Mastering Optimal Control: An ODE Challenge

Hey guys! Today, we're diving deep into a super interesting optimal control problem. You know, the kind that makes your brain do a little workout but is incredibly rewarding when you crack it. We're talking about finding the maximum value of a specific integral, given a differential equation and a constraint on our control input. This isn't just abstract math; understanding these problems is crucial in fields ranging from engineering to economics, where we constantly try to optimize processes. So, buckle up, because we're about to embark on a journey through ordinary differential equations (ODEs) and the fascinating world of control theory!

The Core of the Problem: Maximizing an Integral

Our main quest is to find the maximum value of the integral $\int_{-1}^1 (tx - u^2) dt$. Think of this integral as the 'payoff' we're trying to maximize over the time interval from $t = -1$ to $t = 1$. The integrand, $(tx - u^2)$, is what we're accumulating. It's composed of two parts: $tx$, which depends on time $t$ and the state variable $x$, and $-u^2$, which depends on the control input $u$. We want to make this combined value as large as possible. This is the objective function we're working with, and it’s the heart of our optimization challenge.

The Dynamics: How 'x' Evolves

Now, the state variable $x$ isn't just floating around; its behavior is governed by a differential equation: $\dot{x} = x + u^2$. This equation tells us how $x$ changes over time. Specifically, the rate of change of $x$ (denoted by $\dot{x}$) is equal to the current value of $x$ plus the square of our control input $u$. This means that whatever value $x$ has, it tends to grow exponentially on its own (due to the '$x$' term), and we can further influence its growth by choosing our control input $u$. The $u^2$ term indicates that increasing $u$ will always increase the rate of change of $x$. This relationship is fundamental to how we can steer the system towards our objective.

The Constraint: Keeping 'u' in Check

We can't just pick any value for $u$. The problem states that $u(t) \in [0,1]$ for every $t \in [-1, 1]$. This is a crucial constraint, guys! It means our control input $u$ must always be between 0 and 1, inclusive. We can't use negative inputs, nor can we exceed the value of 1. This limitation is realistic; in many real-world scenarios, control mechanisms have bounds. For instance, a throttle might only go from fully closed to fully open, or a voltage supply might have a maximum limit. This constraint adds a layer of complexity to finding the optimal $u$, as we need to respect these boundaries while still trying to maximize our objective function.

The Boundaries: Where We Start and End

Finally, we have specific conditions for the state variable $x$ at the beginning and end of our time interval. We are given $x(-1) = 0$ and $x(1) = e^2 - e^{1 + \frac{1}{e}}$. These are the boundary conditions. They tell us that our system must start at a state of 0 at time $t = -1$ and must end up at a very specific, non-trivial value at $t = 1$. These conditions are vital because they define the exact path our state variable $x$ must take. Without them, there could be infinitely many solutions; with them, we're looking for a specific trajectory that satisfies both the dynamics and the objective.

Why is This Important, Anyway?

Problems like this, often referred to as optimal control problems, are the backbone of modern system design and analysis. Whether you're designing a rocket's trajectory to reach Mars efficiently, managing an investment portfolio to maximize returns, or even controlling the temperature in a building to minimize energy consumption, you're essentially solving an optimal control problem. The mathematical tools we use, like the calculus of variations and Pontryagin's Maximum Principle, are powerful and widely applicable. Understanding how to formulate and solve these problems gives you a significant edge in tackling complex, real-world optimization challenges. It’s all about finding the best way to do something, given certain rules and limitations. So, when we dive into the specifics of solving this ODE-based problem, remember that you’re learning skills that have tangible impact across a vast array of disciplines. It’s not just about numbers on a page; it’s about intelligent decision-making in dynamic systems.

