If
\begin{equation} \Delta u = (u_{x_1x_1}+4u_{x_2x_2}) \end{equation} what would $\nabla u$ and $u$ become?
u is defined as: \begin{equation} \begin{aligned} u_{t}(x)-\Delta u(x) &=0, & & x \in \Omega, \quad t>0 \\ u(x, t) &=0, & & x \in \partial \Omega, \quad t>0 \\ u(x, 0) &=u_{0}(x), & & x \in \Omega \end{aligned} \end{equation}
$\Omega$ is in $\mathbb{R}^{2}$
I think we would need to have either $$\nabla u = \begin{bmatrix}\frac\partial {\partial x_1}\\2\frac\partial {\partial x_2}\end{bmatrix}u$$
or
$$\nabla u = \begin{bmatrix}\frac\partial {\partial x_1}\\\frac\partial {\partial x_2}\end{bmatrix}u(x_1,2x_2)$$
Maybe because the coordinate system is stretched a factor of $2$ in $x_2$ direction or the partial differential operator itself is.
What will happen with the chain rule?