Consider the smooth solution $u$ of the Dirichlet problem, $$ \begin{cases} - \nabla \cdot (B(x) \nabla u(x)) = 0 & \text{in }\; \Omega \\ \hspace{1.02in} u(x) = g(x) &\text{in }\; \partial \Omega \end{cases} $$ where $\Omega$ is smooth, bounded domain in $\mathbb{R^n}$ and the functions $B: \Omega \to \mathbb{R^+}$ and $g : \Omega \to \mathbb{R}$ are smooth. Suppose there exists a bijective map $\phi : \Omega \to \hat{\Omega}$ satisfying $ D\phi(x) = \frac{1}{B(x)} Q(x),$ where $Q^TQ = I$ and $\det(Q) = 1.$ Show that then $u(x) := \hat{u}(\phi(x))$ is the smooth solution of $$ \begin{cases} - \Delta \hat{u}(\hat{x}) = 0 & \text{in }\; \hat{\Omega} \\ \hspace{0.27in} \hat{u}(\hat{x}) = g(\phi^{-1}(\hat{x})) &\text{in }\; \partial \hat{\Omega} \end{cases} $$
To me, it seems like the problem is trying to say "there exists a change of coordinate system (given by map $\phi$) that simplifies the original PDE to Laplace's equation", and then they specify certain properties such as the full derivative of the map so that we can show the result. I am trying to make my work more rigorous, so I am posting here to see if what I did was correct, and/or if it could be written in a more clear way.
I started by saying $$ x \mapsto \phi(x) := \hat{x}$$ and $$\frac{\partial u(x)}{\partial x_i} \mapsto \sum_j \frac{\partial \hat{u}(\hat{x})}{\partial \hat{x_j}} \frac{\partial \hat{x_j}}{\partial x_i}, $$ meaning $$ \nabla_x u(x) \mapsto D\phi(x) \nabla_{\hat{x}} u(\hat{x}) = \frac{Q(x)}{B(x)}\nabla_{\hat{x}} u(\hat{x}) .$$
Similarly, for any smooth function $f(x),$ $$ \nabla_x \cdot f(x) \mapsto \nabla_{\hat{x}} \cdot f(\hat{x})\frac{Q^T(x)}{B(x)} .$$
Then we have $$ - \nabla_x \cdot (B(x) \nabla_x u(x)) \mapsto -\frac{1}{B(x)} (\nabla_{\hat{x}} \cdot \nabla_{\hat{x}} \hat{u}(\hat{x})) $$ since $Q^TQ = I$ and since $\phi^{-1}(\hat{x}) = x,$ we have our result for the boundary condition as well.
Does this seem correct? Or can this be written in a more convincing way if it is correct? Thanks in advance!