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

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

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

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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    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.2012-10-01

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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 $t$.