A stability estimate for a first-order linear PDE If we have 
$$u_t + u_x =f(x,t)$$
with initial boundary conditions
$u(0,t)=0$ for $t>0$ and $u(x,0)=0$ for $0<x<R$
Can anyone tell me how to prove the stability estimate 
$$\int_0^r (u(x,t))^2 dx \leq e^t \int_0^t \int_0^R f^2(x,s)dx ds$$ where $t>0$
 A: The method of characteristics yields
$$u(x,t)=\int_0 ^x f(s, s + t - x) d s$$ for $t\ge x$
and $$u(x,t)=\int_0 ^t f(s + x - t, s) d s$$ for $x \ge t$.
Then you can use Hölder's inequality to estimate $u^2$ in terms of an integral over $f^2$, so the integral over $u^2$ can be estimated by an integral over a quadrilateral or triangle which is a subset of $[0,R]\times[0,t]$. This shows that the inequality is true with a factor of $t$ instead of $e^t$, so that you get an even stronger estimate.
A: Multiply your equation by $u$ and integrate over $[0,R]$, since $uu_t=\frac{1}{2}(u^2)_t$ we get 
$$
\frac{1}{2}\frac{d}{dt} \int_0^R u^2(x,t)dx + \frac{1}{2}u^2(R,t) = \int_0^R f(x,t)u(x,t)dx \leq \left( \int_0^R f^2 (x,t)dx \right) ^{\frac{1}{2}}\left( \int_0^R u^2 (x,t)dx \right) ^{\frac{1}{2}}
$$
but we have
$$
\frac{d}{dt} \int_0^R u^2(x,t)dx = 2\left( \int_0^R u^2 (x,t)dx \right) ^{\frac{1}{2}} \frac{d}{dt} \left( \int_0^R u^2 (x,t)dx \right) ^{\frac{1}{2}} 
$$
and so we get 
$$
\frac{d}{dt}\left( \int_0^R u^2 (x,t)dx \right) ^{\frac{1}{2}} \leq \left( \int_0^R f^2 (x,t)dx \right) ^{\frac{1}{2}}
$$
and integrate to obtain
$$
\left( \int_0^R u^2(x,t)dx \right)^{\frac{1}{2}} \leq \int_0^t \left( \int_0^R f^2(x,t)dx \right)^{\frac{1}{2}} \leq t^{\frac{1}{2}} \left( \int_0^t  \int_0^R f^2(x,s)dx ds \right)^{\frac{1}{2}}
$$
where we used Jensen's inequality for the last one. This is your result.
