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Chapter 9: Problem 10

Consider the problem \(\max \int_{0}^{1}-u d t, \dot{x}=u^{2}, x(0)=x(1)=0, u \in \mathbb{R}\). (a) Explain why \(u^{*}(t)=x^{*}(t)=0\) solves the problem. (b) Show that the conditions in the maximum principle are satisfied only for \(p_{0}=0\).

Short Answer

Expert verified
The solution \(u^{*}(t) = 0\) and \(x^{*}(t) = 0\) solve the problem because they meet the boundary conditions and integral is maximized at zero. The maximum principle conditions are satisfied only for \(p_{0} = 0\).

Step by step solution

01

State the control problem

The given control problem involves maximizing the integral \[-\int_{0}^{1} u \, dt\], subject to the dynamics \(\frac{dx}{dt} = u^2\) with boundary conditions \(x(0) = x(1) = 0\). The control function \(u(t) \) belongs to \(\mathbb{R}\).
02

Check the candidate solution

The candidate solution is given as \(u^{*}(t) = 0\) and \(x^{*}(t) = 0\). By substituting these into the dynamics, \(\frac{dx}{dt} = 0^2 = 0\), it follows that \(x(t) = 0\) remains constant for all \(t \) in \([0, 1]\).
03

Validate the integral

For \(u^{*}(t) = 0\), the integral to maximize becomes \[\begin{aligned} -\int_{0}^{1} 0 \, dt = 0 \end{aligned}\]. Thus, the integral value is zero, which is the maximum value it can take since it's being subtracted.
04

Maximum principle setup

The Hamiltonian for this problem is defined as \[\begin{aligned} H = -u + p \, u^2 \end{aligned}\] where \(p\) is the costate variable.
05

Costate dynamics

According to the maximum principle, the costate dynamics are given by \(\frac{dp}{dt} = -\frac{\text{鈭倉H}{\text{鈭倉x}\). Since \(H\) does not explicitly depend on \(x\), we have \(\frac{dp}{dt} = 0 \Rightarrow p = p_0\), a constant.
06

Maximize Hamiltonian with respect to u

To satisfy the maximum principle, we need to maximize \(H\) with respect to \(u\). Setting \frac{\text{鈭倉H}{\text{鈭倉u} = -1 + 2p_0 u\ to zero, we get \(-1 + 2p_0 u = 0 \Rightarrow u = \frac{1}{2p_0}\).
07

Analyze the implications of costate

For \(p_0 eq 0\), \(u\) would not be zero everywhere, contradicting our candidate solution. Given \(u^{*}(t) = 0\), we must have \(p_0 = 0\).

