n/2,$ symmetric in $x_1$.
Assume $p$ is nonincreasing in $x_1$ for $x_1\ge 0$ and: $g\in C^{1+\theta
}_{\ell\text{oc}},\;g(\xi )>0$ for $\xi \ge 0$.
\item{\rm (a)} For $\lambda $ small enough, Problem~1 has a
bounded (i.e. $L^\infty )$
positive solution.
\item{\rm (b)} All bounded positive solutions
to {\rm (1)} are symmetric with respect to $x_1,$ and
if moreover $p\in L^{n+\varepsilon }(\Omega )$ then
$\pd {u}{x_1} <0$ for
$0__0,$ with $\Vert
P_\lambda \Vert _{L^{\frac{n+\varepsilon }{2}}(\Omega ^*_\lambda )}$ bounded
for $\lambda \in [\lambda _0,\lambda _1)$.
Note that $\mu _0(\lambda )$ -- the least eigenvalue of $\ell_\lambda $ --
is then continuous in $\lambda ,$
and $p_\lambda (y) \equiv P_\lambda (y)$.
The results in Section~1 then
yield:
\proclaim {Theorem 5} Let {\rm (I), (II)} hold. Then:
\item {\rm (a)} $\mu _0(\lambda )>0$ for $\lambda _0<\lambda <\lambda _1,$
and $u>v_\lambda $ in $\Omega _\lambda $.
\item {\rm (b)} If $\Omega _{\lambda _0} = \Omega \cap \{x_1<\lambda _0\}$
then $u\equiv v_{\lambda _0}$ on $\Omega _{\lambda _0}$.
\item {\rm (c)} If $x_0\in \Omega \cap \{x_1=\lambda \}, \quad\pd
{u}{x_1}\;(x_0)$ exists and $P_\lambda \in L^{n+\varepsilon }(\Omega
^*_\lambda ),$
then $\pd{u}{x_1}\;(x_0) <0$.
\endproclaim
\demo {Proof} We apply the earlier results using $\omega =u-v_\lambda $
in Theorem~1. We first show
that
Theorem~1-d can be used to conclude that $\mu _0(\lambda )>0$.
Specifically, assume otherwise
i.e. $\mu _0(\lambda ^\pr)=0, \; \mu _0(\lambda )>0$ for $\lambda >\lambda
^\pr,$ for some $\lambda ^\pr\in(\lambda _0,\lambda _1),$
and note that $\omega ^-\in H^{1,2}_0(\Omega _{\lambda ^\pr})$ and hence
the min.-max. principle shows $\omega
^-=0,$
since, by continuity, if $\omega ^-\ne 0$ then $(u-v_\lambda )^-\ne 0$ for
some $\lambda >\lambda ^\pr$ and thus $\mu _0(\lambda )\le 0$.
Assume now that
$\omega \in H^{1,2}_0(\wt\Omega _{\lambda^\pr})$.
Choose a
small ball $B\subset \wt\Omega _{\lambda ^\pr}$ and look at the cylinder
$Z=(-a,\wt x_1)\times S$ where: $(\wt x_1,\ol {y^*})$ is the center of $B$
and $S=\{\ol y\vert (\wt x_1,\ol y)\in B\}$.
For notational convenience denote $\lambda ^\pr$ by $\lambda $ henceforth.
Since $v_\lambda $ can be approximated in
$H^{1,2}$ by functions which vanish near $Z\cap \partial\wt\Omega _\lambda$
and $\omega \in H^{1,2}_0,$
then this is also true of $u. $ We may assume $u$ admits in $Z-\wt \Omega
_\lambda $ a trivial extension (also denoted by $u)$ and thus
if $a$ is large enough $u\equiv 0$ on $(-a)\times S$.
Now by a fundamental result employing Fubini's Theorem, [24], there exists
a $\ol y_1\in S$ such that $u(x_1,\ol y_1) = \int^{x_1}_{-a} \;\pd{u(\xi
,\ol y_1)}{x_1}\; d\xi $ for almost all $x_1$.
Since $u$ is continuous in
$Z\cap \Omega _\lambda $ and in $Z-Z\cap \ol \Omega _\lambda $ and clearly
so is the integral, we conclude that equality must actually hold in these
regions. We thus have that $\int^{x_1}_{-a} \;\pd{u}{x_1}\;(\xi ,\ol
y_1)d\xi =0$ if $(x_1,\ol y_1) \in Z-Z\cap \ol\Omega _\lambda $ and
$\int^{x_1}_{-a}\;\pd{u}{x_1}\;(\xi ,\ol y_1)d\xi =u(x_1,\ol y_1)\ge \delta
>0$ if $(x_1,\ol y_1)\in Z\cap K, $ if $K\Subset \Omega ,$ by the positivity of the
solution $u$ in the compacta of $\Omega $.
Now let $\alpha $ be the least number such that we have $C=\{(x_1,\ol
y_1)\vert \alpha 0$ small
enough, and we conclude that $(\alpha ,\ol y_1)\in \Omega $. I.e. $\ol
C\subset \Omega $ and thus $u>\delta >0$ in $\ol C$.
