2$, the case $p> 2$ can be treated analogically. We have $$ f'(t)=\Phi_q(t)-1-\frac{4}{2^\alpha}\Phi_\alpha(t-1),\quad f''(t)= (q-1)|t|^{q-2}-\frac{4(\alpha-1)}{2^\alpha}|t-1|^{\alpha-2}, $$ where $\Phi_q(t)=|t|^{q-2}t$, $\Phi_{\alpha}$ is defined analogically. Hence $f'(-1)=0=f'(1)$, $f(-1)=2-\frac{4}{\alpha}\geq 0$, and $f''(-1)=q-\alpha\geq 0$. Drawing the graphs of the functions $|t|$ and $\left(\frac{4(\alpha-1)}{2^\alpha(q-1)}\right)^{\frac{1}{q-2}}|t-1|^{\frac{\alpha-2}{q-2}}$ shows that the equation $f''(t)=0$ has exactly 2 roots, one positive in the interval $(0,1)$, and one negative in $[-1,0)$. Hence $f''$ is positive outside of the interval determined by these roots and negative inside of it. This means that $f$ has at both stationary points $t=\pm 1$ nonnegative local minima. This also implies that the equation $f'(t)=0$ may have at most one zero in $(-1,1)$, where the function $f$ attains a positive local maximum. Consequently, summarizing these facts about the graph of the functions $f$ we obtain that $f(t)\geq 0$ for $t\in \mathbb{R}$. \end{proof} Observe also that substituting $t\to-t$ gives for $q\geq 2$ the inequality \begin{equation} \label{inequality+} \frac{|t|^q}{q}+t+\frac{1}{p}-\frac{4}{\alpha 2^\alpha}|t+1|^\alpha\geq 0, \quad t\in \mathbb{R} \end{equation} and the opposite inequality for $q\in (1,2]$. The following lemma is an extension of \cite[Lemma 2.4]{DE} which deals with the scalar case and $\alpha=2$. \begin{lemma} \label{lemma:P} \begin{itemize} \item[(i)] Let $p\geq 2$ and $\norm{ u_1}\neq 0$. Then for every $\alpha\in[q,2]$ there exists a number $\gamma(\alpha,p)$ such that \begin{equation} \label{eq:ner-vec1} P( u_1, u_2)\leq \gamma(\alpha,p)\norm{ u_1}^{(p-1)(q-\alpha)}\norm{ u_2-\norm{ u_1}^{p-2} u_1}^\alpha. \end{equation} \item[(ii)] Let $p\in(1,2]$. Then for every $\alpha\in[2,q]$ there exists a number $\gamma(\alpha,p)$ such that \begin{equation} \label{eq:ner-vec2} P( u_1, u_2)\geq \gamma(\alpha,p)\norm{u_1}^{(p-1)(q-\alpha)} \norm{ u_2-\norm{ u_1}^{p-2} u_1}^\alpha. \end{equation} \end{itemize} \end{lemma} \begin{remark} \label{rmk1} \rm In the proof we will show that we can take $\gamma(\alpha,p)=4/\alpha 2^\alpha$. However, the numerical computations show that this constant is not optimal and can be improved. To find this optimal constant is a subject of the present investigation. \end{remark} \begin{proof}[Proof of Lemma \ref{lemma:P}] Observe that \eqref{eq:ner-vec2} trivially holds for $\norm{ u_1}=0$. Therefore, in the remaining part of the proof we suppose $\norm{ u_1}\neq 0$. We will prove the first statement of lemma ($p\geq 2$), the proof of the second part is analogical. By dividing both sides of \eqref{eq:ner-vec1} with the factor $\norm{u_1}^p$ we get the inequality \begin{equation} \label{uu-inequality} \frac 1p-\big\langle\frac{u_1}{\norm{u_1}}, \frac{ u_2}{\norm{ u_1}^{p-1}}\big\rangle + \frac{\norm{\frac{ u_2}{\norm{ u_1}^{p-1}}}^q}{q}\leq \gamma(\alpha,p)\norm{\frac{ u_2}{\norm{ u_1}^{p-1}}-\frac{ u_1}{\norm{ u_1}}}^\alpha.