Partial Differential Equation Change of Variables

Question

Problem Statement

I have a partial differential equation given by

$$\frac{\partial f}{\partial x}=0\tag{1}\label{1}$$

where $f=f(x,y)$. If we introduce a new set of coordinates $q_1$ and $q_2$ such that

$$ \begin{array}{c} q_1=q_1(x,y) \\ q_2=q_2(x,y) \\ \end{array}\tag{2}\label{2} $$

and $g(q_1,q_2)$ corresponds to $f(x,y)$, how do I convert Eq. $\ref{1}$ which is in terms of $f$, $x$, and $y$ to an equation in terms of $g$, $q_1$, and $q_2$?

Solution Attempt

Taking

$$f(x,y) = f(x(q_1,q_2),y(q_1,q_2))=g(q_1,q_2)\tag{3}\label{3}$$

I can substitute equation $\ref{3}$ into equation $\ref{1}$. I think my problem starts when I try to take $\frac{\partial g}{\partial x}$ because the following seems wrong:

$$ \frac{\partial f}{\partial x}=\frac{\partial g}{\partial x}=\frac{\partial g}{\partial q_1}\frac{\partial q_1}{\partial x}\tag{4}\label{4}$$

I think I'm missing some form of the chain rule and believe $\ref{4}$ should really be something like

$$ \frac{\partial f}{\partial x}=\frac{\partial g}{\partial x}=\frac{\partial g}{\partial q_1}\frac{\partial q_1}{\partial x}+\frac{\partial g}{\partial q_2}\frac{\partial q_2}{\partial x}\tag{5}\label{5}$$

because if I make a small change in x, it may cause small changes in $q_1$ and $q_2$, but I don't know how to prove this or if it is true.

Hopefully, this question here will help me answer this question I asked previously. Any help appreciated and thanks in advance!

Answer 1

Yes, $g$ depends on $q_1$ and $q_2$, which both depend on $x, y.$

The chain rule says:

$$\frac{\partial}{\partial x} g(q_1,q_2) = \frac{\partial g}{\partial q_1}\frac{\partial q_1}{\partial x} + \frac{\partial g}{\partial q_2}\frac{\partial q_2}{\partial x},$$

which is what you need.

Answer 2

I think I am trying to find notation that made sense and I think I figured out how to do it using this post.

Start by defining the functions we are using

$$f: \Bbb{R}^2 \to \Bbb{R}$$ $$g: \Bbb{R}^2 \to \Bbb{R}$$

and write Eq. 1 more precisely as

$$(\partial_1f)(x,y) = 0$$

This means we want to take the derivative of $f$ with respect to the first argument, $x$. After substituting $g$ in for $f$ in the above equation, we now wish to find how $g$ changes with changes in $x$. This requires us to use the chain rule as shown below:

$$ \begin{array}{c} (\partial_1f)(x,y)= [(\partial_1g)(q_1(x,y),q_2(x,y))]\cdot(\partial_1q_1)(x,y) \\ + [(\partial_2g)(q_1(x,y),q_2(x,y))]\cdot(\partial_1q_2)(x,y)\\ \end{array}$$

Or using sloppy notation:

$$\frac{\partial f}{\partial x}=\frac{d f}{d x}=\frac{d g}{d x}=\frac{\partial g}{\partial q_1}\frac{d q_1}{d x}+\frac{\partial g}{\partial q_2}\frac{d q_2}{d x}$$

Note how I purposely used $d$ instead of $\partial$ in the second and third terms and did the same for $g$ because in each case, we want to know how each function varies with changes in $x$.