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\headline={\ifnum\pageno=1 \hfill\else%
{\tenrm\ifodd\pageno\rightheadline \else
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\def\rightheadline{EJDE--1994/04\hfil Non-autonomous EllipticBoundary 
Value Problems\hfil\folio} 
\def\leftheadline{\folio\hfil V. Anuradha, S. Dickens, and R. Shivaji
 \hfil EJDE--1994/04}

\def\pretitle{\vbox{\eightrm\noindent\baselineskip 9pt %
 Electronic Journal of Differential Equations\hfil\break
Vol. {\eightbf 1994}(1994), No. 04, pp. 1-10. Published July 8, 1994.\hfil\break
ISSN 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\hfil\break ftp (login: ftp) 147.26.103.110 or 129.120.3.113
\bigskip} }

\topmatter

\title Existence Results for Non-autonomous Elliptic Boundary Value
Problems \endtitle
\thanks \noindent
Submitted: January 23, 1994.\hfil\break
1991 AMS {\eighti Subject  Classification:} 35J65.\hfil\break 
Partially supported by NSF Grants DMS - 8905936 and DMS - 9215027.
\hfil\break
This work was completed while R. Shivaji was a Visiting Research 
Scientist at Georgia Institute of Technology in 1993.\hfil\break
\copyright 1994 Southwest Texas State University  and 
University of North Texas.
\endthanks

\author V. Anuradha, S. Dickens, and R. Shivaji \endauthor
\address V. Anuradha, Department of Mathematics and Statistics, University
of Arkansas at Little Rock, Little Rock, AR 72204-1099\endaddress
\address S. Dickens, Department of Mathematics and Statistics, Mississippi State
 University,
Mississippi State, MS  39762 \endaddress
\address R. Shivaji, Department of Mathematics and Statistics, Mississippi State
 University,
Mississippi State, MS  39762. E-mail rs1\@ra.msstate.edu \endaddress


\abstract 
We study solutions to the boundary value problems
$$-\Delta u(x) = \lambda f(x, u);\quad x \in \Omega$$


$$u(x) + \alpha(x) \frac{\partial u(x)}{\partial n} = 0;\quad x \in
\partial \Omega$$
where $\lambda > 0$, $\Omega$ is a bounded region in $\Bbb{R}^{N}$;
$N \geq 1$ with smooth
boundary $\partial \Omega$, $\alpha(x) \geq 0$, $n$ is the outward
unit normal, and $f$ is
a smooth function such that it has either sublinear or restricted
linear growth
in $u$ at infinity, uniformly in $x$.  We also consider $f$ such
that $f(x, u) u \leq 0$
uniformly in $x$, when $|u|$ is large.  Without requiring any sign
condition on $f(x, 0)$,
thus allowing for both positone as well as
semipositone structure, we discuss the existence of at least three
solutions
for given $\lambda \in (\lambda_{n}, \lambda_{n + 1})$
where $\lambda_{k}$ is the $k$-th eigenvalue of $-\Delta$ subject
to the above boundary conditions.  In
particular, one of the solutions we obtain has non-zero
positive part, while another has non-zero negative part.
We also discuss the existence of three solutions where one of them is
positive, while another is negative, for
$\lambda$ near $\lambda_{1}$, and for $\lambda$
large when $f$ is sublinear.  We use the method of sub-super
solutions to establish our
existence results.  We further discuss non-existence results
for $\lambda$ small.
\endabstract