Setting Up the Solution: The Hamiltonian Approach

To tackle this beast, we'll lean on a powerful tool in control theory: the Hamiltonian. The Hamiltonian, often denoted by $H$, is a function that combines the objective function's integrand with the system's dynamics using a costate variable (let's call it $\lambda$). For our problem, the Hamiltonian is defined as:

$H(t, x, u, \lambda) = f(t, x, u) \lambda - L(t, x, u)$

where $L(t, x, u)$ is the integrand of our objective function and $f(t, x, u)$ is the right-hand side of our state equation (i.e., $\dot{x}$).

In our case, $L(t, x, u) = tx - u^2$ and $f(t, x, u) = x + u^2$. So, our Hamiltonian becomes:

$H(t, x, u, \lambda) = (x + u^2) \lambda - (tx - u^2)$

$H(t, x, u, \lambda) = \lambda x + \lambda u^2 - tx + u^2$

$H(t, x, u, \lambda) = \lambda x - tx + (\lambda + 1)u^2$

This Hamiltonian is key because Pontryagin's Maximum Principle states that for an optimal control $u^*(t)$, it must maximize the Hamiltonian with respect to $u$ at almost every time $t$, subject to the control constraints.

Maximizing the Hamiltonian

Our goal now is to find the value of $u \in [0, 1]$ that maximizes $H$. Let's look at the term involving $u$: $(\lambda + 1)u^2$. To maximize this term, we need to consider the coefficient $(\lambda + 1)$.

• Case 1: $\lambda + 1 > 0$ If $(\lambda + 1)$ is positive, then $u^2$ is maximized when $u$ is as large as possible in magnitude. Since $u \in [0, 1]$, the largest possible value for $u$ is $u=1$. So, $u^*(t) = 1$.

• Case 2: $\lambda + 1 < 0$ If $(\lambda + 1)$ is negative, then $u^2$ is maximized when $u$ is as small as possible in magnitude. Since $u \in [0, 1]$, the smallest possible value for $u$ is $u=0$. So, $u^*(t) = 0$.

• Case 3: $\lambda + 1 = 0$ If $(\lambda + 1) = 0$, meaning $\lambda = -1$, then the term $(\lambda + 1)u^2$ becomes $0 imes u^2 = 0$. In this situation, the value of $u$ doesn't affect the Hamiltonian. This is a singular case, and we might need further analysis, but for now, we focus on the non-singular cases.

So, the optimal control $u^*(t)$ will switch between 0 and 1 depending on the sign of $(\lambda + 1)$. This switching behavior is very common in optimal control problems!

The Costate Equation

Besides maximizing the Hamiltonian, Pontryagin's Maximum Principle also gives us the dynamics for the costate variable $\lambda$. The costate equation is given by:

$\dot{\lambda} = -\frac{\partial H}{\partial x}$

Let's calculate this derivative:

$\frac{\partial H}{\partial x} = \frac{\partial}{\partial x} (\lambda x - tx + (\lambda + 1)u^2) = \lambda$

So, the costate equation is simply:

$\dot{\lambda} = -\lambda$

This is a first-order linear ODE. The solution is straightforward: $\lambda(t) = C e^{-t}$ for some constant $C$. This tells us how the costate variable evolves over time.

Connecting the Dots: The Transversality Condition

We have the state equation ($\dot{x} = x + u^2$), the costate equation ($\dot{\lambda} = -\lambda$), and the rule for finding the optimal control $u^*(t)$ based on $\lambda$. However, we still have unknown constants in the solutions for $x(t)$ and $\lambda(t)$, and we need to determine the exact form of $u^*(t)$. This is where the boundary conditions and transversality conditions come into play.

We know $x(-1) = 0$ and $x(1) = e^2 - e^{1 + \frac{1}{e}}$.

The general solution for $x(t)$ from $\dot{x} = x + u^2$ is $x(t) = e^t \left( \int_{-1}^t e^{-\tau} u^2(\tau) d\tau + x(-1) \right)$. Since $x(-1)=0$, we have:

$x(t) = e^t \int_{-1}^t e^{-\tau} u^2(\tau) d\tau$

The general solution for $\lambda(t)$ is $\lambda(t) = C e^{-t}$.

