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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 09, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/??\hfil Nonexistence of solutions]
{Nonexistence of solutions to KPP-type equations
 of dimension greater than or equal to one}
\author[J. Engl\"{a}nder, P. L. Simon\hfil EJDE-2006/??\hfilneg]
{J\'{a}nos Engl\"{a}nder, P\'{e}ter L. Simon}  % in alphabetical order

\address{J\'{a}nos Engl\"{a}nder \hfill\break
Department of Statistics and Applied Probability\\
University of California, Santa Barbara\\
CA 93106-3110, USA}
\email{englander@pstat.ucsb.edu}
\urladdr{http://www.pstat.ucsb.edu/faculty/englander}

\address{P\'{e}ter L. Simon \hfill\break
Department of Applied Analysis,  E\"otv\"os Lor\'and University\\
 P\'{a}zm\'{a}ny P\'{e}ter S\'{e}t\'{a}ny 1/C, H-1117 Budapest,
Hungary}
\email{simonp@cs.elte.hu}
\urladdr{http://www.cs.elte.hu/$\sim$simonp}

\date{}
\thanks{Submitted September 19, 2005. Published January 24, 2006.}
\subjclass[2000]{35J60, 35J65, 60J80}
\keywords{KPP-equation; semilinear elliptic equations;
 \hfill\break\indent
positive bounded solutions; branching Brownian-motion.}

\begin{abstract}
 In this article, we consider a semilinear elliptic equations of
 the form $\Delta u+f(u)=0$, where $f$ is a concave function. We
 prove for arbitrary dimensions that there is no solution bounded
 in $(0,1)$. The significance of this result in probability theory
 is also discussed.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction and statement of main result}

In this article, we study semilinear elliptic equations of the
form $\Delta u+f(u)=0$. On the nonlinear term  $f:[0,1] \to \mathbb{R}$
we assume that
\begin{itemize}
\item[(i)] $f$ is continuous,
\item[(ii)] $f$ is positive on $(0,1)$,
\item[(iii)] the mapping $z\mapsto f(z)/z$ is strictly decreasing.
\end{itemize}
Under these three conditions, we consider the Kolmogorov Petrovskii
Piscunov-type (KPP-type) equation
\begin{gather}
\Delta u+f(u)=0  \label{eqn} \\
0<u<1 , \quad  \text{in } \mathbb{R}^d.\label{cond}
\end{gather}
Our main result is as follows.

\begin{theorem}\label{mainthm}
Problem \eqref{eqn}-\eqref{cond} has no solution for dimension $d\ge 1$.
\end{theorem}

Semilinear elliptic equations of the form \eqref{eqn} have been
widely studied. We mention here only two reviews \cite{Ouyang,
Tang}, where the exact number of positive solutions with different
nonlinearities are studied. In \cite{Ouyang} the differential
equation is considered on a bounded domain, in \cite{Tang} the
equation is studied in the whole space $\mathbb{R}$, however, it is
subject to the boundary condition $u \to 0$ as $|x| \to \infty$. The
case of concave $f$ has also been studied by several authors. In
\cite{Brezis} the assumption on $f$ is similar to ours, however, the
problem is given on a bounded domain with Dirichlet boundary
condition. In that paper the existence and uniqueness of the
positive solution is proved. Castro et al. studied the case of
concave nonlinearities in a series of papers, see e.g.
\cite{Castro1, Castro2}. In these works the problem is given on a
bounded domain with Dirichlet boundary condition. A generalized
logistic equation, with $f(u)=mu-qu^p$ is studied in
\cite{Hernandez} on a bounded domain with Dirichlet boundary
condition again.

