Given an equation on $[0,1]^2$
$\left\{ \begin{gathered} {u_t} - {u_{xx}} - {u_{xxt}} = \frac{u}{2} \hfill \\ u(0,t) = u(1,t) = 0 \hfill \\ u(x,0) = \varphi (x),{u_x}(x,0) = 0 \hfill \\ \end{gathered} \right.$
I want to prove this inequality
$\begin{gathered} \hfill \\ \int_0^1 {{u^2}(x,t) + u_x^2(x,t)dx} + 2\int_0^1 {\int_0^1 {u_x^2(x,s)dxds} } \leqslant 2e\int_0^1 {{\varphi ^2}(x)dx} \hfill \\ \end{gathered} $
I tried to multiply the equation by $u$ and then integrate it, but I can't see how to use Gronwall inequality to go further... I am stupid in doing this kind of estimates and I really need someone to show me how to deal with it. Please... Thank you in advance:)
If you can assume sufficient regularity, what's easier than using Gronwall is to define the energy $E(t)=\frac12\int_0^1(u^2+u_x^2)\,dx$, and differentiate under the integral sign: $$ \dot E=\int_0^1(uu_t+u_xu_{xt})\,dx=\int_0^1(uu_t-uu_{xxt})\,dx$$ using integration by parts, then use the equation: $$ \dot E=\int_0^1(uu_{xx}+u^2/2)\,dx=\int_0^1(-u_x^2+u^2/2)\,dx\le E, $$ leading to $e^{-t}\ln E$ being non-increasing, and so $$ E(t)\le e^t E(0) = \frac{e^t}2\int_0^1 \phi(x)^2\,dx.$$
The Gronwall method would indeed involve multiplying the equation by $u$ and integrating wrt $x$ first, which after some partial integration should lead to $$ \int_0^1(uu_t+uu_x+u_xu_{xt}\,dx=\tfrac12 \int_0^1 u^2\,dx, $$ which we can rewrite as $$ \int_0^1((u^2)_t+(u^2)_x+(u_x^2)_t\,dx=\int_0^1 u^2\,dx, $$ carefully noting that $u_x^2$ means the square of $u_x$! Now notice that the integral of $(u^2)_x$ vanishes due to the boundary conditions, integrate the rest from $0$ to $t$, and get $$\int_0^1(u^2+u_x^2)\,dx-\int_0^1\phi(x)^2\,dx=\int_0^t\int_0^x u(x,s)^2\,dx\,ds,$$ and so, with $E$ defined as above, this yields $$ E(t)-E(0)\le \int_0^t E(s)\,ds, $$ and you're all set to use Gronwall.