0$,
problem (1.1) possesses no global positive solutions;
(b) when
$p>1+\frac{2}{n}$ and $u_{0}$ is smaller than a small Gaussian, then (1.1)
has global positive solutions. So $1+\frac{2}{n}$ is the critical exponent.
For a reference to the rich literature of the subsequent development on
the topic, we refer the reader to the survey paper [Le].
In recent years, many authors have undertaken research on
semilinear elliptic operators on manifolds, including the well-known Yamabe problem
(see [Sc] and [Yau]). The study of those elliptic problems and
others such as Ricci flow lead naturally to semilinear
or quasi-linear parabolic problems (see [H] and [Sh]).
The first goal of the paper is to study when blow up of solutions
occur and when
global positive solutions exist for equation (1.1) on manifolds. Such an undertaking requires
some new techniques.
The prevailing methods of treating the above semi-linear problems i.e. variational
and comparison method seem difficult to apply.
The method we are using
is based on new inequalities ( see section 3 and Lemma 7.2) involving the heat kernels.
We are able to find an explicit relation between the size of the critical exponent and geometric
properties of the manifold such as the growth rate of geodesic balls (see Theorem B).
It is interesting to note that this technique also leads to a non-existence result of the
well-known non-compact Yamabe problem of prescribing positive constant
scalar curvatures; i.e. whether the following elliptic problem has
``suitable" positive solutions on $\bold{M}^{n}$ with nonnegative scalar curvature (see Theorems C and D).
$$
\Delta u - \frac{n-2}{4(n-1)}R u
+ u^{(n+2)/(n-2)} = 0.
\tag 1.1'
$$This problem has been asked by J. Kazdan [K] and S. T. Yau [Yau]. The compact
version of the problem was proposed by Yamabe [Yam], proved by Trudinger [Tr] and
Aubin [Au1] in some cases and eventually proved by R. Schoen [Sc] completely.
In the non-compact case,
Aviles and McOwen [AM] obtained some existence results for the problem of prescribing
constant negative scalar curvature. Jin [Jin] gave a nonexistence result when $R$ is
negative somewhere. Some existence result when $R$ is positive was obtained
in [Ki]. Recently in [Zhan5] (Theorem A, B), we constructed complete noncompact manifolds
with positive
$R$, which do not have any conformal metric with positive constant scalar curvature. However
those manifolds, obtained by deforming $\bold{R}^3 \times S^1$, do not have nonnegative Ricci
curvature everywhere.
Since manifolds with nonnegative Ricci curvature are one of the most basic objects in geometry. It is the most natural to ask whether the Yamabe equation can be solved in this case. However, as far as we know, there has been no nonexistence
result on the problem
of prescribing constant positive scalar curvature when the Ricci curvature is nonnegative and
$R$ is positive.
When $\bold{M}^{n}$ is $\bold{R}^{n}$ with the Euclidean metric, then problem (1.1')
becomes $\Delta u + u^{(n+2)/(n-2)}=0$, which does have positive solutions (see [Ni]),
that induce incomplete metric of constant positive scalar curvature.
$$
u_{\lambda}(x) = [n(n-2) \lambda^{2}]^{(n-2)/4}/(\lambda^{2}
+ |x|^{2})^{(n-2)/2}, \quad
\lambda>0.
$$So it is reasonable to expect that (1.1') at least has a positive solution which gives
rise to a incomplete metric of scalar curvature one.
However Theorem D
below asserts that unlike the compact Yamabe problem, equation (1.1') can not be solved
in general, regardless of whether one requires the resulting metric is
complete or not. In fact if the existing scalar curvature decays ``too fast" and the
volume of geodesic balls does not increase ``fast enough", then (1.1') does not have
any positive solution at all. This of course rules out the existence of complete
or incomplete metric with constant positive scalar curvature.
Before stating the results precisely, let us list the basic assumptions and some notations to be used
frequently in the paper. For theorems A, B and C in the paper we make the following assumptions
(i), (ii) and (iii)
unless stated otherwise. Instead, Theorem D is exclusively about manifolds with nonnegative Ricci
curvature.
\medskip
(i). There are positive constants $b$, $C$, $K$, $q$ and $Q$ such
that
$$
|B(x, r)| \le C r^{Q}; \quad |B(x, 2r)| \le C 2^{q} |B(x, r)|, \ r>0;
\quad \text{Ricci} \ge -K.
\tag 1.2
$$
(ii). $G$, the fundamental solution
of the linear operator $H_0= \Delta - R - \partial_{t} $ in (1.1), has global Gaussian upper bound. i.e.
$$
0 \le G(x,t;y,s) \le \frac{C}{|B(x, (t-s)^{1/2})|} e^{- b
\frac{d(x,y)^{2}}{t-s}},
\tag 1.3
$$for
all $x, y \in \bold{M}^{n}$ and all $t>s$.
(iii). When $t-s \ge d(x, y)^{2}$, $G$ satisfies
$$
G(x, t; y, s) \ge \min \{ \frac{1}{C |B(x, (t-s)^{1/2})|}, \frac{1}{C |B(y, (t-s)^{1/2})|}\}.
\tag 1.4
$$
We mention that (1.4) actually is equivalent to $G$ having global Gaussian lower bound by a well known
argument in [FS].