Now, for problems with fixed final time and free final state (which is NOT our case, as $x(1)$ is fixed), we often use a terminal condition on $\lambda$. However, when the final state is fixed, the transversality condition relates the costate variable at the final time to the gradient of the objective function with respect to the final state. In essence, it tells us how sensitive the optimal value of the integral is to the final state $x(1)$.

For a problem of the form $\max \int_a^b L(t, x, u) dt$ with $\dot{x} = f(t, x, u)$, and fixed boundary conditions $x(a)=x_a, x(b)=x_b$, the transversality condition at the final time $t=b$ is:

$\lambda(b) = \frac{\partial \phi(x(b))}{\partial x(b)}$

where $\phi(x(b))$ is the final value function. In our case, the objective is $\int_{-1}^1 (tx - u^2) dt$. The structure of the problem is a bit different because the state $x$ appears in the integrand, and the final state $x(1)$ is specified. Let's consider the standard formulation where we maximize $\phi(x(b)) + \int_a^b L(t, x, u) dt$. Here, $\phi(x(1))$ would represent any terminal cost or bonus. In our problem, the final state $x(1)$ is fixed, but it’s not explicitly part of a terminal cost function $\phi$. Instead, the final state is a constraint. For problems with fixed endpoints, the costate variable $\lambda(t)$ does not typically have a direct boundary condition from the objective function itself, other than what is implicitly imposed by the necessity of meeting the final state condition.

However, if we were to define a terminal cost $\phi(x(1))$ that forces $x(1)$ to be its specified value, we could think of it as $\phi(x(1)) = 0$ if $x(1)$ is at the target, and $\infty$ otherwise. A more standard approach is to use the condition that the Hamiltonian should be constant along the optimal trajectory if it doesn't explicitly depend on time. In our case, $H$ does depend on $t$ through the $-tx$ term. Let's re-evaluate our approach.

Revisiting the Problem Structure and Strategy

This problem has fixed boundary conditions for $x$, i.e., $x(-1)=0$ and $x(1)=e^2 - e^{1 + \frac{1}{e}}$. The standard Pontryagin's Maximum Principle (PMP) is designed for problems where either the initial state is fixed and the final state is free, or the initial and final states are fixed, but the objective might include a terminal cost. Our objective function is an integral, and the final state is fixed to a specific value. This means we are looking for a trajectory $x(t)$ that satisfies the ODE, the control constraint, and the boundary conditions, while maximizing the integral.

Let's consider the implications of the fixed final state $x(1)$. The costate variable $\lambda(t)$ at the final time $t=1$ is related to the sensitivity of the optimal value of the objective functional to the final state $x(1)$. For a problem like $\max \int_a^b L(t,x,u)dt$ with $\dot{x} = f(t,x,u)$, $x(a)=x_a$ and $x(b)$ free, the transversality condition is $\lambda(b) = \frac{\partial \phi}{\partial x(b)}$ where $\phi$ is a terminal cost. If there is no terminal cost, $\phi=0$, so $\lambda(b)=0$. In our case, $x(1)$ is fixed.

A common technique when the final state is fixed is to use a Lagrange multiplier approach in conjunction with the PMP. However, the costate equation $\dot{\lambda} = -\lambda$ with $\lambda(t) = C e^{-t}$ and the condition $u^*(t) = 1$ if $\lambda > -1$ and $u^*(t) = 0$ if $\lambda < -1$ still hold.

Let's assume $u^*(t)$ is a function of time. We need to determine which intervals $u^*(t)=1$ and which $u^*(t)=0$. This depends on $\lambda(t) = C e^{-t}$.

• If $C e^{-t} > -1$ for some $t$, then $u^*(t)=1$. This happens when $e^{-t} > -1/C$ (if $C>0$) or $e^{-t} < -1/C$ (if $C<0$). Note that $e^{-t}$ is always positive.
• If $C e^{-t} < -1$ for some $t$, then $u^*(t)=0$. This happens when $e^{-t} < -1/C$ (if $C>0$) or $e^{-t} > -1/C$ (if $C<0$).