Summarizing, we can say that our equation \eqref{eqn} has been
widely studied, however, in the papers where it is considered in the
whole space $\mathbb{R}$, it is always subject to the boundary
condition $u \to 0$ as $|x| \to \infty$. In these publications the
aim is to determine the exact number of the so-called \emph{fast and
slow decay} solutions. Hence, according to the authors knowledge,
there is no result available concerning problem
\eqref{eqn}-\eqref{cond} under the assumptions given on $f$.

\begin{remark}[Low dimensions]\rm
Our theorem can be proved very  easily for $d\le 2$.
To see this, recall that $\Delta$ is a so-called \emph{critical}
operator in $\mathbb{R}^d$ when $d=1,2$. Second order elliptic
operators $L$ with no zeroth order term are classified as being
\emph{subcritical} or critical according to whether the operator
possesses or does not possess a minimal positive Green's function.
In probabilistic terms criticality/subcriticality is captured by
the \emph{recurrence/transience} of the corresponding diffusion
process (see \cite[Chapter 4]{P}).

Another equivalent condition for $L$  to be critical is that all
positive functions $h$ that are superharmonic (i.e. $Lh\le 0$) are
in fact harmonic (i.e. $Lh\equiv 0$).
(See again \cite[Chapter 4]{P})

Now, observe that \eqref{eqn}-\eqref{cond} and the positivity of
$f$ on $(0,1)$ implies
\begin{equation}\label{eqn.lowdim}
\Delta u=-f(u)< 0  \quad  \text{in } \mathbb{R}^d. %\label{cond.lowdim}
\end{equation}
By the above criterion for critical operators, this  is impossible
in dimension one or two.
\end{remark}

The most important model case is the classical KPP equation, when
\begin{equation}\label{specialNL}
 f(u):=\beta u(1-u)
\end{equation}
with $\beta>0$. (In fact this particular nonlinearity is intimately
related to the distribution of a \emph{branching Brownian motion};
see more on the subject in the next paragraph.) We will present a
proof for our result that works basically for concave functions; in
fact, (iii) of Assumption 1  is related to the concaveness of the
function.

The connection between the KPP equation and branching Brownian
motion has already been discovered by H. P. McKean  --- it first
appeared in the classic work \cite{McK1975,McK1976}.

Let $Z=(Z(t))_{t\geq 0}$ be the $d$-dimensional binary branching
Brownian motion  with a spatially and temporally constant branching
rate $\beta>0$. The informal description of this process is as
follows. A single particle starts at the origin, performs a Brownian
motion on $\mathbb{R}^d$, after a mean--$1/\beta$ exponential time
dies and produces two offspring, the two offspring perform
independent Brownian motions from their birth location, die and
produce two offspring after independent mean--$1/\beta$ exponential
times, etc. Think of $Z(t)$ as the subset of $\mathbb{R}^d$
indicating the locations of the particles $z_1^t,\dots ,z^{N_t}_t$
alive at time $t$ (where $N_t$ denote the number of particles at
$t$). Write $P_{x}$ to denote the law of $Z$ when the initial
particle starts at $x$. The natural filtration is denoted by
$\{\mathcal {F}_t,\; t\ge 0\}$.

Then, as is well known (see e.g. \cite[Chapter 1]{D02}), the law
of the process can be described via its Laplace functional as
follows. If $f$ is a positive measurable function, then
\begin{equation}\label{Laplace.func}
E_x\exp \Big(-\sum_{i=1}^{N_t} f(z_i^t)\Big)=1-u(x,t),
\end{equation}
where $u$ solves the initial value problem
\begin{equation}  \label{par.eqn}
\begin{gathered}
\dot{u}=\frac{1}{2}\Delta u+f(u) \quad  \text{in }
\mathbb{R}^d\times\mathbb{R}_+  \\
u(\cdot,0)=1-e^{-f(\cdot)} \quad  \text{in }\mathbb{R}^d \\ %\label{IC}\\
0\le u\le 1 \quad  \text{in } \mathbb{R}^d\times\mathbb{R}_+,%\label{same.cond}
\end{gathered}
\end{equation}
with $f$ of the form \eqref{specialNL}.