\medskip
At the first glance, the relation between conditions (1.3), (1.4) and the function
$R$ does not seem transparent. However we emphasize that in general (1.3)
and (1.4) only require $R$ to satisfy some decay conditions such as
$R^-(x) \le \epsilon/[1+d^{2+\delta}(x, x_{0})]$ for $\delta>0$ and some $\epsilon>0$ and
$R^+(x) \le c/[1+d^{2+\delta}(x, x_{0})]$ for $\delta>0$ and an arbitrary $c>0$ as
indicated in Lemma 6.1 below and Theorem C in [Zhan5].
More specifically we have
\medskip
{\it {\bf Theorem} (Theorem C [Zhan5]). Suppose $\bold{M}$ is a complete noncompact manifold with
nonnegative Ricci curvature outside a compact set and $R \ge 0$, then (1.4) and (1.3) hold if and
only if $\sup_x \int \Gamma_0(x, y) R(y) dy < \infty$. Here $\Gamma_0$ is the fundamental solution
of the free Laplacian $\Delta$. In particular (1.4) and (1.3) hold
if $0 \le R \le c/[1+d^{2+\delta}(x, x_{0})]$ for $\delta>0$ and an arbitrary $c>0$ }
\medskip
In the context of the Yamabe equation,
the relation is even more direct. If the Ricci curvature of $\bold{M}^{n}$ is non-negative
and $R$ is the scalar curvature, then the upper bound (1.3) automatically holds by [LY]
and the maximum principle since $R^-=0$.
\medskip
{\it {\bf Definition} 1.1. A function $u=u(x, t)$ such that
$u \in L^{2}_{loc}(\bold{M}^{n} \times (0, \infty))$ is called a solution of
(1.1) if
$$
u(x, t) = \int_{\bold{M}^{n}} G(x, t; y, 0) u_{0}(y) dy +
\int^{t}_{0}\int_{\bold{M}^{n}} G(x, t; y, s) u^{p}(y, s) dyds
$$for all $(x, t) \in \bold{M}^{n} \times (0, \infty)$. }
\medskip
$G=G(x, t; y, s)$ will denote the fundamental solution of the
linear operator $H_{0}$ in (1.1).
For any $c>0$, we write
$$
G_{c}(x, t;y, s)=
\cases
\frac{1}
{|B(x, (t-s)^{1/2})|}\exp(-c \frac{d(x, y)^{2}}{t-s}), \quad t>s,\\
0, \quad t 0$. But this contradicts the assumption (2.2).
q.e.d.
\medskip
{\bf Remark 2.1.} When the function $R=R(x)$ is not zero, we do not know whether Theorem A
is still valid, except for the case when $\bold{M}^{n}$ has bounded geometry
(see Corollary 1.2).
\medskip
\heading
3. Two inequalities
\endheading
\medskip
In this section we present some inequalities for the heat kernel, which will
play a key role in both the existence and blow up results.
The proof, almost identical to that of Lemma 4.1 in [Zhan2], is
given for completeness.
\medskip
{\it {\bf Lemma 3.1}. Suppose $0\tau>s$,
$$
(i). \ G_{a}(x, t; z, \tau) G_{b}(z, \tau; y, s) \le C [G_{c}(x, t; z, \tau) + G_{c}(z, \tau; y, s)] G_{a}(x, t; y, s),
$$
$$
(ii). \ G_{b}(x, t; z, \tau) G_{a}(z, \tau; y, s) \le C [G_{c}(x, t; z, \tau) + G_{c}(z, \tau; y, s)] G_{a}(x, t; y, s).
$$}
\medskip
$Remark $ 3.1. The condition $a**0$, let
$$
h_{a}(x, t) = \int_{\bold{M}^{n}}G_{a}(x,t;y,0) u_{0}(y)dy,
\tag 2.1
$$where $u_{0}$ is a bounded non-negative function. Suppose $\lim_{d(x, 0) \rightarrow \infty } u_{0}(x) =0$, then
$\lim_{d(x, 0) \rightarrow \infty } h_{a}(x, t) =0$
uniformly with respect to $t>0$.
}
\medskip
Proof. For any $\delta >0$, let $R>0$ be such that
$u_{0}(y) < \delta /2$ when $d(y) \ge R$. When $d(x, 0) \ge 2R$ we have
$$
\aligned
h_{a}(x, t) &= \int_{d(y) > R} \frac{e^{- a \ d(x, y)^{2}/t}}{|B(x, t^{1/2})|} \ u_{0}(y)dy +
\int_{d(y) \le R} \frac{e^{- a \ d(x, y)^{2}/t}}{|B(x, t^{1/2})|} \ u_{0}(y)dy \\
&\le \frac{\delta}{2} \int_{d(y) > R} \frac{1}{|B(x, t^{1/2})|}
e^{- a \ d(x, y)^{2}/t}dy\\
& \quad + \int_{d(y) \le R} \frac{1}{|B(x, t^{1/2})|} e^{- a \ d(x, y)^{2}/(2 t)} e^{- a \ d(x, y)^{2}/(2 t)} u_{0}(y) dy.
\endaligned
$$Note that
$$
\frac{1}{|B(x, t^{1/2})|} e^{- a \ d(x, y)^{2}/(2 t)} \le C/|B(x, d(x, y))|
$$by Proposition 2.1 and $d(x, y) \ge d(x, 0)-R$ when $d(y, 0) \le R$, we have
$$
\aligned
&h_{a}(x, t) \le C \delta /2 + \frac{C}{|B(x, d(x, 0)-R)|}
\int_{d(y) \le R}e^{- a \ d(x, y)^{2}/(2 t)} u_{0}(y) dy \\
&\le C \delta /2 + C \frac{|B(0, R)|}{|B(x, d(x, 0)-R)|} ||u_{0}||_{L^{\infty}}\\
&\le C \delta,
\endaligned
$$when $d(x, 0)$ is sufficiently large. This proves (b).
q.e.d.