\endtopmatter
\document

\subhead{1.\quad Introduction}\endsubhead
We first consider the boundary value problem 
$$-\Delta u(x) = \lambda f(x, u);\quad x \in \Omega \tag{1.1}$$
$$u(x) + \alpha(x) \frac{\partial u(x)}{\partial n} = 0;\quad x \in
\partial \Omega \tag{1.2}$$
where $\Delta$ is the Laplacian operator, $\Omega$ is a bounded
region in $\Bbb{R}^{N}$; $N \geq 1$ with smooth boundary $\partial
\Omega$, 
$\alpha(x) \geq 0$, $n$ is the unit outward normal, and $f$ is a
smooth function such that
$$\lim_{u \to \infty} \frac{f(x, u)}{u} = \sigma \text{ uniformly
in }\, x, \tag{1.3}$$
$$\lim_{u \to - \infty} \frac{f(x, u)}{u} = \beta \text{ uniformly in }\, x,
\tag{1.4}$$
and 
$$\frac{\partial f}{\partial u} (x, u) \geq 0. \tag{1.5}$$
Let $\lambda_{k}$, $\phi_{k}$ be the eigenvalues and corresponding 
eigenfunctions of the boundary value problem 
$$-\Delta \phi_{k} = \lambda_{k} \phi_{k};\quad x \in \Omega
\tag{1.6}$$
$$\phi_{k}(x) + \alpha(x) \frac{\partial \phi_{k}(x)}{\partial n}
= 0;\quad x \in \partial \Omega. \tag{1.7}$$
Let $I = [a, b] \subset (\lambda_{n}, \lambda_{n + 1})$ and for 
$\lambda \in I$ consider the unique solution $Z_{\lambda}$ of the 
boundary value problem 
$$-\Delta Z_{\lambda} - \lambda Z_{\lambda} = -1;\quad x \in \Omega
\tag{1.8}$$
$$Z_{\lambda} (x) + \alpha (x) \frac{\partial Z_{\lambda}
(x)}{\partial n} = 0.;\quad x \in \partial \Omega \tag{1.9}$$
Let $\mu_{1} = \inf_{x \in \bar{\Omega},\, \lambda \in I} Z_{\lambda}(x)$, $\mu_{2}
= \sup_{x \in \bar{\Omega},\, \lambda \in I} Z_{\lambda}(x)$ and 
$\mu = \max \{|\mu_{1}|, \mu_{2}\}$.  Note that 
$Z_{\lambda}^{+} \not\equiv 0$ (see Appendix 1).  Also, $\exists$
$\delta (\Omega) > 0$ such that if $I \subset (\lambda_{1},
\lambda_1 +\delta)$
then $Z_{\lambda} > 0$; $x \in \Omega$ (anti-maximum principle: see
Cl\'em\'ent and 
Peletier in [1]). 

Now assume that 
$$\exists\,\, m > 0 \text{ such that } y + m > f(x, y) > y - m \,\,\, \forall\,\, y \in [-bm\mu, b m\mu]
\tag{1.10}$$
and 
$$\beta, \sigma < \frac{1}{b||w_{\alpha}||_{\infty}} \tag{1.11}$$
where $w_{\alpha}$ is the unique positive solution to 
$$-\Delta w_{\alpha} = 1; \quad x \in \Omega \tag{1.12}$$
$$w_{\alpha}(x) + \alpha(x) \frac{\partial w_{\alpha}
(x)}{\partial n} = 0;\quad x \in \partial \Omega. \tag{1.13}$$

Then we prove:

\proclaim{Theorem 1.1}
Let $I = [a, b] \subset (\lambda_{n}, \lambda_{n + 1})$ and
(1.3)--(1.5), (1.10)--(1.11) 
hold.  Then these exists at least three solutions to (1.1)--(1.2)
for 
$\lambda \in I$.  One of these solutions is a non-negative
or sign changing solution, 
while another is a non-positive or sign changing solution.
\endproclaim

\remark{Remark 1.1}
Note that unlike in the literature of ``jumping nonlinearities"
(see [2]--[3]), our results do not require 
$(\beta, \sigma)$ to include a part of the spectrum of $-\Delta$, 
nor require, a part 
of the spectrum of $-\Delta$ to lie between $f'(0)$ and 
$f'(\pm\infty)$ for autonomous nonlinearities as in [4] and the reference within.
\endremark

\proclaim{Theorem 1.2} 
Let $I = [a, b] \subset (\lambda_1, \lambda_{1} + \delta)$ and
(1.3)--(1.5), 
(1.10)--(1.11) hold.  Then there exists at least three solutions to
(1.1)--(1.2) for 
$\lambda \in I$, where one is a positive solution while
another is a negative solution. 
\endproclaim

Next we consider the particular case when $\sigma = \beta = 0$
(sublinear case).  We further 
assume that there exists 
$f_{1}(u) < f(x, u); \forall\, x \in \bar{\Omega}, u \geq 0$
 such that 
$$f_{1}(r_{1})=0,\; f_{1}'(r_{1}) < 0,\; \int_{0}^{r_{1}} f(s)\, ds
 >0 \tag{1.14}$$
for some $r_{1} > 0$, and that there exists
$f_{2}(u) > f(x, u); \forall\; x \in \bar{\Omega}, u \leq 0$
such that
$$f_{2}(r_{2})=0,\; f_{2}'(r_{2}) < 0,\; \int_{r_{2}}^{0} f(s)\, ds
 <0 \tag{1.15}$$
for some $r_{2} < 0$.  Then we prove:

\proclaim{Theorem 1.3}
Assume (1.14)--(1.15) and let (1.3)--(1.5) hold with $\sigma = \beta
= 0$.  Then there exists at least three solutions to (1.1)--(1.2) 
for $\lambda$ large, where one is a positive solution while
another is a negative solution. 
\endproclaim 

