Find $\lim_{t\to\infty}u(1,t)$, where $u(x,t)$ is a solution of $\frac{\partial u}{\partial t}-\frac{\partial^2u}{\partial x^2}=0$ Let $u(x,t)$ be a solution of $$\frac{\partial u}{\partial t}-\dfrac{\partial^{2}u}{\partial x^{2}}=0\text{ with}\\ u(x,0)=\frac{e^{2x}-1}{e^{2x}+1}.$$Then $\lim\limits_{t\to\infty} u(1,t)$ is equal to


*

*$-1/2$

*$1/2$

*$-1$

*$1. $ 
How to find this limit? No boundary conditions are given.
 A: The question is not really clear about what boundary conditions should be assumed or even what the domain is. One possible interpretation is to consider homogeneous Neumann boundary conditions on $[-M,M]$ and consider $M \to \infty$. In this case, the total integral over the domain is preserved, and the limiting solution in time is constant (to be stationary a function must be linear and a linear function satisfying homogeneous Neumann boundary conditions is constant).
Thus the density would be $\frac{1}{2M} \int_{-M}^M \frac{e^{2x}-1}{e^{2x}+1} dx$. Although the integral does not converge, with the division outside it does, and so you could consider sending $M \to \infty$ here to answer the question. Under these assumptions the answer is actually $0$, so presumably these are not your desired assumptions. A different assumption would be that the domain is $[0,\infty)$ with a homogeneous Neumann condition at zero; in this case the limit would be $1$, however.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
Set ${\rm y} = x/\root{t}$. Then,
\begin{align}
\partiald{{\rm u}}{t} & = \totald{{\rm u}}{y}\,\partiald{y}{t} =
-\,\half\,{x \over t^{3/2}}\,\totald{{\rm u}}{y} =
-\,\half\,{y \over t}\,\totald{{\rm u}}{y}
\\[3mm]
\partiald{{\rm u}}{x} & = \totald{{\rm u}}{y}\,\partiald{y}{x} =
{1 \over t^{1/2}}\,\totald{{\rm u}}{y}\,,\quad\imp\quad
\partiald[2]{{\rm u}}{x} = {1 \over t^{1/2}}\,\totald[2]{{\rm u}}{y}\,\partiald{y}{x} =
{1 \over t}\,\partiald[2]{{\rm u}}{y}\,,\quad
\end{align}
${\rm u}$ satisfies
$$
\totald[2]{{\rm u}}{y} + \half\,y\,\totald{{\rm u}}{y} = 0
\quad\imp\quad
\totald{}{y}\bracks{\expo{y^{2}/2}\,\totald{{\rm u}}{y}} = 0
\quad\imp\quad\totald{{\rm u}}{y} = A\expo{-y^{2}/2}
$$

\begin{align}
{\rm u} & =
\left.{\rm u}\right\vert_{y\ \to\ \infty} +
A\int_{\infty}^{y}\expo{-\xi^{2}/2}\,\dd\xi
\\[3mm]
{\rm u}\pars{x,t} & =
{\rm u}\pars{x,0^{+}} +
A\int_{\infty}^{x/\root{t}}\expo{-\xi^{2}/2}\,\dd\xi
\end{align}

Then,
$$
\lim_{t \to \infty}{\rm u}\pars{1,t} = {\expo{2} - 1 \over \expo{2} + 1}
-A\ \overbrace{\int_{0}^{\infty}\expo{-\xi^{2}/2}\,\dd\xi}
^{\ds{=\ \root{{\pi \over 2}}}}
$$

You still need another condition ( boundary ? ) to get the $A$ constant.
A: From the absence of the boundary conditions I assume it's a Cauchy problem. Note that $u(x,0)\to1$, $x\to+\infty$ and  $u(x,0)\to-1$, $x\to-\infty$. And the convergence is exponential in speed. So it's straightforward to check that $v(x)=u(x,0)-\text{sign}(x-1)\in L_1(\mathbb R)$. Denote $\Gamma(x,t)=(4\pi t)^{-1/2}e^{-x^2/(4t)}$ the fundamental solution of the heat equation. It's even w.r.t. $x$ so
$$
u(1,t)=\int_{-\infty}^\infty \Gamma(1-y,t)u(y,0)\,dy=
$$
$$
\int_{-\infty}^\infty \Gamma(1-y,t)\text{sign}(y-1)\,dy+
\int_{-\infty}^\infty \Gamma(1-y,t)v(y)\,dy=
\int_{-\infty}^\infty \Gamma(1-y,t)v(y)\,dy.
$$
Young's inequality gives
$$
\left|\int_{-\infty}^\infty \Gamma(1-y,t)v(y)\,dy\right|\le
\int_{-\infty}^\infty \Gamma(1-y,t)|v(y)|\,dy\le
$$
$$
\sup_{x\in\mathbb R}|\Gamma(x,t)|\|v\|_{L_1(\mathbb R)}=
(4\pi t)^{-1/2}|\|v\|_{L_1(\mathbb R)}\to0\quad  \text{ as } t\to\infty.
$$
So the answer is $0$ and it doesn't depend on $x$. Actually it is equal to $(u(+\infty,0)+u(-\infty,0))/2$.
