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\def\rightheadline{EJDE--1998/11\hfil Barriers on cones
\hfil\folio}
\def\leftheadline{\folio\hfil Michail Borsuk \& Dmitriy Portnyagin
 \hfil EJDE--1998/11}

\def\pretitle{\vbox{\eightrm\noindent\baselineskip 9pt %
 Electronic Journal of Differential Equations,
Vol.\ {\eightbf 1998}(1998), No.~11, pp.~1--8.\hfil\break
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\hfill\break 
ftp (login: ftp) 147.26.103.110 or 129.120.3.113\bigskip} }

\topmatter
\title
Barriers on cones for degenerate quasilinear elliptic operators
\endtitle

\thanks \noindent
{\it 1991 Mathematics Subject Classifications:} 35J65, 35J70, 35B05, 35B45, 
35B65.\hfil\break
{\it Key words and phrases:} quasilinear elliptic equations, barrier functions,
conical points.
\hfil\break
\copyright 1998 Southwest Texas State University  and
University of North Texas.\hfil\break
Submitted May 15, 1997. Published April 17, 1998.
\endthanks
\author Michail Borsuk \\ Dmitriy Portnyagin  \endauthor

\address Michail Borsuk \hfil\break
Department of Applied Mathematics \hfil\break
Olsztyn University of Agriculture and  Technology \hfil\break 
10-957 Olsztyn-Kortowo, Poland \hfil\break
\endaddress
\email borsuk\@art.olsztyn.pl
\endemail

\address Dmitriy Portnyagin \hfil\break
Department of Physics \hfil\break
Lvov State University \hfil\break
290602 Lvov, Ukraine 
\endaddress

\abstract
Barrier functions  $w=|x|^\lambda \Phi(\omega)$ are constructed
for the first boundary value problem  as well as for the mixed
boundary value problem for quasilinear elliptic second order
equation of divergent form with triple degeneracy on the $n$-dimensional 
convex circular cone:
$$ \frac{d}{dx_i}(|x|^\tau|u|^q|\nabla u|^{m-2}u_{x_i})=
    \mu |x|^\tau{|u|}^{q-1}\text{sgn }{u}{|\nabla u|}^{m}\,,$$
where $-1<\mu \leq 0$, $q\geq 0$, $m>1$, $\tau>m-n$. 
\endabstract
\endtopmatter

\document
\head Introduction \endhead

Lately many mathematicians have been considering nonlinear problems for
elliptic degenerate equations; see e.g.\ [1] and its extensive bibliography. 
In the present paper, we take a first step on the investigation of the
behaviour of solutions of boundary value problems for quasilinear elliptic
second-order equations with triple degeneracy. We study the problem
 $$\gathered 
 Lu\equiv\frac{d}{dx_i}(|x|^\tau|u|^q|\nabla u|^{m-2}u_{x_i})=
    \mu |x|^\tau{|u|}^{q-1}\text{sgn }{u}{|\nabla u|}^{m},\quad x\in G_0\,,\\
    -1<\mu \le0,\quad q\ge0,\quad m>1,\quad\tau>m-n\,, \endgathered\tag1
 $$
where $G_0$ is an $n$-dimensional {\it convex} circular cone with
its vertex at  
the
origin $O$, having $\Gamma_0$ as lateral area; and $\Omega $ is a domain, on
the unit sphere, with a smooth boundary $\partial \Omega$.  We shall construct
functions playing a fundamental role in the study  of the
behaviour of solutions 
to elliptic boundary value
problems in the neighbourhood of the irregular boundary point;
see e.g.\ [2--6]. 
The special structure of the solution near a conical point is of particular
interest in physical applications, [7--9]. It is also used for improving
numerical algorithms, [10--12].