\remark{Remark 1.2}
We refer to [4] where a positive solution for $\lambda \in I = [a,
b] \subset (\lambda_{1}, \lambda_{1} + \delta)$ 
was discussed in the case of \underbar{autonomous},
\underbar{sublinear} $(\sigma = 0)$, and 
\underbar{semipositone} $(f(0) < 0)$ problems with
\underbar{Dirichlet} boundary conditions. 
In [5] it was assumed that there exists $m > 0$ such that $f(y)
\geq y - m$ $\forall y \in [0, bm\mu]$ to obtain a positive solution for
$\lambda \in I$.  Further by 
constructing a function $f_{1}(u)$ satisfying (1.14) a positive
solution for $\lambda$ large was discussed. 
See also [6] where the authors study the existence of a positive
solution via
\underbar{degree theory} arguments.  Hence they allow non-autonomous problems
with Robin boundary condition, but consider 
only the \underbar{sublinear} case, and study the existence of
positive solutions 
for \underbar{$\lambda$ large} with the assumption $f(x, 0) \leq
0$ $\forall x \in \bar{\Omega}$. 
\endremark 

We also consider $f$ such that 
$$\exists\,\, r > 0 \text{ for which } f(x, r) \leq 0 \text{ while } f(x, -r) \geq 0 \text{ for every } x \in \bar{\Omega}.
\tag{1.16}$$

Then we prove:

\proclaim{Theorem 1.4} 
Let $I = [a, b] \subset (\lambda_{n}, \lambda_{n + 1})$ and (1.10),
(1.16) 
hold.  Assume $r \geq bm\mu$.  Then there exists at least three
solutions to 
(1.1)--(1.2) for $\lambda \in I$.
One of these solutions is a non-negative or sign changing solution,
while another is a non-positive or sign changing solution. 
\endproclaim


\proclaim{Theorem 1.5}
Let $I = [a, b] \subset (\lambda_{1}, \lambda_{1} + \delta)$ and
(1.10), (1.16) 
hold.  Assume $r \geq bm\mu$.  Then there exists at least three
solutions 
to (1.1)--(1.2) for $\lambda \in I$.  One of these is 
a positive solution, while another is a negative solution. 
\endproclaim


\proclaim{Theorem 1.6}
Let (1.16) hold, and assume that (1.14) holds for all $0 \leq u \leq r$ with
$r_{1} \leq r$ while (1.15) holds for all $-r \leq u \leq 0$ with 
$r_{2} \geq -r$.  Then there exists at least three solutions to (1.1)--(1.2) 
for $\lambda$ large, where one is a positive solution while another is a 
negative solution. 
\endproclaim


\remark{Remark 1.3}
Consider the boundary value problem 
$$-\Delta u(x) = \lambda f(||x||, u);\quad x \in B_{N}$$
$$u(x) + \alpha \frac{\partial u(x)}{\partial n} = 0;\quad x \in
\partial B_{N}$$
where $\alpha \geq 0$ is a constant and $B_{N}$ is the unit ball in
$\Bbb{R}^{N}$.  Then under 
the corresponding hypothesis on $f(||x||, u)$ instead of $f(x, u)$,
one can generate all the solutions obtained in Theorems 1.1--1.6 to
be \underbar{radial}. 
This follows from the fact that all the sub and super solutions we
will 
use in the proofs of these theorems will  turn out to be radial. 
\endremark 

We next discuss non-existence results under the assumption that
there exists 
$\gamma > 0$ such that
$$\frac{\partial f}{\partial u} \leq \gamma. \tag{1.17}$$
We recall that $\lambda_{1}$ is the principal eigenvalue and 
$\phi_1 > 0$ is a corresponding eigenfunction of $-\Delta$ subject to
boundary Robin 
conditions (1.2). 
Then we prove:

\proclaim{Theorem 1.7}
Assume $\int_{\Omega} f(x, 0) \phi_1 (x) \leq 0$ and $\lambda <
\lambda_{1}/\gamma$.  
If $u(\not\equiv 0)$ is any solution of (1.1)--(1.2) then $u^{-}
\not\equiv 0$ (i.e. there
does not exist any solution $u(\not\equiv 0)$ which is
non-negative).
\endproclaim

\proclaim{Theorem 1.8} 
Assume $\int_{\Omega} f(x, 0) \phi_1(x) \geq 0$ and $\lambda <
\lambda_{1}/\gamma$. 
If $u(\not\equiv 0)$ is any solution of (1.1)--(1.2) then $u^{+}
\not\equiv 0$ 
(i.e. there does not exist any solution $u(\not\equiv 0)$ which is
non-positive). 
\endproclaim

\proclaim{Theorem 1.9} 
Assume $f(x, 0) \leq 0$  and $\lambda < \lambda_{1}/\gamma$.  If $u$
is any solution of (1.1)--(1.2) then $u \leq 0$. 
\endproclaim 

\proclaim{Theorem 1.10} 
Assume $f(x, 0) \geq 0$ and $\lambda < \lambda_{1}/\gamma$.  If $u$
is any 
solution of (1.1)--(1.2) then $u \geq 0$. \endproclaim

We give detailed proofs of our results in section 2. 
For literature on autonomous semipositone problems with 
Dirichlet boundary conditions see [7], while for the positone case 
see [8] and the references cited in [8].

