text:
\begin{document}

\title[A ``Mean Value'' Approach to Asymptotic Behavior]{Exponential Decay Results for Semilinear Parabolic PDE with $C^0$ Potentials: A ``Mean Value'' Approach}

\author{Joseph L. Shomberg}
\address{Providence College \\ Department of Mathematics and Computer Science \\ Providence, RI 02918 \\ USA}
\email{jshomber@providence.edu}

\subjclass[2010]{35K58, 35B45, 35B40}

\keywords{Semilinear parabolic PDE, {\em{a priori}} estimates, asymptotic behavior, exponential decay, finite time blow-up}

\date{}

\begin{abstract}
The asymptotic behavior of some semilinear parabolic PDEs is analyzed by means of a ``mean value'' property.
This property allows us to determine, by means of appropriate {\em{a priori}} estimates, some exponential decay results for suitable global solutions.
We also apply the method to investigate a well-known finite time blow-up result.
An application is given to a one-dimensional semilinear parabolic PDE with boundary degeneracy.
Our results shed further light onto the problem of determining initial data for which the corresponding solution is guaranteed to exponentially decay to zero or blow-up in finite time.
\end{abstract}

\maketitle

\section{Introduction}

Let $\Omega$ be a bounded domain (open and connected) subset of $\mathbb{R}^N$, for some positive integer $N$. Assume the boundary $\Gamma:=\partial\Omega$ is sufficiently smooth.
For $x\in\Omega$ and $t>0$, we study the asymptotic behavior of solutions $u=u(x,t)$ to semilinear parabolic equations for the form,
\begin{equation}
\label{eqn}
\frac{\partial u}{\partial t} - \nu \Delta u + f(u)u = 0,
\end{equation}
$\nu>0$, with the Dirichlet boundary conditions,
\begin{equation}
\label{bndry}
u_{\mid\Gamma}(x,\cdot)=0,
\end{equation}
and given the initial state,
\begin{equation}
\label{ic}
u(\cdot,0) = u_0(\cdot).
\end{equation}
(Of course the results could be suitably adjusted to incorporate other boundary conditions such as Neumann, mixed Neumann/Dirichlet, periodic, or Robin.) We only assume $f$ is a $C^0$ function on $\mathbb{R}$. Note that Hadamard well-posedness for problem (\ref{eqn})-(\ref{ic}) is not known because such with such minimal assumptions on $f$, uniqueness of solutions is not guaranteed. Typically for equations such as (\ref{eqn}), it is assumed that $f\in C^1(\mathbb{R})$ satisfy $f'(s)\geq \ell$, for some $\ell>0$ (cf. e.g. \cite[p. 213]{Robinson01}).
Additionally, we cannot assume that the solutions are instantaneously regularizing.

The goal of this article is the provide a better description to the criteria that surrounds, not the well-posedness of problem (\ref{eqn})-(\ref{ic}), but rather the long-term behavior of the solutions to problem (\ref{eqn})-(\ref{ic}). The asymptotic behavior of solutions to PDE is a rich subject whose development we will only briefly mention. The study of dissipative dynamical systems is motivated by defining a solution operator for a given PDE, possibly posed abstractly as an ODE in a suitable Banach space, where the first task often is to demonstrate, besides global well-posedness, the existence of an absorbing set in the phase space.
After that, one may demonstrate the solutions hold certain properties, like asymptotically smoothing.
In many efforts, the culmination of the study peaks with the existence of a global attractor, the maximal invariant subset of the phase space that attracts all trajectories.
This attractor is typically defined as the omega-limit set of a bounded absorbing set, and consists of smooth solutions.
Some PDE also admit finite dimensional attractors whose rate of attraction is exponential.

The study of dissipative dynamical systems and the development of attractors has flourished since the seminal work of  \cite{Babin&Vishik92,Ladyzhenskaya91,Temam88}. Furthermore, largely due to the permanent importance of the Navier-Stokes equations, attractors for PDE {\em{without unique solutions}} were also developed in \cite{Ball00} and \cite{Melnik&Valero98}. Indeed, generalized semiflows were employed in \cite{Ball04} and \cite{Segatti06} (just to name two applications).
So-called trajectory dynamical systems were developed in \cite{Chepyzhov&Vishik02}.
Also, in the context of supercritical wave equations, the notion of trajectory dynamical systems appears in \cite{Zelik04}.

We will analyze the behavior of solutions for the above class of PDE in rather different terms: for guaranteed exponential convergence to zero. We will find conditions on the nonlinear term $f$ that guarantees the corresponding solutions exponentially decay to zero; hence, rendering some global solutions. Each of our decay results holds for all initial data $u_0\in H^1_0(\Omega)$ and all $\nu>0$.
The criteria we use for each result depends on $f$ through a property we call the ``mean value'' of $f$ through $u$; named after the Mean Value Theorem for Integrals.
It is important to note that we assume a solution $u$ (in the sense defined below) already exists, at least local in time.
Hence, the estimates that follow are {\em{a priori}}, but insure a strict qualitative behavior for all nontrivial solutions.
The method is used to investigate a well-known blow-up result for semilinear parabolic PDE. For certain initial data, we find a positive time, using the mean value of the very solution, at which existence of the solution is no longer guaranteed.
In addition, an application of our method is given.
This concerns a semilinear parabolic PDE with boundary degeneracy recently studied by \cite{Wang13} (surely an extension of problem (\ref{eqn})-(\ref{ic}) described above).

Notation: $u_t=\frac{\partial u}{\partial t}$ and $\Delta$ denotes the Laplace differential operator on $\Omega$ with domain $H^2(\Omega)\cap H^1_0(\Omega)$. The symbols $|u|_p$ and $\|u\|$ denote the norm of $u$ in, respectively, $L^p(\Omega)$ and $L^2(\Omega)$, and $\langle u,u \rangle = \|u\|^2$.
The measure of $\Omega$ is denoted by $|\Omega|:=\int_\Omega \diff x$.
Throughout, $\lambda_1>0$ will denote the Poincar\'{e} constant; $\|u\|^2 \leq \frac{1}{\lambda_1} \|\nabla u\|^2$. We write $H^{-1}(\Omega)$ to denote the dual of the space $H^1(\Omega)$.
Finally, we will commonly identify $u(x,t)\equiv u(t)(x)$; e.g., $u(t)\in H^1_0(\Omega)$, $t>0$, and in many instances we will abbreviate $u(x,t)$ by simply $u$.

\section{The {\em{a priori}} results}  \label{s:2}

The following is the usual notion of a (weak) solution to problem (\ref{eqn})-(\ref{ic}).

\begin{definition}
\label{d:weak-solution}
Let $0<T\leq+\infty$. The function $u$ satisfying
\begin{eqnarray}
u &\in& L^{\infty }(0,T;L^2(\Omega)) \cap L^{2}(0,T;H^1_0(\Omega)), \label{defn-1} \\
\partial_t u &\in& L^2(0,T;H^{-1}(\Omega)), \label{defn-2}
\end{eqnarray}
is said to be a {\sc{weak solution}} to problem (\ref{eqn})-(\ref{ic}) if, for all $\varphi\in H^1_0(\Omega)$, and for almost all $t\in [0,T]$, there holds,
\begin{equation*}
\label{weak-solution-1}
\langle u_t,\varphi \rangle + \nu
\langle \nabla u,\nabla \varphi \rangle + \langle f(u)u,\varphi \rangle = 0.
\end{equation*}
In addition,
\begin{equation*}
\label{weak-solution-3}
u(0)=u_{0}.
\end{equation*}
The function $[0,T]\ni t\mapsto u(t)\in H^1_0(\Omega)$ is called a {\sc{global weak solution}} if it is a weak solution for every $T>0$.
\end{definition}

