Help in proof of $\frac{\partial(u,v)}{\partial(x,y)}.\frac{\partial(x,y)}{\partial(u,v)} = 1$

Question

The following is a proof of $$\frac{\partial(u,v)}{\partial(x,y)}.\frac{\partial(x,y)}{\partial(u,v)} = 1$$

In the above proof, I cannot understand why $\frac{\partial u}{\partial v} = 0$ and $\frac{\partial v}{\partial u} = 0$. Since, $u$ and $v$ are independent variables, $\frac{\partial v}{\partial u}$ and $\frac{\partial u}{\partial v}$ seem to have no meaning. $ $ $ $ $ $

Answer 1

Your book's exposition is a little confusing, since it has $x, y, u, v$ being used as both variables and functions.

To make it clearer, I will try to re-derive the result using $x, y, u, v$ as variables and $X, Y, U, V$ as functions.

Suppose we have the functions $U(x, y), V(x, y), X(u, v), Y(u, v)$ such that the following relations hold:

$$X(U(x, y), V(x, y)) = x$$ $$Y(U(x, y), V(x, y)) = y$$ $$U(X(u, v), Y(u, v)) = u$$ $$V(X(u, v), Y(u, v)) = u$$

These are exactly the conditions that hold when the functions $U, V, X, Y$ specify two alternate coordinate systems. Here is a diagram of the situation where the action of $X,Y$ on points $u,v$ is depicted as a black arrow and the action of $U,V$ on points $x,y$ is depicted as a purple arrow.

Now the expressions $\partial[X|Y|U|V]/\partial[x|y|u|v]$ will make sense for all choices of "numerator" and "denominator" where the functions $X, Y, U, V$ must do double duty. In the expressions $\partial X/\partial x$ and $\partial X/\partial y$, the function $X$ stands for $X(U(x,y), V(x,y))$, and in the expressions $\partial X/\partial u$ and $\partial X/\partial v$, $X$ stands for $X(u,v)$.

Taking derivative of the first relation with respect to $x$ and applying the chain rule, you will arrive at $d_1X (U(x,y), V(x,y)) d_1U(x, y) + d_2X (U(x, y), V(x,y)) d_1V(x,y) = 1$ where $d_i$ means derivative with respect to the $i^{th}$ coordinate.

Writing this in the more terse and implicit notation we will get $\partial X/\partial u \, \partial U/\partial x + \partial X/ \partial v \, \partial V/\partial x = 1$.

Writing it even more confusingly, we get your book's presentation $\partial x/\partial u \, \partial u/\partial x + \partial x/\partial v \, \partial v/\partial x = 1$.

It should be straightforward applications of the chain rule to verify the other 3 equations.

Alternatively, a one liner proof:
In general if you have two functions $A : S_1 \subset \mathbb R^n \rightarrow S_2 \subset \mathbb R^n$, $B : S_2 \subset \mathbb R^n \rightarrow S_1 \subset \mathbb R^n$ such that $A(B(x)) = x$ and $B(A(x)) = x$ for all $x \in S_2, S_1$ respectively.

We have immediately $A'(B(x))B'(x) = I$ by the higher dimensional chain rule.

Answer 2

The expressions $\frac{\mathrm{d}v}{\mathrm{d}u}$ and $\frac{\mathrm{d}u}{\mathrm{d}v}$ are indeed meaningless because the differentials $\mathrm{d}u$ and ${\mathrm{d}v}$ are not ratios of one another.

But those are not the expressions appearing in the formulas.

When given a system of coordinates such as $\{ u, v \}$, the equations

$$ \frac{\partial v}{\partial u} = 0 = \frac{\partial u}{\partial v}$$

are basically (half of) the definition of the notation. For reductions to other familiar notions:

• $\frac{\partial}{\partial u}$ is notation that means to take the directional derivative in the direction where $v$ is held constant and $u$ increases with unit rate. That is, $\frac{\partial}{\partial u} = \nabla_{\vec{s}}$ where $\vec{s}$ is chosen to be the vector such that $\nabla_{\vec{s}} u = 1$ and $\nabla_{\vec{s}} v = 0$
• $\frac{\partial}{\partial u}$ is the linear operator on differential forms that sends $\mathrm{d}u \mapsto 1$ and $\mathrm{d}v \mapsto 0$. The notation $\frac{\partial z}{\partial u}$ means to apply $\frac{\partial}{\partial u}$ to the differential $\mathrm{d} z$.

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Aside: do take care to note that despite only one of the two variables appearing in $\frac{\partial}{\partial u}$, the operation depends on the entire choice of coordinate system $\{ u, v \}$. If $\{ u, x \}$ also happened to be a coordinate system, the operator $\frac{\partial}{\partial u}$ you would write for that coordinate system is different from the operator by the same name you would write for the $\{ u, v \}$ system.

If you ever find yourself in such a situation and still want to retain the use of this notation, you could do something like

$$ \left.\frac{\partial}{\partial u}\right|_{v \mathrm{\ const}} $$

to refer to the operator that is relative to the $\{ u, v \}$ coordinate system.