Our existence results are based on sub-super solutions.
Namely, a \underbar{super solution} is defined as a smooth
function $\phi$ such that 
$$-\Delta \phi \geq \lambda f(x, \phi);\quad x \in \Omega
\tag{1.18}$$
$$\phi(x) + \alpha(x) \frac{\partial \phi(x)}{\partial n} \geq
0;\quad x \in \partial \Omega \tag{1.19}$$
and a \underbar{subsolution} is a smooth function $\psi$ that
satisfies (1.18)--(1.19) with the inequalities reversed.  If $\psi
\leq \phi$, 
then it follows that (1.1)--(1.2) has a solution $u$ such that
$\psi \leq u \leq \phi$ (see [9]--[10]).  

Further if $\psi_{1}$ is a subsolution, $\psi_{2}$ is a strict 
subsolution, $\phi_{1}$ is a strict supersolution and $\phi_{2}$ is
a supersolution such that $\psi_{1} \leq \psi_{2} \leq \phi_{2}$, $\psi_{1} \leq
\phi_{1} \leq \phi_{2}$, 
$\psi_{2}(x_{0}) > \phi_{1}(x_{0})$ for some $x_{0} \in \bar{\Omega}$,
then (1.1)--(1.2) has at least \underbar{three} 
distinct solutions $u_{1}$, $u_{2}$, $u_{3}$ such that $u_{1} \leq
u_{2} \leq u_{3}$.  See [11] for this
multiplicity result which 
was proved for Dirichlet boundary conditions.  However, it is easy
to see that this result holds for the Robin boundary conditions
(1.2) as well. 

\subhead{2.\quad Proofs of Theorems 1.1--1.10}\endsubhead

\demo{\bf Proof of Theorem 1.1}
Let $v_{1}(x) := bmZ_{\lambda}(x)$ and $u_{2}(x) = -v_{1}(x)$. 
Then 
$-\Delta v_{1} = bm(-\Delta Z_{\lambda}) = bm (\lambda Z_{\lambda}
- 1)$
$\leq bm(\lambda Z_{\lambda} - \lambda/b) (\text{ since } \lambda \leq b)
= \lambda [bm Z_{\lambda} - m]$ $= \lambda [v_{1} - m] < 
\lambda f(x, v_{1})$ by (1.10), and $-\Delta u_{2} = \Delta v_{1}
\geq \lambda [-v_{1} + m] = \lambda[u_{2} + m]$ 
$> \lambda f(x, u_{2})$ again by (1.10).  Thus $v_{1}$ is a strict 
subsolution while $u_{2}$ is a strict supersolution.  Note that 
$v_{1}^{+} \not \equiv 0$ and $u_{2}^{-} \not \equiv 0$.  Now let 
$u_{1} (x) := J w_{\alpha}(x)$ where $J > 0$ is large enough so
that 
$$\frac{1}{b||w_{\alpha}||_{\infty}} \geq \frac{f(x,
J||w_{\alpha}||_{\infty})}{J||w_{\alpha}||_{\infty}} \tag{2.1}$$
and 
$$u_{1} \geq v_{1},\quad u_{1} \geq u_{2}. \tag{2.2}$$
Here (2.1) is possible since $\sigma < \frac{1}{||w_{\alpha}||_{\infty}b}$,
and (2.2) is possible 
by the Hopf's maximum principle.  Then $-\Delta u_{1} = J \geq
\lambda f(x, J||w_{\alpha}||_{\infty}); \,\, x \in \Omega$ (by
(2.1) since 
$\lambda \leq b)$ $\geq \lambda f(x, J w_{\alpha})$ (since
$\frac{\partial f}{\partial u} \geq 0) = \lambda f(x, u_{1})$. 
Next 
let $v_{2}(x) = - \tilde{J} w_{\alpha} (x)$ where $\tilde{J} > 0$ is
large enough so that 
$$\frac{1}{b||w_{\alpha}||_{\infty}} \geq \frac{f(x, - \tilde{J}
||w_{\alpha}||_{\infty})}{- \tilde{J} ||w_{\alpha}||_{\infty}} \tag{2.3}$$
and
$$v_{2} \leq v_{1},\quad v_{2} \leq u_{2}. \tag{2.4}$$
Here again (2.3) is possible since $\beta <
\frac{1}{||w_{\alpha}||_{\infty}b}$ and 
(2.4) is possible by the Hopf's maximum principle.  Then 
$-\Delta v_{2} = -\tilde{J} \leq \lambda f(x,
-\tilde{J}||w_{\alpha}||_{\infty});\quad x \in \Omega$ 
(by (2.3) since $\lambda \leq b) \leq \lambda f(x, -
\tilde{J} w_{\alpha})$ 
(since $\frac{\partial f}{\partial u} \geq 0) = \lambda f(x, v_{2})$. 
Thus $u_{1}$ is a supersolution 
while $v_{2}$ is a subsolution such that $v_{2} \leq v_{1} \leq
u_{1}$ and 
$v_{2} \leq u_{2} \leq u_{1}$, where $v_{1}$ is a strict
subsolution, $u_{2}$ 
is a strict supersolution with $v_{2} \leq 0$, $v_{1}^{+} \not
\equiv 0$, $u_{2}^{-} \not \equiv 0$ and $u_{1} \geq 0$.  Hence the result.
\qed\enddemo

