Asymptotic behaviour of the solution of a PDE I have the following PDE:
\begin{align}
\frac{d}{dt}u(t,r)&=\alpha(r-\beta)\frac{d}{dr}u(t,r)+\alpha u(t,r)\\
u(0,r)=&u_0(r)
\end{align}
where $r\in [0,1]$ and $\alpha$ and $\beta$ are fixed constants in $[0,1]$.
I would like to prove that $\lim_{t\to  +\infty}u(t, r)\equiv \beta$. 
What I know about the initial condition is that $u_0(0)=u_0(1)=0$, $\int_0^1u_0(r)dr=1$ and that $\int_0^1 u_0(r)rdr=\beta$.
The equation can be solved trough the characteristic method and the solution is given by
$u(t,r)=u_0((r-\beta)e^{\alpha t})+\beta)e^{\alpha  t}$.
How to prove that $\lim_{t\to +\infty}u(t,r)=\beta, \forall r\in (0,1)$? Could someone help me?
 A: The solution by characteristics that you give, is only valid when $(r-\beta)e^{\alpha t}+\beta\in[0,1]$, that is, for
$$ \beta-\beta e^{-\alpha t} \le r \le \beta+(1-\beta) e^{-\alpha t}, $$
an interval of width $e^{-\alpha t}$. Outside that interval, the characteristic will hit one of the boundary conditions at $r=0$ or $r=1$, and so $u(t,r)=0$ at those points.
You should find that $\int_0^1 u(t,r) \,dr = \int_0^1 u_0(r)\,dr=\beta$ (a constant).
Conclusion: The asymptotic limit of the solution as $t\to\infty$ is $\beta\delta_\beta$, where $\delta_\beta$ is a delta function located at $\beta$ (sometimes written $\delta_\beta(r)=\delta(r-\beta)$).
Edited to add:
This analysis assumes $0<\beta<1$ and $\alpha>0$. If $\alpha=0$, the solution is of course independent of $t$. And if $\beta=0$ or $\beta=1$, the analysis needs to be changed a bit, but much the same will still hold
Edit the second:
Here is a picture of the solution. I chose $u_0(r)=4(r-r^2)$, $\beta=2/3$, and $\alpha>0$. The red graph is the initial condition $u_0$,
blue is the solution for $e^{\alpha t}=2$, and the tallest one (khaki) is for $e^{\alpha t}=4$. The area under each curve is the same in each case.
Also, I removed my totally misguided “consistency check” from the answer. Sorry about that; not enough caffeine, I suppose.

