Hi!
Great question! This problem combines economic intuition with differential equations. The main idea is that a perfectly competitive firm gradually adjusts its output depending on whether the market price is above or below marginal cost.
Let's work through each part.
(a) Derive the nonlinear differential equation
We're given the adjustment rule:
[
\frac{dq}{dt}=\alpha[\beta^2-MC(q)]
]
and the marginal-cost function:
[
MC(q)=\frac12q^2
]
Substituting marginal cost into the adjustment equation gives:
[
\boxed{\frac{dq}{dt}=\alpha\left(\beta^2-\frac12q^2\right)}
]
This is a first-order, autonomous, nonlinear differential equation.
It is nonlinear because it contains (q^2).
Economically, if price exceeds marginal cost, output increases. If marginal cost exceeds price, output decreases.
(b) Draw the phase diagram
To identify the equilibrium points, set the rate of change equal to zero:
[
\frac{dq}{dt}=0
]
Therefore:
[
\alpha\left(\beta^2-\frac12q^2\right)=0
]
Assuming (\alpha>0):
[
q^2=2\beta^2
]
Thus:
[
q^*=\pm\sqrt2\beta
]
Because output cannot be negative, the economically relevant equilibrium is:
[
\boxed{q^*=\sqrt2\beta}
]
Now examine the direction of output adjustment.
When:
[
0\le q<\sqrt2\beta
]
we have:
[
\frac{dq}{dt}>0
]
Therefore, output increases.
When:
[
q>\sqrt2\beta
]
we have:
[
\frac{dq}{dt}<0
]
Therefore, output decreases.
The economically relevant phase line is:
[
0\qquad\longrightarrow\qquad
\boxed{\sqrt2\beta}
\qquad\longleftarrow
]
The arrows point toward equilibrium from both directions.
This suggests that the equilibrium is stable.
(c) Solve the differential equation
Starting with:
[
\frac{dq}{dt}
=\alpha\left(\beta^2-\frac12q^2\right)
]
Separate the variables:
[
\frac{dq}{\beta^2-\frac12q^2}=\alpha,dt
]
Define:
[
K=\sqrt2\beta
]
Then:
[
\frac{2,dq}{K^2-q^2}=\alpha,dt
]
Integrating both sides:
[
\frac1K\ln\left|\frac{K+q}{K-q}\right|
=\alpha t+C
]
Multiplying by (K):
[
\ln\left|\frac{K+q}{K-q}\right|
=\alpha Kt+C_1
]
Exponentiating and rearranging:
[
\frac{K+q}{K-q}=Ce^{\alpha Kt}
]
Solving for (q):
[
q(t)=K\frac{Ce^{\alpha Kt}-1}{Ce^{\alpha Kt}+1}
]
Substituting (K=\sqrt2\beta):
[
\boxed{
q(t)=\sqrt2\beta
\frac{Ce^{\alpha\sqrt2\beta t}-1}
{Ce^{\alpha\sqrt2\beta t}+1}
}
]
This gives the nonconstant solutions wherever the denominator is nonzero.
If the initial output is (q(0)=q_0), then:
[
C=\frac{K+q_0}{K-q_0}
]
for (q_0\ne K).
The equilibrium solution (q(t)=K) must also be included separately.
(d) Determine equilibrium stability
We can verify stability mathematically using the derivative test.
Define:
[
f(q)=\alpha\left(\beta^2-\frac12q^2\right)
]
Differentiate with respect to (q):
[
f'(q)=-\alpha q
]
The standard stability theorem for a one-dimensional autonomous differential equation states:
If (f'(q^*)<0), the equilibrium is locally asymptotically stable.
If (f'(q^*)>0), the equilibrium is unstable.
At the economically relevant equilibrium:
[
q^*=\sqrt2\beta
]
we obtain:
[
f'(q^*)=-\alpha\sqrt2\beta
]
Since (\alpha>0) and (\beta>0):
[
\boxed{f'(q^*)<0}
]
Therefore:
[
\boxed{\text{The positive equilibrium is asymptotically stable.}}
]
The phase diagram confirms this result because output increases below equilibrium and decreases above equilibrium.
In fact, for every nonnegative initial output, the solution approaches the positive equilibrium as time increases.
Final Answers
(a) Differential equation:
[
\boxed{\frac{dq}{dt}=\alpha\left(\beta^2-\frac12q^2\right)}
]
(b) Phase diagram:
Output moves toward (q^*=\sqrt2\beta) from either side.
(c) General nonconstant solution:
[
\boxed{
q(t)=\sqrt2\beta
\frac{Ce^{\alpha\sqrt2\beta t}-1}
{Ce^{\alpha\sqrt2\beta t}+1}
}
]
(d) Stability:
[
\boxed{\text{The positive equilibrium is asymptotically stable.}}
]
Economic interpretation: The firm adjusts its production toward the quantity where price equals marginal cost. If it produces too little, price exceeds marginal cost, encouraging greater production. If it produces too much, marginal cost exceeds price, encouraging reduced production.
This adjustment process naturally pushes the firm toward its profit-maximizing output.
Hope this helps!
Michael
