Find $\lim_{t\to\infty}u(1,t)$, where $u(x,t)$ is a solution of $\frac{\partial u}{\partial t}-\frac{\partial^2u}{\partial x^2}=0$

Question

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. $-1/2$

2. $1/2$

3. $-1$

4. $1. $

How to find this limit? No boundary conditions are given.

Answer 1

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.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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.

Answer 3

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 $(u(+\infty,0)+u(-\infty,0))/2$.