\medskip
%{\it {\bf Lemma 2.2.}
% Suppose $0 \le u_{0}(x) \le C/(1+|B(0, d(x, 0))|)$ and $u_{0} \in L^{1}(
%\bold{M}^{n})$, then
%$$
%h_{a}(x, t) \le C(1+||u_{0}||_{L^{1}})/(1+|B(0, d(x, 0))|),
%$$for all $t>0$ and $x \in \bold{M}^{n}$.}
%Proof.
%For $d(x, 0) >R$, we have
%$$
%\aligned
%&h_{a}(x, t) =\int_{d(y) > R} \frac{e^{- a \ d(x, y)^{2}/t}}{|B(x, t^{1/2})|} %\ u_{0}(y)dy +
%\int_{d(y) \le R} \frac{e^{- a \ d(x, y)^{2}/t}}{|B(x, t^{1/2})|} \ %u_{0}(y)dy \\
%&\le \frac{C}{1+|B(0, R)|} + \int_{d(y) \le R} \frac{1}{|B(x, t^{1/2})|} e^{- %a \ d(x, y)^{2}/2t} e^{- a \ d(x, y)^{2}/2t} \ u_{0}(y)dy \\
%&\le \frac{C}{1+|B(0, R)|} +
%\frac{C}{|B(x, d(x, y))|} \int_{d(y) \le R}e^{- a \
%d(x, y)^{2}/(2 t)} u_{0}(y) dy\\
%&\frac{C}{1+|B(0, R)|} +
%\frac{C}{|B(x, d(x, 0)-R)|} \int_{d(y) \le R}e^{- a \
%d(x, y)^{2}/(2 t)} u_{0}(y) dy.
%\endaligned
%$$Taking $R$ so that $d(x, 0) = 2R$ and using the doubling condition we have
%$$
%h_{a}(x, t) \le \frac{C}{1+|B(0, R)|} + %\frac{||u_{0}||_{L^{1}(\bold{M}^{n})}}{1+|B(x, d(x, 0))|} \le %\frac{||u_{0}||_{L^{1}(\bold{M}^{n})}}{1+|B(0, d(x, 0))|}.
%$$Part (b) is thus proved.
%q.e.d.
%\medskip
\medskip
{\bf Proof of Theorem A.} Since the proof follows the lines of Theorem 1 in [Me],
we will be sketchy.
(a). We first show that if $p>p^{*}$ then (1.1) has global positive solutions for some $u_{0}$. As in the case of [Me],
it is enough to show that if there is a non-trivial positive solution $W$ of $H_{0}W=0$ such that
$$
\int^{\infty}_{0} ||W(., t)||^{p-1}_{L^{\infty}} dt< \infty,
$$then (1.1) has global positive solutions for some $u_{0}$.
Let $a(t)$ be the solution of the initial value problem
$$
a'(t) = ||W(., t)||^{p-1} a(t)^{p}, \quad a(0) = a_{0}>0.
$$Then $u(x, t) \equiv a(t) W(x, t)$ is a super solution of (1.1) with $u_{0}(x) = a_{0}W(x, 0).$ Thus it is enough to show that choice of $a_{0}$ can ensure that $a(t)$ exists for all $t>0$ and is uniformly bounded. As
$$
a(t) = [a^{1-p}_{0}-(p-1) \int^{t}_{0}||W(., s)||^{p-1} ds]^{-1/(p-1)},
$$this follows from the condition. Since $0$ is always a subsolution, the standard comparison
theorem shows that
(1.1) has a global positive solution.
(b). We want to show that if $p****s \ge 0$,
$$
(i). \int^{t}_{s}\int_{\bold{M}^{n}}G_{a}(x, t; z, \tau) \ |V(z, \tau)| \
G_{b}(z, \tau; y, s) dz d\tau \le C_{a, b} \
N_{c, \infty}(V) G_{a}(x, t; y, s);
$$
$$
(ii). \int^{t}_{s}\int_{\bold{M}^{n}}G_{b}(x, t; z, \tau) \ |V(z, \tau)| \
G_{a}(z, \tau; y, s) dz d\tau \le C_{a, b} \
N_{c, \infty}(V) G_{a}(x, t; y, s).
$$}
\medskip
{\it Proof of Lemma 3.2.} We will only give a proof of (i) since (ii) can be
handled similarly. For simplicity we write
$$
J(x, t; y, s) = \int^{t}_{s}\int_{\bold{M}^{n}}G_{a}(x, t; z, \tau) \ |V(z, \tau)| \
G_{b}(z, \tau; y, s) dz d\tau.
$$The desired estimate follows from the main inequality in Lemma 3.1 i.e.
$$
G_{a}(x, t; z, \tau) G_{b}(z, \tau; y, s) \le C [G_{c}(x, t; z, \tau) +
G_{c}(z, \tau; y, s)] G_{a}(x, t; y, s)
$$where $c$ is a suitable positive constant. This is because the last inequality implies
$$
J(x, t; y, s) \le C \int^{t}_{s}\int_{\bold{M}^{n}}[G_{c}(x, t; z, \tau) +
G_{c}(z, \tau; y, s)] |V(z, \tau)| dzd\tau G_{a}(x, t; y, s).
$$Taking the maximum of the integral the right hand side, we finish the proof
of the lemma.
q.e.d.