\demo{\bf Proof of Theorem 1.2} 
If $\lambda \in (\lambda_{1}, \lambda_{1} + \delta)$ then
$Z_{\lambda} > 0$; $x \in \Omega$.  
Then $v_{1}(x) > 0$; $x \in \Omega$ while $u_{2} < 0$; $x \in
\Omega$ and hence the 
result follows by the proof in Theorem~1.1. 
\qed\enddemo

\demo{\bf Proof of Theorem 1.3} 
Consider the autonomous Dirichlet problem 
$$-\Delta v = \lambda f_{1}(v);\quad x \in \Omega \tag{2.5}$$
$$v = 0;\quad x \in \partial \Omega. \tag{2.6}$$

Then by (1.14) it follows that there exists $\bar{\lambda}_{1} > 0$
such that 
for $\lambda \geq \bar{\lambda}_{1}$ (2.5)-(2.6) has a positive
solution $v_{\lambda}$ 
(see [12]).  Clearly since $\frac{\partial v_{\lambda}}{\partial n}
\leq 0$; $x \in \partial \Omega$ and 
$f_{1}(S) < f(x, S)$; $\forall x \in \bar{\Omega}$, $S \geq 0$
$v_{\lambda}$ is a strict subsolution to (1.1)--(1.2) for 
$\lambda \geq \bar{\lambda}_{1}$. 

Next consider 
$$-\Delta u = \lambda f_{2} (u);\quad x \in \Omega \tag{2.7}$$
$$u = 0;\quad x \in \partial \Omega. \tag{2.8}$$
Then setting $w = -u$ we see that $w$ satisfies 
$$-\Delta w = \lambda [-f_{2}(-w)];\quad x \in \Omega \tag{2.9}$$
$$w = 0;\quad x \in \partial \Omega. \tag{2.10}$$ 

Let $g(w) = -f_{2}(-w)$.  Then $g(-r_{2}) = 0$, 
$g'(-r_{2}) = f_{2}'(r_{2}) < 0$ and $\int_{0}^{-r_{2}} g(s) ds$ 
$= \int_{0}^{-r_{2}} - f_{2}(-s) ds = -\int_{\gamma_{2}}^{0}
f_{2}(s) ds > 0$.  Hence 
again by [12], there exists $\bar{\lambda}_{2} > 0$ such that for 
$\lambda \geq \bar{\lambda_{2}}$, (2.9)--(2.10) has a positive
solution $w_{\lambda}$.  
Equivalently, (2.7)--(2.8) has a negative solution 
$u_{\lambda} = -w_{\lambda}$ for $\lambda \geq \bar{\lambda}_{2}$. 
Also since 
$\frac{\partial u_{\lambda}}{\partial n} \geq 0$;  $x \in \partial \Omega$
and $f_{2}(S) > f(x, S)$; $\forall x \in \bar{\Omega}$, $S \geq 0$,
$u_{\lambda}$ is a strict super solution to (1.1)--(1.2) for 
$\lambda \geq \bar{\lambda_{2}}$.  Let $\bar{\lambda} = \max
\{\bar{\lambda_{1}}, \bar{\lambda}_{2}\}$ and 
$\lambda > \bar{\lambda}$ be fixed.  Consider $u_{1}(x) := J
w_{\alpha} (x)$
where $J > 0$ is large enough so that 
$$\frac{1}{\lambda ||w_{\alpha}||_{\infty}} \geq \frac{f(x,
J||w_{\alpha}||_{\infty})}{J||w_{\alpha}||_{\infty}} \tag{2.11}$$
and
$$u_{1} \geq v_{\lambda}. \tag{2.12}$$
Here (2.11) is possible since $\sigma = 0$ and (2.12) is possible
by the Hopf's maximum 
principle.  Then $-\Delta u_{1} = J \geq \lambda f(x,
J||w_{\alpha}||_{\infty}) \geq \lambda f(x, Jw_{\alpha})$ (since
$\frac{\partial f}{\partial u} \geq 0) = \lambda f(x, u_{1})$.  
Next consider $v_{2}(x) := - \tilde{J} w_{\alpha} (x)$ when $\tilde{J}
> 0$ is large enough so 
that 
$$\frac{1}{\lambda ||w_{\alpha}||_{\infty}} \geq \frac{f(x, -
\tilde{J} ||w_{\alpha}||_{\infty})}{-\tilde{J} ||w_{\alpha}||_{\infty}}
\tag{2.13}$$
and
$$v_{2} \leq u_{\lambda}. \tag{2.14}$$
Here again (2.13) is possible since $\beta = 0$ and (2.14) is
possible 
by the Hopf's maximum principle.  Then $-\Delta v_{2} = - \tilde{J}
\leq \lambda f(x, -\tilde{J}||w_{\alpha}||_{\infty})$               
$\leq \lambda f(x_{1}, - \tilde{J} w_{\alpha})$ (since $\frac{\partial
f}{\partial u} \geq 0) = \lambda f(x, v_{2})$.  
Thus $u_{1}$ is a supersolution while $v_{2}$ is a subsolution such
that 
$v_{2} \leq u_{\lambda} \leq 0 \leq v_{\lambda} \leq u_{1}$, where 
$u_{\lambda}$ is a strict supersolution and $v_{\lambda}$ is a
strict subsolution.  Hence the result. 
\qed\enddemo 

