A-priori estimate help (something to do with Gronwall lemma)

Here $\Gamma(t)$ is a family of smooth compact connected and oriented surfaces in $\mathbb{R}^n$. $u$ solves the PDE $$\dot{u} + u\nabla_\Gamma \cdot v - \Delta_\Gamma u = 0$$ where $\dot{u} = u_t + v \cdot \nabla u$ is the material derivative and $v$ is a velocity field on $\Gamma$. The $\nabla_{\Gamma(t)}$ is the surface gradient (for fixed $t$) obtained by projecting the ordinary gradient onto the tangent space.

Multiplying this PDE by a test function, integrating by parts, setting the test function to $u$ and doing some manipulations, we can get $$\frac{1}{2}\frac{d}{dt}\int_{\Gamma(t)} u^2 + \int_{\Gamma(t)} |\nabla_\Gamma u|^2 + \frac{1}{2}\int_{\Gamma(t)} u^2 \nabla_{\Gamma(t)} \cdot v = 0$$ How can we get the following estimate from the above: $$\sup_{t \in [0,T]} \lVert u(t) \rVert^2_{L^2(\Gamma(t))} + \int_0^T\lVert \nabla_{\Gamma(t)} u(t) \rVert^2_{L^2(\Gamma(t))} < c\lVert u_0 \rVert^2_{L^2(\Gamma_0)}\;?$$

Here, $u(t=0) = u_0$ is known and we can assume that $\nabla_\Gamma \cdot v$ is in $L^\infty(\cup (\{t\}\times \Gamma(t)))$ (and the constant $c$ depends on $v$, $T$ and $\Gamma$).

Clearly integrating in time between 0 and $T$ is what we need to do but I don't know how to deal with the third term in the first equation.

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If you really want someone to look into your problem I think you should explain a bit more. There is not much of a background nor notational explanation. – AD. Jun 4 '12 at 19:15
@AD. I just edited it. I think (but maybe I'm wrong) one doesn't need too much background and that there is some standard PDE trick that I'm missing. – soup Jun 4 '12 at 19:33

$$\frac{1}{2}\frac{d}{dt}\int_{\Gamma(t)} u^2 + \lVert \nabla_\Gamma u \rVert_{L^2}^2 - \frac{c}{2}\int_{\Gamma(t)} u^2 \leq 0$$ for a constant $c$. This is $$\frac{1}{2}\frac{d}{dt}\lVert u \rVert^2_{L^2(\Gamma(t))} + \lVert \nabla_\Gamma u \rVert_{L^2}^2 \leq \frac{c}{2}\lVert u \rVert_{L^2}^2.\tag{1}$$ The LHS is greater than the LHS without the second term, and using Gronwall on this "new" LHS: $$\lVert u \rVert^2_{L^2(\Gamma(t))} \leq C(t)\lVert u_0 \rVert_{L^2}^2 \tag{2}.$$
Integrate (1) with respect to time: $$\frac{1}{2}\lVert u \rVert^2_{L^2(\Gamma(t))} + \int_0^T\lVert \nabla_\Gamma u \rVert_{L^2}^2 \leq \int_0^T\frac{c}{2}\lVert u \rVert_{L^2}^2 + \lVert u_0 \rVert_{L^2(\Gamma(0))}^2$$ and using the Gronwall bound (2) gives you $$\frac{1}{2}\lVert u \rVert^2_{L^2(\Gamma(t))} + \int_0^T\lVert \nabla_\Gamma u \rVert_{L^2}^2 \leq \int_0^TC_1(t)\lVert u_0 \rVert_{L^2}^2 + \lVert u_0 \rVert_{L^2(\Gamma(0))}^2$$ The right hand side is data and the factor of a half on the LHS can be removed. I think the supremum can be incorporated in (2) and carried through.