\medskip
\heading
4. Proof of Theorem B, part (a)
\endheading
This section is divided into three parts. In the first part we list a number
of notations and symbols, which include an integral operator and an appropriate
function space. In the next part we will prove lemma 4.1 which states that
the integral operator has a fixed point in the function space. Theorem B,
part (a),
will be proved in the end of the section.
\medskip
First we recall and define a number of notations.
Given a positive $u_{0} \in L^{\infty}(\bold{M}^{n})$, write
$$
h(x, t) = \int_{\bold{M}^{n}} G(x, t; y, 0) u_{0}(y) dy.
\tag 4.1
$$Here $G$ is the fundamental solution of the operator $H_{0}$ in (1.1).
By (1.3), there are positive
constants $C$ and $b$ such that
$$
G(x, t; y, s) \le \frac{C}{|B(x, (t-s)^{1/2})|} \exp (- b \frac{d(x, y)^{2}}{t-s}) =
C G_{b}(x, t; y, s),
$$for all $t>s$ and $x, y \in \bold{M}^{n}$.
\medskip
For $u \in L^{\infty}(\bold{M}^{n} \times [0, \infty) )$, we
define $T$ to be the integral operator:
$$
T u\ (x, t) = h(x, t) + \int^{t}_{0}\int_{\bold{M}^{n}} G(x, t; y, s)
u^{p}(y, s) dyds.
\tag 4.2
$$For any constants $a>0$, $d>1$ and $M>1$,
the space $S_{d}$ is defined by
$$
S_{d} = \{ u(x, t) \in C(\bold{M}^{n} \times [0, d] ) \ |
\ 0
\le u(x, t) \le M h_{a}(x, t) \}
$$where the function $h_{a}$ is given by (2.1).
From this moment, we fix the number
$a$ to be a positive number strictly
less than $b$, which is the constant in the Gaussian upper bound for $G$.
This choice of $a$ is crucial when we prove Lemma 3.1 below. Since $a 1$
and $b_{0} >0$ independent of $d$ such that
the integral operator (4.2) has a fixed point in $S_{d}$, provided that
$$u_{0} \in C^{2}(\bold{M}^{n}),
\quad \lim_{d(x, 0) \rightarrow \infty} u_{0}(x)=0, \quad \text{and}, \quad ||u_{0}||_{L^{\infty}(\bold{M}^{n})} + ||u_{0}||_{L^{1}(\bold{M}^{n})} \le b_{0}.$$}
\medskip
Proof.
$step$ 1. We want to use the Schauder fixed point theorem. To this end we
need to check the following conditions.
(i). $S_{d}$ is nonempty, closed, bounded and convex.
(ii). $T S_{d} \subset S_{d}.$
(iii). $T S_{d}$ is a compact subset of $S_{d}$ in $L^{\infty}$ norm.
(iv). $T$ is continuous.
\medskip
$step$ 2.
Condition (i) is obviously true. So let's verify (ii), which requires us
to show that $0 \le Tu \le M h_{a}$ when $0 \le u \le M h_{a}$.
Since $0 \le u \le M h_{a}$ we have
$$
u^{p}(y, s) \le M^{p} h^{p-1}_{a}(y, s) h_{a}(y, s) = M^{p} h^{p-1}_{a}(y, s)
\int_{\bold{M}^{n}} G_{a}(y, s; z, 0) u_{0}(z) dz.
\tag 4.3
$$Recalling the definition of $h_{a}$ and using the fact that $u_{0} \in L^{1}(\bold{M}^{n})$ we obtain
$$
\aligned
&h^{p-1}(y, s) = [\int_{\bold{M}^{n}}G_{a}(y, s ; z, 0)
u_{0}(z) dz]^{p-1} \\
&= \frac{1}{|B(y, s^{1/2})|^{p-1}} [\int_{\bold{M}^{n}} u_{0}(z) dz]^{p-1} \le
\frac{1}{|B(y, s^{1/2})|^{p-1}} ||u_{0}||^{p-1}_{L^{1}(\bold{M}^{n})}.
\endaligned
\tag 4.4
$$Therefore
$$
u^{p}(y, s) \le M^{p} ||u_{0}||^{p-1}_{L^{1}(\bold{M}^{n})}\frac{1}{|B(y, s^{1/2})|^{p-1}}.
$$ Substituting (4.3) into (4.2) and using Fubini's theorem we obtain
$$
\aligned
&T u (x, t) \le h(x, t)\\
&\quad + C M^{p} \int_{\bold{M}^{n}}
\int^{t}_{0}
\int_{\bold{M}^{n}} G(x, t; y, s) h^{p-1}_{a}(y, s) G_{a}(y, s; z, 0) dyds
\ u_{0}(z) dz,
\endaligned
\tag 4.5
$$
Remembering that
$$G(x, t;y, s) \le \frac{C}{|B(x, (t-s)^{1/2})|}
\exp (- b \ d(x, y)^{2}/ (t-s) ) = C G_{b}(x, t;y, s),
$$
we have
$$
\aligned
\int^{t}_{0}
\int_{\bold{M}^{n}}& G(x, t; y, s) h^{p-1}_{a}(y, s) G_{a}(y, s; z, 0) dyds\\
&\le C \ \int^{t}_{0}
\int_{\bold{M}^{n}} G_{b}(x, t; y, s) h^{p-1}_{a}(y, s) G_{a}(y, s; z, 0) dyds\\
\endaligned
$$
At this stage we quote Lemma 3.2 which was first proved in
[Zhan3] for the Euclidean case. Given $b>a$,
$$
\int^{t}_{0}
\int_{\bold{M}^{n}} G_{b}(x, t; y, s) h^{p-1}_{a}(y, s) G_{a}(y, s; z, 0) dyds
\le C C_{a, b} N_{c, \infty}(h^{p-1}_{a}) G_{a}(x, t; z, 0),
\tag 4.6
$$for all $t>0$ and some positive $c$ and $C_{a, b}$. We claim that
$N_{c, \infty}(h^{p-1}_{a})$, which is defined in (1.7), is a finite number.