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Key Concepts

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

Maximum Principle
Optimal Control Theory often revolves around the **Maximum Principle**. This principle allows us to find the best control to either maximize or minimize a given performance criterion, usually defined by an integral.
In our problem, we aim to maximize the integral \[-\int_{0}^{1} u \, dt\]. The dynamics are given by \(\frac{dx}{dt} = u^2\), with boundary conditions \(x(0) = x(1) = 0\). The candidate solutions are \(u^{*}(t) = 0\) and \(x^{*}(t) = 0.\)
The Maximum Principle aids in forming **Hamiltonian Dynamics** which integrates the original problem and its constraints.
Hamiltonian Dynamics
A key component in the Maximum Principle is the Hamiltonian. For our control problem, the Hamiltonian \(H\) combines the performance criterion with the system dynamics through a costate variable \(p\). Here, the Hamiltonian is defined as:
\[H = -u + p \, u^2\]
In this equation, \(p\) represents the costate variable, a Lagrange multiplier ensuring the constraints hold. The Hamiltonian helps transfer our integral maximization into a manageable form involving ordinary differential equations for both **state variables** and **costate variables**.
For this problem, the next step involves defining the evolution of the costate variable through the system's dynamics.
Costate Variable
The **Costate Variable** \(p\) plays a pivotal role in Optimal Control. It provides necessary conditions to reach the optimal solution. The dynamics of this variable are given by:
\[\frac{dp}{dt} = -\frac{\text{鈭倉H}{\text{鈭倉x}\]
In our case, since the Hamiltonian \(H\) does not directly depend on \(x\), we find that \( \frac{dp}{dt} = 0 \). This shows that \(p\) is constant. Naming this constant as \(p_0\), we proceed to the next part of the Maximum Principle process: maximizing the Hamiltonian with regards to the control variable \(u\).
Integral Maximization
The ultimate goal involves **Integral Maximization** under given constraints. In our specific problem, the integral in question is:
\[-\text{鈭珆_{0}^{1} 0 \, dt = 0\]
Given our candidate solution \(u^{*}(t) = 0\), the integral evaluates to zero. This shows that it meets the requirement for the maximum value due to subtraction.
To ensure the Maximum Principle is satisfied, we adjust \(u\) to maximize the Hamiltonian. By differentiating the Hamiltonian and setting it to zero, we found that \( u = \frac{1}{2p_0} \), leading us back to the condition that \(p_0 = 0 \) for our candidate solution \(u^{*}(t) = 0\) to hold true. Through these steps, we affirm that our problem is correctly solved and satisfies optimal criteria.

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Most popular questions from this chapter

$$ \min _{u(t) \in(-\infty, \infty)} \int_{0}^{1}\left[x(t)+u(t)^{2}\right] d t, \quad \dot{x}(t)=-u(t), \quad x(0)=0, \quad x(1) \text { free } $$

$$ \max _{u(t) \in(-\infty, \infty)} \int_{0}^{1}\left[1-u(t)^{2}\right] d t, \dot{x}(t)=x(t)+u(t), \quad x(0)=1, \quad x(1) \text { free } $$

Find the only possible solution to the following problem by using both the calculus of variations and control theory: $$ \max \int_{0}^{1}\left(2 x e^{-t}-2 x \dot{x}-\dot{x}^{2}\right) d t, \quad x(0)=0, \quad x(1)=1 $$

3\. Consider the optimal extraction problem over a fixed extraction period, $$ \max _{u(t) \geq 0} \int_{0}^{T}\left[a e^{\alpha t} u(t)-(u(t))^{2} e^{\beta t}-c\right] e^{-r t} d t, \dot{x}(t)=-u(t), x(0)=K, x(T)=0 $$ Here \(x(t)\) and \(u(t)\) have the same interpretation as in Example 1, with \(q(t)=a e^{\alpha t}\) as the world market price, and \((u(t))^{2} e^{\beta t}-c\) as the cost of extraction, with \(c>0\) (a) One can prove that if \(u^{*}(t)\) is optimal, then \(u^{*}(t)>0\) for all \(t .\) (You are not required to show this.) The adjoint function is a constant \(\bar{p}\). Find \(u^{*}(t)\) expressed in terms of \(\bar{p}\). Then find \(x^{*}(t)\) and \(\bar{p}\) for the case \(\alpha=\beta=0, r \neq 0\) (b) Let \(T>0\) be subject to choice (keeping the assumptions \(\alpha=\beta=0, r \neq 0\) ). Prove that the necessary conditions lead to an equation for determining the optimal \(T^{*}\) which has a unique positive solution. Assume that \(\max _{u}\left(a u-u^{2}-c\right)>0\), i.e. \(a^{2}>4 c\)

Solve the problem (where \(T, \alpha\), and \(\beta\) are positive constants, \(\alpha \neq 2 \beta\) ) $$ \max \int_{0}^{T} e^{-\beta t} \sqrt{u} d t \text { when } \dot{x}(t)=\alpha x(t)-u(t), x(0)=1, x(T)=0, u(t) \geq 0 $$ What happens if the terminal condition \(x(T)=0\) is changed to \(x(T) \geq 0\) ?
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