Question

Solve the following exhaustible resource problem for the optimal extraction path: $$\begin{array}{ll}\text { Maximize } & \int_{0}^{T} \ln (q) e^{-s t} d t \\\\\text { subject to } & s^{\prime}=-q \\ \text { and } & s(0)=s_{0} \quad s(t) \geq 0\end{array}$$

Short answer

The optimal extraction path is \( q(t) = \frac{1}{\lambda(0)} e^{-(s+1)t} \), adjusting for initial conditions and ensuring \( s(t) \geq 0 \).

Step-by-step solution

Understanding the Problem Setup

The problem aims to maximize the integral \( \int_{0}^{T} \ln(q) e^{-st} dt \) subject to constraints related to the extraction rate of an exhaustible resource: \( s' = -q \) and initial stock \( s(0) = s_0 \), with the constraint \( s(t) \geq 0 \). This involves determining the optimal path for \( q(t) \) such that the integral is maximized under these constraints.

Setting Up the Hamiltonian

For optimal control problems, we use the Hamiltonian function which in this problem is defined as: \[ H = \ln(q) e^{-st} + \lambda (-q) \] where \( \lambda \) is the co-state variable associated with the stock of resources \( s(t) \).

Deriving the Optimality Conditions

The first order conditions for an optimal solution involve taking the derivative of the Hamiltonian with respect to \( q \) and setting it to zero:\[ \frac{\partial H}{\partial q} = \frac{1}{q} e^{-st} - \lambda = 0 \]Solving for \( q \), we get:\[ q = \frac{1}{\lambda} e^{-st} \]

Differential Equation for Co-state Variable

The co-state equation is derived from the Hamiltonian by using \( \frac{d\lambda}{dt} = s \lambda - \frac{\partial H}{\partial s} \). Given \( \frac{\partial H}{\partial s} = -t \ln(q) e^{-st} \) we have:\[ \frac{d\lambda}{dt} = \lambda \]

Solution of Co-state Variable Differential Equation

Since \( \frac{d\lambda}{dt} = \lambda \), it satisfies the differential equation of the form \( \lambda(t) = \lambda(0) e^t \) with \( \lambda(0) \) being a constant of integration.

Applying Conditions and Solving for \( q(t) \)

Substitute \( \lambda(t) = \lambda(0) e^{t} \) into the equation for \( q \):\[ q(t) = \frac{1}{\lambda(0) e^{t}} e^{-st} = \frac{1}{\lambda(0)} e^{-(s+1)t} \]This is the extraction path, but since \( s' = -q \) and initial conditions must be satisfied, further integration and evaluation are required.

Key concepts

Exhaustible Resources

Exhaustible resources are natural resources that have a finite supply. They are not replenished in a short time period. These resources include fossil fuels like coal, oil, and natural gas, as well as certain minerals and metals.
Understanding how to manage the extraction of these resources is crucial, as their depletion affects economic sustainability and environmental balance. In economics, a key problem is to determine the rate of extraction that maximizes utility over a given time period.
In the problem setup we're exploring, the initial quantity of the resource, denoted by \( s_0 \), dictates the availability over time. The challenge is maximizing the benefit derived from extracting the unit quantity of resource \( q(t) \), considering its declining stock over time, \( s'(t) = -q(t) \). Managing these constraints effectively is at the core of optimal control theory applied to exhaustible resources.

Hamiltonian Function

The Hamiltonian function is a pivotal concept in optimal control theory. It combines the objective function and the constraints into one comprehensive equation, helping us find the optimal path for control variables. In this context, it incorporates both the economic benefit of resource extraction and the dynamics of resource stock change.
In our problem, the Hamiltonian is expressed as:
\[ H = \ln(q) e^{-st} + \lambda (-q) \] - The term \( \ln(q) e^{-st} \) represents the instantaneous benefit from extraction, discounted over time with the exponential function. - The term \( \lambda (-q) \) involves the co-state variable, which signifies the shadow price or the marginal value of the resource stock.
The Hamiltonian concept aids in aligning short-term extraction benefits with long-term resource availability, offering a structured pathway to find the optimal solution for the extraction sequence.

Optimal Extraction Path

The optimal extraction path refers to the sequence of extraction rates over time that maximizes the overall utility from the resource.
For our scenario, this involves solving the Hamiltonian's first-order condition:
\[ \frac{\partial H}{\partial q} = \frac{1}{q} e^{-st} - \lambda = 0 \]
Solving for \( q \), we get the extraction path as
\[ q = \frac{1}{\lambda} e^{-st} \]
This equation signifies how the extraction rate, \( q(t) \), should adjust in response to the resource's remaining quantity and the time passage, factoring into the discount rate and co-state variable. The ultimate extraction path balances the marginal cost and benefit of extraction over time.
Understanding and calculating this path is crucial for the optimal use of exhaustible resources.

Co-state Variable

The co-state variable, \( \lambda(t) \), serves as the shadow price or the marginal valuation of an additional unit of the resource stock over time in optimal control problems.
It represents the value of having one more unit of the stock, indicating how much the objective would improve with a slight increase in stock. In our exhaustible resources problem, \( \lambda(t) \) directly influences the optimal extraction rate.
The differential equation governing \( \lambda(t) \) is:
\[ \frac{d\lambda}{dt} = \lambda \]
This equation implies exponential growth or decay, depending on its sign. The constant \( \lambda(0) \) is determined by boundary conditions and integration.
In our solution, we found \( \lambda(t) = \lambda(0) e^{t} \), which helped define the extraction path by establishing a relationship between the resource's marginal value and its extraction pace. The co-state variable plays a fundamental role in mathematically linking resource stock dynamics with economic decisions in optimal control theory.