Change of variables involving Poisson's Equation Consider Poisson's equation on $\Omega = [0,3] \times[0,3]$
$$\nabla^2 u(x_1, x_2) = f(x_1, x_2)$$
Suppose we had a map $M: = (x_1, x_2) \to (3x_1, 3x_2)$ from the domain $\Omega_0 = [0,1] \times[0,1]$ to $\Omega$. And suppose we had an equation that when the map $M$ was applied to it we obtained Poisson's equation above. What would this equation look like? I assume it would be similar to the equation above but I am not sure how the $\nabla$ would be affected. I believe the Jacobian is involved somewhere to account for the change of variables. But I am not sure what the form of the equation should be before the map $M$ is applied to it?
 A: How to relate the Laplacian of composition $u\circ \phi$ to the Laplacian of $u$ and some derivatives of map $\phi$? In general this is a messy formula involving the second derivatives of $\phi$; it does not preserve the structure of Poisson's equation. But if the Jacobian matrix of $\phi$ is a multiple of an orthogonal matrix, i.e., $(D\phi)^T (D\phi) =c I$, then the computation simplifies somewhat. Note that $c$ must be equal to $(\det D\phi)^{2/n}$ above, where $n$ is the dimension of the space. When $n=2$, the exponent is $1$, which simplifies the computation  a lot more: see below.   
By the chain rule,
$$
\nabla (u\circ \phi) = ((\nabla u)\circ \phi) D\phi
$$ 
where $D\phi$ is the (square) matrix of the derivatives of $\phi$, and $\nabla u$ is a row vector. 
The Laplacian is the divergence of the gradient, but applying the divergence to the above seems boring. Instead, consider a test  function $v$ with compact support and use integration by parts:
$$
\int \Delta (u\circ \phi) (v\circ \phi) 
=-
\int \nabla  (u\circ \phi) (\nabla (v\circ \phi))^T \\
= -\int [(\nabla  u)\circ \phi] (D\phi) (D\phi)^T [(\nabla v)^T\circ \phi ]
$$
Here is where orthogonality of $\phi$ can be applied $(D\phi) (D\phi)^T = \det D\phi$. The Jacobian determinant is exactly what we need for  the change of variables:
$$
-\int [(\nabla  u)\circ \phi] [(\nabla v)^T\circ \phi ] \,\det D\phi
= -\int  (\nabla  u) (\nabla v)^T =   \int (\Delta u )v$$
Thus, for every test function $v$ 
$$
\int \Delta (u\circ \phi) (v\circ \phi) 
= \int (\Delta u )v \tag{1}$$
On the other hand, 
$$
\int \Delta (u\circ \phi) (v\circ \phi)\det D\phi = \int (\Delta u )v\tag{2}$$ by the change of variables. Comparing $(1)$ and $(2)$ and recalling that $v$ was an arbitrary test function, we conclude that 
$$\Delta (u\circ \phi)  = (\Delta u )\det D\phi \tag{3}$$
In your specific case, $\det D\phi=9$, so $(3)$ simplifies a bit more. 
