How Is Equilibrium Price Stability Determined in Market Dynamics?

[TheBestMilk]

Homework Statement

In the market for a certain good, the price p(t) adjusts continuously in the presence of excess supply or demand:
$\frac{dp}{dt}$ = F(D(p)-S(p)) where F(0) = 0, F'>0.
Obtain the condition for stability of the equilibrium price p* in terms of the slopes D'(p*) and S'(p*), and illustrate stable and unstable equilibria graphically for straight-line supply and demand schedules.

Homework Equations

General I suppose, p' = F(p)

The Attempt at a Solution

This is what I've come up with so far, but I'm not sure if it's correct or delves deeply enough:
p'=F(D(p)-S(p))
We know there would be an equilibrium point where p'=0 and therefore where F(D(p)-S(p))=0.
Since F(0) = 0, we know that at least one equilibrium exists at p=0, so p*=0. From that, we can see that p'=0=F(D(0)-S(0)) and therefore at p*, D(0) = S(0).

To test stability we would look at the sign of p'' at p*.
p'' = F'(D(p*)-S(p*))(D'(p*)-S'(p*))
Since we're given F'>0, we know that the D'(p*)-S'(p*) dictates stability.
So if D'(p*) > S'(p*) then p'' >0 and p* is unstable.
Else, if D'(p*) < S'(p*) then p'' <0 and p* is stable.

That's what I've come up with. It seems a bit simple which leads me to believe I may be missing something. Any help or insight would be greatly appreciated.

Thanks!

 

[mighty2000]

Your solution is correct and provides a good understanding of the stability condition for equilibrium price. However, to provide a more in-depth analysis, I would like to suggest considering the second derivative of p with respect to time, p''.

As you have correctly stated, p'' = F'(D(p*)-S(p*))(D'(p*)-S'(p*)). This means that the stability of the equilibrium price p* also depends on the slope of the function F'(p). If F'(p) is positive, then p* will be stable only if D'(p*) > S'(p*), as you have mentioned. However, if F'(p) is negative, then p* will be stable only if D'(p*) < S'(p*).

Moreover, it is also important to consider the magnitude of the second derivative p''. If p'' is close to zero, then p* will be stable for a small range of values of D'(p*) and S'(p*). However, if p'' is significantly positive or negative, then p* will be stable for a wider range of values of D'(p*) and S'(p*).

To illustrate this, you can plot graphs of p'' vs D'(p*) and p'' vs S'(p*). This will give you a better understanding of how the stability of p* changes with respect to the slopes of D(p) and S(p).

Additionally, you can also consider the case when F'(p) = 0. In this case, the stability of p* will depend solely on the slopes D'(p*) and S'(p*). If D'(p*) > S'(p*), then p* will be stable, and if D'(p*) < S'(p*), then p* will be unstable.

I hope this helps in providing a deeper understanding of the stability condition for equilibrium price. Keep up the good work in analyzing and solving scientific problems!