Prove an identity of the type "integration by parts" How can I apply integration by parts to prove that 
$$- \int_0^T \int_\Omega (\nabla \cdot (u^\alpha \nabla \Delta u)) \phi \ dx dt = \int_0^T\int_{\Omega} u^\alpha \nabla u \cdot \nabla \Delta \phi \ dx dt + \alpha \int_0^T\int_{\Omega} u^{\alpha-1}\nabla u \cdot D^2\phi \cdot \nabla u \ dx dt \\ + \frac{\alpha}{2} \int_0^T\int_{\Omega} u^{\alpha-1}|\nabla u|^2\Delta \phi \ dx dt  + \frac{\alpha(\alpha-1)}{2} \int_0^T\int_{\Omega} u^{\alpha-2}|\nabla u|^2\nabla u \cdot \nabla \phi \ dx dt,$$
where $T>0$, $\alpha > 1$, $\Omega$ is bounded, $D^2\phi$ denotes the Hessian matrix. All the differential operators are with respect to the space variable. We assume that $\nabla u = u^\alpha\nabla \Delta u = u = 0$ on $\partial \Omega$. We also assume that $u(x,t)$ and $\phi(x,t)$ are sufficiently regular and integrable so that the terms make sense.
 A: Notation : $\nabla\cdot \nabla f=\Delta f$
Green Formula : $\int_\Omega \ \nabla f\cdot \nabla g + f\Delta
g=\int_{\partial \Omega}\ f \nabla g\cdot n $ where $n$ is out unit
normal of $\partial \Omega$
Divergence Theorem : $\int_\Omega\ \nabla\cdot X
=\int_{\partial\Omega} \ X\cdot n $ where $X$ is a vector field.
EXE : $\nabla \cdot (f\nabla g)=\nabla f\cdot \nabla g + f\Delta g$
EXE : \begin{align*} &\ \nabla u\cdot \nabla (\nabla u\cdot \nabla
\phi ) = \nabla u\cdot \nabla (\sum_i u_{x_i}\phi_{x_i})
\\&=
 u_{x_i} u_{x_jx_i}\phi_{x_j} + u_{x_i} u_{x_j}
 \phi_{x_j x_i}
 \\&=D^2u(\nabla u,\nabla \phi ) +
 D^2\phi (\nabla u,\nabla u) \end{align*}
So
\begin{align*}
\int_0^T\int_\Omega \ \nabla\cdot \bigg( u^\alpha \nabla \Delta u\
\phi \bigg) &= \int_{\partial ([0,T] \times \Omega )}
 \ n\cdot \bigg( u^\alpha \nabla \Delta u\ \phi \bigg)
 =0 \end{align*}
if $\phi$ has a compact support in $ [0,T]\times
 \Omega$.
Hence \begin{align*} &-
 \int_0^T\int_\Omega \ \nabla\cdot \bigg( u^\alpha \nabla \Delta u\
 \bigg) \phi \\&=
 \int_0^T\int_\Omega
  \ u^\alpha \nabla\Delta u \cdot \nabla \phi \\&=
  \int_0^T\int_\Omega
\ -\Delta u \nabla\cdot \bigg( u^\alpha \nabla \phi \bigg)\\&=
 \int_0^T\int_\Omega
 \ -\Delta u \{ \alpha u^{\alpha-1}\nabla u\cdot
 \nabla \phi +u^\alpha \Delta \phi \}
 \\&= \int_0^T\int_\Omega
\ \nabla u\cdot \nabla \{ \alpha u^{\alpha-1}\nabla u\cdot
 \nabla \phi +u^\alpha \Delta \phi \}\\&=
  \int_0^T\int_\Omega \
 \alpha (\alpha -1)u^{\alpha-2} |\nabla u|^2\ \nabla u\cdot \nabla\phi +
\alpha u^{\alpha -1} \nabla u\cdot \nabla(\nabla u\cdot \nabla \phi
)
 \\&+\alpha u^{\alpha-1} |\nabla u|^2\Delta \phi + u^\alpha
 \nabla u\cdot \nabla \Delta\phi
\\&=
   \int_0^T\int_\Omega \
 \alpha (\alpha -1)u^{\alpha-2} |\nabla u|^2\ \nabla u\cdot \nabla\phi +
\alpha u^{\alpha-1} |\nabla u|^2\Delta \phi + u^\alpha
 \nabla u\cdot \nabla \Delta\phi
\\&+ \alpha u^{\alpha -1} \{ D^2u(\nabla u, \nabla\phi ) +D^2\phi
(\nabla u,  \nabla u) \}
\end{align*}