%\\ \end{equation} Define $x=\frac{ u_2}{\norm{ u_1}^{p-1}}$ and $a=\frac{ u_1}{\norm{ u_1}}$. Then $\Vert a\Vert=1$ and \eqref{uu-inequality} can be written in the form \begin{equation} \label{inequality} \frac{\Vert x\Vert^q}{q}-\langle a,x\rangle +\frac{1}{p}\leq \gamma(\alpha,p)\Vert x-a\Vert^{\alpha}. \end{equation} As mentioned above, we show that this inequality holds with $\gamma(\alpha,p)=\frac{4}{\alpha 2^\alpha}$. Let $g(x)=\langle x,a\rangle +\frac{4}{\alpha 2^\alpha} \Vert x-a\Vert^{\alpha}$. We will examine the minimal value of this function over the sphere $\Vert x\Vert=t$, $t\geq 0$. Any $x\in \mathbb{R}^n$ can be written in the form $x=\mu a+ \nu a^{\perp}$ for some unit vector $a^{\perp}$ with $\langle a,a^{\perp}\rangle=0$. Then $$ t^2=\Vert x\Vert ^2=\langle \mu a+\nu a^\perp,\mu a+\nu a^\perp \rangle =\mu^2 + \nu^2. $$ We have \begin{align*} g(x)=&\langle \mu a+\nu a^\perp, a\rangle + \frac{4}{\alpha 2^\alpha} \langle \mu a + \nu a^\perp - a, \mu a + \nu a^\perp - a\rangle^{\alpha/2} \\ =& \mu +\frac{4}{\alpha 2^\alpha}(t^2-2\mu +1)^{\alpha/2}. \end{align*} Now we solve the extremal problem $g(x)\to \min$, $\Vert x \Vert=t$ which can be written in the form $$ \mu +\frac{4}{\alpha 2^\alpha} (t^2-2\mu +1)^{\alpha/2}\to \min, \quad \mu \in [-t,t]. $$ Since $\alpha/2\leq 1$, the minimized function is concave and hence it attains its minimum over $[-t,t]$ at the boundary point of this interval; i. e., $$ g(x)\bigr|_{\Vert x\Vert=t}\geq \min\big\{-t+\frac{4}{\alpha 2^\alpha}|t+1|^{\alpha}, t+\frac{4}{\alpha 2^\alpha}|t-1|^\alpha\big\}. $$ Consequently, inequality \eqref{inequality} holds if $$ \frac{|t|^q}{q}+\frac{1}{p}\mp t-\frac{4}{\alpha 2^\alpha}|t\mp 1|^2 \leq 0. $$ But this is just the inequality from Lemma \ref{L1} for the sign ``$-$'' or its equivalent reformulation after the substitution $t\to -t$ (see \eqref{inequality+} in case $q\in (1,2]$) . The Lemma is proved. \end{proof} The next lemma presents a link between two Riccati type operators, namely the operator which corresponds to half-linear equation \eqref{eq:E} (the power at the dependent variable is $q$) and the Riccati operator with $\alpha$-degree nonlinearity, where $\alpha\in [\min\{q,2\},\max\{q,2\}]$. \begin{lemma}\label{lemma:ineq} Let $h\in C^2(\Omega, \mathbb{R}^+)$. Define $ G=h\norm{\nabla h}^{p-2}\nabla h$ and $ v=h^p w- G$. Further, let $\alpha\in [\min\{q,2\},\max\{q,2\}]$ and $\gamma(\alpha,p)$ be the number from Lemma \ref{lemma:P}. \begin{itemize} \item[(i)] If $1< p\leq 2$, then \begin{equation} \label{eq:ner1} h^pR[ w]\geq \operatorname{div} v+hL[h]+\gamma(\alpha,p) p h^{-\alpha}\norm{\nabla h}^{(p-1)(q-\alpha)}\norm{ v}^\alpha \end{equation} holds on $\Omega$. \item[(ii)] If $p\geq 2$ and $\norm {\nabla h}\neq 0$ on $\Omega$, then \begin{equation} \label{eq:ner2} h^pR[ w]\leq \operatorname{div} v+hL[h]+\gamma(\alpha,p) p h^{-\alpha}\norm{\nabla h}^{(p-1)(q-\alpha)}\norm{ v}^\alpha \end{equation} holds on $\Omega$. \end{itemize} \end{lemma} \begin{proof} We start with the following