\demo{\bf Proof of Theorem 1.4} 
Let $v_{1}(x)$ and $u_{2}(x)$ be the strict sub and strict super
solutions to (1.1)--(1.2) as in the 
proof of Theorem 1.1.  Then since $r \geq bm\mu$, both $v_{1}(x)$
and $u_{2}(x)$ satisfy $-r \leq v_{1}(x) \leq r$, $-r \leq u_{2}(x)
\leq r$.  But 
by (1.16) $u_1(x) = r$ and $v_{2}(x) \equiv -r$ are 
super and sub solutions respectively to (1.1)--(1.2).  Hence the
result. 
\qed\enddemo 

\demo{\bf Proof of Theorem 1.5} 
Let $v_{1}(x)$ and $u_{2}(x)$ be as in the proof of Theorem 1.4. 
But $\lambda \in (\lambda_{1}, \lambda_{1} + \delta)$.  Hence 
$v_{1}(x) \geq 0$ while $u_{2}(x) \leq 0$.  The rest of the proof
is 
identical to the proof of Theorem 1.4. 
\qed\enddemo

\demo{\bf Proof of Theorem 1.6}
Let $u_{\lambda}$ and $v_{\lambda}$ be respectively the strict super and strict
subsolutions to (1.1)--(1.2) as in the proof of Theorem 1.3.  But $-r_{2} \leq u_{\lambda} \leq 0 \leq v_{\lambda} \leq r_{1}$ 
(see [12]) while $u_{1}(x) \equiv r$ are $v_{2}(x) = -r$ are 
super and subsolutions respectively to (1.1)--(1.2).  Hence the result.
\qed\enddemo


\demo{\bf Proof of Theorem 1.7} 
Assume $u \geq 0$, $u \not \equiv 0$.  Then 
$$\align -\Delta u &= \lambda f(x, u)\\
&= \lambda [f(x, u) - f(x, 0)] + \lambda f(x, 0)\\
&= \lambda \frac{\partial f}{\partial u} (x, \eta) u + \lambda f(x,
0) (\text{ where } 0 \leq \eta \leq u)\\
&\leq \lambda \gamma u + \lambda f(x, 0) (\text{ by } (1.17)).\endalign$$
Thus $u$ satisfies 
$$-\Delta u - \lambda \gamma u \leq \lambda f(x, 0);\quad x \in
\Omega \tag{2.15}$$
$$u(x) + \alpha(x) \frac{\partial u(x)}{\partial n} = 0;\quad x \in \partial
\Omega. \tag{2.16}$$
Multiplying (2.15) by $\phi_1$ and integrating we obtain 
$$\int_{\Omega} - \Delta u \phi_1\, dx - \int_{\Omega} \lambda \gamma
u \phi_1\, dx \leq \int_{\Omega} \lambda f(x, 0) \phi_1\, dx \leq 0.$$
Applying Green's second identity we obtain 
$$\int_{\partial \Omega} \{-\phi_1 \frac{\partial u}{\partial n} + u
\frac{\partial \phi_1}{\partial n} \}\, ds + \int_{\Omega} u
\lambda_{1} \phi_1\, dx - \int_{\Omega} \lambda \gamma u \phi_1\, dx \leq
0.$$
But applying the boundary conditions we see that 
$$\int_{\partial \Omega} \{-\phi_1 \frac{\partial u}{\partial n} + u
\frac{\partial \phi_1}{\partial n}\}\, ds = 0.$$
Thus we obtain 
$$\int_{\Omega} u \phi_1 (\lambda_{1} - \lambda \gamma)\, dx \leq 0.$$
But $u \geq 0$, $\phi_1 > 0$ in $\Omega$ and this is a
contradiction if $\lambda < \lambda_{1}/\gamma$.  Hence 
the result. 
\qed\enddemo 

