We got two functions given as:
$\alpha(x,y)=a_1 + a_2\Delta y + a_3\Delta x$
$\beta (y,z)= b_1 + b_2\Delta y +b_3\Delta z$
and I need to figure out if the coefficients $a_i$ and $b_i$ are somehow dependent.
We also know that the $\alpha$ and $\beta$ functions satistify two following relationships:
$\alpha (x,y) = \frac {1}{z} (\frac{\partial z}{\partial y})_x$ $\;$(meaning $x$ is held constant)
$\beta (y,z) = - \frac {1}{z} (\frac{\partial z}{\partial x})_y$ $\;$(meaning $y$ is held constant)
And we also have that $\Delta x = (x - x_0)$,$\;$ $\Delta y = (y - y_0)$$\;$ and$\;$ $\Delta z = (z - z_0)$ and we examine the behaviour around the point $(x_0, y_0, z_0)$.
Attempt of the solution:
I wanted to express $z$ in both functions by the integration and then sum them.
$\frac {dz}{z} = \alpha (x,y) dy $ $\;$ and $\;$ $-\frac {dz}{z} = \beta (y,z) dx$
$\alpha (x,y) dy + \beta (y,z) dx = 0$
(So we got 2-form equal to a zero function, if it is possible to view it like this)
$\int_{y_0}^{y} \alpha (x,y) dy + \int_{x_0}^{x} \beta (y,z) dx = 0$
$ \alpha (x,y)(y-y_0) + A(x) + \beta (y,z) (x-x_0) + B(y,z) = 0$
where $A$ and $B$ are integration constants.
(My first question is, if I can integrate it like the $\alpha$ and $\beta$ functions first and after that put the expression of them in it. And how can I get the $A$, $B$)
Then I would study the dependence of the coefficients $a_i$, $b_i$:
I was thinking about putting $(x_0,y_0,z_0) = (0,0,0)$ for simplicity, if I can. Then It would look like:
$ \alpha (x,y)(y) + A(x) + \beta (y,z) (x) + B(y,z) = T(a_1 + a_2y + a_3x) + x(b_1 + b_2 y +b_3z)=0$
OR
$ \alpha (x,y)(y-y_0) + A(x) + \beta (y,z) (x-x_0) + B(y,z) = (y-y_0)(a_1 + a_2(y-y_0) + a_3(x-x_0)) + (x-x_0)(b_1 + b_2 (y-y_0) +b_3(z-z_0))=0$
and try what would happen If $x=x_0$ and so on for example.
This is a part of the thermodynamics problem actually, where the math is maybe a little bit advanced for me. I have an idea what to do, but I struggle how to make it correctly mathematically. So I would appreciate any help! $(x,y,z)$ actually corresponds to $(p,V,T)$.