The proof of the estimates for the solution is based on the 
observation that the function $r^{\lambda}\Phi(\omega)$ is usable as barrier 
in the above problem. By the weak comparison principle in [2, chapt.\ 10], it
is possible to verify that the assumptions of this principle are fulfilled. 
Since (1) is equivalent to
 $$\gathered 
 \frac{d}{dx_i}\left({|\nabla u|}^{m-2}u_{x_i}\right)+\tau |x|^{-2}{|\nabla 
u|}^{m-2}
(x \nabla u)+(q- \mu){|u|}^{-1}\text{sgn }{u}{|\nabla u|}^{m}=0,\\ x\in G_0, 
\quad
   -1<\mu \le0,\quad q\ge0,\quad m>1,\quad\tau>m-n \endgathered 
 $$
on the set where $u \neq 0$,
one can obtain  the bound for solution near conical boundary point. In
this setting, finding of exact value of the exponent $\lambda$ is very
important and very difficult.
For the case of a planar bounded domain with corner boundary points, the exact 
value
of the exponent $\lambda$ will be calculated explicitly.

Let us transfer the above problem to spherical coordinates with the pole at 
the  point $O$.
 $$\align x_1&=r\cos\omega_1, \\
           x_2&=r\cos\omega_2 \sin\omega_1, \\
               &\quad\vdots       \\
           x_{n-1}&=r\cos\omega_{n-1}
\sin\omega_{n-2}\dots\sin\omega_1, \\                  
           x_{n}&=r\sin\omega_{n-1}\dots \sin \omega_1, \endalign
 $$                 
where $r=|x|>0$,  $0<\omega_i<\pi$  for $i=1,\dots,n-2$, and
$0<\omega_{n-1}<2\pi$.

The differential operator $L$ takes the form         
$$  Lu=\frac{1}{J}\sum_{i=1}^{n}\frac{d}{d \xi_i}\left(r^\tau |u|^q|\nabla u|^
           {m-2}\frac{J}{H_i^2}\frac{\partial u}{\partial \xi_i}\right),$$
where $ J=r^{n-1}\sin^{n-2}\omega_1,\dots,\sin\omega_{n-2}$; 
$H_1=1$; $\xi_1=r$; $\xi_{i+1}=\omega_{i}$ and
  $H_{i+1}=r\sqrt{q_i}$, for $i=\{\overline{1,n-1}\}$;
  $q_1=1$; $q_i=(\sin\omega_1\dots\sin\omega_{i-1})^2$ for
  $i=\{\overline{2,n-1}\}$.

 We shall seek the solution of (1) as a function of the form
$u=r^\lambda\Phi(\omega)$
with  $\Phi(\omega) \ge0$. Then $\Phi(\omega)$ satisfies the equation
  $$\multline
\frac{1}{j(\omega)}\sum_{k=1}^{n-1} \frac{d}{d \omega_k}\left(
 \frac{j(\omega)}{q_k}(\lambda^2 \Phi^2
 +|\nabla_{\omega} \Phi|^2)^{(m-2)/2}|\Phi|^q
 \frac{\partial \Phi}{\partial \omega_k}\right)\\
 +\lambda[\lambda(q+m-1)+ \tau+n-m](\lambda^2 \Phi^2+|\nabla_{\omega}
 \Phi|^2)^{(m-2)/2}\Phi|\Phi|^q\\
 =\mu \Phi|\Phi|^{q-2}(\lambda^2 \Phi^2+|\nabla_{\omega} \Phi|^2)
 ^{m/2}, \omega\in \Omega, \endmultline\tag2
 $$
where $|\nabla_{\omega}\Phi|^2=\sum_{j=1}^{n-1}\frac{1}{q_j}
(\frac{\partial \Phi}{\partial \omega_j})^2$, and 
  $j(\omega)=\sin^{n-2}\omega_1\dots\sin\omega_{n-2}$.

