I have $u(x,t)$ - solution of Cauchy problem $$u_t=u_{xx},~~~u(x,0)=e^{-x^2},$$ where $t>0, x\in\mathbb{R}.$
Is there a way to find such a limit? I have seen several theorems which help to find value of $\lim\limits_{t \to \infty} u(x,t),$ but this integral realy spoils everything.
Thank you.
The $L_p$ solution to the problem can be obtained by convolution with Gauss-Wierstrass kernel:
$$ u(x,t)= W_t*f(x)=\frac{1}{\sqrt{4\pi t}}\int f(y)e^{\frac{-|x-y|^2}{4t}}\,dy $$
where $f(x)=e^{-x^2}$. The kernel $W_t$ is a good approximation to the identity and so, $W_t*f\rightarrow f$ in $L_p$ ($1\leq p<\infty$ ) and point wise to $f$ (it should be at any Lebesgue point of $f$ fut $f(x)=e^{-x^2}$ is continuous so every point in $\mathbb{R}$ is a Lebesgue point of $f$. In fact, the problem is simple here since $f(x)$ is uniformly continuous.
Here are some generalities of approximation to identity. You have to show that $W_t(x)=t^{-1}W_1(x/t)$, where $W_1(y)=\frac{1}{\sqrt{4\pi}}e^{-y^2}$ satisfies the conditions (1)--(3) described below:
Consider a collection $\{K_\varepsilon:\varepsilon>0\}\subset\mathcal{L}_1(\mathbb{R}^n,\lambda_n)$ that satisfy the following properties:
Theorem. Suppose $\{K_\varepsilon:\varepsilon>0\}\subset\mathcal{L}_1(\mathbb{R}^n,\lambda_n)$ satisfy (1)--(3) above. Then, for any $f\in\mathcal{L}_p(\mathbb{R}^n,\lambda_n)$, $1\leq p<\infty$, \begin{aligned} \lim_{\varepsilon\rightarrow0}\|f* K_{\varepsilon} - a\,f\|_p = 0. \end{aligned}
If $f\in\mathcal{L}_\infty(\mathbb{R}^n,\lambda_n)$ is continuous at a point $x$, then $\lim_{\varepsilon\rightarrow0}f*K_\varepsilon(x)=f(x)$. If $f$ is bounded and uniformly continuous, then $f*K_\varepsilon$ converges to $f$ uniformly as $\varepsilon\rightarrow0$.
Proof:
Let $M=\sup_{\varepsilon>0}\|K_\varepsilon\|_1$. If $f\in\mathcal{L}_p(\mathbb{R}^n,\lambda_n)$, $1\leq p<\infty$, and application of the generalized Minkowski's inequality gives
\begin{aligned}
\|f* K_\varepsilon -af\|_p &\leq \Big(\int\Big(\int_{\mathbb{R}^n}|f(x-y)-f(x)|
|K_\varepsilon(y)|\,dy\Big)^p\,dx\Big)^{1/p}\\
& \leq \int_{\mathbb{R}^d}\Big(\int|f(x-y)-f(x)|^p\,dx\Big)^{1/p}|K_\varepsilon(y)|\,dy\\
&=\int\|\tau_yf-f\|_p|K_\varepsilon(y)|\,dy.
\end{aligned}
where $\tau_y$ is the translation operator. Recall that $\lim_{y\rightarrow0}\|\tau_yf-f\|_p=0$ for all $f\in\mathcal{L}_p$. Assumption (2) implies that for any $\eta>0$, there exists $\delta>0$ such that $M\|\tau_yf-f\|_p<\eta/2$ whenever $|y|\leq\delta$. By assumption (3), for some $\varepsilon'>0$ we have
$2\|f\|_p\int_{|y|>\delta}|K_\varepsilon(x)|\,dx<\eta/2$ whenever $\varepsilon<\varepsilon'$. Combining these facts, we obtain
\begin{aligned}
\|f*K_\varepsilon-af\|_p\leq&
\int_{|y|\leq\delta}\|\tau_yf-f\|_p|K_\varepsilon(y)|\,dy\\
&\quad + \int_{|y|>\delta}\|\tau_yf-f\|_p|K_\varepsilon(y)|\,dy
\leq \frac{\eta}{2} + \frac{\eta}{2}
\end{aligned}
whenever $0<\varepsilon<\varepsilon'$.
The second statement follows similarly. Let $\eta>0$ be fixed. If $f$ is continuous at $x$, then for some $\delta>0$, $|x-u|\leq\delta$ implies that $|f(x)-f(u)|<\frac{\eta}{2 M}.$ For such $\delta>0$, there is $\varepsilon'>0$ such that $2\|f\|_{\infty}\int_{|x|>\delta}|K_\varepsilon(x)|\,dx<\frac{\eta}{2}$ whenever $0<\varepsilon<\varepsilon'$. Putting these statements together gives \begin{aligned} |f*K_\varepsilon(x)&-af(x)|\leq \int |f(x-y)-f(x)||K_\varepsilon|(y)\,dy\\ &\leq \int_{|y|\leq\delta}+\int_{|y|>\delta}|f(x-y)-f(x)||K_\varepsilon|(y)\,dy \leq \frac{\eta}{2}+\frac{\eta}{2} \end{aligned} If $f$ is bounded and uniformly continuous, then $\delta>0$ can be chosen so that $$\sup_{|v-u|<\delta}|f(u)-f(u)|<\frac{\eta}{2M}.$$ Uniform convergence follows immediately.
Solution 1. Since $(x, t) \mapsto u(-x, t)$ solves the same Cauchy problem, it follows that $u(x, t) = u(-x, t)$. So,
$$ \int_{0}^{\infty} u(x, t) \, \mathrm{d}x = \frac{1}{2} \int_{-\infty}^{\infty} u(x, t) \, \mathrm{d}x. $$
Now, the heat equation preserves the total mass, hence the integral is equal to
$$ \frac{1}{2} \int_{-\infty}^{\infty} u(x, 0) \, \mathrm{d}x = \frac{\sqrt{\pi}}{2}. $$
Since the integral does not depend on $t$, the answer is still $\sqrt{\pi}/2$.
Solution 2. Alternatively, note that the heat kernel
$$ \Phi_t(x) = \frac{1}{\sqrt{4\pi t}} e^{-x^2/4t} $$
solves the heat equation $\partial_t \Phi_t = \partial_{xx} \Phi_t$ and that $u(x, 0) = \sqrt{\pi} \Phi_{1/4}(x)$. So, it follows that
$$ u(x, t) = \sqrt{\pi} \Phi_{(1/4)+t}(x). $$
From this, it is easy to deduce that $\int_{0}^{\infty} u(x, t) \, \mathrm{d}x = \sqrt{\pi}/2$ and hence the limit is $\sqrt{\pi}/2$ as well.
I believe you have that: $$u(x,t)=e^{-x^2}f(t)$$ and so you want: $$\lim_{t\to\infty}f(t)\int_0^\infty e^{-x^2}dx$$ assuming that this is interchangeable goes to: $$\frac{\sqrt{\pi}}{2}\lim_{t\to\infty}f(t)$$