\demo{\bf Proof of Theorem 1.8} 
Assume $u \leq 0$, $u \not \equiv 0$.  Proceeding as in the proof
of Theorem 1.7 we obtain 
$$\align -\Delta u &= \lambda \frac{\partial f}{\partial u} (x, z)
u + \lambda f(x, 0)\\
& \geq \lambda \gamma u + \lambda f(x, 0) (\text{ by } (1.17)). \endalign$$
Thus $u$ satisfies 
$$-\Delta u - \lambda \gamma u \geq \lambda f(x, 0);\quad x \in
\Omega \tag{2.17}$$
$$u(x) + \alpha(x) \frac{\partial u(x)}{\partial n} = 0;\quad x \in \partial
\Omega. \tag{2.18}$$
Multiplying (2.17) by $\phi_1$, using $\int_{\Omega} f(x, 0) \phi_1 dx
\geq 0$, and proceeding 
as in the proof of Theorem 1.7 we obtain 
$$\int_{\Omega} u\phi_1 (\lambda_{1} - \lambda \gamma)\, dx \geq 0.$$
But $u \leq 0$ while $\phi_1 > 0$ for $x \in \Omega$, thus this is a
contradiction if $\lambda < \lambda_{1}/\gamma$. 
Hence the result. 
\qed\enddemo

\demo{\bf Proof of Theorem 1.9} 
Suppose $u > 0$ somewhere in $\bar{\Omega}$.  Then $\exists$ some
$\Omega_{1} \subseteq \Omega$ such that 
$u > 0$ in $\Omega$, and either 
$$u(x) = 0 \text{ or } u(x) + \alpha(x) \frac{\partial u(x)}{\partial n} =
0;\quad x \in \partial \Omega. \tag{2.19}$$
Let $\tilde{\lambda}_{1}$ be the principal eigenvalue and 
$\tilde{\phi}_1 > 0$ be a corresponding eigenfunction of $-\Delta$ on
the region $\Omega_{1}$ subject to Robin conditions (1.2) on $\partial \Omega_1$.  
Note that $\tilde{\lambda}_{1} \geq \lambda_{1}$.  Now since $f(x, 0)
\leq 0$ we have 
$\int_{\Omega} f(x, 0) \tilde{\phi}_1\, dx \leq 0$.  Thus following 
the steps in the proof of Theorem 1.7 we obtain 
$$\int_{\Omega_{1}} - \Delta u \tilde{\phi}_1\, dx - \int_{\Omega_{1}}
\lambda \gamma u \tilde{\phi}_1\, dx \leq 0$$ 
and then 
$$\int_{\partial \Omega_{1}} [-\tilde{\phi}_1 \frac{\partial
u}{\partial n} + u \frac{\partial \tilde{\phi}_1}{\partial n}]\, ds +
\int_{\Omega_{1}} [\tilde{\lambda}_{1} - \lambda \gamma] u \tilde{\phi}_1
\, dx \leq 0.$$
But if $u = 0$ on $\partial \Omega_{1}$, then $\frac{\partial
u}{\partial n} \leq 0$ on 
$\partial \Omega_{1}$, while if $u + \alpha \frac{\partial
u}{\partial n} = 0$ on $\partial \Omega_{1}$, since 
$\tilde{\phi}_1 + \alpha \frac{\partial \tilde{\phi}_1}{\partial n} = 0$, we
have 
$-\tilde{\phi}_1 \frac{\partial u}{\partial n} + u \frac{\partial
\tilde{\phi}_1}{\partial n} = 0$ on 
$\partial \Omega_{1}$.  Thus in any case (see (2.19)) $-\tilde{\phi}_1
\frac{\partial u}{\partial n} 
+ u \frac{\partial \tilde{\phi}_1}{\partial n} \geq 0$.  Hence 
$$\int_{\Omega_{1}} [\tilde{\lambda}_{1} - \lambda \gamma] u
\tilde{\phi}_1\, dx \leq 0$$
which is a contradiction since $u > 0$, $\tilde{\phi}_1 > 0$ for 
$x \in  \Omega_{1}$ while $\lambda < \lambda_{1}/\gamma \leq
\tilde{\lambda_{1}}/\gamma$. 
Hence the result. 
\qed\enddemo