Here it is well known that (cf. e.g. \cite[Lemma II.3.2]{Temam88}),
\[
u\in C([0,T];L^2(\Omega)),
\]
and for almost all $t\in[0,T]$,
\begin{equation}
\label{Temam}
\frac{\diff}{\diff t}\|u\|^2 = 2\langle u_t,u \rangle.
\end{equation}

\begin{proposition}
\label{MV-lemma}
Let $T>0$. Suppose $u\in C([0,T];L^2(\Omega))$ and $f\in C(\mathbb{R})$.
Then there is $\xi\in C([0,T];\Omega)$ in which
\begin{equation}
\label{xi}
\int_\Omega f(u(x,t))u^2(x,t) \diff x = f(u(\xi(t),t))\int_\Omega u^{2}(x,t) \diff x.
\end{equation}
\end{proposition}

\begin{proof}
The existence of the function $\xi\in C([0,T];\Omega)$ in (\ref{xi}) follows from the Mean Value Theorem for Integrals after some straightforward generalizations. See Theorem \ref{trick}.
\end{proof}

The following theorem provides the general result.
It concerns the behavior of $f$ on solutions that are restricted to the path $(\xi(t),t)\in\Omega\times[0,T]$, where the function $\xi\in C([0,T];\Omega)$ is due to Proposition \ref{MV-lemma}.
Recall that, when $u$ is a solution to problem (\ref{eqn})-(\ref{ic}), then $u(\xi(t),t)\in\mathbb{R}$ for each $t\in[0,T]$.

\begin{theorem}
\label{general-result}
Let $\nu>0$, $f\in C(\mathbb{R})$ and $u_0\in H^1_0(\Omega)$. Suppose $u=u(x,t)$ is a weak solution to problem (\ref{eqn})-(\ref{ic}) for $t\in[0,T]$, for some $T>0$.
Then $u$ satisfies, for all $t\in[0,T]$,
\begin{equation}
\label{general-condition}
\|u(t)\| \leq \exp\left[-\int_0^t f(u(\xi(\tau),\tau))\diff\tau \right] \|u_0\|.
\end{equation}
\end{theorem}

\begin{proof}
Let $u$ be a solution to problem (\ref{eqn})-(\ref{ic}) according to Definition \ref{d:weak-solution}.
Since we are only interested in estimates at the {\em{a priori}} level, we are allowed to formally multiply (\ref{eqn}) by $u=u(x,t)$ in $L^2(\Omega)$ to obtain the identity, which holds for almost all $t\in(0,T)$,
\begin{equation}
\label{id-1}
\langle u_t,u \rangle - \nu \langle \Delta u,u \rangle + \langle f(u),u^2 \rangle = 0.
\end{equation}
First we recall (\ref{Temam}), then Green's first identity to see that there holds,
\[
\nu\langle -\Delta u, u \rangle = \nu\|\nabla u\|^2.
\]
Next, by Proposition \ref{MV-lemma}, we now know that there is $\xi\in C([0,T],\Omega)$ in which,
\begin{equation}
\label{id-2-f}
\langle f(u)u,u \rangle = f(u(\xi(t),t)) \|u\|^2.
\end{equation}
Together, (\ref{id-1}) becomes, for almost all $t\in(0,T)$,
\begin{equation}
\label{id-2}
\frac{1}{2}\frac{\diff}{\diff t}\|u\|^2 + \nu\|\nabla u\|^2 + f(u(\xi(t),t)) \|u\|^2 = 0.
\end{equation}
Omitting the term $\nu\|\nabla u\|^2$ produces the differential inequality,
\begin{equation*}
\frac{1}{2}\frac{\diff}{\diff t}\|u\|^2 + f(u(\xi(t),t)) \|u\|^2 \leq 0,
\end{equation*}
and from this we find (\ref{general-condition}).
\end{proof}

\begin{remark}
After multiplying identity (\ref{id-2}) by $\exp[\int_0^t 2f(u(\xi(\tau),\tau))\diff\tau]$ and integrating with respect to $t$ on $[0,T]$, we arrive at the new identity,
\[\begin{aligned}
& \|u\|^2 + 2\nu\int_0^t \exp\left[-\int_s^t 2f(u(\xi(\tau),\tau))\diff\tau \right]\|\nabla u(x,s)\|^2 \diff s \\
& = \exp\left[-\int_0^t 2f(u(\xi(\tau),\tau))\diff\tau \right]\|u_0\|^2.
\end{aligned}\]
Notice that for any $f\in C(\mathbb{R})$, we recover the bounds
\[
u\in L^\infty(0,T;L^2(\Omega))\cap L^2(0,T;H^1_0(\Omega)),
\]
with dependence on $T>0$.
\end{remark}

With Theorem \ref{general-result}, we may now move onto the consideration of the case when solutions are guaranteed to exponentially decay to zero. Obviously, one immeditely read from (\ref{general-condition}) that exponential decay for solutions $u$ in $L^2(\Omega)$ occurs on the time intervals where
\[
F(t):=\int_0^t f(u(\xi(\tau),\tau)) \diff\tau < 0.
\]
However, given an arbitrary continuous function $f$, we seek conditions, with pragmatic assumptions on $f$, which guarantee solutions $u$ decay to zero exponentially, in $L^2(\Omega)$, as $t\rightarrow+\infty$.
We encounter the first assumption we can make that insures solutions to problem (\ref{eqn})-(\ref{ic}) decay exponentially to zero; when $f$ is {\em{positive}}.

\begin{corollary}
\label{exp-zero-result}
Let $f\in C(\mathbb{R})$ be a positive function on $\mathbb{R}$. Then for any $u_0\in H^1_0(\Omega)$ and $\nu>0$, the corresponding solutions $u$ to problem (\ref{eqn})-(\ref{ic}) are global ones (i.e., $T=+\infty$) and $\|u(t)\|$ exponentially decays to zero as $t\rightarrow +\infty$.
\end{corollary}

Recall that, for all $\varphi\in H^1_0(\Omega)$,
\begin{equation}
\label{Poincare}
\|\varphi\| \leq \frac{1}{\sqrt{\lambda_1}}\|\varphi_x\|,
\end{equation}
where $\lambda_1$ is the fist eigenvalue of the Laplacian with respect to homogeneous Dirichlet boundary conditions. Of course, (\ref{Poincare}) is known as the Poincar\'{e} inequality.

Under certain conditions, solutions will exponentially decay to zero when $f$ is not necessarily positive on all of $\mathbb{R}$. Because of the Poincar\'{e} inequality, $f$ may be allowed to assume some negative values, and we may still guarantee that solutions exponentially decay to zero.
The following extensions are for when $f\in C(\mathbb{R})$ is (eventually) {\em{bounded below}} by $-\nu\lambda_1$.

