# Change of variables in multivariable differential equations

This is a very easy question about how to justify the change of variables. Let $f$ be a $C^1$ function of two variables $x,y$. Introduce the variables $s,t$ as: $$\begin{cases} s=x+y \\ t=x-y \end{cases}$$ Then $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial f}{\partial t}\frac{\partial t}{\partial x}$$ $$\\ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial s}\frac{\partial s}{\partial y} + \frac{\partial f}{\partial t}\frac{\partial t}{\partial y}$$ Say that (for example) $$\frac{\partial f}{\partial t} = 0$$ This would imply that $f$ is only a function of $s$, i.e. $$f(s,t)=\phi(s)$$ Now to my question. How does this mean that $f(x,y)=\phi(x+y)$? Shouldn't it mean that $f(x,y)=\phi(x)$ (just by looking at the pattern, which is that if you plug in two variables $s,t$ in to $f$ you are also plugging in only $s$ in to $\phi$ yielding the same output).

The error here is that you're defining $f(s,t)$ and then pretending that you can substitute any variable inside, like $f(x,y)$. However $s,t,x,y$ are not free anymore and they're bound altogether by the very relations you set in the first place. You can't substitute them anymore like they're only meaningless letters.

The problem with the notation of partial derivatives is, even though it's practical, it hides the fact that you're actually defining functions of other variables: $s$ and $t$ are not variables per se but they're functions: $$s(x,y) = x+y \\ t(x,y)= x-y$$

I believe a clearer way would be to put aside function notation for a while. Instead of $f$, let's call it $u$, and keep the variable notation:

$$u = \text{something which depends on } s \text{ and } t$$

$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial x}$$ $$\\ \frac{\partial u}{\partial y} = \frac{\partial u}{\partial s}\frac{\partial s}{\partial y} + \frac{\partial u}{\partial t}\frac{\partial t}{\partial y}$$

So, if we say that $$\frac{\partial u}{\partial t} = 0$$

you're right here; it does imply that $u$ is only a function of $s$, i.e. $$u=\phi(s)$$

As you can see, with the variable notation I can't make the next step of your reasoning, where you replace $f(x,y)$ by $f(s,t)$, because I can't "plug" or "substitute" other variables in $u$.

There is slight yet common abuse of notation. To be more specific, we should write

Assume $f:\mathbb{R}^2\to\mathbb{R}$ is a functions, and $\begin{bmatrix} s\\ t\end{bmatrix}:\mathbb{R}^2\to\mathbb{R}^2$ are parametric variables. Then for $(x,y) \in \mathbb{R}^2$ we write $f = f(x,y)$.

Now, let us make change of variables in several steps:

1. Rename variables $(x,y)$ into $(s,t)$ and write $f = f(s,t)$. Note that both domain and codomain of $f$ are the same; the only thing we changed is labeling of variables.
2. Assume that $(s,t)$ variables actually depend on two parameters $(u,v)$, e.g. $$\begin{bmatrix} s\\ t\end{bmatrix} = \begin{bmatrix} \hat s\,(u,v)\\ \hat t(u,v)\end{bmatrix} \iff \begin{cases} s= \hat s\,(u,v),& \hat s : \mathbb{R}^2\to\mathbb{R} \\ t = \hat t(u,v), & \hat t : \mathbb{R}^2\to\mathbb{R} \end{cases}$$ Then $$f = f(s,t) = f\big(\hat{s}\,(u,v), \hat{t}(u,v)\big) = \hat f(u,v),$$ where $f$ and $\hat f$ are, strictly speaking, two different functions, as they have different domain.

3. Recycle the old variable labels by renaming $(u,v)$ into $(x,y)$. Then we will have $f(s,t) = \hat f(x,y)$.

Note that $(x,y)$ here are not the same as in the very first definition of $f$. The labels are reused, but the variables are different. People like to abuse notation in this way because it allows to reserve variable names for other needs; however this practice does introduce some confusion.

4. Omit hats in $\hat f$, obtaining $f(x,y)$ again.

Note that, just like before, $f(x,y)$ here is technically not the same as $f(x,y) \longleftrightarrow f(s,t)$ we started from. Actually, behind this $f(x,y)$ hides the old $\hat f (u,v) = f \big( \hat s\,(u,v), \hat t(u,v)\big)$

Therefore ultimately we have

$$f(x,y) \stackrel{\substack{x\leftrightarrow s \\y\leftrightarrow t}}{=} f(s,t) \xrightarrow[]{\substack{s\leftrightarrow \hat s(u,v)\\t\leftrightarrow \hat t(u,v)}} f \big(\hat s\,(u,v), \hat t(u,v)\big) \stackrel{\substack{u\leftrightarrow x \\v\leftrightarrow y}}{=}f \big(\hat s\,(x,y), \hat t(x,y)\big) =:\hat{f}(x,y).$$

Then, when you write $f(s,t) = \phi(s)$ you actually mean $f\big(\hat s\,(u,v), \hat t(u,v)\big) = \phi \big(\hat s\,(u,v)\big)$, or, replacing $(u,v)$ with $(x,y)$ labels again, $$f(s,t) = f\big(\hat s\,(x,y), \hat t(x,y)\big) = \phi \big(\hat s\,(x,y)\big) = \phi(x+y),$$ since for your case $$s = \hat s\,(x,y) = x+y \\ t = \hat t(x,y)= x-y$$

• I do not understand this: "Note that $(x,y)$ here are not the same as in the very first definition of $f$. The labels are reused, but the variables are different. " How are the variables different? Is it because $s$ and $t$ depend on the parameters $u$ and $v$? Also can you clarify the distinction between $\leftrightarrow$ and $\rightarrow$? – Lozansky May 18 '15 at 13:57
• Yes, it is. What we have is, basically, $s$ and $t$ depending on $u$ and $v$, and the old $x,y$, in turn, depend on $s,t$. But since we do not need the old $x$ and $y$ which depend on $s$ and $t$, we simply reuse these labels. Also $\to$ is used when we have some kind of mapping, e.g. one set of variables is mapped into another. On the other hand, I used $\leftrightarrow$ to mark the substitution or relabeling of the same set of variables. – Vlad May 23 '15 at 1:37