Show $(\frac{\partial f}{\partial x})^2-(\frac{\partial f}{\partial y})^2=(\frac{\partial f}{\partial r})^2-1/{r^2}(\frac{\partial f}{\partial t})^2$

Question

I have been asked to prove the following relation: $$(\frac{\partial f}{\partial x})^2-(\frac{\partial f}{\partial y})^2=(\frac{\partial f}{\partial r})^2-\frac 1{r^2}(\frac{\partial f}{\partial t})^2$$

Given that $x=r\cosh (t)$ and $y=r\sinh(t)$

From this it can be deduced that: $f=f(x,y)$, $f=f(r,t)$, $x=x(r,t)$, $y=y(r,t)$, $r=r(x,y)$ and $t=t(x,y)$

Each of the above ($df,dx,dy,dt...$) will have a total derivative associated with them. Thus it if possible to use the chain rule to obtain expressions for the partial derivatives given in the question.

$$(\frac{\partial f}{\partial r})=(\frac{\partial f}{\partial x})(\frac{\partial x}{\partial r})+(\frac{\partial f}{\partial y})(\frac{\partial y}{\partial r})$$ $$(\frac{\partial f}{\partial t})=(\frac{\partial f}{\partial x})(\frac{\partial x}{\partial t})+(\frac{\partial f}{\partial y})(\frac{\partial y}{\partial t})$$ $$(\frac{\partial f}{\partial x})=(\frac{\partial f}{\partial r})(\frac{\partial r}{\partial x})+(\frac{\partial f}{\partial t})(\frac{\partial t}{\partial x})$$ $$(\frac{\partial f}{\partial y})=(\frac{\partial f}{\partial r})(\frac{\partial r}{\partial y})+(\frac{\partial f}{\partial t})(\frac{\partial t}{\partial y})$$

Now here's my issue. The problem should be possible to solve by finding equations for the squares of $(\frac{\partial f}{\partial r})$ and $(\frac{\partial f}{\partial t})$ and finding thier difference. Or by doing the same with $(\frac{\partial f}{\partial x})$ and $(\frac{\partial f}{\partial y})$. I found it very simple to do the former but I could not find the relation using the expressions for the derivatives with respect to $x$ and $y$.

I used the following derivatives in my failed attempt using the expressions for the derivatives with respect to $x$ and $y$:

$(\frac{\partial r}{\partial x})=\cosh(t)$, $(\frac{\partial r}{\partial y})=\sinh(t)$, $(\frac{\partial t}{\partial x})=-\frac{\sinh(t)}r$ and$(\frac{\partial t}{\partial y})=\frac{\cosh(t)}r$

How is it possible?

Answer 1

to go the other way round,

$$ f_x=f_r r_x + f_t t_x \\ f_y=f_r r_y + f_t t_y $$ since $$ r^2 = x^2-y^2 $$ we have $$ r_x= \frac{x}{r} \\ r_y= -\frac{y}{r} $$ also $$ t = \tanh^{-1}\frac{y}{x} $$ so $$ t_x = \frac{y}{r^2} \\ t_y = - \frac{x}{r^2} $$ this gives: $$ (f_x)^2 - (f_y)^2 = (\frac{x}{r}f_r +\frac{y}{r^2}f_t)^2 -(-\frac{y}{r}f_r -\frac{x}{r^2}f_t)^2 \\ = (f_r)^2 - \frac1{r^2} (f_t)^2 $$ using the same identity as before, and again with cancellation of the mixed terms, which this time are $\frac{2xyf_rf_t}{r^3}$

it is worth noting that the symmetry shown by the relations: $$ \begin{pmatrix} f_r \\f_t \end{pmatrix}=\begin{pmatrix}x_r & y_r \\ x_t & y_t \end{pmatrix} \begin{pmatrix} f_x \\ f_y \end{pmatrix} $$

$$ \begin{pmatrix} f_x \\f_y \end{pmatrix}=\begin{pmatrix}r_x & t_x \\ r_y & t_y \end{pmatrix} \begin{pmatrix} f_r \\ f_t \end{pmatrix} $$ is expressed in terms of the Jacobian determinants: $$ \frac{\partial(x,y)}{\partial(r,t)} \frac{\partial(r,t)}{\partial(x,y)} = 1 $$

Answer 2

use the abbreviations $c=\cosh t, s = \sinh t$ and denote differentiation by a subscript.

then, as you have remarked: $$ f_r = f_x x_r + f_y y_r = f_x c + f_y s \\ f_t = f_x x_t + f_y y_t = r(f_x s + f_y c) $$ this gives: $$ (f_r)^2 - \frac1{r^2}(f_t)^2 = (f_x c + f_y s)^2 - (f_x s + f_y c)^2 $$ which simplifies to $(f_x)^2 - (f_y)^2$ on account of the identity $c^2-s^2 =1$ and the cancellation of the mixed terms $2f_xf_y c s$