Apply Gronwall's inequality to the following inequality Assume the norms in the following inequality make sense, $U\subset \Bbb{R}^n$ open and $u: U\times [0,T] \to \Bbb{R}$ , with the relation holds:
$$\frac{d}{d t}\left\|{u}\right\|_{L^{2}(U)}^{2}+\theta\left\|{u}\right\|_{H_{0}^{1}(U)}^{2} \leq C\left(\left\|f(t)\right\|_{L^{2}(U)}^{2}+\left\|{u}\right\|_{L^{2}(U)}^{2}\right) \tag{*}$$ with $\theta >0,C>0$ being constant.
I want to apply Gronwall's inequality to it to get the estimate:
$$\max_{t\le T}\|{u}(t)\|^2_{L^2(U)} + \int_0^T \|u\|_{H_0^1}^2 dt \le C(\|f(t)\|^2_{L^2(0,T,U)} + \|u(0)\|^2_{L^2(U)})$$
I do as follows:
denote $$\eta(t) = \|u\|_{L^2(U)}^2 + \int_0^t \theta\|u\|^2_{H^1_0}dt$$
Then we get from (*):
$$\eta'(t) \le C(\|f(t)\|^2_{L^2(U)} + \|u\|_{L^2(U)}^2) \le C(\|f(t)\|_{L^2} + \eta(t))$$
Then apply Gronwall's inequality we have :
$$\eta(t) = \|u\|_{L^2(U)}^2 + \int_0^t \theta\|u\|^2_{H^1_0}dt \le C(\|f(t)\|^2_{L^2(0,T,U)} + \eta(0))$$
with $\eta(0) = \|u(0)\|^2_{L^2(U)}$.Taking sup on both side over $0 \le t \le T$ gets the result.
$$\max_{0\le t\le T}\|{u}(t)\|^2_{L^2(U)} + \int_0^T \theta \|u\|_{H_0^1}^2 dt \le C(\|f(t)\|^2_{L^2(0,T,U)} + \|u(0)\|^2_{L^2(U)})$$
wihch is equivalent to :
$$\max_{0\le t\le T}\|{u}(t)\|^2_{L^2(U)} + \int_0^T \|u\|_{H_0^1}^2 dt \le C(\|f(t)\|^2_{L^2(0,T,U)} + \|u(0)\|^2_{L^2(U)})$$
Is my proof correct?
 A: I don't know what version of Grönwall you are trying to apply; most versions end up with an exponential on the right hand side. Unless the $C$ in the final result depends on time (in particular different from the first line's $C$), I think your inequality is wrong.
What you can do is use the main idea of any proof of Grönwall, which is to use an integrating factor. Put $$g(t)= C\|f(t)\|_{L^2}^2-\theta \|u(t)\|_{H^1_0}^2,\quad X(t)=\|u(t)\|_{L^2}^2 e^{-Ct}.$$
Then the given inequality can be written
$$ X'(t) \le g(t) e^{-Ct}$$
Integrating both sides gives
$$ X(t)-X(0) \le \int_0^t g(s) e^{-Cs}ds.$$
Undoing the substitutions
$$ \|u(t)\|_{L^2}^2 + \theta\int_0^t \|u(s)\|_{H^1_0}^2 e^{C(t-s)}ds \le \|u(0)\|e^{Ct} + \int_0^t \|f(s)\|_{L^2}^2 e^{C(t-s)}ds. $$
using $1\le e^{C(t-s)}$ on the LHS and $e^{C(t-s)}\le e^{Ct}$ on the RHS, we get
$$ \|u(t)\|_{L^2}^2 + \theta\int_0^t \|u(s)\|_{H^1_0}^2 ds \le e^{Ct}(\|u(0)\|_{L^2}+\|f\|_{L^2((0,T)\times U)}). $$
Hence,
$$ \|u(t)\|_{L^2}^2 + \int_0^t \|u(s)\|_{H^1_0}^2 ds \le \max(1,\theta^{-1}) e^{Ct}(\|u(0)\|_{L^2}+\|f\|_{L^2((0,T)\times U)}). $$
