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 solve this equation with the characteristic's method.
If my computations are correct the equations of the characeristics are given by \begin{align} &\frac{dt}{ds}=-1,\\ &\frac{dr}{ds}=\alpha({r-\beta}) \end{align} while the equation along the characteristics is \begin{align} \frac{du}{ds}=-\alpha u \end{align} and consequently the total solution is given by $u(t,r)=u_0((r-\beta)e^{\alpha t})+\beta)e^{\alpha t}$.
My problem is that, under the hypotesis $u_0(r)\in[0,1]$ I should have $u(t,r)\in[0,1]$ for every $t\geq 0$, but according to the solution that I get, it doesen't seem to be in this way. There is something wrong in my argument?
Could someone help me?
$$\frac{d}{dt}u(t,r)-\alpha(r-\beta)\frac{d}{dr}u(t,r)=\alpha u(t,r)$$ $$\frac{dt}{1}=\frac{dr}{\alpha(\beta-r)}=\frac{du}{\alpha u}=-ds$$ It doesn't mater the coefficient of the parameter $s$. I put $-1$ to be consistent with the meaning of $s$ in your notations. So, your characteristic ODEs are consistent with mine. Your mistake is later. Compare your further calculus with the following :
First characteristic curves, from $\quad \frac{dt}{1}=\frac{du}{\alpha u}$ $$u e^{-\alpha t}=c_1$$ Second characteristic curves, from $\quad \frac{dt}{1}=\frac{dr}{\alpha(\beta-r)}$ $$(r-\beta)e^{\alpha t}=c_2$$ The general solution on implicit form is $\quad u e^{-\alpha t}=F\left((r-\beta)e^{\alpha t} \right)\quad$ where $F$ is an arbitrary function.
The general solution on explicit form is : $$u(r,t)=e^{\alpha t}F\left((r-\beta)e^{\alpha t} \right)$$ Condition :
$u(0,r)=u_0(r)=e^{\alpha 0}F\left((r-\beta)e^{\alpha 0} \right)=F\left((r-\beta) \right)$
Let $x=r-\beta$. $\quad r=x+\beta\quad;\quad u_0(r)=u_0(x+\beta)=F(x).$ $$F(x)=u_0(x+\beta)$$
Now, the function $F(x)$ is determined. We put it into the above general solution where $x=(r-\beta)e^{\alpha t}\quad$ thus $\quad F(x)=u_0(x+\beta)=u_0\left((r-\beta)e^{\alpha t}+\beta \right)$.
$$u(r,t)=e^{\alpha t}u_0\left((r-\beta)e^{\alpha t}+\beta \right)$$