Transforming differential equations to new coordinate system (using Jacobian)

Question

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.

Answer 1

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.

Answer 2

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?