obvious identities \begin{equation} \operatorname{div} G=\norm{\nabla h}^p+h\Delta_ph \end{equation} and \begin{equation} \begin{aligned} h^p\operatorname{div} w &=h^p\operatorname{div}(h^{-p}( v+ G))\\ &=\operatorname{div} v+\operatorname{div} G-ph^{-1}\inner{ v+ G}{\nabla h}\\ &=\operatorname{div} v+\norm{\nabla h}^p+h\Delta_p h-ph^{-1} \inner{v+ G}{\nabla h}. \end{aligned} \end{equation} Now a direct computation shows \begin{align*} h^p R[ w]&= h^p\operatorname{div} w+h^p c(x)+(p-1)h^p \norm{ w}^q\\ &=\operatorname{div} v+\norm{\nabla h}^p+h\Delta_p h-ph^{-1}\inner{ v+ G}{\nabla h}+ h^p c(x)+(p-1)h^{-q}\norm{h^p w}^q\\ &=\operatorname{div} v+hL[h]+ph^{-q}\Big(\frac{\norm{h^{q-1}\nabla h}^p}p-\inner{ v+ G}{h^{q-1}\nabla h}+\frac{\norm{ v+ G}^q}q\Big) \\ &=\operatorname{div} v+hL[h]+ph^{-q}P(h^{q-1}\nabla h, v+ G). \end{align*} For $ u_1=h^{q-1}\nabla h$ and $ u_2= v+ G$ we have \begin{align*} \norm{ u_1}^{(p-1)(q-\alpha)}&\norm{ u_2-\norm{ u_1}^{p-2} u_1}^\alpha\\ &= h^{(q-1)(p-1)(q-\alpha)}\norm{\nabla h}^{(p-1)(q-\alpha)} \norm{ v+ G-h\norm{\nabla h}^{p-2}\nabla h}^\alpha\\ &=h^{q-\alpha}\norm{\nabla h}^{(p-1)(q-\alpha)}\norm{ v}^\alpha. \end{align*} Now the lemma follows from the estimates in Lemma \ref{lemma:P}. \end{proof} \section{Main results} In this section we introduce the main results of the paper. Since most of the work has been already done in the previous section, the proofs are short and straightforward. Our first theorem certifies the importance of Riccati type inequality \eqref{eq:R[w]leq} in the theory of half-linear differential equations \eqref{eq:E}. \begin{theorem}\label{th1} The following statements are equivalent: \begin{itemize} \item[(i)] The equation $L[u]=0$ has a positive $C^2$ solution on $\Omega$. \item[(ii)] The inequality $L[u]\leq 0$ has a positive $C^2$ solution on $\Omega$. \item[(iii)] The equation $R[ w]=0$ has a $C^1$ solution $ w$ on $\Omega$ such that the vector field $\norm{ w}^{q-2} w$ is conservative. \item[(iv)] The inequality $R[ w]\leq 0$ has a $C^1$ solution $ w$ on $\Omega$ such that the vector field $\norm{ w}^{q-2} w$ is conservative. \end{itemize} \end{theorem} \begin{proof} Define \begin{equation} w=\frac{\norm{\nabla u}^{p-2}\nabla u}{|u|^{p-2}u}.\label{eq:def_w} \end{equation} By a direct calculation, the $i$-th component of the vector $ w$ satisfies \[ \frac{\partial w_i}{\partial x_i}= \frac{\frac{\partial }{\partial x_i}\left(\norm{\nabla u}^{p-2}\frac{\partial u}{\partial x_i}\right)}{|u|^{p-2}u} -(p-1)\frac{\norm{\nabla u}^{p-2}\frac{\partial u}{\partial x_i}}{|u|^{p}}\frac{\partial u}{\partial x_i} \] and summing up over all independent variables we get \begin{align*} \operatorname{div} w&=\frac{\operatorname{div}\left(\norm{\nabla u}^{p-2}\nabla u\right)}{|u|^{p-2}u} -(p-1)\frac{\norm{\nabla u}^{p-2}}{|u|^{p}}\norm{\nabla u}^2\\&= \frac{\operatorname{div}\left(\norm{\nabla u}^{p-2}\nabla u\right)}{|u|^{p-2}u} -(p-1)\norm{ w}^q. \end{align*} Using this computation we easily observe that \begin{equation} \label{eq:RL} R[ w]=\frac{L[u]}{|u|^{p-2}u} \end{equation} holds. (i)$\implies$(iii). Follows from \eqref{eq:RL} and from the fact that if $ w$ is defined by \eqref{eq:def_w}, then $\norm{w}^{q-2} w=\frac{\nabla u}{u}=\nabla(\ln u)$ and $\ln u$ is a scalar potential to $\norm{ w}^{q-2} w$. (iii)$\implies$(iv). Clearly holds. (iv)$\implies$(ii). Since $\norm{ w}^{q-2} w$ has a scalar potential, there exists a scalar function $\varphi$, such that $\nabla \varphi= \norm{ w}^{q-2} w$. Define function $u=e^\varphi$. The function $u$ satisfies \eqref{eq:def_w} and in view of \eqref{eq:RL} the implication holds. (ii)$\implies$(i). Follows from Theorem \ref{th:Allegretto}. \end{proof} Our second theorem relates two Riccati type inequalities. One of them is inequality \eqref{eq:R[w]leq} which is associated to the half-linear equation with $p$-Laplacian \eqref{eq:E} (the dependent variable appears in the inequality in the power $q$), while the second one contains the dependent variable in the power $\alpha$; i. e., the equation is associated with a half-linear PDE with $\beta$-degree Laplacian, where $\beta$ is the conjugate number to the number $\alpha$ (see also Remark \ref{rem} below). \begin{theorem}\label{th2} Let $h\in C^2(\Omega,\mathbb{R}^+)$. \begin{itemize} \item [(i)] Let $p\in(1,2]$, \eqref{eq:R[w]leq} has a $C^1$ solution on $\Omega$, $\alpha\in[2,q]$ be arbitrary number and $\gamma(\alpha,p)$ be the number from Lemma \ref{lemma:P}. Then \begin{equation} \label{eq:lin_rce} \operatorname{div} v+h(x)L[h(x)]+p\gamma(\alpha,p) h^{-\alpha}(x)\norm{\nabla h(x)}^{(p-1)(q-\alpha)}\norm{ v}^\alpha \leq 0 \end{equation} has also a $C^1$ solution on $\Omega$. \item[(ii)] Let $p\geq 2$, $\alpha\in[q,2]$ be arbitrary number, $\gamma(\alpha,p)$ be the number from Lemma \ref{lemma:P} and let $h$ satisfy $\norm{\nabla h}\neq 0$ on $\Omega$. If \eqref{eq:lin_rce} has a $C^1$ solution on $\Omega$, then \eqref{eq:R[w]leq} has also a $C^1$ solution on $\Omega$. \end{itemize} \end{theorem} The proof of the above theorem is a direct consequence of the inequalities from Lemma \ref{lemma:ineq}. \begin{remark}\label{rem} \rm Suppose that both $h$ and $\norm{\nabla h}$ do not vanish in $\Omega$. The equation \[ \operatorname{div} v+h(x)L[h](x)+p\gamma(\alpha,p) h^{-\alpha}(x)\norm{\nabla h(x)}^{(p-1)(q-\alpha)}\norm{ v}^\alpha =0 \] is the Riccati equation for the second order partial differential equation \begin{equation} \operatorname{div}\Bigl(A(x)\norm{\nabla u}^{\beta-2}\nabla u\Bigr)+ h(x)L[h](x)|u|^{\beta -2}u=0, \label{eq:Emod} \end{equation} where $\beta=\frac{\alpha}{\alpha-1}$ and $A(x)=\left[\frac{p\gamma(\alpha,p)}{\beta-1}\right]^{1-\beta}h^{\beta} (x)\norm{\nabla h(x)}^{p-\beta}$. Thus if \eqref{eq:Emod} has a positive solution on $\Omega$, then \eqref{eq:lin_rce} has a solution. Conversely, if \eqref{eq:lin_rce} has a $C^1$ solution $ v$ on $\Omega$ and $\norm{ v}^{\alpha-2} v$ is conservative, then \eqref{eq:Emod} has a positive $C^2$ solution on $\Omega$. 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