Challenge. Solve for $y$ the expression $$\Delta \log\frac{y}{x_{1}}=s.\Delta \log\frac{x_{2}}{x_{1}}$$ That's it, that's my question. But you will probably need some background.
Background. This is from my research in economics, but my math is rusty so I am posting here. More precisely, $y$ measures how much income (or GDP) there is in an economy at a given time. I am interested in finding an equation for $y$, which if I am correct can be derived from the equation above. Some further points:
- My notations are not the best. $\Delta:=\frac{d}{dt}$, and $\Delta \log=\frac{dx/dt}x$ i.e. the rate of change over time. Furthermore, every variable $y,x_{1},x_{2}$ as well as $s$ are all indexed in time. Feel free to rewrite the equation above if needed.
- In addition I know that the following relationship holds true: $\frac{x_{2}}{x_{1}}=\frac{s}{1-s}\frac{p_{1}}{p_{2}}$
What I've done. Below I provide some steps I have already taken, hoping that they are correct:
First I integrated both with respect to time, i.e. $$ \int\Delta \log\frac{y}{x_{1}}dt=\int s.\Delta \log\frac{x_{2}}{x_{1}}dt$$
Then I note that I have a product of two functions in the right-hand side integral, so I proceeded by integrating by parts. Let $u=s, du=ds$ and $dv=\Delta \log\frac{x_{2}}{x_{1}}, v=\log\frac{x_{2}}{x_{1}}$. Again every variable is indexed in time. The integration by part follows $\int u.dv = u.v -\int v.du$. This leads to $$\log\frac{y}{x_{1}}+C_{1}=s\log\frac{x_{2}}{x_{1}}-\int \log\frac{x_{2}}{x_{1}}ds$$
This is already a nice and very interesting first step for my research. But I am not done as I still have to calculate the integral $\int \log\frac{x_{2}}{x_{1}}ds$. As I understand it, I have to integrate $\log\frac{x_{2}}{x_{1}}$ with respect to $s$. This is where my problem starts, as $\log\frac{x_{2}}{x_{1}}$ is not an expression in $s$. But I can use the relationship in point #2 above, which relates $\frac{x_{2}}{x_{1}}$ to an expression in $s$ times $\frac{p_{1}}{p_{2}}$. Using this substitution I get $$\log\frac{y}{x_{1}}+C_{1}=s\log\frac{x_{2}}{x_{1}}-\int \log\left(\frac{s}{1-s}\frac{p_{1}}{p_{2}}\right)ds$$ or equivalently: $$\log\frac{y}{x_{1}}+C_{1}=s\log\frac{x_{2}}{x_{1}}-\int \log\frac{s}{1-s}ds-\int \log\frac{p_{1}}{p_{2}}ds$$
Now I have an expression consisting of four terms. The first three do not cause any problem. It is the last term I have trouble with. Specifically, what is $$\int \log\frac{p_{1}}{p_{2}}ds$$ equal to? Maybe I should end with some more background. $\frac{p_{1}}{p_{2}}$ does not seem equal to anything useful. Those two $p$'s are the prices of two products, and the ratio of those two prices is what it is. There is no expression I can substitute for this ratio, except using the relationship in point #2 above again, which probably means going backwards, not forwards.
Maybe there is something to do with $s$ in that last integral? From relationship #2 again, we find that $s=\frac{p_{2}x_{2}}{p_{1}x_{1}+p_{2}x_{2}}$, which means that the last integral is $$\int \log\frac{p_{1}}{p_{2}}.d\left(\frac{p_{2}x_{2}}{p_{1}x_{1}+p_{2}x_{2}}\right)$$ ... but this is far from helping me.
So I will conclude with one word: help. Help with solving the very first equation directly, or better yet, help with solving the expressions in #6 or #7.
Start with the original question, to solve $$\Delta log \frac{y}{x_1} = s \Delta log \frac{x_2}{x_1}.$$ We can re-write that as $$\frac{d}{dt}[log\ y -log\ x_1]=s\frac{d}{dt}[log\ x_2 - log\ x_1]$$ or $$\frac{d}{dt}[log\ y ]=s\frac{d}{dt}[log\ x_2 - log\ x_1] +\frac{d}{dt}log\ x_1 $$ or $$\frac{d}{dt}log\ y =s\frac{d}{dt}log\ x_2 +(1-s)\frac{d}{dt}log\ x_1 $$
Now we recall the useful fact, which is the point of the log formulation, that $$\frac{d}{dt}log\ y = \frac{1}{y}\frac{dy}{dt}$$ thus the derivative of the log is the percentage change. In econ jargon the original problem is a question about the relation of two elasticities.
So now we have a differential equation of the following form (I stick with the one variable case to make the logic more clear): $$\frac{1}{y}\frac{dy}{dt}=k\frac{1}{x}\frac{dx}{dt}$$ and using separation of variables, that reduces to $$\frac{1}{y}dy=k\frac{1}{x}dx.$$ Integrating both sides gives the solution $log \ y=k\ log\ x$ or $$y=x^k.$$
Now by analogy, the solution to the original problem becomes $$y=x_1^{1-s}x_2^s$$ which looks a lot like a Cobb-Douglas production function. It pays to remember that $x_1, x_2$ and $y$ are all functions of $t$. This says nothing about the prices $p_1,p_2$, but you can substitute out for either one of the x's and just have a few more constants floating around.
Separately, is this a growth accounting exercise? Have you looked at chapter 12 of Glaister's Mathematical Methods of Economists? or the More advanced The Theory of Equilibrium Growth by Dixit?