\begin{theorem}
\label{bounded}
Let $\nu>0$ and $f\in C(\mathbb{R})$.
Suppose $f$ satisfies the lower-bound,
\begin{equation}
\label{below}
\inf_{s\in\mathbb{R}} f(s) > -\nu\lambda_1.
\end{equation}
For any $u_0\in H^1_0(\Omega)$, the corresponding solutions $u$ to problem (\ref{eqn})-(\ref{ic}) are global ones (i.e., $T=+\infty$) and $\|u(t)\|$ exponentially decays to zero as $t\rightarrow +\infty$.
\end{theorem}

\begin{proof}
Let $u_0\in H^1_0(\Omega)$ and assume $u$ is a solution to problem (\ref{eqn})-(\ref{ic}) according to Definition \ref{d:weak-solution}.
After applying the Poincar\'{e} inequality (\ref{Poincare}) to the identity (\ref{id-2}), we arrive at the differential inequality, which holds, for almost all $t\in(0,T)$,
\begin{equation}
\label{bnd-ineq-1}
\frac{\diff}{\diff t}\|u\|^2 + 2\left( \nu\lambda_1 + f(u(\xi(t),t)) \right)\|u\|^2 \leq 0.
\end{equation}
Integrating (\ref{bnd-ineq-1}) with respect to $t$ on $[0,T]$ yields,
\begin{equation}
\label{bnd-ineq-2}
\|u(t)\| \leq \exp \left[ - \int_0^t \left( \nu\lambda_1 + f(u(\xi(\tau),\tau)) \right) \diff \tau \right] \|u_0\|.
\end{equation}
At this point we recall assumption (\ref{below}) which together implies,
\begin{equation}
\label{bound-2}
f(u(\xi(t),t)) > -\nu\lambda_1, \quad\forall t\in[0,T].
\end{equation}
Thus, together (\ref{bnd-ineq-2}) and (\ref{bound-2}) show that $u$ is globally defined and exponentially decays to zero as $t\rightarrow +\infty$.
This completes the proof.
\end{proof}

Of course when $f$ is strictly bounded below by $-\nu\lambda_1$, we are allowed the take any initial data.
For the results that follow, solutions $u$ are assumed to be {\em{positive}}. Similar results for negative solutions can be show with minor modifications.
We will now show that for any initial data, positive solutions $u$ converge to zero exponentially (in $L^2(\Omega)$) when $f(s)$, $s\geq 0$, is bounded below in an appropriate manor.

\begin{theorem}
\label{eventual}
Let $\nu>0$ and $f\in C([0,+\infty))$. Suppose that the average value of $f(\sigma)$ on $(0,s)$ satisfies the lower-bound, for all $s>0$,
\begin{equation}
\label{mvp}
f_{\rm{avg}}(s) = \frac{1}{s} \int_0^s f(\sigma)\diff\sigma > -\nu\lambda_1.
\end{equation}
Then for any $u_0\in H^1_0(\Omega)$, the corresponding solutions $u$ to problem (\ref{eqn})-(\ref{ic}) are global ones (i.e., $T=+\infty$) and $\|u(t)\|$ exponentially decays to zero as $t\rightarrow +\infty$.
\end{theorem}

\begin{proof}
Let $u_0\in H^1_0(\Omega)$ and assume $u$ is a positive solution to problem (\ref{eqn})-(\ref{ic}) according to Definition \ref{d:weak-solution}.
The claim follows directly by applying the assumption (\ref{mvp}) to (\ref{bnd-ineq-2}). Indeed, it suffices to show that,
\[
\int_0^t \left( \nu\lambda_1 + f(u(\xi(\tau),\tau)) \right) \diff \tau >0, \quad\forall t\geq 0,
\]
or
\[
-\nu\lambda_1 t < \int_0^t f(u(\xi(\tau),\tau))\diff\tau.
\]
Thanks to assumption (\ref{mvp}), the proof is complete.
\end{proof}

\subsection*{Examples}

1. First, we note that the well-known problem, the Chafee--Infante reaction diffusion equation,
\[
u_t - \nu\Delta u + u^3 - (\nu\lambda_1) u = 0,
\]
satisfies condition (\ref{below}) because $f_1(s)=s^2-\nu\lambda_1$. (We will consider the finite time blow-up problem motivated by the nonlinear term $-f_1(s)s$ below.)

2. We also give an example of a function satisfying the condition (\ref{mvp}). Let us take $\nu=1$ and $\Omega=(0,1)$ so that $\lambda_1=1$ (cf. e.g. \cite{Hunter&Nachtergaele01} or \cite{Robinson01}). Define
\[
f_2(s) = \left\{ \begin{array}{ll}
1-s & \text{for $0 \leq s < 3$} \\
\frac{1-s}{s-2} & \text{for $3 \leq s$.}
\end{array} \right.
\]
Clearly, $f_2\in C^0([0,+\infty))$, and
\[
f_{2\rm{avg}}(s):=\frac{1}{s}\int_0^s f_2(\sigma)\diff\sigma = \left\{ \begin{array}{ll}
1-\frac{s}{2} & \text{for $0\leq s <3$} \\
-1 -\frac{1}{s}\ln|s-2| +\frac{1}{s}\ln 2 & \text{for $3 \leq s$.}
\end{array} \right.
\]
See Figure \ref{fig3} below for the graphs of $f_2(s)$ and $f_{2\rm{avg}}(s)$.

Some important properties of this function are:
\begin{enumerate}
\item[(i)] Here there holds,
\[
\liminf_{s\rightarrow+\infty}f(s) = -\nu\lambda_1,
\]
and notably $f_2(s)$ converges to $-\nu\lambda=-1$ from {\em{below}}; also

\item[(ii)] the minimum average value of $f_2(s)$, on $(0,+\infty)$, is $-1$ and it is reached asymptotically, when $s\rightarrow+\infty$.
\end{enumerate}

\begin{figure}[htbp]
\centering
\includegraphics[scale=0.6]{fig3}
\caption{The plot of $f_2(s)$ (blue) and $f_{2\rm{avg}}(s)$ (magenta) for $s\in(0,15]$.}
\label{fig3}
\end{figure}

Finally, we give an example of a nonlinear term $f$ satisfying the assumptions of Theorem \ref{eventual} where the nonlinear term $f$ is unbounded below. Consider the singular potential $f_3(s) = -(1-s)^{-p}$ on $s\in [0,1)$ for any $0 < p < 1 - \frac{1}{\nu\lambda_1}$. Then $f_3(s)$ is strictly decreasing on $[0,1)$, and
\[
f_{3{\rm{avg}}}(s) = \int_0^1 -(1-s)^{-p} \diff s = \frac{1}{p-1} > -\nu\lambda_1.
\]

\begin{remark}
Notice that each assumption (\ref{below}) and (\ref{mvp}) implies that the same necessary condition on $f$ and the data $u_0$ holds:
\[
f(u(\xi(0),0)) = \frac{\langle f(u_0),u_0^2 \rangle}{\|u_0\|^2} > -\nu\lambda_1.
\]
\end{remark}

The final result in this section is motivated by the blow-up result in \cite[p. 176]{Zheng04}. Indeed, we give a description of a blow-up condition in terms of the ``mean value'' technique developed thus far.

\begin{theorem}  \label{blow-up-result}
Let $\nu>0$ and $f\in C((0,+\infty))$. Suppose $f$ satisfies the upper- and lower-bounds, for all $s\in (0,+\infty)$,
\begin{equation}  \label{blow-up}
c_1 - \frac{c_2(r+2)}{4} s^r \leq f(s) \leq c_1 - c_2 s^{r},
\end{equation}
for some $c_1\geq 0$, $c_2>0$ and some $r > 2$.
If $u_0\in H^1_0(\Omega)$ ($u_0>0$) satisfies the condition,
\begin{equation}
\label{criteria-blow-up}
\nu\|\nabla u_0\|^2 + c_1\|u_0\|^2 < \frac{c_2}{2} |u_0|^{r+2}_{r+2},
\end{equation}
then the corresponding positive solutions $u$ to problem (\ref{eqn})-(\ref{ic}) possess the interval of existence $(0,t')$, where,
\begin{equation}
\label{time}
t' = \frac{2|\Omega|^{(r-2)/2}}{c_2(r-2) u^2(\xi(t^*),t^*) \|u_0\|^{r-2}},
\end{equation}
for some $t^*=t^*(t)>0$.
\end{theorem}

