Problem 6 - Thomas Calculus

Question

An economic model Consider the following economic model. Let \(P\) be the price of a single item on the market. Let \(Q\) be the quantity of the item available on the market. Both \(P\) and \(Q\) are functions of time. If one considers price and quantity as two inter- acting species, the following model might be proposed: $$ \begin{aligned} \frac{d P}{d t} &=a P\left(\frac{b}{Q}-P\right) \\ \frac{d Q}{d t} &=c Q(f P-Q) \end{aligned} $$ where \(a, b, c,\) and \(f\) are positive constants. Justify and discuss the adequacy of the model. $$ \begin{array}{l}{\text { a. If } a=1, b=20,000, c=1, \text { and } f=30 \text { , find the equilibrium }} \\ {\text { points of this system. If possible, classify each equilibrium }} \\ {\text { point with respect to its stability. If a point cannot be }} \\ {\text { readily classified, give some explanation. }}\\\\{\text { b. Perform a graphical stability analysis to determine what will }} \\ {\text { happen to the levels of } P \text { and } Q \text { as time increases. }} \\ {\text { c. Give an economic interpretation of the curves that determine }} \\ {\text { the equilibrium points. }}\end{array} $$

Step-by-Step Solution

Answer

The stable equilibrium is at \((600, 18000)\), representing market balance.

Step 1: Define Equilibrium Points

To find the equilibrium points, set \( \frac{dP}{dt} = 0 \) and \( \frac{dQ}{dt} = 0 \) and solve for \( P \) and \( Q \). This leads to the equations: \( P \left( \frac{b}{Q} - P \right) = 0 \) and \( Q(fP - Q) = 0 \).

Step 2: Solve for Equilibrium Points

From the first equation, \( P = 0 \) or \( P = \frac{b}{Q} \). From the second equation, \( Q = 0 \) or \( Q = fP \). Using these values, we find potential equilibrium points are \((0,0)\), \((0, 20000)\), and \((600, 18000)\).

Step 3: Analyze Stability of Equilibrium Points

To classify the stability, use Jacobian matrix analysis. The Jacobian is calculated from the partial derivatives of the system. Evaluating the Jacobian at each equilibrium point, \((0,0)\) is a saddle point, \((0, 20000)\) is a saddle point, and \((600, 18000)\) is an asymptotically stable node.

Step 4: Graphical Stability Analysis

Graph the nullclines \( \frac{dP}{dt} = 0 \) and \( \frac{dQ}{dt} = 0 \). The nullclines divide the space into regions where \( P \) and \( Q \) increase or decrease. Around \( (600, 18000) \), trajectories converge, indicating stability, while near the other points, trajectories diverge, confirming instability.

Step 5: Economic Interpretation

The curves represent supply and demand balance. The point \((600, 18000)\) represents a stable market equilibrium where supply meets demand. Points \((0,0)\) and \((0, 20000)\) suggest conditions of no supply, which are unstable as they can't sustain market operations.

Key Concepts

Economic Model

An economic model is a simplified framework designed to illustrate complex processes in the economy. This model specifically looks at the interactions between price \((P)\) and quantity \((Q)\) of a market item over time.
The given model treats price and quantity like interacting species, where changes in one influence the other. The differential equations used reflect how growth in each aspect depends on both internal conditions and their interaction probabilities. For example:

• The price change \( \frac{dP}{dt} = aP(\frac{b}{Q} - P) \) describes how price increases depend on the available quantity and decrease with saturation.
• Similarly, \( \frac{dQ}{dt} = cQ(fP - Q) \) suggests that quantity increases with price but reduces if overproduction occurs.

This model helps study market dynamics, forecasting how prices and quantities adjust over time under different conditions.

Equilibrium Points

Equilibrium points in this context are where price \(P\) and quantity \(Q\) no longer change over time. To find them, set the rate of change equations to zero. This leads to solving for values of \(P\) and \(Q\).In our model:

• Setting \( \frac{dP}{dt} = 0 \) yields solutions \(P = 0\) or \(P = \frac{b}{Q}\).
• Setting \( \frac{dQ}{dt} = 0 \) gives \(Q = 0\) or \(Q = fP\).

Finding common solutions gives us equilibrium points like \((0,0)\), \((0, 20000)\), and \((600, 18000)\).Each equilibrium point represents a different market condition:

• \((0,0)\) indicates a market with no activity.
• \((0, 20000)\) shows a large supply with zero pricing power.
• \((600, 18000)\) represents a balanced, sustainable market state.

Stability Analysis

Stability analysis helps in determining how the economic model behaves near equilibrium points over time. Specifically, it asks whether small changes will return to equilibrium or move away.To examine stability, we:

• Use the Jacobian matrix, which is a matrix of first-order partial derivatives of our system.
• Evaluate it at each equilibrium point to study behavior.

From our solution, it’s found that:

• The point \((600, 18000)\) is asymptotically stable, meaning small deviations will converge back to it.
• Both \((0,0)\) and \((0, 20000)\) are saddle points, in which any slight deviation leads to instability.

This analysis informs us which market conditions are self-sustaining and which are prone to collapse.

Jacobian Matrix

A Jacobian matrix provides us insight into the local stability of the system around equilibrium points.
In our economic model, it is derived from the linearization of the differential equations around these points:For each equilibrium, the Jacobian consists of partial derivatives:

• The derivative of \(\frac{dP}{dt}\) with respect to \(P\) and \(Q\)
• The derivative of \(\frac{dQ}{dt}\) with respect to \(P\) and \(Q\)

Evaluating this matrix around the equilibrium points helps determine their behavior:

• If all eigenvalues are negative, the point is stable.
• If any are positive, it indicates possible divergence.

In our specific example, calculating and analyzing the Jacobian allows us to predict the stability and nature of equilibrium points, verifying our earlier conclusions.