Question

Let D: {(x,y)| x>0, y>0}. Consider two function f and g from D to R, defined by:

f(x,y) = Inx - Iny and g(x,y)= x^2+ 3y^2/(2xy)

Show that the necessary condition for the functional dependence of f and g is satisfied. Also find a functional relation between f and g

Answer 1

As we know, two functions $f(x,y)$ and $g(x,y)$ are dependent if the Jacobi matrix is equal to zero :

$$J=\frac{\partial\left(f,g\right)}{\partial\left(x,y\right)}= \left|\begin{array}{cc} f'_x&f'_y\\[0.3cm] g'_x&g'_y \end{array}\right|$$

In our case,

$$f(x,y)=\ln x-\ln y\longrightarrow\left\{ \begin{array}{l} f'_x=\displaystyle\frac{1}{x}\\[0.3cm] f'_y=-\displaystyle\frac{1}{y} \end{array}\right.\\[0.3cm] g(x,y)=\frac{x^2+3y^2}{2xy}\equiv\frac{x}{2y}+\frac{3y}{2x}\longrightarrow\left\{ \begin{array}{l} g'_x=\displaystyle\frac{1}{2y}-\displaystyle\frac{3y}{2x^2}\\[0.3cm] g'_y=-\displaystyle\frac{x}{2y^2}+\displaystyle\frac{3}{2x} \end{array}\right.\\[0.3cm]$$

Then,

$$J=\left|\begin{array}{cc} f'_x&f'_y\\[0.3cm] g'_x&g'_y \end{array}\right|=\left|\begin{array}{cc} \displaystyle\frac{1}{x} &-\displaystyle\frac{1}{y}\\[0.5cm] \displaystyle\frac{1}{2y}-\displaystyle\frac{3y}{2x^2}& -\displaystyle\frac{x}{2y^2}+\displaystyle\frac{3}{2x} \end{array}\right|=\\[0.3cm] =\frac{1}{x}\cdot\left(-\frac{x}{2y^2}+\frac{3}{2x}\right)- \left(-\frac{1}{y}\right)\cdot\left(\frac{1}{2y}-\frac{3y}{2x^2}\right)=\\[0.3cm] =-\frac{1}{2y^2}+\frac{3}{2x^2}+\frac{1}{2x^2}-\frac{3}{2x^2}\equiv0\longrightarrow\\[0.3cm] \boxed{\text{Conclusion,}\,\,\,J\left(f,g\right)\equiv0,\,\,\forall (x,y)\in D=\left\{(x,y)\left|x>0,y>0\right.\right\}}$$

To find the functional relationship between $f(x,y)$ and $g(x,y)$ we will do the following :

$$z=f(x,y)=\ln x-\ln y\equiv\ln\left(x/y\right)\\[0.3cm] u=g(x,y)=\frac{x^2+3y^2}{2xy}=\frac{1}{2}\cdot\left(\frac{x}{y}+3\cdot\frac{y}{x}\right)\\[0.3cm] u=\frac{1}{2}\cdot\left(\frac{x}{y}+3\cdot\left(\frac{x}{y}\right)^{-1}\right)$$

Our task is reduced to finding a function $\varphi$ between variables $z$ and $u$ .

As we know

$$e^z=e^{\ln\left(x/y\right)}=\frac{x}{y}\\[0.3cm] e^{-z}=e^{-\ln\left(x/y\right)}=e^{\ln\left(y/x\right)}=\frac{y}{x}\\[0.3cm] \varphi(z)=\frac{1}{2}\cdot\left(e^z+3e^{-z}\right)$$

Or in the original notation

$$\boxed{\varphi\left(f(x,y)\right)=\frac{1}{2}\cdot\left(e^{f(x,y)}+3e^{-f(x,y)}\right)\equiv g(x,y)}$$