\begin{proof}
Let $u_0\in H^1_0(\Omega)$ satisfy (\ref{criteria-blow-up}) and assume $u$ is a solution to problem (\ref{eqn})-(\ref{ic}) according to Definition \ref{d:weak-solution}.
This time we apply (\ref{blow-up}) to (\ref{id-2-f}) to find that there holds,
\[\begin{aligned}
\langle f(u)u,u \rangle & = \int_\Omega f(u)u^2 \diff x \\
& \leq \int_\Omega \left( c_1 - c_2 u^r \right) u^2 \diff x \\
& \leq \int_\Omega c_1 u^2 \diff x - \frac{1}{2} \int_\Omega c_2 u^2 u^{r} \diff x - \frac{1}{2} \int_\Omega c_2 u^2 u^{r} \diff x \\
& = c_1 \|u\|^2 - \frac{c_2}{2} |u|^{r+2}_{r+2} - \frac{c_2}{2} u^2(\xi(t),t) |u|^r_r.
\end{aligned}\]
With this, instead of (\ref{id-2}), we find that there holds, for almost all $t\in(0,T)$,
\begin{equation}  \label{blow-up-2}
\frac{1}{2}\frac{\diff}{\diff t}\|u\|^2 + \nu\|\nabla u\|^2 + c_1 \|u\|^2 - \frac{c_2}{2} |u|^{r+2}_{r+2} \geq \frac{c_2}{2} u^2(\xi(t),t) |u|^r_r.
\end{equation}
Define the functional $\mathcal{E}:H^1_0(\Omega)\rightarrow\mathbb{R}$, for all $\varphi\in H^1_0(\Omega)\cap L^{r+2}(\Omega)$, by
\[
\mathcal{E}(\varphi) := \nu\|\nabla \varphi\|^2 + c_1\|\varphi\|^2 - \frac{c_2}{2} |\varphi|^{r+2}_{r+2}.
\]
Let $\mathcal{E}(t)$ denote the functional $\mathcal{E}$ along trajectories $\varphi=u(t)$.
We observe that
\[\begin{aligned}
\frac{1}{2}\frac{\diff\mathcal{E}(t)}{\diff t} & = \langle -\nu\Delta u + \left( c_1 - \frac{c_2}{4}(r+2) u^r \right)u, u_t \rangle \\
& \leq -\|u_t\|^2.
\end{aligned}\]
Thus, for all $t\in(0,T)$,
\[
\mathcal{E}(0) \geq \mathcal{E}(t),
\]
and since (\ref{criteria-blow-up}) holds, $\mathcal{E}(0)<0$.
With this, (\ref{blow-up-2}) becomes
\[
\frac{\diff}{\diff t}\|u\|^2 \geq c_2 u^2(\xi(t),t) |u|^r_r,
\]
whereby with H\"{o}lder's inequality,
\[
|u|^r_r \geq |\Omega|^{(2-r)/2} \|u\|^r,
\]
so we now have, for almost all $t\in(0,T)$,
\begin{equation}  \label{blow-up-3}
\frac{\diff}{\diff t}\|u\|^2 \geq c_2 |\Omega|^{(2-r)/2} u^2(\xi(t),t) \|u\|^r.
\end{equation}
Integrating (\ref{blow-up-3}) with respect to $t$ over $[0,T]$ yields,
\begin{equation}
\label{time-2}
\|u(t)\|^2 \geq \left( \|u_0\|^{2-r} + \frac{c_2(2-r)}{2} |\Omega|^{(2-r)/2} \int_0^t u^2(\xi(\tau),\tau) \diff\tau \right)^{2/(2-r)}.
\end{equation}
Recall that $r>2$; whence, finite time blow-up occurs when
\[
\int_0^t u^2(\xi(\tau),\tau) \diff\tau = \frac{2|\Omega|^{(r-2)/2}}{c_2(r-2) \|u_0\|^{r-2}}.
\]
We now appeal to the Mean Value Theorem for Integrals once again; there is $t^*\in (0,t)$ in which
\[
\int_0^t u^2(\xi(\tau),\tau) \diff\tau = u^2(\xi(t^*),t^*)t.
\]
Thus, the right-hand side of (\ref{time-2}) is singular whenever (note, $t$ is positive),
\[
t = \frac{2|\Omega|^{(r-2)/2}}{c_2(r-2) u^2(\xi(t^*),t^*) \|u_0\|^{r-2}}.
\]
This shows (\ref{time}) as claimed.
\end{proof}

\begin{remark}
The preceding result guarantees the local existence of a solution to problem \eqref{eqn}-\eqref{ic}.
In the proof, $t^*$ depends on $t$, so we cannot claim that we know for which $t$ a finite time blow-up occurs.
Nevertheless, this shows that the ``MVT method'' can be used to show that certain problems cannot possess global solutions.
Below, we will see how the method can be used to show finite time blow-up.
\end{remark}

\section{Application}

In this section we encounter a type of perturbation for the problem considered in \cite{Fujita-66}. The study of the equation,
\[
u_t -\Delta u - u^{\alpha+1} = 0, \quad\alpha>0,
\]
has led to the development of the study of blow-up solutions of PDE.
We will consider, for $x\in(0,1)$ and $t>0$, the asymptotic behavior of positive solutions $u=u(x,t)$ of the semilinear parabolic equation with boundary degeneracy,
\begin{equation}
\label{ap-eqn}
u_t = \left( x^d u_x \right)_x + u^p,
\end{equation}
$d>0$, $p>1$, with the (mixed weighted Neumann and Dirichlet) boundary conditions,
\begin{equation}
\label{ap-bndry}
x^d u_x(0,\cdot)=0, \quad u(1,\cdot)=0,
\end{equation}
and
\begin{equation}
\label{ap-ic}
u(\cdot,0) = u_0(\cdot).
\end{equation}

Of particular interest is the recent work by \cite{Wang13}, which we now report.
First a definition (cf. \cite[Definition 2.1]{Wang13}).

\begin{definition}
\label{def-Wang}
Assume that $0<T\leq +\infty$. A nonnegative function $u$ is called a {\sc{solution}} to problem (\ref{ap-eqn})-(\ref{ap-ic}) if
\begin{enumerate}
\item[(i)] For any $0<\tilde T<T$, $u\in L^\infty((0,1)\times(0,\tilde T))$ with $u_t\in L^2((0,1)\times(0,\tilde T))$ and $x^{d/2}u_x\in L^2((0,1)\times(0,\tilde T))$.

\item[(ii)] For any $0<\tilde T<T$ and any nonnegative $\varphi\in C^1([0,1]\times[0,\tilde T])$ vanishing at $x=1$,
\[\begin{aligned}
\int_0^{\tilde T} \int_0^1 & \left( u_t(x,t)\varphi(x,t) + x^d u_x(x,t)\varphi(x,t) \right) \diff x \diff t \\
& = \int_0^{\tilde T} \int_0^1 u^p(x,t)\varphi(x,t)\diff x \diff t.
\end{aligned}\]

\item[(iii)] $u(1,\cdot) = 0$ on $(0,T)$ and $u(\cdot,0) = u_0(\cdot)$ on $(0,1)$ (in the sense of trace).
\end{enumerate}
\end{definition}

The results we investigate are (cf. \cite[Theorems 2.1, 2.2]{Wang13}).

\begin{theorem}  \label{thm}
Assume $0< d <2$. Then there exists both nontrivial global and blowing-up solutions to problem (\ref{ap-eqn})-(\ref{ap-ic}).
\end{theorem}

\begin{theorem}  \label{blow-up-thm}
Assume $d \ge 2$.
The solution to problem (\ref{ap-eqn})-(\ref{ap-ic}) must blow up in a finite time for any nontrivial $0 \le u_0 \in L^\infty(0,1)$ with $x^{d/2}\partial_x u_0 \in L^2(0,1)$.
\end{theorem}

The purpose of this application is to provide a better description to the states $u_0$ for which the corresponding solution converges exponentially to zero, or leads to finite time blow-up.
This is done in terms of the structural parameters $d$ and $p$.
Our idea employs a similar argument that produces the results in the previous section; i.e., we find an {\em{a priori}} estimate which exploits a generalization of the Mean Value Theorem for Integrals to derive a condition that guarantees solutions $u$ (to a comparable problem) exponentially converge to zero.
Because of the nature of the equation examined here, we are able to explicitly determine a finite time blow-up result as well.