Since $e^{-t}$ is always positive, the condition $C e^{-t} > -1$ is always satisfied if $C > 0$. If $C < 0$, then $C e^{-t}$ is always negative. The condition $C e^{-t} < -1$ requires $e^{-t} > -1/C$ (since $C$ is negative, $-1/C$ is positive). The condition $C e^{-t} > -1$ requires $e^{-t} < -1/C$. The value $e^{-t}$ decreases as $t$ increases.

Let's consider the possibility that $u^*(t)$ switches. This would happen if $\lambda(t) = -1$ at some point. $C e^{-t} = -1 \\implies e^{-t} = -1/C$. This requires $C < 0$. Let $t_0$ be the time such that $e^{-t_0} = -1/C$. Then for $t < t_0$, $e^{-t} > e^{-t_0} = -1/C$, which means $C e^{-t} < -1$, so $u^*(t)=0$. For $t > t_0$, $e^{-t} < e^{-t_0} = -1/C$, which means $C e^{-t} > -1$, so $u^*(t)=1$. This implies a switch from $u=0$ to $u=1$ at $t_0$.

Alternatively, if $C > 0$, then $\lambda(t) = C e^{-t} > 0 > -1$ for all $t$. In this case, $u^*(t) = 1$ for all $t$. Let's test this simpler scenario first.

Scenario 1: $u^*(t) = 1$ for all $t \in [-1, 1]$

If $u^*(t)=1$, the state equation is $\dot{x} = x + 1^2 = x + 1$. The solution with $x(-1)=0$ is:

$x(t) = e^t \int_{-1}^t e^{-\tau}(1) d\tau = e^t [-e^{-\tau}]_{-1}^t = e^t (-e^{-t} - (-e^{1})) = e^t (-e^{-t} + e) = -1 + e^{t+1}$.

Let's check the final state: $x(1) = -1 + e^{1+1} = e^2 - 1$.

Our required final state is $x(1) = e^2 - e^{1 + \frac{1}{e}}$. Since $e^{1 + \frac{1}{e}} > 1$, our calculated $x(1)$ is not equal to the required $x(1)$. So $u^*(t)=1$ is not the optimal control.

Scenario 2: $u^*(t) = 0$ for all $t \in [-1, 1]$

If $u^*(t)=0$, the state equation is $\dot{x} = x + 0^2 = x$. The solution with $x(-1)=0$ is:

$x(t) = e^t \int_{-1}^t e^{-\tau}(0) d\tau = e^t (0) = 0$.

This gives $x(1)=0$, which is not our required final state. So $u^*(t)=0$ is not the optimal control.

Scenario 3: $u^*(t)$ switches.

This means we must have $\lambda(t_0) = -1$ for some $t_0 \in (-1, 1)$, and $\lambda(t) = C e^{-t}$.

If $\lambda(t_0) = -1$, then $C e^{-t_0} = -1$, so $C = -e^{t_0}$.

Then $\lambda(t) = -e^{t_0} e^{-t} = -e^{t_0 - t}$.

We have $\lambda(t) < -1$ if $-e^{t_0 - t} < -1 \\implies e^{t_0 - t} > 1 \\implies t_0 - t > 0 \\implies t < t_0$. So $u^*(t)=0$ for $t < t_0$.

We have $\lambda(t) > -1$ if $-e^{t_0 - t} > -1 \\implies e^{t_0 - t} < 1 \\implies t_0 - t < 0 \\implies t > t_0$. So $u^*(t)=1$ for $t > t_0$.

Thus, the optimal control candidate is $u^*(t) = \begin{cases} 0 & \text{if } t < t_0 \\ 1 & \text{if } t > t_0 \end{cases}$ for some $t_0 \in (-1, 1)$.

Now we need to find $t_0$ such that $x(1) = e^2 - e^{1 + \frac{1}{e}}$.

