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$u_t +au_x = f(x,t)$ for $ 0<x<R$ and $t>0$

$u(0,t)=0$ for $t>0$

$u(x,0)=0$ for $0<x<R$

I have manage to get:

$\int_0^R \! u^2(x,t) \, \mathrm{d} x \leq e^t \int_0^t \int_0^R \! f^2(x,t) \, \mathrm{d} x \mathrm{d} s$

How does that implies stability with respect to initial condition with $L^2 norm$ ?

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Please don't post the same question twice. Also, since you've been using the site to ask several questions by now, it makes perfect sense to register. Being an Unregistered user for a prolonged period of time is a sure way to mess up you questions/answers and create more work for moderators. – user31373 Oct 1 '12 at 4:42
sorry about it, I'm already registered ^^. I got problems with the old one because I was not registered that why I double post, sorry for that. – Porufes Oct 1 '12 at 5:33
up vote 2 down vote accepted

I understand the stability with respect to initial condition as: if $u$ and $v$ solve the above equation with the same $f$ and the same boundary condition $u(0,t )=0=v(0,t)$, then $\|u(\cdot , T)-v(\cdot , T)\|_{L^2}$ is bounded in terms of $\|u(\cdot , 0)-v(\cdot , 0)\|_{L^2}$. In this case, we don't have to worry about $f$ at all because the difference $w=u-v$ solves the homogeneous transport equation. The solution $w$ is constant on every line with slope $1/a$. Taking into account the boundary data $0$, you should be able to see that the $L^2$ norm of $w$ with respect to the space variable $x$ does not increase with $t$.

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