It was crucial to the development of the results in section \ref{s:2} that we explicitly know the value of the first eigenvalue of the Laplacian with respect the the imposed boundary conditions, we now introduce the comparable problem which we will use to illustrate Theorem \ref{thm}.
Observe that any solution to \eqref{ap-eqn} satisfying the homogeneous mixed Neumann--Dirichlet boundary condition
\begin{equation}
\label{ap-bndry2}
u_x(0,\cdot)=0, \quad \text{and} \quad u(1,\cdot)=0,
\end{equation}
also satisfies \eqref{ap-bndry} when $d>0$.
Then recall that the Laplacian on $(0,1)$ subject to the Neumann--Dirichlet boundary conditions \eqref{ap-bndry2} admits the following system of eigenvalues and eigenvectors, \footnote{https://en.wikipedia.org/wiki/Eigenvalues\_and\_eigenvectors\_of\_the\_second\_derivative\#Mixed\_Neumann-Dirichlet\_boundary\_conditions}
\[
\lambda_k=\frac{(2k-1)^2\pi^2}{4}, \quad \phi_k(x)=\sqrt{2}\cos\left( \frac{(2k-1)\pi x}{2} \right), \quad k=1,2,3,\dots.
\]
Whence, on $(0,1)$, the Poincar\'{e} constant (first eigenvalue of the Laplacian subject to \eqref{ap-bndry2}) is $\lambda_1=\frac{\pi^2}{4}$.

Our first result in response to Theorem \ref{thm} is the following

\begin{theorem}
\label{ap-result}
Let $T>0$, $0 < d < 2$ and $p>1$. Suppose $u=u(x,t)$ is a positive solution to problem (\ref{ap-eqn}), (\ref{ap-ic}), and \eqref{ap-bndry2}.
There are $\xi,\chi\in C([0,T];(0,1))$ in which
\begin{equation}  \label{ap-xi}
\int_0^1 x^d u_x^2(x,t) \diff x = \xi^d(t) \int_0^1 u_x^2(x,t) \diff x,
\end{equation}
and
\begin{equation}
\label{ap-chi}
\int_0^1 u^{p+1}(x,t) \diff x = u^{p-1}(\chi(t),t)\int_0^1 u^{2}(x,t) \diff x.
\end{equation}
Furthermore, for any $u_0\in H^1(\Omega)$, $0<d<2$ and $p>1$ satisfying the condition,
\begin{equation}
\label{Wang-cond}
\frac{|u_0|^{p+1}_{p+1}}{\|u_0\|^2} < \frac{\pi^2}{4}\frac{\langle x^d,(\partial_x u_0)^2 \rangle}{\|\partial_x u_0\|^2},
\end{equation}
then the solution of problem (\ref{ap-eqn}), (\ref{ap-ic}), and \eqref{ap-bndry2} exponentially decays to zero for all $t\in(0,T)$.
\end{theorem}

\begin{proof}
Suppose $u$ is a solution to problem (\ref{ap-eqn}), (\ref{ap-ic}), and \eqref{ap-bndry2} in the sense of Definition \ref{def-Wang}.
The existence of the functions $\xi,\chi\in C([0,T];(0,1))$ in (\ref{ap-xi}) and (\ref{ap-chi}) again follows from the Mean Value Theorem for Integrals (see Theorem \ref{trick}). It remains to show the condition which guarantees the solution's exponential decay to zero.
As usual, we begin by multiplying (\ref{ap-eqn}) by $u=u(x,t)$ in $L^2(0,1)$ to obtain the identity,
\begin{equation}
\label{ap-id-1}
\frac{1}{2}\frac{\diff}{\diff t}\|u\|^2 = \left\langle \partial_x\left( x^d u_x \right), u \right\rangle + \left\langle u^{p+1},1 \right\rangle.
\end{equation}
Applying the Mean Value Theorem for Integrals, we obtain,
\[\begin{aligned}
\left\langle \partial_x\left( x^d u_x \right), u \right\rangle & = \int_0^1 \partial_x\left( x^d u_x(x,t) \right) u \diff x = -\int_0^1 x^d u_x^2(x,t) \diff x \\
& = -\xi^d(t) \int_0^1 u_x^2(x,t) \diff x = -\xi^d(t) \|u_x\|^2,
\end{aligned}\]
for some $\xi(t)\in C([0,T],(0,1))$.  For some $\chi(t)\in C([0,T],(0,1))$, we also find
\[\begin{aligned}
\left\langle u^{p+1}, 1 \right\rangle & = \int_0^1 u^{p+1}(x,t) \diff x = \int_0^1 u^{p-1}(x,t) u^2(x,t) \diff x \\
& = u^{p-1}(\chi(t),t) \int_0^1 u^2(x,t) \diff x = u^{p-1}(\chi(t),t)\|u\|^2.
\end{aligned}\]
So (\ref{ap-id-1}) becomes
\begin{equation}
\label{ap-id-2}
\frac{\diff}{\diff t}\|u\|^2 + 2\xi^d(t) \|u_x\|^2 = 2u^{p-1}(\chi(t),t)\|u\|^2.
\end{equation}

Recall $\lambda_1=\frac{\pi^2}{4}$.
Hence, from (\ref{id-2}) we arrive at the differential inequality
\begin{equation}
\label{ap-ineq-1}
\frac{\diff}{\diff t}\|u\|^2 + 2\left( \frac{\pi^2}{4}\xi^d(t) - u^{p-1}(\chi(t),t) \right)\|u\|^2 \leq 0.
\end{equation}
Thus, integrating (\ref{ap-ineq-1}) with respect to $t$ on $[0,T]$ yields,
\begin{equation}
\label{ap-ineq-2}
\|u(t)\| \leq \exp \left[ - \int_0^t \left( \frac{\pi^2}{4}\xi^d(\tau)  - u^{p-1}(\chi(\tau),\tau) \right) \diff \tau \right] \|u_0\|,
\end{equation}
Hence, exponential decay to zero is guaranteed when,
\[
u^{p-1}(\chi(t),t) < \frac{\pi^2}{4}\xi^d(t), \quad \forall t\in(0,T),
\]
and by monotonicity (in this case the decreasing function is $f(s)= -s^{p-1}$ for $s>0$ and $p>1$),
\begin{equation}
\label{ap-cond}
u^{p-1}_0(\chi(0)) < \frac{\pi^2}{4}\xi^d(0).
\end{equation}
Thus, in terms of the initial data $u_0$, thanks to (\ref{ap-xi}) and (\ref{ap-chi}), the condition (\ref{ap-cond}) becomes (\ref{Wang-cond}) as claimed.
This finishes the proof.
\end{proof}

\begin{remark}
The conditions that guarantee finite time blow-up when $0<d<2$ are not as clear. However, there is the partial result. After applying the Cauchy-Schwarz and Poincar\'{e} inequalities to the condition (\ref{Wang-cond}) we see that, when
\[
\frac{|\partial_x u_0|^2_4}{|u_0|^{p+1}_{p+1}} \leq \frac{\pi^2}{4},
\]
then blow-up in finite time is {\em{possible}} for any $0<d<2$ and $p>1$.
\end{remark}

