Solving a 2-variable differential equation using logarthmic differentiation

Question

Let $f(x,k): \mathbb{R}^2_+\rightarrow \mathbb{R}_+$ be the following function,

$$\dfrac{2kx}{x+c},$$

where $c$ is some positive constant. I'm essentially trying to figure out the "elasticity" of this equation between variables $k$ and $x$ (this is just context, and understanding of the concept is not required for the maths).

Let $f_x$ be $df(x,k)/dx$, and $f_k$ be $df(x,k)/dk$. Then what is,

$$\frac{dln(x/k)}{dln(f_k/f_x)}?$$

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I know that this becomes the following,

$$\frac{d(x/k)}{x/k}/\frac{d(f_k/f_b)}{f_k/f_x}$$

Given that $f_x=\frac{2kc}{(x+c)^2}$, and $f_k=\frac{2x}{x+c}$, this is

$$\frac{d(x/k)}{x/k}/\bigg(\frac{d(f_k/f_b)}{x(x+c)/kc}\bigg).$$ But how do I evaluate what's left? I'm used to derivatives, and not so much to differentiation expressions.

Answer 1

You made an error in computing $f_k$. \begin{align} f_x &= \frac{2kc}{(x+c)^2} \\ f_k &= \frac{2x}{x+c} \\ \frac{f_k}{f_x} &= \frac{x(x+c)}{kc} \end{align} Let $k = v(x)$ where $v(x)$ is some function of x. \begin{align} \Rightarrow \frac{x}{k} &= \frac{x}{v(x)} \text{ and } \frac{f_k}{f_x} = \frac{x(x+c)}{v(x)c} \\ \Rightarrow E_{xk} &= \frac{d\ln(\frac{x}{k})}{d\ln(\frac{f_k}{f_x})} = \frac{d\ln(\frac{x}{v(x)})}{d\ln(\frac{x(x+c)}{v(x)c})} \\ \Rightarrow E_{xk} &= \frac{\frac{d\ln(\frac{x}{v(x)})}{dx}}{\frac{d\ln(\frac{x(x+c)}{v(x)c})}{dx}} = \frac{(x+c)(v(x) - xv'(x))}{x(2v(x) - xv'(x)) + c(v(x) - xv'(x))} \\ &= \frac{(x+c)u'(x)}{x(k + u'(x)) + cu'(x))} = \frac{(x+c)u'(x)}{xk + (x+c)u'(x)} \end{align} where $u(x) = \frac{v(x)}{x} = \frac{k}{x} \Rightarrow u'(x) = \frac{\partial \frac{k}{x}}{\partial x}$.

Hence, in such a case you can't report the elasticity without knowing rate of change of one input w.r.t to the other.

In the case where there is no functional relation w.r.t k and x, then simply equate $u'(x) = -\frac{k}{x^2}$ in the above expression.

Answer 2

Since the dependence or independence of the variables $x$ and $k$ is not evident a priori I suggest considering both of them as functions of a hidden variable $t$ ("time").

Given $f(u,v):={2uv\over u+c}$ we can compute $f_u(u,v)={2vc\over (u+c)^2}$, $\ f_v(u,v)={2u\over u+c}$. You are interested in the quantity $$q(t):={{d\over dt}\log\bigl(x(t)/k(t)\bigr)\over{d\over dt}\log\bigl(f_v\bigl(x(t),k(t)\bigr)/f_u\big(x(t),k(t)\bigr)\bigr)}\ .$$ After performing the necessary calculations the result can be expressed as a function of $x=x(t)$, $k=k(t)$, and the quotient ${k'(t)\over x'(t)}$ which encodes a possible dependence between the variables $x$ and $k$.