$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$ with $u(x,0)=\frac{e^{2x}-1}{e^{2x}+1}$. Then $\lim_{t \to \infty}u(1,t)$ equals Let $u(x,t)$ be a bounded solution of $\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$ with $u(x,0)=\frac{e^{2x}-1}{e^{2x}+1}$. Then $\lim_{t\to \infty}u(1,t)$ equals:
In the above problem, the boundary is not given but let's assume that the boundary conditions are as follows: $u(0,t)=u(\pi,t)=0$. Then how to solve for the bounded equation and how to find the above limit?
I will show my working below: with the given (assumed) initial conditions,
$$u(x,t)=\sum_{n=1}^{\infty}a_n e^{-\frac{n^2\pi^2t}{\pi^2}}\sin(\frac{n\pi x}{\pi})=\sum_{n=1}^{\infty}a_n e^{-{n^2t}}\sin(nx)$$
Now,  $u(x,0)=f(x)=\frac{e^{2x}-1}{e^{2x}+1}=\sum_{n=1}^{\infty}a_n \sin(nx)$, which is a fourier sine series, thus the constants $a_n$ are given by:
$$a_n=\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx$$
But I am unable to determine $a_n$ or solve the problem. Can someone please help me solve this problem?
Thanks in advance.
The ans is :

$ -\frac{1}{2}$

 A: Case $x\in \mathbb{R}$
Here I suppose that the domain is $(x,t)\in \mathbb{R}\times \mathbb{R}^+$ and compute the solution as convolution of the initial datum $u(x,0)=\frac{e^{2x}-1}{e^{2x}+1}=\tanh(x)$ with the fundamental solution:
$$
u(1,t)=\frac{1}{\sqrt{4\pi t}}\int_\mathbb{R} e^{-\frac{|y-1|^2}{4t}} \tanh(y) \,dy. 
$$
With the change of variable $w=\frac{y-1}{2\sqrt{t}}$ the integral becomes
$$
2\sqrt{t}\int_\mathbb{R} e^{-w^2} \underbrace{\frac{e^{4\sqrt{t}w+2}-1}{e^{4\sqrt{t}w+2}+1}}_{=:f(w,t)} \,dw.
$$
Now we split the domain in $(-\infty,0)$ and $[0,\infty)$. The integrand satisfies $|f(w,t)|\leq e^{-w^2}\in L^1(\mathbb{R})$ hence we can apply the Dominate convergence theorem and get
$$
\int_{-\infty}^0 e^{-w^2} \frac{e^{4\sqrt{t}w+2}-1}{e^{4\sqrt{t}w+2}+1} \,dw\to \int_{-\infty}^0 e^{-w^2}(-1) \,dw=-\frac{\pi}{2},\qquad
\int_0^{+\infty} e^{-w^2} \frac{e^{4\sqrt{t}w+2}-1}{e^{4\sqrt{t}w+2}+1} \,dw\to \int_{0}^{+\infty} e^{-w^2} \,dw=\frac{\pi}{2}, \quad \text{as} \quad t \to +\infty
$$
Hence we obtain $u(1,t) \to 0$ as $t\to +\infty$.

$\lim\limits_{t\to \infty}u(1,t)=-1/2$ if $u(x,0)\equiv 0$ for $x> 0$.
The only way I got $\lim_{t\to \infty}u(1,t)=-1/2$ as answer is by using the initial condition $u(x,0)=\tanh(x)$ in $(-\infty,0)$ and defining it $u(x,0)\equiv 0$ for $x\geq 0$. Use the computation above only for $x<0$.


Case $x\in \mathbb{R}^+$
Some computations using the domain $(0,\infty)$ with the additional boundary condition $u(0,t)=0$, which is consistent with $u(x,0)=\tanh(x)$, gives the same result since $u(x,0)$ is odd.
A: I don't know if this is what you want. It is easy to see
$$ |a_n|\le\frac{2}{\pi}\int_{0}^{\pi}\frac{e^{2x}-1}{e^{2x}+1}dx\le2.$$
So for $t\ge N>0$,
$$\sum_{n=1}^{\infty}\bigg|a_n e^{-{n^2t}}\sin(nx)\bigg|\le2\sum_{n=1}^{\infty} e^{-{n^2N}}$$
converges and hence $\sum_{n=1}^{\infty}a_n e^{-{n^2t}}\sin(nx)$ converges uniformly for any $t\ge N$ and $x\in[0,\pi]$. Thus
$$ \lim_{t\to\infty}\sum_{n=1}^{\infty}a_n e^{-{n^2t}}\sin(nx)=\sum_{n=1}^{\infty}\lim_{t\to\infty}a_n e^{-{n^2t}}\sin(nx)=0. $$
From this, you can see $\lim_{t\to\infty}u(1,t)=0$.