We are now in position to provide a better description for the solutions that converge exponentially to zero.
Using (\ref{Wang-cond}), we are motivated to define the set on which solutions to problem (\ref{ap-eqn}), (\ref{ap-ic}), and \eqref{ap-bndry2} are guaranteed to exponentially decay to zero,
\[
\mathcal{E} = \left\{ u_0\in H^1_0(\Omega) : u_0\not\equiv 0, \frac{\|u_0\|^2 |\partial_x u_0|^2_4}{|u_0|^{p+1}_{p+1} \|\partial_x u_0\|^2} > \frac{4}{\pi^2}\sqrt{1+2d} \right\}.
\]

It is possible to further illustrate this application with an inverse problem. Given $u_0(x)$, one computes $|\partial_x u_0|^2_4$, $\|u_0\|^2$, $|u_0|^{p+1}_{p+1}$, and $\|\partial_x u_0\|^2$.
Since $0<d<2$, the set of $p>1$ which guarantee the corresponding solution $u$ decays exponentially to zero can be implicitly determined from the inequality
\[
\frac{\sqrt{1+2d} |u_0|^{p+1}_{p+1} \|\partial_x u_0\|^2} {\|u_0\|^2 |\partial_x u_0|^2_4} < \frac{4}{\pi^2} \approx 0.405285.
\]
Now the left hand side also defines a function of the two variables $(d,p)$. Hence, when we are given $u_0(x)$, the region containing the value of $(d,p)$ which guarantees the solution converges exponentially to zero can be found on the contour map of the surface described by that function. We illustrate this with three different initial conditions, $u_0(x)=x\sin(\pi x)$ and $u_0(x)=e^{1-x^2}-1$ (see Figures \ref{fig1} and \ref{fig2} below).

\begin{figure}[htbp]
\centering
\includegraphics[scale=0.5]{fig1}
\caption{Given $u_0(x)=x\sin(\pi x)$. Plot of the parameter space: $d\in[0,2]$, $p\in[1,3]$. Parameters $(d,p)$ above the critical contour (the $4/\pi^2$ level curve shown in red) guarantee the solution exponentially decays to zero.}
\label{fig1}
\end{figure}

\begin{figure}[htpb]
\centering
\includegraphics[scale=0.5]{fig2}
\caption{Given $u_0(x)=e^{1-x^2}-1$. Plot of the parameter space: $d\in[0,2]$, $p\in[1,3]$. All parameters $(d,p)$ in this space guarantee the solution exponentially decays to zero.}
\label{fig2}
\end{figure}

Finally, our last result is in response to Theorem \ref{blow-up-thm}.
In this case we consider the original problem \eqref{ap-eqn}-\eqref{ap-ic} described in the application (no knowledge of the best Poincar\'{e} constant is needed here).

\begin{theorem}  \label{ap-result2}
Let $T>0$, $d>0$ and $p>1$. Suppose $u=u(x,t)$ is a positive solution to problem (\ref{ap-eqn})-(\ref{ap-ic}).
There is $\xi\in C([0,T];(0,1))$ in which
\begin{equation}
\label{ap-xi2}
\int_0^1 x^d u_x^2(x,t) \diff x = \xi^d(t) \int_0^1 u_x^2(x,t) \diff x,
\end{equation}
Furthermore, for any $u_0\in H^1(\Omega)$, $d>0$ and $p>1$ satisfying the condition,
\begin{equation}  \label{Wang-cond2}
\langle x^d,(\partial_x u_0)^2 \rangle < \frac{1}{p+1} |u_0|^{p+1}_{p+1},
\end{equation}
then the corresponding positive solutions $u$ to problem (\ref{ap-eqn})-(\ref{ap-ic}) blow-up at the finite time
\begin{equation}  \label{time2}
t' = \frac{p+1}{2p} \frac{1}{\|u_0\|^\frac{p-1}{2}},
\end{equation}
that is, $\lim_{t\rightarrow t'^-} \|u(t)\| = +\infty$.
\end{theorem}

\begin{proof}
The assertion in \eqref{ap-xi2} was already shown in the proof of the previous theorem.

We proceed as in the proof of Theorem \ref{blow-up-result}.
Multiplying \eqref{ap-eqn} by $u$ in $L^2(0,1)$ and applying \eqref{ap-xi} we obtain,
\begin{equation}  \label{ap-poi-0}
\frac{1}{2}\frac{\diff}{\diff t}\|u\|^2 + \langle -\partial_x(x^du_x),u \rangle - |u|^{p+1}_{p+1} = 0.
\end{equation}
A further multiplication of \eqref{ap-eqn} by $u_t$ produces,
\[
\frac{\diff}{\diff t} \left\{ \langle -\partial_x(x^du_x),u \rangle - \frac{1}{p+1}|u|^{p+1}_{p+1} \right\} + \|u_t\|^2 = 0.
\]
We define the functional
\begin{equation}  \label{ap-poi-1}
E(t):=\langle -\partial_x(x^du_x(t)),u(t) \rangle - \frac{1}{p+1}|u(t)|^{p+1}_{p+1},
\end{equation}
and thanks to \eqref{ap-poi-1}, we immediately know that
\[
E(t)\le E(0).
\]
From \eqref{ap-poi-0} we also have,
\[
\frac{1}{2}\frac{\diff}{\diff t}\|u\|^2 + E - \frac{p}{p+1} |u|^{p+1}_{p+1} = 0,
\]
and if $E(0)<0$, then
\[
\frac{1}{2}\frac{\diff}{\diff t}\|u\|^2 \ge \frac{p}{p+1} |u|^{p+1}_{p+1} \ge \frac{p}{p+1} \|u\|^{2\cdot\frac{p+1}{2}}.
\]
We now observe that, due to the complicated degenerate boundary conditions imposed on the solution in \eqref{ap-bndry}, we find following integration by parts and recalling \eqref{ap-xi2},  the condition \eqref{Wang-cond2}.
Indeed, consider
\begin{align*}
E(0) & = \langle -\partial_x(x^d \partial_x u_0),u_0 \rangle - \frac{1}{p+1}|u_0|^{p+1}_{p+1}  \\
& = \langle x^d \partial_x u_0,\partial_xu_0 \rangle - \frac{1}{p+1}|u_0|^{p+1}_{p+1}  \\
& = \xi^d(0)\|\partial_x u_0\|^2 - \frac{1}{p+1}|u_0|^{p+1}_{p+1}  \\
& = \langle x^d,(\partial_x u_0)^2 \rangle - \frac{1}{p+1}|u_0|^{p+1}_{p+1}.  \\
\end{align*}
Now following some of the calculations as in the proof of Theorem \ref{blow-up-result}, we readily find the blow-up time \eqref{time2} as claimed.
This finishes the proof.
\end{proof}

\section*{Acknowledgments}

The author would like to thank the referees for their generous comments when reading the manuscript.