Let's calculate $x(t)$ with this piecewise control.

For $t \in [-1, t_0]$, $\dot{x} = x$, with $x(-1)=0$. This gives $x(t) = 0$ for $t \in [-1, t_0]$. So $x(t_0) = 0$.

For $t \in [t_0, 1]$, $\dot{x} = x + 1$, with initial condition $x(t_0) = 0$. The solution is $x(t) = e^t \int_{t_0}^t e^{-\tau}(1) d\tau = e^t [-e^{-\tau}]_{t_0}^t = e^t (-e^{-t} - (-e^{-t_0})) = -1 + e^{t-t_0}$.

Now, let's evaluate $x(1)$ using this expression:

$x(1) = -1 + e^{1 - t_0}$.

We need this to equal the required final state: $e^2 - e^{1 + \frac{1}{e}}$.

So, $-1 + e^{1 - t_0} = e^2 - e^{1 + \frac{1}{e}}$.

$e^{1 - t_0} = e^2 - e^{1 + \frac{1}{e}} + 1$.

This equation looks complicated to solve for $t_0$ analytically. Let's double-check the transversality condition for fixed final state.

For a fixed final state $x(T)=x_T$, the PMP requires that there exists a $\lambda(t)$ such that

1. $\dot{x} = \partial H / \partial \lambda$
2. $\dot{\lambda} = -\partial H / \partial x$
3. $H(x^*(t), u^*(t), \lambda(t), t) \ge H(x(t), u, \lambda(t), t)$ for all admissible $u$.
4. $\lambda(T) = \partial \phi / \partial x(T)$ (terminal cost)
5. Hamiltonian is constant if not explicitly time-dependent.

In our problem, the objective is $\int_{-1}^1 (tx - u^2) dt$. The final state $x(1)$ is fixed. This setup can be viewed as maximizing $\int_{-1}^1 (tx - u^2) dt + \mu(x(1) - (e^2 - e^{1 + \frac{1}{e}}))$, where $\mu$ is a Lagrange multiplier. However, the standard PMP formulation handles terminal costs $\phi(x(T))$. If there is no explicit terminal cost function $\phi$, but just a fixed value $x(T)$, the transversality condition on $\lambda(T)$ is often taken to be related to the gradient of the value function $V(x(T))$, which is implicitly defined by the problem. For problems with fixed endpoints, the costate variable $\lambda(T)$ isn't necessarily zero.

Let's reconsider the value of $u^*(t)$. The Hamiltonian is $H = \lambda x - tx + (\lambda + 1)u^2$. We found $u^*(t)=1$ if $\lambda > -1$ and $u^*(t)=0$ if $\lambda < -1$. $\lambda(t) = C e^{-t}$.

Could there be a scenario where $u^*(t)=1$ always, but it's not the one we calculated? No, because the final condition was not met.

Let's re-evaluate the boundary condition for $\lambda$. The condition $\lambda(t_0)=-1$ implies $u$ switches at $t_0$. This is called a chattering control if the switch happens infinitely often, but here it's a single switch. A single switch occurs when $\lambda(t)$ passes through $-1$. The value of $C$ determines where this crossing happens.

Consider the final state: $x(1) = e^2 - e^{1 + \frac{1}{e}}$. We have $x(t) = e^t (x(a) + extrm{integral})$.

If $u(t)=1$ for $t \in [t_1, t_2]$, then $\dot{x}=x+1$, $x(t) = C_1 e^t - 1$. If $u(t)=0$ for $t \in [t_1, t_2]$, then $\dot{x}=x$, $x(t) = C_2 e^t$.

Let's try to guess the structure of the control. The term $tx$ in the integral encourages larger $x$ values, especially for positive $t$. The term $-u^2$ penalizes using $u$. The dynamics $\dot{x} = x + u^2$ mean that using $u$ increases $x$. To maximize the integral, we want $x$ to be large, especially for positive $t$. This suggests we'd prefer $u=1$ for larger $t$. However, $u=1$ also drives $x$ up faster, which could lead to a large $x(1)$ that we don't want if it means violating the constraint. The constraint is $x(1) = e^2 - e^{1 + \frac{1}{e}}$.

