p-2}y'\right)'+q(t)|y|^{p-2}y=0, \] where $s>0$, $q$ are real-valued continuous functions on $t\in [t_0,\infty)$, is said to be oscillatory if any nontrivial solution of this equation has a sequence of zeros tending to $\infty$. Concerning linear and half-linear partial equations, there are two types of oscillation: \textit{oscillation} (sometimes also weak oscillation) and \textit{nodal oscillation} (strong oscillation). Denote $$ \Omega(r_0)=\{x\in \mathbb{R}^n: \norm{x}\geq r_0\} $$ and assume that the coefficients of the operator $L$ satisfy $A\in C^1(\Omega(r_0),\mathbb{R}^{n\times n})$, $C\in C^{0,\alpha}(\Omega(r_0))$. We say that a solution $v$ of $L(v)=0$ is \textit{oscillatory\/} if it has a zero in $\Omega(r)$ for every $r\geq r_0$. Equation $L(v)=0$ is said to be \textit{oscillatory\/} if every solution of this equation is oscillatory. The equation $L(v)=0$ is said to be \textit{nonoscillatory\/} if it is not oscillatory. Similarly, the equation $L(v)=0$ is said to be \textit{nodally oscillatory}, if every its solution has a nodal domain outside of every ball in $\mathbb{R}^n$ and \textit{nodally nonoscillatory} in the opposite case. It is known that nodal oscillation implies oscillation. The opposite implication is known to be valid only in the linear case $p=2$ (see \cite{Moss}) and remains an open question in the half-linear multidimensional case (the case $n=1$ is trivial). While Riccati technique is suitable to study weak oscillation, Picone identity and variational technique are suitable to study both types of oscillation (and hence both techniques overlap for weak oscillation). In the remaining part of this paper we deal (for simplicity) with the weak oscillation (referred to as oscillation) and show one simple but important application of Picone inequality. The following oscillation theorem compares oscillation of the PDE $L(u)=0$ with oscillation of a certain ordinary differential equation. This enables to extend many oscillation criteria from theory of ordinary equations to partial differential equations. Note that the statement we present has been proved using the Riccati technique in \cite{M2007JMAA}. Using Picone identity the proof is simple and straightforward. \begin{theorem} \label{th:osc-int} Suppose that the half-linear ordinary differential equation \begin{equation} \label{eq:ODR-int} \tilde l(y):=\left(\tilde K(r)|y'|^{p-2}y'\right)'+\tilde C(r)|y|^{p-2}y=0, \end{equation} where \[ \tilde K(r):=\int_{\norm{x}=r}K(x)\d S,\quad \tilde C(r):=\int_{\norm{x}=r}C(x)\d S, \] is oscillatory. Then the equation $L(v)=0$ is also oscillatory. \end{theorem} \begin{proof} Let $y=y(r)$ be an (oscillatory) solution of \eqref{eq:ODR-int} and let $r_1