This contradicts the
absolute continuity of the integral
and it follows that $\omega \notin H^{1,2}_0(\wt\Omega _{\lambda ^\pr})$
and thus $\mu _0(\lambda ^\pr)>0$ by Theorem~1-d.
Since $\mu _0(\lambda )>0$ and
$(u-v_\lambda )^- \in H^{1,2}_0(\Omega _\lambda ),$ then the min.-max. principle
again implies $(u-v_\lambda )^-=0,$ whence $u\ge v_\lambda $. The earlier arguments
show that $u\not\equiv v_\lambda $ in $\Omega _\lambda ,$ and then
$u>v_\lambda $.
This shows part (a).
As for part (b), we have $u(x)\ge u(x^{\lambda _0})$ in $\Omega
_{\lambda _0}$ by continuity, since $u>v_\lambda $ on $\Omega _\lambda
,\;\lambda >\lambda _0$. However $\Omega $ is symmetric in this case
about $x_1=\lambda _0$. If we first perform a reflection about
$x_1=\lambda _0,$ and then repeat the above procedure we would conclude
$u(x^{\lambda _0})\ge u\big((x^{\lambda _0})^{\lambda _0}\big) = u(x)$ and
the result.
Finally, for part (c), note that $x_0\in \Omega $
and $P_\lambda \in L^{n+\varepsilon },$
imply that $\omega $ is,
without loss of generality, in $C^{1+\theta} (\ol B)$ for some
ball $B\subset \Omega _\lambda $ with $x_0\in \partial B,$ [16]. Choose
a function $z\in C^{1+\theta }(\ol B)$ such that $-\Delta z-P_\lambda z=0$
in $B,\; z>0$ in $\ol B$. Then by considering the equation $\omega /z$
satisfies we conclude, again by [16], that $\omega /z\in C^2(B)\cap
C^{1+\theta }(\ol B)$ and $\pd{}{x_1}\;(\omega /z)(x_0)<0,$ i.e.
$\pd{\omega }{x_1}\;(x_0) = 2\;\pd{u}{x_1}\;(x_0) <0$.
Next assume that $f$ depends on $x$ as well, i.e. $f\equiv f(x,u)$. As
was the case in the previous references, this situation can also be dealt
with if we assume $f$ is monotone in $x_1: \;f(x,\xi )\ge f(x^\lambda ,\xi
)$ for $\lambda >0$.
There is no significant change in the proofs. Note that we could
thus deal with some cases where $f$ had
singularities with respect to $x$ on $\{x_1=\lambda
_0\}\cap \Omega ,$
or with
singularities along planes $\{x_m =c\}\cap \Omega ,\quad m\ne 1,$
as examples with $f(x,u) = p(x)g(u)$ easily show, as long as the resulting
$P_\lambda $ was in $L^\alpha (\Omega _\lambda )$.
One limitation in the applicability of Theorem~5 is given by
condition~(II).
Observe that since $\Vert P_\lambda \Vert _{L^{\frac{n+\varepsilon }{2}}}$
is assumed bounded, then it suffices that $\chi _\lambda P_\lambda (x)$ be
pointwise continuous, a.e. in $\lambda ,$ which will be immediately the
case if $f$ is smooth in $u$. This follows from the observation that
if $\{g_n\}$ is a sequence of functions bounded in $L^\alpha $ and $g_n\to
g\in L^\alpha $ pointwise then $g_n\to g$ in $L^{\alpha -\varepsilon }$ for
any $\varepsilon >0,$ and this is because we can find a constant
$c$ --independent of $n\text{--}$ such that $g_n =\bar g^c_n +g^\pr_n$
with $\bar g^c_n = g_n$ if $\vert g_n\vert \le c,\;\vert \bar g^c_n\vert =c$
otherwise, and $g^\pr_n$ of small $L^{\alpha-\varepsilon } $ norm. We thus only need to
check the boundedness of $\chi _\lambda P_\lambda $.
Recall that we may also express $P_\lambda $ as
$P_\lambda =\int^1_0 f^\pr\big(tu+(1-t)v_\lambda \big)dt$ and thus if, for example,
$u\in L^\infty (\Omega )$ and $f\in C^{1+\theta }_{\ell\text{oc}},$
then Theorem~5 applies.
Finally, observe that
the results apply if $f$ is Lipschitz (locally Lipschitz if $u$ is in
$L^\infty )$. This follows by setting $p_\lambda =[f(u)-f(v_\lambda
)]/[u-v_\lambda ]$ if $u(x)\ne v_\lambda(x) ;\;p_\lambda =0$ otherwise,
and this example also shows that the continuity of $\mu _0$ is really
only needed from the left:
If we
let $\lambda ^\pr$ be given by $\mu _0(\lambda )>0$ if $\lambda >\lambda
^\pr,$ as before, then we repeat the above arguments, in particular
Theorem~1-c, and conclude that both $\mu _0(\lambda ^\pr)>0$ and
$u>v_{\lambda ^\pr}$. We next observe that $\chi _\mu P_\mu \to
\chi _{\lambda
^\pr}P_{\lambda ^\pr}$ pointwise, and that $\vert P_\mu \vert __