Transforming differential equations to new coordinate system (using Jacobian) I am trying to understand how to express any (partial) differential equation in a new coordinate system. As an example I have used the 2D Laplace equation: $$\Delta f=\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}$$ and the transformation given by the following expressions:
$$u=x$$
$$v=\frac{y-b(x)}{t(x)-b(x)}$$
where $u$,$v$ are the new coordinates, $b(x)$ is the function describing bottom of the domain, $t(x)$ is the function describing top of the domain.

To transform the laplacian I have used the following expression using summation notation (taken from here):
$$\Delta f=\frac{1}{J}\frac{\partial}{\partial \zeta_j}\left[J\left(\frac{\partial f}{\partial \zeta_k}\frac{\partial \zeta_k}{\partial x_i}\frac{\partial \zeta_j}{\partial x_i}\right)\right]$$ where $x_1=x$, $x_2=y$ and $\zeta_1=u$, $\zeta_2=v$.
After expansion of the indexes I've got:
$$\Delta f=\frac{1}{J}\left\{\frac{\partial}{\partial u}\left[J\left(\frac{\partial f}{\partial u} \frac{\partial u}{\partial x}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}\frac{\partial u}{\partial y}\right) \right]+ \frac{\partial }{\partial v}\left[J\left( \frac{\partial f}{\partial u} \frac{\partial u}{\partial x}\frac{\partial v}{\partial x}+\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}\frac{\partial v}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}\frac{\partial v}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}\frac{\partial v}{\partial y}\right) \right] \right\}$$.
The derivatives are:
$\frac{\partial u}{\partial x}=1$; $\frac{\partial u}{\partial y}=0$;
$\frac{\partial v}{\partial x}=\frac{-b'(x)\left(t(x)-b(x)\right)-(y-b(x))(t'(x)-b'(x))}{\left(t(x)-b(x)\right)^2}$; $\frac{\partial v}{\partial y}=\frac{1}{t(x)-b(x)}$;
The Jacobian is  $J=t(x)-b(x)$.
After substituting the derivatives the laplacian takes the following form
$$\Delta f=\frac{1}{J}\left\{\frac{\partial}{\partial u}\left[J\left(\frac{\partial f}{\partial u} +\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}\right) \right]+ \frac{\partial }{\partial v}\left[J\left( \frac{\partial f}{\partial u}\frac{\partial v}{\partial x}+\frac{\partial f}{\partial v}\left(\frac{\partial v}{\partial x}\right)^2+\frac{\partial f}{\partial v}\left( \frac{\partial v}{\partial y}\right)^2\right) \right] \right\}$$
As the Jacobian depends only on $x$ its derivatives with respect to $u$ and $v$ are equal to zero. So after calculating the $\frac{\partial}{\partial u}$ and $\frac{\partial}{\partial v}$ the laplacian takes the following form which is wrong (first order derivatives are not present):
$$\Delta f_{u,v}=\frac{\partial^2f}{\partial u^2}+2\frac{\partial v}{\partial x}\frac{\partial^2f}{\partial u\partial v}+\left[\left(\frac{\partial v}{\partial x}\right)^2+\left(\frac{\partial v}{\partial y}\right)^2\right]\frac{\partial^2f}{\partial v^2}=0$$
How can I use this way of transformation to get the appropriate result (given below)?
$$\Delta f_{u,v}=\frac{\partial^2f}{\partial u^2}
+2\frac{\partial^2f}{\partial u\partial v}\frac{\partial v}{\partial x}+
\frac{\partial^2f}{\partial v^2}\left(\left(\frac{\partial v}{\partial x}\right)^2+ \left(\frac{\partial v}{\partial y}\right)^2\right )+
\frac{\partial f}{\partial v}\left(\frac{\partial^2v}{\partial x^2} \right )$$
I'd be very grateful for pointing out what I've made wrong as well for any general rules, tips, materials and examples on the topic.
 A: Lets try it from first principles then look for the mismatches with the given derivation.
Chain rule:
$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u}  \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x} \tag{1}$$
$$\frac{\partial^2 f}{\partial x^2} = \frac{\partial }{\partial x} \left( \frac{\partial f}{\partial u}  \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x} \right) \tag{2}$$
$$\frac{\partial^2 f}{\partial x^2} = \left(  \frac{\partial^2 f}{\partial u^2} \frac{\partial u}{\partial x} +   \frac{\partial^2 f}{\partial v \partial u} \frac{\partial v}{\partial x}  \right) \frac{\partial u}{\partial x} + 
\frac{\partial{f}}{\partial u} \frac{\partial^2{u}}{\partial x^2} +
\left(  \frac{\partial^2 f}{\partial v^2} \frac{\partial v}{\partial x} +   \frac{\partial^2 f}{\partial u \partial v} \frac{\partial u}{\partial x}  \right) \frac{\partial v}{\partial x} +
\frac{\partial{f}}{\partial v} \frac{\partial^2{v}}{\partial x^2}
\tag{4}$$
$$\frac{\partial^2 f}{\partial x^2} = 
\frac{\partial^2 f}{\partial u^2} \left(\frac{\partial u}{\partial x } \right)^2 + 
\frac{\partial^2 f}{\partial v^2} \left(\frac{\partial v}{\partial x } \right)^2 + 
2 \frac{\partial^2 f}{\partial u \partial v}\frac{\partial u}{\partial x } \frac{\partial v}{\partial x }  +
\frac{\partial{f}}{\partial u} \frac{\partial^2{u}}{\partial x^2} +
\frac{\partial{f}}{\partial v} \frac{\partial^2{v}}{\partial x^2} 
\tag{5}$$
Similarly for $y$.
$$\frac{\partial^2 f}{\partial y^2} = 
\frac{\partial^2 f}{\partial u^2} \left(\frac{\partial u}{\partial y } \right)^2 + 
\frac{\partial^2 f}{\partial v^2} \left(\frac{\partial v}{\partial y } \right)^2 + 
2 \frac{\partial^2 f}{\partial u \partial v}\frac{\partial u}{\partial y } \frac{\partial v}{\partial y }  +
\frac{\partial{f}}{\partial u} \frac{\partial^2{u}}{\partial y^2} +
\frac{\partial{f}}{\partial v} \frac{\partial^2{v}}{\partial y^2} 
\tag{6}$$
$$\Delta f = \frac{\partial^2 f}{\partial x^2 } + \frac{\partial^2 f}{\partial y^2 } \tag{7}$$
Substitute $(5)$ and $(6)$ into $(7)$.
\begin{equation}
\begin{split}
\Delta f =&\frac{\partial^2 f}{\partial u^2} \left[ \left(\frac{\partial u}{\partial x } \right)^2 +  \left(\frac{\partial u}{\partial y } \right)^2\right]
 +  \frac{\partial^2 f}{\partial v^2} \left[ \left(\frac{\partial v}{\partial x } \right)^2 +  \left(\frac{\partial v}{\partial y } \right)^2\right]\\
&+ 2 \frac{\partial^2 f}{\partial u \partial v} \left[ \frac{\partial u}{\partial x } \frac{\partial v}{\partial x }  + \frac{\partial u}{\partial y } \frac{\partial v}{\partial y }  \right] \\
 &+ \frac{\partial f}{\partial u } \left[ \frac{\partial^2 u}{\partial x^2 }  + \frac{\partial^2 u}{\partial y^2 } \right]  +
\frac{\partial f}{\partial v } \left[ \frac{\partial^2 v}{\partial x^2 }  + \frac{\partial^2 v}{\partial y^2 }  \right] \\
\end{split}  \tag{8} 
\end{equation}