Let $x_{target} = e^2 - e^{1 + \frac{1}{e}}$. This value is approximately $e^2 - e^{1.367} \approx 7.389 - 3.92 \\approx 3.469$.

If $u(t)=1$ for all $t$, $x(1) = e^2 - 1 \approx 6.389$. This is larger than $x_{target}$.

If $u(t)=0$ for all $t$, $x(1) = 0$. This is smaller than $x_{target}$.

This suggests that we need a control that results in a final $x(1)$ smaller than $e^2-1$ but larger than 0. This supports the idea of a switching control.

If $u(t)$ switches from 1 to 0 at $t_0$, then for $t < t_0$, $\dot{x} = x+1$, and for $t > t_0$, $\dot{x} = x$. This would lead to a lower $x(1)$ than if $u(t)=1$ always, which is what we need. The switch would be from $u=1$ to $u=0$ if $\lambda$ goes from $>-1$ to $<-1$. This requires $C<0$. So $\lambda(t) = C e^{-t}$. If $\lambda(t_0) = -1$, then $C = -e^{t_0}$. $\lambda(t) = -e^{t_0-t}$. For $t < t_0$, $t_0-t > 0$, $e^{t_0-t} > 1$, so $\lambda(t) < -1$, $u^*(t)=0$. For $t > t_0$, $t_0-t < 0$, $e^{t_0-t} < 1$, so $\lambda(t) > -1$, $u^*(t)=1$. This leads to $u^*(t)$ switching from 0 to 1, which increases $x(1)$. This is the opposite of what we need.

Let's reconsider the condition $\lambda+1$. If $\lambda+1>0$, $u=1$. If $\lambda+1<0$, $u=0$. This means $\lambda > -1 \\implies u=1$ and $\lambda < -1 \\implies u=0$. With $\lambda(t) = C e^{-t}$.

If $C>0$, $\lambda(t) = C e^{-t} > 0 > -1$ for all $t$. So $u^*(t)=1$ for all $t$. We already showed this doesn't work.

If $C<0$, let $C = -A$ where $A>0$. Then $\lambda(t) = -A e^{-t}$.

• We need $\lambda(t) > -1$, so $-A e^{-t} > -1 \\implies A e^{-t} < 1 \\implies e^{-t} < 1/A \\implies -t < \ln(1/A) = -\ln(A) e t > \ln(A)$.
• We need $\lambda(t) < -1$, so $-A e^{-t} < -1 \\implies A e^{-t} > 1 \\implies e^{-t} > 1/A \\implies -t > \ln(1/A) = -\ln(A) e t < \ln(A)$.

Let $t_s = \ln(A)$. So, if $t < t_s$, $\lambda(t) < -1$ and $u^*(t)=0$. If $t > t_s$, $\lambda(t) > -1$ and $u^*(t)=1$. This implies $u^*(t)$ switches from 0 to 1 at $t_s$. We need $t_s \in (-1, 1)$.

Let's calculate $x(1)$ with $u^*(t) = \begin{cases} 0 & \text{if } t < t_s \\ 1 & \text{if } t > t_s \end{cases}$.

For $t \in [-1, t_s]$, $\dot{x}=x$, $x(-1)=0$. So $x(t)=0$ for $t \in [-1, t_s]$. Thus $x(t_s)=0$.

For $t \in [t_s, 1]$, $\dot{x}=x+1$, $x(t_s)=0$. Solution is $x(t) = e^t \int_{t_s}^t e^{-\tau}(1) d\tau = e^t [-e^{-\tau}]_{t_s}^t = e^t (-e^{-t} - (-e^{-t_s})) = -1 + e^{t-t_s}$.

So $x(1) = -1 + e^{1-t_s}$.