\head The Dirichlet problem\endhead 

First we consider the Dirichlet problem for (1) when 
$u|_{\Gamma_0}=0$. 
On multiplying (2) by $\Phi(\omega)$ and integrating by parts over $\Omega$ we
obtain
   $$\align
&\int\limits_{\Omega}\left(\lambda^2\Phi^2+|\nabla_\omega\Phi|^2
\right)^{(m-2)/2}|\Phi|^q|\nabla_\omega\Phi|^2d\Omega\\
&=\lambda[\lambda(q+m-1)+\tau+n-m]\int\limits_{\Omega}
\left(\lambda^2\Phi^2+|\nabla_\omega\Phi|^2\right)^{(m-2)/2}
|\Phi|^{q+2}d\Omega\\
&\quad -\mu\int\limits_{\Omega}
|\Phi|^q\left(\lambda^2\Phi^2+|\nabla_\omega\Phi|^2\right)^{m/2}
d\Omega\\
&\equiv\int\limits_{\Omega}\left(\lambda^2\Phi^2
+|\nabla_\omega\Phi|^2\right)^{(m-2)/2}|\Phi|^q \times \\
&\quad\left\{\lambda[\lambda(q+m-1)+\tau+n-m]\Phi^2
-\mu(\lambda^2\Phi^2+|\nabla_\omega\Phi|^2)\right\}
d\Omega\,.\endalign
$$
Hence, it follows that
$$\multline
(1+\mu)\int\limits_{\Omega}\left(\lambda^2\Phi^2+|\nabla_\omega\Phi|^2
\right)^{(m-2)/2}|\Phi|^q|\nabla_\omega\Phi|^2d\Omega\\
=\lambda[\lambda(q+m-1-\mu)+\tau+n-m]\int\limits_{\Omega}
\left(\lambda^2\Phi^2+|\nabla_\omega\Phi|^2\right)^{(m-2)/2}
|\Phi|^{q+2}d\Omega\,.\endmultline
$$
Since $\Phi(\omega)\not \equiv 0$ and $\mu>-1$, we have
   $$\lambda[\lambda(q+m-1-\mu)+\tau+n-m]>0\,.\tag *$$
We shall consider the case of $\Phi(\omega)$ not depending on  
$\omega_2,\dots ,\omega_{n-1}$;
so that $\Phi$ is a function of a single angular coordinate
$\omega_1=\omega\in(-\omega_0/2,\omega_0/2)$, $0<\omega_0<\pi$. Such 
function $\Phi(\omega)$ satisfies the
boundary value problem for ordinary differential equation
 $$\aligned
  &[(m-1){\Phi'}^2+\lambda^2 \Phi^2] \Phi \Phi'' 
  +(\lambda^2 \Phi^2+{\Phi'}^2) \times \\  & \left\{
(q-\mu){\Phi'}^2+\lambda[\lambda(q+m-1-\mu)+\tau+n-m]\Phi^2 +(n-2)\Phi\Phi'
\cot \omega \right\}\\
& +(m-2)\lambda^2\Phi^2{\Phi'}^2=0,
\qquad \omega\in (-\omega_0/2,\omega_0/2) \\
&  \Phi(-\omega_0/2)=\Phi(\omega_0/2)=0\,.\endaligned \tag ODE
$$
By making the substitution $y=\Phi'/\Phi$ and
$y'+y^2=\Phi''/\Phi$, we arrive to 
 $$\multline 
[(m-1)y^2+\lambda^2]y'+(m-1+q-\mu)(y^2+\lambda^2)^2\\
      +[\lambda(\tau+n-m)+(n-2)y\cot \o](y^2+\lambda^2)=0;
\quad \omega\in (-\frac{\omega_0}{2},\frac{\omega_0}{2}). \endmultline \tag3
$$
 Let us now verify that
  $$\gather \Phi(-\omega)=\Phi(\omega),\quad y(-\omega)=-y(\omega),\quad 
y'(-\omega)=y'(\omega), \\
            \Phi'(-\omega)=-\Phi'(\omega),\quad\forall
            \omega\in (-\frac{\omega_0}{2},\frac{\omega_0}{2})\,. \endgather$$
                                                                            