The proof is as follows.
Since $s>0$ and $h_{a}(x, t) = 0$ when $t<0$, we have
$$
\aligned
N_{c, \infty}(h^{p-1}_{a}) \equiv &\text{sup}_{x,t>0}\int^{t}_{0}\int_{\bold{M}^{n}}
|h^{p-1}_{a}(y,s)|
G_{c}(x,t;y,s) dyds\\
& + \text{sup}_{y,s \ge 0}\int^{\infty}_{s}\int_{
\bold{M}^{n}}|h^{p-1}_{a}(x,t)|
G_{c}(x,t;y,s) dxdt.
\endaligned
$$By (4.4) and the fact that $h_{a}(x, t) \le C ||u_{0}||_{L^{\infty}}$ we have
$$
\aligned
&\int^{t}_{0}\int_{\bold{M}^{n}}
|h^{p-1}_{a}(y,s)|
G_{c}(x,t;y,s) dyds \\
&= \int^{t}_{r_{0}}\int_{\bold{M}^{n}}
|h^{p-1}_{a}(y,s)|
G_{c}(x,t;y,s) dyds + \int^{r_{0}}_{0}\int_{\bold{M}^{n}}
|h^{p-1}_{a}(y,s)|
G_{c}(x,t;y,s) dyds\\
&\le ||u_{0}||^{p-1}_{L^{1}(\bold{M})} \int^{t}_{r_{0}}\int_{\bold{M}^{n}}\frac{1}{|B(y, s^{1/2})|^{p-1}}G_{c}(x, t; y, s) dyds+ C ||u_{0}||^{p-1}_{L^{\infty}} r_{0}.\\
\endaligned
$$Similarly
$$
\aligned
&\int^{\infty}_{s}\int_{
\bold{M}^{n}}|h^{p-1}_{a}(x,t)|
G_{c}(x,t;y,s) dxdt\\
&\le C ||u_{0}||^{p-1}_{L^{1}(\bold{M})} \int^{\infty}_{r_{0}}\int_{\bold{M}^{n}} \frac{G_{c}(x, t; y, s)}{|B(x, t^{1/2})|^{p-1}} dxdt + C ||u_{0}||^{p-1}_{L^{\infty}} r_{0}.
\endaligned
$$Using $C_{0}$ to denote
$$
\sup_{x \in \bold{M}^{n}, t>0} \int^{t}_{r_{0}}\int_{\bold{M}^{n}}\frac{G_{c}(x, t;y,s)}{|B(y, s^{1/2})|^{p-1}} dyds + \sup_{y \in \bold{M}^{n}, s>0} \int^{\infty}_{r_{0}}\int_{\bold{M}^{n}} \frac{G_{c}(x, t; y, s)}{|B(x, t^{1/2})|^{p-1}} dxdt,
$$we have
$$
N_{c, \infty}(h^{p-1}_{a}) \le C C_{0} ||u_{0}||^{p-1}_{L^{1}(\bold{M})} + C ||u_{0}||^{p-1}_{L^{\infty}} r_{0}< \infty,
$$which proves the claim.
Combining (4.6) with (4.5), we reach
$$
T u(x, t) \le h(x, t) + C C_{a, b} M^{p} N_{c, \infty}(h^{p-1}_{a})
\int_{\bold{M}^{n}} G_{a}(x, t; z, 0)
u_{0}(z)dz,
$$
which yields
$$
T u(x, t)
\le (C+ C C_{a, b} M^{p} N_{c, \infty}(h^{p-1}_{a})) h_{a}(x, t)
\tag 4.7
$$Since $p>1$, by taking $M > 2C$ and $||u_{0}||_{L^{1}(M)} + ||u_{0}||_{L^{\infty}(M)}$ suitably small we find that
$$
0 \le T u(x, t) \le M h_{a}(x, t).
\tag 4.8
$$Thus condition (ii) is satisfied.
\medskip
$step$ 3.
Now we need to check condition (iii). We note that the local regularity theory
for solutions of uniformly parabolic equations can be transplanted
to the operator $H_{0}$ in (1.1).
By our choice, functions $u$ in $S_{d}$
are uniformly bounded and therefore,
$Tu$ is equicontinuous and in fact H\"older continuous. This is because
$Tu$ actually satisfies, in the weak sense,
$H_{0}(Tu) = - u^{p}$ in $\bold{M}^{n} \times (0,
d)$ and $Tu(x, 0) = u_{0}(x)$ and $u_{0} \in C^{2}(\bold{M}^{n})$. Taking
into account that
$$
0 \le \lim_{d(x, 0) \rightarrow \infty}Tu(x, t)
\le C \lim_{d(x, 0) \rightarrow \infty} h_{a}(x, t) = 0
$$uniformly (by Lemma 2.1 ), we know that
$$
\lim_{d(x, 0) \rightarrow \infty}Tu(x, t) = 0
$$Hence
$T S_{d}$ is a relatively compact subset of $S_{d}$.
