Estimate involving Transport Equation and Gronwall Inequality Let the constant $\alpha > 0$ be the problem
$$\left\{\begin{array}{cll}
u_t + \alpha u_x & = & f(x,t); \ \ 0 < x < L; \ t > 0\\
u(0,t) & = & 0; \ \ t > 0;\\
u(x,0) & = & 0; \ \ 0 < x < L.
\end{array}\right.$$
Prove that, for every $t > 0$, the following applies:
$$\int_{0}^L |u(x,t)|^2dx \leq \int_{0}^L \int_{0}^t |f(x,s)|^2 dsdx.$$
TIP: Uses Gronwall Inequality.
Outline: I tried to use the fact that
$$u(x,t) = \int_{0}^t f(x + (s-t)\alpha,s)ds$$
is the solution to the above problem when $u(x,0) = 0$. Then I used Holder inequality to get to
$$|u(x,t)|^2 \leq t\int_{0}^t |f(x + (s-t)\alpha,s)|^2ds.$$
Then I got stuck because I couldn't apply the Gronwall inequality satisfactorily...
 A: It would be best to include in your question, what you have tried so far, so you get more helpful answers. But here are a couple of suggestions.
You have something linear, and need to prove something quadratic, so some multiplication is needed at some point. If for example you multiply the PDE by $u$, you have
$$ uu_t + \alpha uu_x = uf. $$
Now to relate bilinear things like $uf$ to quadratic things there are a couple of ways to go. One is the pointwise
$$ uf \le \tfrac{1}{2}(u^2+f^2) $$
and another is
$$
 \int_0^L uf dx \le
 \sqrt{\int_0^L u^2dx}\sqrt{\int_0^L f^2 dx}.
$$
So you try from one of those to see where is leads. Sometimes tips are helpful and sometimes they are distracting, and this applies to mine as well as what was given, but these are some ideas anyway.
A: Here's a way that uses Gronwall's inequality and gets close to your desired result, maybe you can use the ideas to improve on it. I'm assuming that $u$ is at least continuously differentiable which allows us to change the order of integration at some point:
\begin{alignedat}{2}
&&u_t + \alpha u_x &= f \\
&\implies 
&uu_t + \alpha uu_x &= uf \\
&\implies
&\frac{1}{2} (u^2)_t + \frac{ \alpha}{2} (u^2)_x &= uf \\
&\implies
&\frac{1}{2}\left(\int_0^L \int_0^t (u^2)_s \mathrm{d}s\mathrm{d}x + \alpha \int_0^L \int_0^t (u^2)_x \mathrm{d}s\mathrm{d}x\right) &= \int_0^L \int_0^t uf \mathrm{d}s\mathrm{d}x\\ 
&\implies
&\frac{1}{2}\left(\int_0^L u^2\mathrm{d}x + \alpha \int_0^t \int_0^L (u^2)_x \mathrm{d}x\mathrm{d}s\right)
&\leq \int_0^L \int_0^t \frac{1}{2} \left(u^2 +  f^2 \right)\mathrm{d}s\mathrm{d}x\\
&\implies
&\int_0^L u^2\mathrm{d}x + \alpha \int_0^t u^2(L,s)\mathrm{d}s
&\leq \int_0^L \int_0^t u^2 \mathrm{d}s\mathrm{d}x + \int_0^L \int_0^t  f^2 \mathrm{d}s\mathrm{d}x\\
&\implies
&\int_0^L u^2\mathrm{d}x
&\leq \int_0^t \int_0^L u^2 \mathrm{d}x\mathrm{d}s + \int_0^L \int_0^t  f^2 \mathrm{d}s\mathrm{d}x.\\
\end{alignedat}
Applying Grönwall's lemma to the function $t \mapsto \int_0^L u^2\mathrm{d}x$ in this inequality gives us
\begin{aligned}
\int_0^L u^2(x,t)\mathrm{d}x
&\leq e^t \int_0^L \int_0^t  f^2(x,t) \mathrm{d}s\mathrm{d}x,
\end{aligned}
which is your inequality with an extra factor of $e^t$ on the right hand side.
