I have such an equation; When I mutliply -2Δut, I get energy inequality.
\begin{align}
& u_{tt}- \Delta u + a \Delta^{2} u = b f (-\Delta u), \quad (t,x) \in (0,T) \times (\Omega ∪ \partial \omega) \tag{1} \\
& \Delta u = 0,\alpha (x)\dfrac{\partial u_{t}}{\partial \nu} - \Delta^{2} u = 0, \quad (t,x) \in (0,T) \times \partial \Omega \tag{2} \\
& u(x,0) = u_{0}(x), u_{t}(x,0) = u_{1}(x), \quad x \in \Omega \tag{3}
\end{align}
where $\Omega$ is a bounded domain in $\mathbb{R}^{n}$ with a sufficiently smooth boundary $\partial \Omega := \Gamma, T>0$ is an arbitrary real number, and $\nu$ is the outward normal of the boundary $\Gamma$.
Let the function $f(u)$ with its primitive. $F(u)=\int_{u}f(s)ds$ have the following properties
$$f(0) = 0, sf(s) \geq 2(2\gamma + 1)F(s), \forall s \in \mathbb R^{1},\gamma > 0 $$
after some calculations, we get
$$\phi '(t) \leq 2(b+1)\int_{\Omega} - \Delta u_{t} f(-\Delta u) \tag{4}$$
By using Green-Gauss, find out
$$\phi '(t) \leq \int_{\Omega} \triangledown u_{t} \triangledown (f(- \Delta u)) \tag{5}$$
and finally, obtain:
$$\phi ′(t) \leq \int_{\Omega} \vert \triangledown u_{t} \vert \, \vert f'(-\Delta u)\vert \vert \triangledown \Delta u\vert \tag{6}$$
but our claims are that:
- Claim 1 : $\phi '(t) \leq c\cdot \phi^{k}+c_{2};k>1$
- Claim 2 : $\phi '(t)\leq c_{1}(\left\Vert \triangledown u_{t}\right\Vert^{2}+ \left\Vert \Delta u \right\Vert + a \left\Vert \triangledown \Delta u \right\Vert^{2} + 2 \langle 1,F(-\Delta u)\rangle)^{k}$
How can we get Claim 1 or Claim 2 by using (6) inequality.
Thank you so much.