This is an easy modification of the classical Ascoli-Arzela theorem (see
[Zhao]). Hence we have verified (iii).
\medskip
$step$ 4. Finally we need to check condition (iv).
Given $u_{1}$ and $u_{2}$ in $S_{d}$, we have, by (3.2),
$$
( T u_{1}-T u_{2}) (x, t) = \int^{t}_{0}\int_{\bold{M}^{n}}G(x, t; y, s)
) \ [u^{p}_{1}(y, s)-u^{p}_{2}(y, s)] dyds.
\tag 4.9
$$Next we notice that
$$|u^{p}_{1}(y, s) - u^{p}_{2}(y, s)| \le
p \max\{u^{p-1}_{1}(y, s), u^{p-1}_{2}(y, s)\} |u_{1}(y, s) - u_{2}(y, s)|.
$$Since $u_{1}$ and $u_{2}$ are bounded from above by $M h_{a}$, we have
$$
|u^{p}_{1}(y, s) - u^{p}_{2}(y, s)| \le C M^{p-1} h^{p-1}_{a}(y, s) |u_{1}(y, s) - u_{2}(y, s)|.
$$
Substituting the last inequality to (4.9) we obtain
$$
\aligned
||Tu_{1}-T u_{2}||_{L^{\infty}} &\le C M^{p-1}
||u_{1}-u_{2}||_{L^{\infty}} \int^{t}_{0}\int_{\bold{M}^{n}}G(x, t; y, s)
h^{p-1}_{a}(y, s) dyds\\
&\le C M^{p-1} ||u_{1}-u_{2}||_{L^{\infty}} N_{c, \infty} (h^{p-1}_{a}).
\endaligned
$$Here $c$ is a suitable constant. Since
$ N_{c, \infty} (h^{p-1}_{a})$ is a finite constant
we have proved the continuity of $T$ and the lemma. q.e.d.
\medskip
Now we are ready to give the
\medskip
{\bf Proof of Theorem B, part (a).}
For any $d>1$, let $u_{d}$ be a fixed point
of $T$ in the space $S_{d}$ as given in Lemma 3.1. Define
$$
U_{d}(x, t)=
\cases
u_{d}(x, t), \quad t \le d;\\
u_{d}(x, d), \quad t >d.
\endcases
$$Then from the proof of Lemma 4.1 ( (4.8) e.g.), we know that
$\{U_{d}\}$ is uniformly bounded and equicontinuous. Hence there is a
subsequence $\{U_{d_{m}} | m=1, 2, ..\}$ which converges uniformly
to a function
$u$ in any compact region of $\bold{M}^{n} \times [0, \infty)$. For any
fixed $(x, t) \in \bold{M}^{n} \times [0, \infty)$ and $m$ sufficiently large,
we know that
$$
U_{d_{m}}(x, t) = h(x, t) + \int^{t}_{0}\int_{\bold{M}^{n}} G(x, t;y,s)
U^{p}_{d_{m}}(y,s) dyds.
$$This is because $U_{d_{m}}$ is a fixed point of $T$ in $S_{d_{m}}$. Now
by the
dominated convergence theorem, $u$ satisfies
$$
u(x, t) = h(x, t) + \int^{t}_{0}\int_{\bold{M}^{n}} G(x, t;y,s)
u^{p}(y,s) dyds,
$$for all $(x, t) \in \bold{M}^{n} \times [0, \infty).$ Moreover, by (4.8)
We know
$$0 < u(x, t) \le M h_{a}(x, t)
$$for all $(x, t) \in \bold{M}^{n} \times [0, \infty).$ Clearly $u$ is a
global positive solution of (1.1). This finishes the proof.
q.e.d.
\medskip
Next we give
\medskip
{\it Proof of Corollary 1.1.} Note that
$$
\int_{\bold{M}^{n}} G_{c}(x, t; y, s) dx, \quad \int_{\bold{M}^{n}} G_{c}(x, t; y, s) dy \le C,
$$for all $t>s$, we have
$$
\aligned
&\int^{\infty}_{r_{0}}\int_{\bold{M}^{n}}\frac{G_{c}(x, t; y, s)}{|B(y, s^{1/2})|^{p-1}} dyds + \int^{\infty}_{r_{0}}\int_{\bold{M}^{n}}\frac{G_{c}(x, t; y, s)}{|B(x, t^{1/2})|^{p-1}} dxdt\\
&\le C \int^{\infty}_{r_{0}} \sup_{y \in \bold{M}} \frac{1}{|B(y, r^{1/2})|^{p-1}} dr<\infty.
\endaligned
$$
Now suppose $\inf_{x \in \bold{M}^{n}} |B(x, r)| \ge C r^{\alpha}$ when $r > r_{0}$, then, for $p>1 + \frac{2}{\alpha}$,
$$
\int^{\infty}_{r_{0}} \sup_{y \in \bold{M}} \frac{1}{|B(y, r^{1/2})|^{p-1}} dr
\le C \int^{\infty}_{r_{0}} \frac{1}{r^{\alpha (p-1)/2}} < \infty,
$$since $\alpha (p-1)/2>1$. Therefore (1.1) has a global positive solution for some $u_{0}>0$ by Theorem B, part (a). This proves the corollary. q.e.d.