Reductions for the specific example:
$$\frac{\partial u}{\partial x} = 1, \frac{\partial u}{\partial y} = 0,
\frac{\partial^2 u}{\partial x^2} = 0, \frac{\partial^2 u}{\partial y^2} = 0,\frac{\partial^2 v}{\partial y^2} = 0 \tag{9}$$
Substituting $(9)$ into $(8)$:
$$\Delta f = \frac{\partial^2 f}{\partial u^2}  + 2 \frac{\partial^2 f}{\partial u \partial v}  \frac{\partial v}{\partial x } 
+ \frac{\partial^2 f}{\partial v^2} \left[ \left(\frac{\partial v}{\partial x } \right)^2 +  \left(\frac{\partial v}{\partial y } \right)^2\right]
+ \frac{\partial f}{\partial v } \frac{\partial^2 v}{\partial x^2 } \tag{10}$$

Mismatches with the given derivation.
As the Jacobian depends only on x its derivatives with respect to u and v are equal to zero.

This statement is false $u = x$ and $v$ is a function of $x$. So $\frac{\partial J}{\partial u}$ and $\frac{\partial J}{\partial v}$ are not necessarily zero.
Correction: By the chain rule:
$\frac{\partial J}{\partial u} = 
\frac{\partial J}{\partial x}\frac{\partial x}{\partial u} +
\frac{\partial J}{\partial y}\frac{\partial y}{\partial u}
= \frac{\partial J}{\partial x}$, since $\frac{\partial x}{\partial u} = 1$ and $\frac{\partial y}{\partial u} = 0$
$\frac{\partial J}{\partial v} = 
\frac{\partial J}{\partial x}\frac{\partial x}{\partial v} 
+ \frac{\partial J}{\partial y}\frac{\partial y}{\partial v} = 
\frac{\partial J}{\partial x}\frac{\partial x}{\partial v} $ , since $\frac{\partial J}{\partial y} = 0$
It looks like a false assumption threw off the solution.
A: Given
$$u=x\tag{1}$$
$$v=\frac{y-b(x)}{t(x)-b(x)} \tag{2}$$
Substitute $u$ for $x$ into $(2)$:
$$v=\frac{y-b(u)}{t(u)-b(u)} \tag{3}$$
$v$ is dependent on $u$.
$u$ and $v$ are not independent variables.
The other argument is that if $u$ and $v$ are independent variables then  $\frac{\partial v}{\partial u} = 0$ and $\frac{\partial u}{\partial v} = 0$
Since $u=x$ then $\frac{\partial v}{\partial x} = 0$ and $\frac{\partial x}{\partial v} = 0$.
e.g. Let $b(x) = x$ and $t(x) = 2x$ then $v = \frac{y}{x} - 1$ and $\frac{\partial v}{\partial x} = -\frac{y}{x^2}$
This implies that the assumption of independance of $u$ and $v$ is false.
The equations for the Jacobian and the change of variables for the Laplacian are based on the independence of $x$ with respect to $y$ and $u$ wrt $v$.
Perhaps someone could shed more light on this?
