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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!

Community
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eball
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    From equation $(3)$ you know that

    $$f(x,y)=g(q_1(x,y),q_2(x,y))=g(q_1,q_2)$$

    where both $q_1(x,y)$ and $q_2(x,y)$ are functions of the variable $x$. This implies by the chain rule that

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

    therefore

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

     – Axion004 Mar 18 '20 at 23:30

2 Answers2

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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.

suckplectomorphic
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  • Thanks for the reply! I was looking for something more detailed and pedantic about notation. – eball Mar 18 '20 at 22:12
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    I see! It's called the chain rule. You could prove it using mean value theorem - I'm sure a proof is on Wikipedia. The idea is what you said: a small change in x causes small changes in q1 and q2, and the change is exactly like this. The notation in this answer uses correct deltas, but it seems the answer below treats f, q1 and q2 and functions of one variable. What more can I clarify? – Maxim Gilula Mar 24 '20 at 23:47
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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 $x$.

eball
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