\appendix
\section{}

Here we report the Mean Value Theorem for Integrals that is used throughout this paper.
The interested reader can also see \cite[Chapter 7]{SR}.

\begin{theorem}
\label{trick}
Let $T>0$ and $\Omega$ be a bounded domain (open and connected) in $\mathbb{R}^N$.
Let $f,\varphi\in C([0,T];L^1(\Omega))$. Suppose $\varphi(t)$ is nonnegative a.e. on $\Omega$.
Then there is $\xi(t)\in C([0,T];\Omega)$ in which, for all $t\in[0,T]$,
\[
\int_\Omega f(x,t)\varphi(x,t) \diff x = f(\xi(t),t)\int_\Omega\varphi(x,t) \diff x,
\]
that is,
\[
\langle f(t),\varphi(t) \rangle_{L^2(\Omega)} = f(\xi(t),t) |\varphi(t)|_{L^1(\Omega)}.
\]
\end{theorem}

\begin{proof}
Denote by $C_c(\Omega)$ the continuous functions on $\Omega$ that are compactly supported in $\Omega$ (recall that such functions are dense in $L^p(\Omega)$, $1\leq p<\infty$).
Fix $t'\in[0,T]$ and let $\tilde f(t'),\tilde\varphi(t')\in C_c(\Omega)$, where $\tilde\varphi(t')$ is nonnegative on $\Omega$.
By the Extreme Value Theorem for continuous functions, there exist $x_m$ and $x_M$ in $\Omega$ so that
\[
m(t'):=\min_{x\in\Omega}\left\{\tilde f(x,t')\right\} = \tilde f(x_m,t')
\]
and
\[
M(t'):=\max_{x\in\Omega}\left\{\tilde f(x,t')\right\} = \tilde f(x_M,t').
\]
Hence, there holds,
\[
m(t') \int_\Omega \tilde\varphi(x,t') \diff x \leq \int_\Omega \tilde f(x,t')\tilde\varphi(x,t') \diff x \leq M(t') \int_\Omega \tilde\varphi(x,t') \diff x.
\]
Since $\tilde\varphi(t')$ is nonnegative, $|\tilde\varphi(t')|_{L^1(\Omega)}=\int_\Omega \tilde\varphi(x,t') \diff x$. In the case when $\tilde\varphi(t')=0$, we obtain equality and identity. So in the case when $\tilde\varphi(t')$ is nonzero, we obtain
\[
m(t') \leq \frac{1}{|\tilde\varphi(t')|_{L^1(\Omega)}}\int_\Omega \tilde f(x,t')\tilde\varphi(x,t') \diff x \leq M(t').
\]
Since $\tilde f(x,t')$ is continuous and $\tilde f(x,t')\in[m(t'),M(t')]$, then the Intermediate Value Theorem provides an $\xi(t')\in\Omega$ in which
\[
\tilde f(\xi(t'),t') = \frac{1}{|\tilde\varphi(t')|_{L^1(\Omega)}}\int_\Omega \tilde f(x,t')\tilde\varphi(x,t') \diff x.
\]
Thus,
\[
\int_\Omega \tilde f(x,t')\tilde\varphi(x,t') \diff x = \tilde f(\xi(t'),t') \int_\Omega \tilde\varphi(x,t') \diff x.
\]
By the density of $C_c(\Omega)\subset L^1(\Omega)$, we also have for $f(t'),\varphi(t')\in L^1(\Omega)$,
\[
\int_\Omega f(x,t')\varphi(x,t') \diff x = f(\xi(t'),t') \int_\Omega \varphi(x,t') \diff x,
\]
or
\begin{equation}
\label{mvt-1}
\langle f(t')\varphi(t') \rangle_{L^2(\Omega)} = f(\xi(t'),t') |\varphi(t')|_{L^1(\Omega)}.
\end{equation}
So far we have shown that, for each fixed $t'\in[0,T]$, there exists $\xi(t') \in \Omega$ in which (\ref{mvt-1}) holds.
It remains to show that the map $[0,T]\ni t\mapsto \xi(t)\in\Omega$ is continuous.
Let $t^*\in[0,T]$ and $(t_n)_{n\in\mathbb{N}_{>0}}$ be such that $t_n\rightarrow t^*$ as $n\rightarrow\infty$.
Then with the continuity of $f$ and $\varphi$ on $[0,T]$,
\[\begin{aligned}
\lim_{n\rightarrow \infty} f(\xi(t_n),t_n) & = \lim_{n\rightarrow \infty} \left( \frac{1}{|\varphi(t_n)|_{L^1(\Omega)}} \langle f(t_n),\varphi(t_n) \rangle_{L^2(\Omega)} \right) \\
& = \frac{1}{|\varphi(t^*)|_{L^1(\Omega)}} \langle f(t^*),\varphi(t^*) \rangle_{L^2(\Omega)} = f(\xi(t^*),t^*).
\end{aligned}\]
Therefore, $\xi\in C([0,T];\Omega)$. This finishes the proof.
\end{proof}

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\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }

\providecommand{\MRhref}[2]{
  \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
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TITLE: Question about open mapping theorem
QUESTION [5 upvotes]: Let, $f:\Omega\to \mathbb C$ be a non constant anlytic function on an open set $\Omega \subset \mathbb C$.For $r>0$ let $\mathbb D_r=\{z\in \mathbb C:|z|<r\}$ and let $\bar{ \mathbb D_r}$ be its closure. Which of the followings are true ?
(A) If  $\bar{ \mathbb D_1}\subset f(\Omega)$ then $\mathbb D_r\subset f(\Omega)$ for some $r>1$.
(B) If $\bar{ \mathbb D_1}\subset f(\Omega)$ then $\mathbb D_r= f(\Omega)$ for some $r>1$.
(C) If $\bar{ \mathbb D_1}\subset f(\Omega)$ then $\bar{\mathbb D_r}\subset f(\Omega)$ for some $r>1$.
(D) $f(\Omega)$ is open.
As, $\Omega$ is an open subset of $\mathbb C$ , so by open mapping theorem $f(\Omega)$ is open. So, option (D) is true.
But I have no idea about the other options.
Please help...

REPLY [1 votes]: First note that (A) and (C) are equivalent.
The implication “(C) ⇒ (A)” is trivial, and the other direction “(A) ⇒ (C)” is almost trivial: If $\mathbb D_r ⊂ f(Ω)$, then for $r’ = r/2$, you have $\overline{\mathbb D_{r’}} ⊂ \mathbb D_r$.
For (B), take a very trivial map $ℂ → ℂ$ and restrict it to a non-disc shape $Ω$ containing $\overline{\mathbb D_1}$.

REPLY [0 votes]: Option b : We can give a counter example , $\Omega= {|z|<2} $  union  $  |z-10|<1$ (not connected) with $f(z)= z$. So, this works only because we don't need to have the domain to be connected.
Option d : If the domain is not connected, then we could have a function that gives different constants on the different components as a counterexample.  If the domain is connected, then we apply the open mapping theorem to imply that it is true.
Option a and c : ( Since thay are equivalent as mentioned in another answer )  Suppose $f(\Omega)$ is open. We have that the closed unit disc is contained in it. For every point on the boundary, there is an open disc centred at that point which lies completely inside $f(\Omega)$. The collection of all such open discs is an open cover for the boundary $|z|=1$. Since this is compact, this open cover has a finite cover. From the finite cover (which is a finite number of open discs) choose the radius of the smallest disc. This radius works for every point on the boundary. From here we can conclude that a ( and therefore c ) are true.


TITLE: notation convention of vectors in differential calculus and integration
QUESTION [0 upvotes]: Recently I've been reviewing my calculus 2 materials and noticed a strange inconsistency in the notation.
So, in differential calculus, a vector is usually noted as a column vector, but in Lebesgue-Integration, they are noted as row vectors. Is this just a convention, or is there some meaning to this im missing?

REPLY [2 votes]: Integration and differential calculus are closely releted topics and there is no reason why one should use row vectors in one and column vectors in the other. The choice between row vectors in one and column vectors  is a personal one and one can use either one in any topic.