\medskip
{\it Proof of Corollary 1.2.} Suppose $p**

p^{*} = 1+ \frac{2}{\alpha^{*}}$, then there exists $\epsilon>0$ such that $\frac{2}{p-1} + \epsilon < \alpha^{*}$ and hence
$$
\lim_{r \rightarrow \infty}\inf \frac{|B(x, r)|}{r^{\frac{2}{p-1} + \epsilon}} = \infty.
$$Therefore, for some $r_{0}>0$ and all $r \ge r_{0}$,
$$
1/|B(x, r^{1/2})|^{p-1} \le C/r^{1+ \epsilon(p-1)/2}.
$$By assumption (1.8), we have
$$
\int^{\infty}_{r_{0}} \sup_{y \in \bold{M}} \frac{1}{|B(y, r^{1/2})|^{p-1}} dr \le \int^{\infty}_{r_{0}} \frac{1}{r^{1+ \epsilon(p-1)/2}} dr <\infty.
$$Theorem B, part (a) shows that (1.1) has global positive solutions for some $u_{0} >0$. q.e.d.
\medskip
\heading
5. Proof of Theorem B, part (b)
\endheading
{\bf Proof. } Without loss of generality we take $x_{0}=0$.
Suppose $u$ is a global positive solution of (1.1), by Definition 1.1,
we know that $u$ solves the integral equation
$$
u(x, t) = \int_{\bold{M}^{n}} G(x, t;y,0) u_{0}(y) dy +
\int^{t}_{0}\int_{\bold{M}^{n}} G(x, t;y,s)
u^{p}(y,s) dyds,
\tag 5.1
$$for all $(x, t) \in \bold{M}^{n} \times [0, \infty).$
Given $t>0$, choosing $T>t$, multiplying $G(x, T; 0, t)$ on both sides
of (5.1) and integrating with respect to $x$, we obtain
$$
\aligned
\int_{\bold{M}^{n}} &G(x, T; 0, t) u(x, t) dx \ge
C \int_{\bold{M}^{n}} \int_{\bold{M}^{n}}G(x, T; 0, t) \ G(x, t;y,0) dx
\ u_{0}(y)dy+\\
& + C \int^{t}_{0}\int_{\bold{M}^{n}} \int_{\bold{M}^{n}}
G(x, T; 0, t)\ G(x, t;y,s)dx\ u^{p}(y,s) dyds.
\endaligned
\tag 5.2
$$Even though $H_{0}$ is an operator with variable coefficients, the
fundamental solution $G$ still enjoys the symmetry
$$
G(x, T; y, t)=G(y, T; x, t)
$$for all $x, y \in \bold{M}^{n}$ and $T>t$ (see [D]).
Therefore by the reproducing
property of the heat kernel, we reach
$$
\int_{\bold{M}^{n}}G(x, T; 0, t) \ G(x, t;y,0) dx = C G(0, T; y, 0)
= C G(y, T; 0, 0),
$$
$$
\int_{\bold{M}^{n}}G(x, T; 0, t)\ G(x, t;y,s)dx
= C G(0, T; y, s) = C G(y, T; 0, s).
$$Substituting the last two equalities into (5.2), we see that
$$
\aligned
\int_{\bold{M}^{n}} &G(x, T; 0, t) u(x, t) dx\\
&\ge
C \int_{\bold{M}^{n}} G(y, T; 0, 0) u_{0}(y)dy +
C \int^{t}_{0}\int_{\bold{M}^{n}} G(y, T; 0, s) u^{p}(y,s) dyds.
\endaligned
\tag 5.3
$$By (1.3), $\int_{\bold{M}^{n}} G(y, T; 0, s)dy \le C$. Using H\"older's inequality, we obtain
$$
\aligned
\int_{\bold{M}^{n}}& G(y, T; 0, s) u(y,s) dy =
\int_{\bold{M}^{n}} G^{1/q}(y, T; 0, s) G^{1/p}(y, T; 0, s) u(y,s) dy
\\
&\le [\int_{\bold{M}^{n}} G(y, T; 0, s) dy]^{1/q} [\int_{\bold{M}^{n}}G(y, T; 0, s) u^{p}(y,s) dy]^{1/p}\\
&\le C [\int_{\bold{M}^{n}}G(y, T; 0, s) u^{p}(y,s) dy]^{1/p},
\endaligned
$$where $1/p + 1/q = 1$. Inequality (5.3) then implies
$$
\aligned
\int_{\bold{M}^{n}} &G(x, T; 0, t) u(x, t) dx\\
&\ge
C \int_{\bold{M}^{n}} G(y, T; 0, 0) u_{0}(y)dy +
C \int^{t}_{0} [\int_{\bold{M}^{n}} G(y, T; 0, s) u(y,s) dy]^{p} ds.
\endaligned
\tag 5.4
$$
Without loss of generality we assume that $u_{0}$ is strictly positive
in a neighborhood of $0$. Using the lower bound in (1.4) for
$G$, we can then find a constant $C>0$ so that, for $T>1$,
$$
\int_{\bold{M}^{n}} G(y, T; 0, 0) u_{0}(y)dy \ge \int_{d(y, 0)^{2} \le 1} \frac{C}{|B(0, T^{1/2})|} u_{0}(y)dy \ge \frac{C}{|B(0, T^{1/2})|}.