 Putting $\omega=0$ we obtain $y(0)=0$. Therefore, it is sufficient to consider
 the equation only on the interval $(0,\omega_0/2)$.
 Since $\cot \omega>0$ on $(0,\omega_0/2)$, from (3) and
 (*) it follows that
$$[(m-1)y^2+\lambda^2]y'+(n-2) y(y^2+\lambda^2)\cot \omega<0\,,
              \quad \omega\in (0,\frac{\omega_0}{2})\,. \tag4$$
 Let us solve the Cauchy problem
  $$\gather 
  [(m-1)\overline{y}^2+\lambda^2]\overline{y}'+(n-2)\overline{y}(\overline{y}^2
+\lambda^2)\cot \omega=0
           ,\quad \omega\in (0,\frac{\omega_0}{2});\\
          \overline{y}(0)=0\,. \endgather $$
 We obtain
  $$\int \frac{(m-1)\overline{y}^2+\lambda^2}{\overline{y}(\overline{y}^2+
\lambda^2)}d\overline{y}=
        -(n-2)\int \cot \omega d \omega + \text{const}$$
which implies 
 $$\gather \overline{y}(\overline{y}^2+\lambda^2)^{(m-2)/2}=C {\text{sin }}^{2-n}\omega \\
         \overline{y}(0)=0\,. \endgather $$
This, in turn, implies   $C=0$ and $\overline{y}\equiv 0$. 
  