Equation \eqref{eqn}-\eqref{cond} appears when one studies certain
`natural' martingales associated with branching Brownian motion (see
e.g. \cite{EK}). To understand this, let $\mathcal{\widehat
F}_t:=\sigma(\bigcup_{s\ge  t}\mathcal{F}_s)$ and consider the tail
$\sigma$-algebra $\mathcal{\widehat F}_{\infty}:=\bigcap_{t\ge
0}\mathcal{\widehat F}_s$. Choosing appropriate (sequences of) $f$'s
one can then express the probabilities of various events $A_t\in
\mathcal{\widehat F}_t$, for $t>0$, in terms of the function $u$ in
\eqref{par.eqn}. Letting $t\to\infty$ then leads to the conclusion
that if $A\in \mathcal{\widehat F}_{\infty}$ denotes a certain tail
event (e.g. having strictly positive limit for a certain nonnegative
`natural' martingale, or local/global extinction) then the function
$u(x):=P_x(A)$ is either constant ($=0$ or $=1$), or it must solve
\eqref{eqn}-\eqref{cond}. Hence, it immediately follows from our
main theorem that \emph{the tail $\sigma$-algebra is trivial}, that
is, all those events $A$ satisfy $P_{\cdot}(A)\equiv 0$ or
$P_{\cdot}(A)\equiv1$.

Note that if $\beta>0$ is replaced by a smooth nonnegative function
$\beta(\cdot)$ that does not vanish everywhere, then this
corresponds to having \emph{spatially dependent} branching rate for
the branching Brownian motion. It would be desirable therefore to
investigate whether our main theorem can be generalized for such
$\beta$'s.

\section{Proof of the theorem}

The proof is based on two ideas: The application of the semilinear
elliptic maximum principle, which is generalized here fore concave
functions, and a comparison between the semilinear and the linear
problems. Using these two ideas we will show that the \emph{minimal
positive solution} of \eqref{eqn} is $u_{\min}\equiv 1$, hence
\eqref{eqn} has no solution satisfying \eqref{cond}.

First we state and prove a semilinear maximum principle.
The results in this form is a
generalization of \cite[Proposition 7.1]{EP99}
for the particular case when the elliptic
operator is $L=\Delta$.

\begin{lemma}[Semilinear elliptic maximum principle]\label{emp}
Let $f:[0,\infty ) \to \mathbb{R}$ be a continuous function, for
which the mapping $z\mapsto f(z)/z$ is strictly decreasing.
Let  $D\subset \mathbb{R}^{d}$ be a bounded domain with smooth boundary.
If $v_{i}\in C^2(D)\cap C(\bar D)$\ satisfy $v_{i}>0$ in $D$,
$\Delta v_{i}+f( v_{i})=0$,
in $D$ for $i=1,2$, and $v_{1}\ge v_{2}$ on $\partial D$, then
$v_{1}\ge v_{2}$ in $\bar D$.
\end{lemma}


\begin{proof}
 Note that the  function $w:=v_1-v_2$ satisfies
\begin{equation}
\Delta w + f(v_1)-f(v_2)=0 . \label{e3}
\end{equation}
We show that $w\ge 0$  in $D$. Suppose to the contrary that there
exists a point $y\in D$ where $w$ is negative. Let $\Omega_0:=\{x\in
D\mid w(x)<0\}$. Let $\Omega$ be the connected component of
$\Omega_0$ containing $y$. Since $w \geq 0$ on $\partial D$, one has
$\Omega \subset \subset D$ and
\begin{equation}
w<0 \quad \mbox{in } \Omega,  \quad  w=0 \quad \mbox{on } \partial \Omega .
\label{e4}
\end{equation}
Let us multiply the
equation $\Delta v_{1}+f( v_{1})=0$ by $w$ and equation \eqref{e3}
by $v_1$, then subtract the second equation from the first, and
integrate on $\Omega$. Using that $w=v_1-v_2$ one obtains
\begin{equation}
I+II:=\int_{\Omega} (w \Delta v_1 - v_1 \Delta w) +
\int_{\Omega} (v_1 f(v_2) -v_2 f(v_1)) =0. \label{5}
\end{equation}
Using Green's second identity and that $w=0$ on $\partial \Omega$,
along with the fact that $\partial_{\nu} w \geq 0$ on $\partial
\Omega$, we obtain
$$
I= -\int_{\partial \Omega} v_1
\partial_{\nu} w \leq 0,
$$
where $\nu$ denotes the unit outward normal to $\partial \Omega$.
Furthermore, since  $v_1<v_2$ in $\Omega$, using (iii) of
Assumption 1, we have
 that also $II<0$:
$$
v_1 f(v_2) - v_2 f(v_1)= v_1 v_2 \left[ \frac{f(v_2)}{v_2} -
\frac{f(v_1)}{v_1} \right] < 0.
$$
It follows that the left hand side of (\ref{5}) is negative, while
its right hand side is zero. This contradiction proves that in fact
$w\ge 0$ in $D$.
\end{proof}

