Function $u$ solves the heat equation $\Longrightarrow$ $\langle x, \nabla u \rangle + 2t u_t$ solves the heat equation I am baffling with this homework problem:
Assume that a smooth function u solves the heat equation $u_t − \Delta u = 0.$ Show that also the function
$v(x, t) = \langle x, ∇u(x, t) \rangle + 2tu_t (x, t)$
solves the heat equation. Here $\langle \cdot, \cdot \rangle$ denotes the innerproduct $ f,g \mapsto \int_{\mathbb{R}^N} f(x) \cdot g(x) dx$ .
I was considering to use integration by parts but, to my understanding, nothing guarantees that the solution $u$ even vanishes on the boundaries. Some clues are welcome.

For $\langle \cdot, \cdot \rangle $ denoting the Euclidian scalar product you get
$$v_t = \nabla u \cdot \boldsymbol{x}_t + \boldsymbol{x} \cdot \partial_t \nabla u + 2 (u_{tt} + u_t) = \boldsymbol{x} \cdot \partial_t \nabla u + 2 (tu_{tt} + u_t) \tag{1}$$
Use Green's Vector identity for the $\Delta \langle \boldsymbol{x}, \nabla u \rangle$ and the fact that $\nabla \times \nabla v \equiv 0 \: \forall \: v \in \mathbb{R}^n$
\begin{align}\Delta v &= \nabla u \cdot \Delta   \boldsymbol{x} - \boldsymbol{x}  \cdot \Delta \nabla u + 2 \nabla \cdot \Big( (\boldsymbol{x} \cdot \nabla) \nabla u + \boldsymbol{x} \times (\nabla \times \nabla u) \Big) + 2 t \Delta u_t \\
&= - \boldsymbol{x}  \cdot \Delta \nabla u + 2 \nabla \cdot \Big( (\boldsymbol{x} \cdot \nabla) \nabla u \Big) + 2 t \Delta u_t \tag{2} \end{align}
The terms including $t$ cancel each other for a smooth $u$ (can interchange differentiations arbitrarily). Formally, you can define the operator $2 t \partial_t$ which is then applied to $u_t - \Delta u_t$.
To simplify the remaining terms, index notation is employed.
\begin{align} - \boldsymbol{x}  \cdot \Delta \nabla u + 2 \nabla \cdot \Big( (\boldsymbol{x} \cdot \nabla) \nabla u \Big) =& -x_i \partial_i \partial_j\partial_j u + 2 \partial_i (x_j \partial_j\partial_i u ) \\
&= -x_i \partial_i \partial_j\partial_j u + 2 \big( \delta_{ij} \partial_j \partial_i u +  x_i \partial_i\partial_j\partial_j u \big) \\
& = 2  \partial_i \partial_i u +  x_i \partial_i\partial_j\partial_j u  \\
&= 2 \Delta u + \boldsymbol{x} \cdot \nabla \Delta u \end{align}
Clearly, $2u_t - 2 \Delta u = 0$. By interchanging the temporal and spatial derivatives and defining the operator $\boldsymbol{x} \cdot \nabla $ also the last remaining terms cancel.