Comparing the solution of (4) with that of the Cauchy
 problem, we deduce that $y(\omega)\le 0$.
 Since   $\cot \omega >0$ \text{and} $y\le0$ on our interval, by (3)
 we have
  $$ \multline [(m-1)y^2+\lambda^2]y'+[(m-1+q-\mu)(\lambda^2+y^2)+
  \lambda(\tau+n-m)](\lambda^2+y^2)\\
       = -(n-2) y(y^2+\lambda^2)\cot \omega\ge0,
              \quad \omega\in (0,\frac{\omega_0}{2})\,.\endmultline
  $$
Thus, for $(0,\omega_0/2)$ we have,
  $$ \gather
[(m-1)y^2+\lambda^2]y' \ge -[(m-1+q-\mu)(\lambda^2+y^2)+
\lambda(\tau+n-m)](\lambda^2+y^2) \\
y(0)=0\,. \endgather
$$
Similarly by the comparison theorem, we obtain $y(\omega)\ge z(\omega)$,
 where $z(\omega)$ with  $\omega \in (0,\omega_0/2)$ is the solution to 
 Cauchy problem
$$\gather
 [(m-1)z^2+\lambda^2]z'= -[(m-1+q-\mu)(\lambda^2+z^2)+
 \lambda(\tau+n-m)](\lambda^2+z^2),\\ 
 z(0)=0\,.\endgather$$
 On solving the latter, we obtain the expression for $z$ in the implicit form
$$\gathered 
\frac{\frac{m-1}{m-1+q-\mu}+\lambda \frac{m-2}{\tau+n-m}}
{\sqrt{\lambda^2+\lambda\frac{\tau+n-m}{m-1+q-\mu}}}
\arctan \frac{z}{\sqrt{\lambda^2+\lambda\frac{\tau+n-m}
{m-1+q-\mu}}}\\
 +\omega+\frac{m-2}{m-n-\tau}\arctan (\frac{z}{\lambda})=0\,.
                                                 \endgathered \tag5$$
 By combining the obtained results, we conclude that
       $$0\ge y(\omega)\ge z(\omega)\,.\tag6$$
 Let us now return to the equation for $y(\omega)$. On making the substitution
 $\varphi = \ln \Phi$, $w(\varphi)=y^2(\varphi)$, 
 $$
w'(\varphi)=2yy'(\varphi)=2y\frac{d \omega}{d 
\varphi}y'(\omega)=2y'(\omega)\,,$$ 
 we obtain
 $$\gather \frac{1}{2}[(m-1)w+\lambda^2]w'+[(m-1+q-\mu)(\lambda^2+w)+
 \lambda(\tau+n-m)](\lambda^2+w)\\
        -(n-2) \sqrt{ w}(w+\lambda^2)\cot \omega=0\,, \endgather $$
where we have used  $y=\pm\sqrt{ w}$ and $y<0$.
As we did above,  we obtain a differential inequality for $w$,
 $$ \frac{1}{2}[(m-1)w+\lambda^2]w'+[(m-1+q-\mu)(\lambda^2+w)+
 \lambda(\tau+n-m)](\lambda^2+w)>0\,.$$
 Integrating the respective differential equation
 $$ \frac{1}{2}[(m-1)\overline{w}+\lambda^2]\overline{w}'+
[(m-1+q-\mu)(\lambda^2+\overline{w})+
 \lambda(\tau+n-m)](\lambda^2+\overline{w})=0$$
 we obtain
 $$\multline
 \lambda\frac{m-2}{m-n-\tau}\ln (\lambda^2+\overline{w})\\
       +\left(\frac{m-1}{m-1+q-\mu}+
\lambda \frac{m-2}{\tau+n-m}\right)\ln ((m-1+q-\mu)(\lambda^2+\overline{w})+
\lambda(\tau+n-m))\\
         +2\ln \Phi=\ln C\,.\endmultline$$
 Solving the latter expression with the respect to $\Phi$ we obtain
 $$\gather 
 {\Phi}^2(\omega)=C^2\left(\frac{(m-1+q-\mu)(\lambda^2+\overline{w})
 +\lambda(\tau+n-m)}{\lambda^2+\overline{w}}\right)^
     {\lambda(m-2)/(m-n-\tau)}\times\\
  [(m-1+q-\mu)(\lambda^2+\overline{w})+\lambda(\tau+n-m)]^{-(m-1)/(m-1+q-\mu)}.
                                                 \endgather$$
 Now it is evident that $\overline{w}=z^2(\varphi)$ and $w=y^2(\varphi)$. 
 From (6) and $w\le \overline{w}$, it follows that
$$\Phi^2(\omega)=C^2(z^2+\lambda^2)^{\frac{1-m}{m-1+q-\mu}}\left(m-1+q-\mu+
 \frac{\lambda(\tau+n-m)}{(z^2+\lambda^2)}\right)^
 {\frac{\lambda(m-2)}{m-n-\tau}-\frac{m-1}{m-1+q-\mu}}\,.
 $$
Whence it follows that
 $${\Phi}(\omega)\sim |z|^{-\frac{(m-1)}{m-1+q-\mu}}
 \quad\text{as}\quad |z|\to+\infty\,.
 $$
Since $y^2\le z^2$, it follows that $1/z^2\le 1/y^2$, and it is now clear 
that 
$$\lim_{\omega\to(\omega_0/2)-0} z(\omega)=-\infty $$
(since $\Phi(\omega_0/2)=0)$. Furthermore, since $y=\frac{\Phi'}{\Phi}<0$ and 
 $\Phi>0$ on $(0,\omega_0/2)$,  $\Phi'<0$. i.e., $\Phi(\omega)$ decreases on 
 $(0,\omega_0/2)$
from the positive value $\Phi(0)$ to $\Phi(\omega_0/2)=0$. $\Phi$ does not
vanish anywhere else in $(0,\omega_0/2)$, otherwise it should increase
 somewhere. From this equation we have
$$\gather y'= -[(m-1+q-\mu)(y^2+\lambda^2)+\lambda(\tau+n-m)]\frac{(y^2+
 \lambda^2)}{(m-1)y^2+\lambda^2}\\
    -(n-2)y\frac{y^2+\lambda^2}{(m-1)y^2+\lambda^2}\cot \omega \to-\infty
     \quad\text{as}\quad y\to-\infty\,.\endgather
$$
That is to say $y(\omega)$ decreases in the vicinity of the point $-\omega$, 
when $y\to-\infty$. It is possible only at $-\omega=\omega_0/2$, (when passing
 $\omega\to\frac{\omega_0}{2}-0)$.
 On performing the passage to the  limit $\omega\to\frac{\omega_0}{2}-0$ 
in $(5)$, and taking into account  that $z\to-\infty$, we obtain
 $$\multline
 \frac{\omega_0}{\pi}+\frac{m-2}{\tau+n-m} \\
 =\left(\frac{m-1}{m-1+q-\mu}+\lambda \frac{m-2}{\tau+n-m}\right)
 \left(\frac{\lambda[\lambda(m-1+q-\mu)
+ \tau+n-m]}{m-1+q-\mu}\right)^{-1/2}\,.\endmultline$$