\begin{remark}[Spatially dependent $f$'s]\rm
One can similarly prove the analogous more general result for the
case, when $f:D\times [0,\infty ) \to \mathbb{R}$ is continuous in
$u$ and bounded in $x$, and $u \mapsto f(x,u)/u$ is strictly
decreasing.
\end{remark}

Let $f:[0,1] \to \mathbb{R}$ be a continuous function which is
positive in $(0,1)$. Based on ideas in \cite{KS} and using the
comparison between the linear and the semilinear equations,  we
prove the following lemma.

\begin{lemma}[Radially symmetric solutions] \label{lemma2}
Assume in addition that $f$ satisfies
$\liminf_{z\downarrow 0} \frac{f(z)}{z}>0$
(this is automatically satisfied under assumption that
the mapping $z\mapsto f(z)/z$ is strictly decreasing).
Then for any $y\in \mathbb{R}^d$ and
$p\in (0,1)$ there exists a ball $\Omega:= B_R(y)$ (with some
$R>0$) and a radially symmetric $C^2$ function $v:\Omega \to
\mathbb{R}$ such that
\begin{gather*}
\Delta v + f(v)= 0  \\
v>0 \quad \mbox{in } \Omega\\
v=0 \quad \mbox{on } \partial\Omega\,\quad v(y)=p \,.
\end{gather*}
\end{lemma}

\begin{proof} We show the existence of  a radially
symmetric solution of the form $v(x)=V(|x-y|)$. Let
$V\in C^2([0,\infty))$ be the solution of the initial value problem
\begin{equation}
\begin{gathered}
(r^{d-1}V'(r))'+r^{d-1}f(V(r))=0  \\
V(0)=p, \ V'(0)=0  .
\end{gathered}\label{e6}
\end{equation}
Writing $\Delta$ in polar coordinates, one sees that it is
sufficient to prove that there exists an $R>0$ such that $V(R)=0$
and $V(r)>0$ for all $r\in [0,R)$.
To this end, consider the \emph{linear} initial value problem
\begin{equation} \label{e7}
\begin{gathered}
(r^{d-1}W'(r))'+r^{d-1}m W(r)=0  \\
W(0)=p, \ W'(0)=0 ,
\end{gathered}
\end{equation}
where $m>0$ is chosen so that $f(u)>mu$ holds for all $u\in (0,p)$.
(Our assumptions on $f$ guarantee the existence of such an $m$.) It
is known that $W$ has a first root, which we denote by $\rho$. Note
that in this case $-m$ is the first eigenvalue of the Laplacian on
the ball $B_{\rho}$. We now show that $V$ has a root in $(0,\rho]$.
In order to do so let us multiply \eqref{e7} by $V$ and \eqref{e6} by
$W$, then subtract one equation from the other, and finally,
integrate on $[0,\rho]$. We obtain
\begin{equation}
\begin{aligned}
I+II&:=\int_0^{\rho} [(r^{d-1}W'(r))'V(r) -
(r^{d-1}V'(r))'W(r)]\, \mbox{d}r \\
&\quad +\int_0^{\rho} r^{d-1}[mW(r)V(r) - W(r)f(V(r))]\, \mbox{d}r =
0\,.
\end{aligned}\label{e8}
\end{equation}
Suppose now that $V$ has no root in $(0,\rho]$. Then, integrating by
parts, $ I = \rho^{d-1}W'(\rho)V(\rho)<0. $

