Let $u \in \mathcal C^\infty (\mathbb R^3 \times (0, \infty); \mathbb R)$ be a smooth function mapping from three spatial dimensions and a positive value for time to the real numbers. $u_t$ shall denote the derivative of $u(\vec x, t)$ with respect to time. $\nabla$ and $\Delta$ shall denote the spatial gradient and Laplacian respectively. In the following, I suppress the arguments $(\vec x, t)$ to remove visual clutter.
I wish to expand the dot product $$\nabla \Delta u_t \cdot \nabla \Delta u$$ so that $u_t$ is left by itself without a spatial derivative. I begin with the identity $$\nabla \cdot (\Delta u_t \nabla \Delta u)=\nabla \Delta u_t\cdot \nabla\Delta u +\Delta u_t \Delta^2u$$ so that we have $$\nabla \Delta u_t\cdot \nabla\Delta u=\nabla \cdot (\Delta u_t \nabla \Delta u)-\Delta u_t \Delta^2u$$
I have already found a way to cancel $\Delta u_t \Delta^2 u$ with another equation, but I am having difficulty expressing $\Delta u_t \nabla \Delta u$ in terms of $u_t$. Largely, my problem stems from the fact that $\nabla \Delta u$ is a vector valued quantity. I was wondering if the Laplacian vector calculus identity $$ \tilde \Delta \vec v = \nabla (\nabla \cdot \vec v) - \nabla \times (\nabla \times \vec v)$$ applied to $\nabla u$ would yield a solution. Taking
$$ \begin{align} \tilde \Delta \nabla u &= \nabla (\nabla \cdot \nabla u) + \nabla \times (\nabla \times u) \\ &= \nabla \Delta u + \nabla \times (\nabla \times \nabla u) \\ &= \nabla \Delta u \end{align} $$
and
$$ \begin{align} \nabla \Delta u_t \cdot \nabla \Delta u &= \tilde \Delta \nabla u_t \cdot \tilde \Delta \nabla u \end{align} $$
Can I express the above quantity in terms of $\tilde \Delta (u_t \nabla u)$, $\Delta (u_t u)$, or something else of the form $\tilde \Delta (u_t D u)$ where $D$ is some spatial derivative operator?