\tag 5.5
$$ Going
back to (5.4) and writing $J(t) \equiv \int_{\bold{M}^{n}}
G(x, T; 0, t) u(x, t) dx$, we have
$$
J(t) \ge C/|B(0, T^{1/2})| + C \int^{t}_{0} J^{p}(s) ds, \quad T>t, T >1.
\tag 5.6
$$Using the notation $g(t) \equiv \int^{t}_{0} J^{p}(s) ds$,
we obtain,
$$
g'(t)/(|B(0, T^{1/2})|^{-1} + g(t))^{p} \ge C.
\tag 5.7
$$Integrating (5.7) from $0$ to $T$ and noticing $g(0) = 0$, we have
$$
-\frac{1}{(|B(0, T^{1/2})|^{-1}+g(t))^{p-1}}\big{|}^{T}_{0} \ge (p-1) C T
\tag 5.8
$$and therefore
$$
|B(0, T^{1/2})|^{p-1} \ge (p-1) C T,
$$for all $T>1$. This is possible only when
$$
|B(0, T)| \ge C T^{2/(p-1)}
\tag 5.9
$$when $T$ is large. Under the assumption of part (b) of Theorem B,
$$\lim_{r \rightarrow \infty} \inf \frac{|B(x, r)|}{r^{\alpha}} < \infty
$$and $p< 1+ \frac{2}{\alpha}$. Therefore
$$
\lim_{r \rightarrow \infty} \inf \frac{|B(x, r)|}{r^{2/(p-1)}} =0.
$$This contradicts (5.9).
Hence no global positive solutions exist for such $p$.
q.e.d.
\medskip
\heading
6. Proof of Theorem C: non-existence result for the non-compact Yamabe problem
\endheading
{\bf Proof of Theorem C.} Since the properties (1.2), (1.3) and (1.4) hold, we can apply part (b) of Theorem B, which says, for any $p< 1+ \frac{2}{\alpha}$, equation (1.1) has no global positive solutions. Since
$$
\alpha< (n-2)/2,
\tag 6.1
$$we know that equation (1.1) has no
global positive solutions for $p=\frac{n+2}{n-2}$ . Therefore the Yamabe equation can not have
solutions. This is so because any solution $u=u(x)$ of the Yamabe equation is a stationary and hence global positive solution of (1.1). Following the arguments in Lemma 1 of [P], which can be easily generalized to manifolds with Ricci bounded from below, we know that
$$
u(x) \ge \int_{\bold{M}^{n}} G(x, t; y, 0) u(y)dy + \int^{t}_{0}\int_{\bold{M}^{n}} G(x, t; y, s) u^{p}(y) dyds.
$$By Theorem B, part (b), such a $u$ does not exist in this case. q.e.d.
\medskip
To prove Corollary 1.3 we need to show that the Green's function $G$ of the operator $\Delta -R -\partial_{t}$ have global bounds (1.3) and (1.4). Due to the presence of $R$, we need much more effort. This is done in the next
\medskip
{\it {\bf Lemma 6.1.}Let $G_{0}$ be the fundamental solution of the unperturbed operator
$\Delta - \partial_{t}$. Suppose there are constants $b,C>0$ such that
$$
\frac{1}{C |B(x, (t-s)^{1/2})|} e^{- \frac{d(x, y)^{2}}{b(t-s)}} \le G_{0}(x, t; y, s)
\le
\frac{C}{ |B(x, (t-s)^{1/2})|} e^{- b\frac{d(x, y)^{2}}{t-s}},
\tag 6.2
$$for all $x, y \in \bold{M}^{n}$ and all $t>s$. Let $G$ be the fundamental solution of
the operator $\Delta - V - \partial_{t}$ where $V=V(x)$ is a bounded function.
Suppose
$N_{c, \infty}(V^-)$ is sufficiently small and $N_{c, \infty}(V^+)$ is
finite then properties (1.3) and (1.4) hold for $G$.}
\medskip
{Proof.} This essentially follows from combining Theorem C in [Zhan5] with
the proof of Theorem A part (b) in [Zhan2].
For the sake of completeness we include the proof here.
First we show that (1.3) holds.
Without
any loss of generality we assume that $V^+=0$ otherwise the maximum principle gives the desired
result.
We pick a number $a$ such that $0T$. The claim
can be checked easily by using the reproducing formula
$$
G(x, t; y, 0) = \int_{\bold{M}^{n}} G(x, t; z, T)
G(z, T; y, 0) dz, \quad t>T,
$$and the fact that $G(x, t; z, T) = G_{0}(x, t; z, T)$ for $t > T$.
The main task is to show that $B$ depends
on $V$ only in the form of $N_{c, \infty}(V)$.
From the Duhamel's principle and (6.3) we have
$$
G(x, t; y, s) \le G_{0}(x, t; y, s) +
B C_{0} \int^{t}_{s}\int_{\bold{M}^{n}}
G_{a}(x, t; z, \tau) \ |V(z, \tau)| \ G_{b}(z, \tau; y, s)dzd\tau,
$$
$$
G(x, t; y, s) \ge G_{0}(x, t; y, s) -
B C_{0} \int^{t}_{s}\int_{\bold{M}^{n}}
G_{a}(x, t; z, \tau) \ |V(z, \tau)| \ G_{b}(z, \tau; y, s)dzd\tau,
$$where $x, y \in \bold{M}^{n}$ and $s