Suppose I have a function $u_0:\mathbb R\to \mathbb R^+$ with compact support contained in the interval $[0,1]$ and such that $\int_0^1u_0(r)dr=1$.
I would like to analyze the function $$v(t,r)=u_0((r-r_0)e^t+r_0)e^t$$ where $r_0=\int_0^1ru_0(r)dr$.
I can prove that $\int_0^1v(t, r)dr=1$ and that since $u_0$ has compact support the function $v(t,r)$ is different from $0$ only in an interval which becomes smaller and smaller as the time increases and this allows to conclude that, as $t\to +\infty$, $$\int_0^1v(t,r)G(r)dr\rightarrow G(r_0)$$, for every test function $G$.
What can I say about the $$\lim_{t\to +\infty}v(t,r)$$
Thank you
The limit depends on the function $u_0$.
Denote the support of $u_0$ as $A$. If $r_0=\int_0^1ru_0(r)\,dr\in A^。$, for example, $u_0(r)=1$ on $[0,1]$ and $=0$ in other places, then $$\lim_{t\rightarrow+\infty}v(t,r)=\left\{ \begin{array}{cl} 0& r\neq r_0,\\ +\infty&r=r_0\\ \end{array}\right.$$ if $u_0(r_0)>0$, since when $r\neq r_0$, for sufficiently large $t$, $(r-r_0)e^t+r_0\notin[0,1]$ and thus $v(t,r)=0$.
If $u_0(r_0)=0$, for example, $$u_0(r)=\left\{ \begin{array}{cl} 2&r\in[0,\frac14]\cup[\frac34,1],\\ 0&\text{otherwise}\\ \end{array}\right.$$ where $r_0=\frac12$ and thus $u_0(r_0)=0$. In this case, $\lim\limits_{t\rightarrow+\infty}v(t,r)=0$.