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Chapter 10: Problem 2

(a) Solve the following problem when \(r>a>0\) : $$ \max _{u \in[0,1]} \int_{0}^{\infty} x_{2} e^{-r t} d t, \quad \begin{cases}\dot{x}_{1}=a u x_{1}, & x_{1}(0)=x_{1}^{0}>0 \\\ \dot{x}_{2}=a(1-u) x_{1}, & x_{2}(0)=x_{2}^{0}=0\end{cases} $$ (b) Show that the problem has no solution when \(r

Short Answer

Expert verified
The optimal control problem can be solved through the Hamiltonian approach and costate equations. For \( r < a \), the problem has no solution because of integrand divergence.

Step by step solution

01

Define the Hamiltonian

Given the system dynamics, the Hamiltonian can be defined as:\[ H = x_2 e^{-r t} + u_1 (a u x_1) + u_2 (a(1 - u) x_1) \]
02

Determine the Costate Equations

To find costate equations, use the Hamiltonian:\[ \frac{du_1}{dt} = -\frac{\text{partial} H}{\text{partial} x_1} \]\[ \frac{du_2}{dt} = -\frac{\text{partial} H}{\text{partial} x_2} \]Derivative of Hamiltonian with respect to state variables gives:\[ \frac{du_1}{dt} = -(u_1 a u + u_2 a(1-u)) \]\[ \frac{du_2}{dt} = -e^{-r t} \]
03

Solve the Second Costate Equation

Solving \( \frac{du_2}{dt} = - e^{-r t} \) gives the costate variable \( u_2 \):\[ u_2 = \frac{1}{r} e^{-r t} + C_2 \]Applying the transversality condition \( u_2(\infty) = 0 \) means that \( C_2 = - \frac{1}{r} \), so\[ u_2 = \frac{1}{r}(1 - e^{-r t}) \]
04

Substitute Back into the Costate Equation for \(u_1\)

Substitute \( u_2 \) back into \( \frac{du_1}{dt} = -(u_1 a u + u_2 a(1-u)) \): \[ \frac{du_1}{dt} = - u_1 a u - \frac{a (1 - u)}{r}(1 - e^{-r t}) - C_1 \]
05

Maximize the Hamiltonian

To maximize the Hamiltonian with respect to control variable \(u \), compute the derivative \( \frac{\text{partial} H}{\text{partial} u} \) and equate to 0:\[ \frac{\text{partial} H}{\text{partial} u} = u_1 a x_1 - u_2 a x_1 = 0 \]This gives \( u_1 = u_2 \). But for maximization, we need boundary conditions to solve completely.
06

Evaluate When \( r < a \)

When \( r < a \), the integrand grows negatively since the costates \( u_i \); thus, the function no longer converges. Hence, the problem is not well defined and has no solution for \( r < a \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hamiltonian
In optimal control theory, the Hamiltonian plays a crucial role. It combines the objective function and the dynamics of the system being studied. In simple terms, it helps us evaluate how well our control strategy works over time. For our exercise, the Hamiltonian is given by: \[ H = x_2 e^{-r t} + u_1 (a u x_1) + u_2 (a(1 - u) x_1) \]Here, the function integrates the immediate reward from the state variable \( x_2 \) with the control effects on \( x_1 \) and \( x_2 \). The Hamiltonian lets us encapsulate both the current state and the control鈥檚 impact on future states into one formula.
Costate Equations
The costate equations are derived from the Hamiltonian and help in finding the optimal control path. They show how the shadow prices or marginal valuations of the state variables evolve over time. Mathematically, we derive them by taking partial derivatives of the Hamiltonian:
  • \(\frac{du_1}{dt} = -\frac{\partial H}{\partial x_1}\)
  • \(\frac{du_2}{dt} = -\frac{\partial H}{\partial x_2}\)
In our problem, these result in:
  • \(\frac{du_1}{dt} = -(u_1 a u + u_2 a(1-u))\)
  • \(\frac{du_2}{dt} = -e^{-r t}\)
These equations describe how the valuation of state variables changes over time, influenced by the control variable \( u \).
Transversality Condition
The transversality condition ensures that the solution to our problem is optimal at the infinite time horizon. For our solution, we use it to determine the constants in our equations:\[ u_2 (t) = \frac{1}{r} e^{-r t} + C_2 \]
Using the transversality condition \( u_2(\infty) = 0 \), we find that \( C_2 = -\frac{1}{r} \), leading to:\[ u_2 = \frac{1}{r}(1 - e^{-r t}) \]
Transversality conditions are essential because they help to ensure that the behavior of costate variables converges appropriately as time goes to infinity.
Control Variable
The control variable \( u \) represents the decisions we make at each point in time to optimize our objective. In our exercise, maximizing the Hamiltonian with respect to the control variable helps us define its optimal path. Maximizing \( H \) results in:\[ \frac{\partial H}{\partial u} = u_1 a x_1 - u_2 a x_1 = 0 \]This simplifies to \( u_1 = u_2 \). However, to identify the boundary conditions and fully solve the problem, additional steps are needed. The control variable is critical since it directly influences the state dynamics and therefore the outcome of our optimization problem.

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Most popular questions from this chapter

Solve the problem $$ \min \int_{0}^{5}(u+x) d t, \quad \dot{x}=u-t, \quad x(0)=1, \quad x(5) \quad \text { free, } \quad x \geq 0, \quad u: \geq $$ (Hint: See if it pays to keep \(x(t)\) as low as possible all the time.)

Solve the problem $$ \max \int_{0}^{10}\left(-u^{2}-x\right) d t, \quad \dot{x}=u, \quad x(0)=1, \quad x(10) \quad \text { free, } \quad x \geq 0, \quad u \in \mathbb{R} $$

Solve the problem $$ \max \int_{0}^{2}\left(x-\frac{1}{2} u^{2}\right) d t, \quad \dot{x}=u, \quad x(0)=1, \quad x(2) \text { free, } \quad x \geq u $$ (Hint: Guess that \(u^{*}(t)=x^{*}(t)\) on some interval \(\left[0, t^{*}\right]\), and \(u^{*}(t)

(a) Solve the problem $$ \max \int_{0}^{T}\left(\frac{1}{2} x_{1}+\frac{1}{5} x_{2}-u_{1}-u_{2}\right) d t, \quad\left\\{\begin{array}{lll} \dot{x}_{1}(t)=u_{1}(t), & x_{1}(0)=0, & x_{1}(T) \text { is free } \\ \dot{x}_{2}(t)=u_{2}(t), & x_{2}(0)=0, & x_{2}(T) \text { is free } \end{array}\right. $$ with \(0 \leq u_{1}(t) \leq 1,0 \leq u_{2}(t) \leq 1\), and with \(T\) as a fixed number greater than \(5 .\) (b) Replace the objective functional by \(\int_{0}^{T}\left(\frac{1}{2} x_{1}+\frac{1}{5} x_{2}-u_{1}-u_{2}\right) d t+3 x_{1}(T)+2 x_{2}(T)\) and find the solution in this case.

Solve the problem $$ \max _{u \in[-1,1]} \int_{0}^{4}\left(10-x_{1}+u\right) d t, \quad\left\\{\begin{array}{lll} \dot{x}_{1}(t)=x_{2}(t), & x_{1}(0)=2, & x_{1}(4) \text { is free } \\ \dot{x}_{2}(t)=u(t), & x_{2}(0)=4, & x_{2}(4) \text { is free } \end{array}\right. $$
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