\demo{\bf Proof of Theorem 1.10} 
Suppose $u < 0$ somewhere in $\bar{\Omega}$.  Then 
$\exists$ some $\Omega_{1} \subseteq \Omega$ such that $u < 0$ in
$\Omega_{1}$ and either 
$$u(x) = 0 \text{ or } u(x) + \alpha(x) \frac{\partial u(x)}{\partial n} =
0;\quad x \in \partial \Omega_{1} \tag{2.20}$$
Let $\tilde{\lambda}_{1}$ and $\tilde{\phi}_1 > 0$ be as in the proof of
Theorem 1.9.  Now $f(x, 0) \geq 0$ and 
hence $\int_{\Omega_{1}} f(x, 0) \tilde{\phi}_1 dx \geq 0$.  Thus following
the steps in the 
proof of Theorem 1.8 we obtain 
$$\int_{\Omega_{1}} - \Delta u \tilde{\phi}_1\, dx - \int_{\Omega_{1}}
\lambda \gamma u \tilde{\phi}_1\, dx \geq 0,$$
and then 
$$\int_{\partial \Omega} [-\tilde{\phi}_1 \frac{\partial u}{\partial n}
+ u \frac{\partial \tilde{\phi}_1}{\partial n}]\, ds + \int_{\Omega_1}
[\tilde{\lambda}_{1} - \lambda \gamma] u \tilde{\phi}_1\, dx \geq 0.$$
But if $u = 0$ on $\partial \Omega_{1}$, then $\frac{\partial
u}{\partial n} \geq 0$ on $\partial \Omega_{1}$, 
while if $u + \alpha \frac{\partial u}{\partial n} = 0$ on
$\partial \Omega_{1}$, since
$\tilde{\phi}_1 + \alpha \frac{\partial \tilde{\phi}_1}{\partial n} = 0$ on 
$\partial \Omega$, we have $-\tilde{\phi}_1 \frac{\partial u}{\partial n}
+ u \frac{\partial \tilde{\phi}_1}{\partial n} = 0$ on $\partial
\Omega_{1}$.  Thus in any case (see (2.20))
$-\tilde{\phi}_1 \frac{\partial u}{\partial n} + u \frac{\partial
\tilde{\phi}_1}{\partial n} \leq 0$. 
Hence
$$\int_{\Omega_{1}} [\tilde{\lambda}_{1} - \lambda \gamma] u \tilde{\phi}_1
\, dx \geq 0$$ 
which is a contradiction since $u < 0$, $\tilde{\phi}_1 > 0$ for $x \in
\Omega$, while 
$\lambda < \lambda_1/\gamma \leq \tilde{\lambda}_{1}/\gamma$.  Hence
the result. 
\qed\enddemo 

\subhead{Appendix 1} \endsubhead
Let $\lambda \in (\lambda_{n}, \lambda_{n + 1})$ and consider the 
unique solution to the boundary value problem 
$$-\Delta Z_{\lambda} - \lambda Z_{\lambda} = -1;\quad x \in
\Omega$$
$$Z_{\lambda}(x) + \alpha(x) \frac{\partial Z_{\lambda}(x)}{\partial n} =
0;\quad x \in \partial \Omega.$$
Let $\phi_{1} > 0$ satisfy (1.6)--(1.7) for $k = 1$.  Then
$$\int_{\Omega} - \Delta Z_{\lambda} \phi_{1}\, dx - \int_{\Omega}
\lambda Z_{\lambda} \phi_{1}\, dx = \int_{\Omega} - \phi_{1}\, dx$$
which implies 
$$\int_{\partial \Omega} \{-\phi_{1} \frac{\partial
Z_{\lambda}}{\partial n} + Z_{\lambda} \frac{\partial
\phi_{1}}{\partial n}\}\, ds + \int_{\Omega} Z_{\lambda} \lambda_{1}
\phi_1\, dx
- \int_{\Omega} \lambda Z_{\lambda}\phi_{1} dx = \int_{\Omega} -
\phi_{1}\, dx\,.$$
But $$\int_{\partial \Omega} \{-\phi_1 \frac{\partial
Z_{\lambda}}{\partial n} + Z_{\lambda} \frac{\partial
\phi_{1}}{\partial n}\}\, ds = \int_{\partial \Omega} \{\alpha
\frac{\partial \phi_1}{\partial n} \frac{\partial
Z_{\lambda}}{\partial n} - \alpha \frac{\partial
Z_{\lambda}}{\partial n} \frac{\partial \phi_{1}}{\partial n}\}
\, ds = 0\,.$$
Thus 
$$\int_{\Omega} (\lambda_{1} - \lambda) Z_{\lambda} \phi_{1}\, dx =
\int_{\Omega} - \phi_{1}\, dx.$$
But $\lambda > \lambda_1$ and $\phi_1 > 0$ for $x \in \Omega$. 
Hence $Z_{\lambda}^{+} \not \equiv 0$. 

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\enddocument
\bye