{\it Hence we obtain an explicit expression for $\lambda$,
$$\gathered\lambda=\frac{\pi}{2\omega_0(m-1+q-\mu)}\times\\
\left\{\frac{m(m-2)-2(m-2)t-t^2}{t+2(m-2)}+\frac{\sqrt{[t^2+2(m-2)t+m^2]
[t^2+2(m-2)t+(m-2)^2]}}{t+2(m-2)}\right\},\endgathered\tag7
$$
where $t=(\tau+n-m)\omega_0/\pi$.

 In the case of $n=2$, $\tau=0$ we obtain 
$$
\lambda=\frac{(m-1)}{(m-1-\mu+q)}+\frac{(\pi-\omega_0)[m(\pi-\omega_0)
+\sqrt{(m-2)^2(\pi-\omega_0)^2+4(m-1){\pi}^2} ]}
   {2\omega_0(2\pi-\omega_0)(m-1-\mu+q)}.\tag8
$$
 In the case of $\tau=\mu=q=0$, $n=2$ we get the result of [13].
If $m=n=2$,  from (7) we get
$$\lambda=\frac
{\sqrt{\left(\frac{2\pi}{\omega_0}\right)^2+\tau^2}-\tau}{2(1+q-\mu)}.\tag9
$$}
\medskip

For the case  $n=3$, we assume that $\tau=0$.
We shall seek a solution of the form 
$u=r^{\lambda}\Phi(\omega)\sin^{\lambda}\Phi$,
with $\Phi\in(0,\pi)$ and $\omega\in(-\omega_0/2,\omega_0/2)$.
Then we obtain for $\Phi(\omega)$ a problem which coincides with
(ODE) with $n=2$, and so for $\lambda$ we have the Expression (8).

\head The mixed boundary value problem \endhead

Now we consider the mixed boundary value problem in the planar
domain $G_0=\{(r,\omega)\mid r>0,\;0<\omega<\omega_0<\pi\}$, with a
corner boundary point,
  $$\gather
 \frac{d}{dx_i}\left({|u|}^q{|\nabla u|}^{m-2}u_{x_i}\right)=
 \mu {|u|}^{q-1}\text{sgn }{u}{|\nabla u|}^{m}, \quad x\in G_0\,,\\
 u\big|_{\omega=\omega_0}=0\,,\quad\left.{\frac{\partial u}{\partial x_2}}
 \right|_{\omega=0}=0\,,
  \endgather $$
where $\omega_0$ is the angle with the vertex at the point $O$.
By a process analogous to the one above, we come to the expression 
$$\multline\lambda=\frac{(m-1)}{(m-1-\mu+q)}\\
+\frac{(\pi-2\omega_0)[m(\pi-2\omega_0)+\sqrt{(m-2)^2(\pi-2\omega_0)^2+4(m-1)
{\pi}^2} ]}
   {8\omega_0(\pi-\omega_0)(m-1-\mu+q)}\,.\endmultline\tag10
$$
Obviously, this expression coincides with (8) for the  Dirichlet
problem, if in the latter we put $2\omega_0$ instead of $\omega_0$.
\medskip

{\it Therefore, barrier functions $w=r^\lambda \Phi(\omega)$ have been constructed
for the first boundary value problem for the equation (1), and 
for the mixed boundary value problem for (1) with $\tau=0$.}

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\ref \no 13  \by Borsuk M.V., Plesha M.I.
\paper Estimates of generalized solutions of
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\jour Ukrainer  Math. journal  \toappear \endref 
\endRefs
\enddocument