Next, observe that by integrating \eqref{e6}, one gets $V'(r)<0$
(i.e. $V$ is decreasing). Hence $V(r)<p$, yielding
$mV(r)-f(V(r))<0.$ Therefore $II$, and thus the whole left hand side
of \eqref{e8} are negative; contradiction. This contradiction proves
that $V$ in fact has a root in $(0,\rho]$.
\end{proof}

\begin{remark}[Spatially dependent f's]\rm
When $f$ depends also on $x$, our method breaks down as it is no
longer possible to use ordinary differential equations to show the
existence of a solution attaining a value close to one at a given
point.

There is one easy case though: it is immediately seen that if
there exists a $g(u)$, with $f(x,u)\ge g(u)$ and $g(u)$ satisfies
the conditions of Theorem \ref{mainthm}, then Theorem
\ref{mainthm} remains valid for $f(x,u)$ as well.

Indeed, we know that $u_{min}\ge 1$, where $u_{min}$ is the minimal
positive solution for the semilinear equation with $g$. Recall (see
e.g. \cite{EP99, EP03}) that one way of constructing the minimal
positive solution is as follows. One takes  large balls $B_R(0)$,
and positive solutions  with zero boundary condition on these balls
(in our case  we know from \cite{KS} that there exist such positive
solutions for arbitrarily large $R$'s), and finally, lets
$R\to\infty$; using the monotonicity in $R$ that follows from the
semilinear elliptic maximum principle (Lemma \ref{emp}), the
limiting function exists and positive. It is standard to prove that
it solves the equation on the whole space, and  by Lemma \ref{emp}
again it must be the \emph{minimal} such solution.

Now suppose that $0<v$ solves the semilinear equation with $f(x,u)$.
Then  $v$ is a \emph{supersolution}: $0\ge \Delta v+ g(v)$; hence by
the above construction of $u_{min}$ and by an obvious modification
of the proof of Lemma \ref{emp}, $v\ge u_{min}\ge 1$.

The general case is harder. For example, when
$f(x,u):=\beta(x)(u-u^2)$ and $\beta$ is a smooth nonnegative
bounded function, the mere existence of positive solutions on large
balls is no problem  as long as the generalized principal eigenvalue
of $\Delta+\beta$ on $\mathbb{R}^d$ is positive. (The method in
\cite{Pconst}, pp. 26-27 goes through for $f(x,u):=\beta(x)(u-u^2)$
even though $\beta$ is constant in \cite{Pconst}.) The problematic
part is to show that the solution is large at the center of the
ball.
\end{remark}

\begin{proof}[Proof of Theorem \ref{mainthm}]
Suppose that problem \eqref{eqn}-\eqref{cond} has a solution. Choose
an arbitrary point $y\in \mathbb{R}^d$ and an arbitrary number
$p\in (0,1)$. Note that by Assumption 1, $f$ satisfies the conditions of
Lemma \ref{lemma2} and consider  the ball $B_R(y)$ and the radially
symmetric function $v$ on it, which are guaranteed by Lemma
\ref{lemma2}. We can apply Lemma \ref{emp} with $D=B_R(y)$, $v_1=u$
and $v_2=v$ and obtain that $u\geq v$. In particular then, $u(y)\geq
v(y)=p$. Since $y$ and $p$ were arbitrary, we obtain that $u\geq 1$,
in contradiction with \eqref{cond}. Consequently,
\eqref{eqn}-\eqref{cond} has no solution.
\end{proof}


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\end{document}