TITLE: Functional Analysis: Folland Problem
QUESTION [3 upvotes]: Folland problem 5.36 (b)
Let $\mathcal{X}$ be a separable Banach space and let $\mu$ be counting measure on $\mathbb{N}$. Suppose that $\{x_{n}\}_{n=1}^{\infty}$ is a countable dense subset of the unit ball of $\mathcal{X}$, and define $T:L^{1}(\mu)\to \mathcal{X}$ by $Tf = \sum_{1}^{\infty}f(n)x_{n} $.
Show that $T$ is surjective.
My approach: Consider any $x\in\mathcal{X}$ and let $x_{n}$ be a sequence such that $x_{n}\to x$ (since $\mathcal{X}$ is separable this can be done). Now consider $f^{k}$ such that $f^{k}(k) = 1$ and $f^{k}(j)= 0$ if $j\neq k$. Clearly $Tf^{k} = x_{k}\to x$. But now the problem is that $f^{k}$ doesn't converge.

REPLY [0 votes]: Here is a complete proof of 36
Problem 36 - Let $\mathcal{X}$ be a separable Banach space and let $\mu$ be counting measure on $\mathbb{N}$. Suppose that $\{x_n\}_{1}^{\infty}$ is a countable dense subset of the unit ball of $\mathcal{X}$, and define $T:L^{1}(\mu)\rightarrow \mathcal{X}$ by $Tf = \sum_{1}^{\infty}f(n)x_n$.
a.) $T$ is bounded.
b.) $T$ is surjective.
c.) $\mathcal{X}$ is isomorphic to a quotient space of $L^{1}(\mu)$ (Use Exercise 35).
Proof a.) - We have that $Tf = \sum_{1}^{\infty}f(n)x_n$, hence  $$\| Tf\|  = \| \sum_{1}^{\infty}f(n)x_n\| \leq \sum_{1}^{\infty}|f(n)|\| x_n\| \leq \sum_{1}^{\infty}|f(n)| = \| f\|$$
Thus $T$ is bounded.
Proof b.) - Since $T$ is linear, it's enough to show that every $x\in X$ with $||x||\leq 1$ is in the image of $T$. To do this, we proceed inductively: there exists $x_{n_1}$ such that $||x-x_{n_1}||<\frac{1}{2}$. If $y=2(x-x_{n_1})$ then $||y||<1$, so there exists $x_{n_2}\neq x_{n_1}$ such that $||y-x_{n_2}||<\frac{1}{2}$, hence $||x-x_{n_1}-\frac{1}{2}x_{n_2}||<\frac{1}{4}$. In general, if $x_{n_1},\dots,x_{n_k}$ have been chosen such that
$$ \Big|\Big|x-\sum_{j=1}^k2^{1-j}x_{n_j}\Big|\Big|<2^{-k}$$
then $y=2^k(x-\sum_{j=1}^k2^{1-j}x_{n_j})$ is in the unit ball, hence there exists $x_{n_{k+1}}\not\in\{x_{n_1},\dots,x_{n_k}\}$ such that $||y-x_{n_{k+1}}||<\frac{1}{2}$, i.e.
$$ \Big|\Big|x-\sum_{j=1}^{k+1}2^{1-j}x_{n_j}\Big|\Big|<2^{-k-1}$$
Finally, define $f\in \ell^1(\mathbb{N})$ by $f(n)=2^{1-k}$ if $n=n_k$, $f(n)=0$ otherwise. Then $Tf=x$.
proof c.) - We have that  $L^{1}(\mu)$ and $\mathcal{X}$  are Banach spaces and $T\in L(L^{1}(\mu),\mathcal{X})$. Moreover,
$T$ is surjective, so $\textrm{range}(T)= \mathcal{X}$ and $\mathcal{X}$ is closed in $\mathcal{X}$. So by exercise 35.
$L^{1}(\mu)/\mathcal{N}(T)$ is isomorphic to $\mathcal{X}$.


TITLE: Probability for intersecting lines
QUESTION [0 upvotes]: 20 lines in the plane intersect with each other at distinct points. Find the probability for 3 such points to form a triangle.
Well, I assume we have 20C2 = 190 such points. Then what? We calculate 190C3 to find how many triangles we can form?
The condition we are examining is for the 3 points not to be collinear.

REPLY [2 votes]: If we assume there are no "accidental" colinearities, you can form a triangle whenever the three points are not all on the same line.  Each line has $19$ points on it, so there are $20 {19 \choose 3}=19380$ sets of three colinear points.  There are ${190 \choose 3}=1125180$ ways to choose three points so the chance you can form a triangle is $1-\frac {19380}{1125180}=\frac{970}{987}\approx 0.983$
Another way to get there is to pick the first point.  To have the three points be colinear the second point needs to be on one of the two lines going through the first point, which is a probability of $\frac {38}{189}$.  The third point needs to be on the same line as the first two, which is a probability of $\frac {18}{188}$.  This gives a chance for no triangle of $\frac {38 \cdot 18}{189 \cdot 188}=\frac {17}{987}$ like the first approach.


TITLE: When is Lebesgue integration useful over Riemann integration in physics?
QUESTION [29 upvotes]: Riemann integration is fine for physics in general because the functions dealt with tend to be differentiable and well behaved. Despite this, it's possible that Lebesque integration can be more powerfully used even in physical situations that can be solved by Riemann integration. So my questions is:
In solving physics problems, when is Lebesque integration useful over Riemann integration?

REPLY [5 votes]: I would like to point out a completely practical example here, too. You can also need to change the order of integration and summation, or integration and derivative is some calculations, i.e., $\int dx\sum_n \to \sum_n \int dx$, or $\int dx\, \partial/\partial t \to \partial/\partial t \int dx$. While this is a problem with Riemann integration, it works for the Lebesgue integral, under certain assumptions, which are, in physical systems, usually fulfilled.


TITLE: Combining functions
QUESTION [0 upvotes]: I want to know how to represent functions by one formula which are defined by multiple formula.
For example:
The function $f(x)$ defined as follows
$$f(x)=x,0\leq x< 1 \text{ and } f(x)=2x-1,1\leq x\leq2$$
could be represented as $$f(x)=\frac{3}{2}(x-\frac{1}{3})+\frac{1}{2}|x-1|$$
Then, I was working on the following function
$$ f(y)=y^2-5y,0<y\leq6, \text{ and } f(y)=-y^2-5y,-6<y \leq0$$
I wanted to know is there a method using which such functions could be combined to a single formula

REPLY [1 votes]: In that particular example
$$
f(y)=|y|(y-5).
$$
With the new formulation, we have
$$
f(y)=y\,|y|-5\,y.
$$


TITLE: Pair production and recombination with a more energetic beta particle
QUESTION [0 upvotes]: Can gamma particles with sufficient energy for pair production gain energy by recombining their positron with a more energetic beta particle than the one they produced? If yes, is this something that would happen to any significant degree in a sample of radioactive material decaying with high energy gammas and betas eg. Na24?

REPLY [1 votes]: Pair production with a gamma ray

Feynman diagram of electron–positron pair production. One must calculate multiple diagrams to get the net cross section

needs a field to  interact with, in the diagram the charged field of a nucleus, for momentum conservation in the center of mass of the final electron-positron pair. A  gamma has mass zero, the pair at least two electron masses, so a single gamma cannot decay to a pair.
Once the pair is produced the original gamma disappears. The positron of the interaction will have an energy momentum vector certainly smaller than the incoming gamma. If the positron interacts with an electron two fresh gammas will be produced (at the low nuclear energy levels your question assumes).  If the electron has high energy, it will be distributed according to the probability of the interaction, on the two new gammas.
At the level of nuclear physics it is not possible energetically to get one of the two  outgoing gammas to have higher energy than the  original incoming one.