We need $x(1) = e^2 - e^{1 + \frac{1}{e}}$.

$-1 + e^{1-t_s} = e^2 - e^{1 + \frac{1}{e}}$

$e^{1-t_s} = e^2 - e^{1 + \frac{1}{e}} + 1$

This is the same equation as before. It seems there might be an issue with my understanding of the transversality condition for fixed endpoints or a numerical value that needs to be calculated.

Let's re-evaluate the problem structure. Is it possible that the optimal control is bang-bang, meaning it only takes values 0 or 1?

The Hamiltonian is $H = \lambda x - tx + (\lambda + 1)u^2$. Maximizing $H$ wrt $u \in [0,1]$ gives $u=1$ if $\lambda+1>0$ and $u=0$ if $\lambda+1<0$. This is bang-bang control.

The costate equation is $\dot{\lambda} = -\lambda$, so $\lambda(t) = C e^{-t}$.

We need to satisfy $x(-1)=0$ and $x(1)=e^2 - e^{1 + \frac{1}{e}}$.

Let's try to find $C$ and potentially a switching time $t_0$ such that these conditions are met.

If $u(t)=1$ always, $x(1) = e^2-1$. Target is $e^2 - e^{1 + 1/e}$. Since $e^{1+1/e} > 1$, target $x(1)$ is smaller than $e^2-1$. This suggests we need a control that reduces $x(1)$ compared to always using $u=1$. This implies we should use $u=0$ for some interval.

If $u(t)=0$ always, $x(1)=0$. Target $x(1)$ is positive. This suggests we need a control that increases $x(1)$ compared to always using $u=0$. This implies we should use $u=1$ for some interval.

This leads to a switching control. For $u$ to switch from 1 to 0, we need $\lambda+1$ to go from positive to negative. This means $\lambda$ goes from $>-1$ to $<-1$. This requires $C>0$. $\lambda(t) = C e^{-t}$. If $\lambda(t_0)=-1$, then $C e^{-t_0} = -1 ightarrow C = -e^{t_0}$. This contradicts $C>0$. So $u$ cannot switch from 1 to 0.

For $u$ to switch from 0 to 1, we need $\lambda+1$ to go from negative to positive. This means $\lambda$ goes from $<-1$ to $>-1$. This requires $C<0$. Let $C = -A$ with $A>0$. $\lambda(t) = -A e^{-t}$. If $\lambda(t_0)=-1$, then $-A e^{-t_0}=-1 ightarrow A = e^{t_0}$. Then $\lambda(t) = -e^{t_0} e^{-t} = -e^{t_0-t}$.

For $t < t_0$, $t_0-t > 0$, $e^{t_0-t} > 1$, so $\lambda(t) < -1$. Thus $u^*(t)=0$. For $t > t_0$, $t_0-t < 0$, $e^{t_0-t} < 1$, so $\lambda(t) > -1$. Thus $u^*(t)=1$.

So the control is $u^*(t) = \begin{cases} 0 & t < t_0 \\ 1 & t > t_0 \end{cases}$ for some $t_0 \in (-1, 1)$.

We calculated $x(1) = -1 + e^{1-t_0}$ with this control.

We need $x(1) = e^2 - e^{1 + \frac{1}{e}}$.

So, $-1 + e^{1-t_0} = e^2 - e^{1 + \frac{1}{e}}$.

$e^{1-t_0} = e^2 - e^{1 + \frac{1}{e}} + 1$.

It seems the problem might be set up such that $t_0$ is not easily solvable or there's a detail missed. Let's check the Hamiltonian at the final time. For fixed endpoints, $\lambda(T)$ is not necessarily zero. The value of $x(1)$ is fixed, so its variation is zero. The condition $\lambda(T) = \partial \phi / \partial x(T)$ becomes tricky.

Let's consider the structure of the objective function and the state equation again. $\max extrm{Integral} = extrm{Integral of } (tx - u^2)$. Dynamics: $\dot{x} = x + u^2$. Boundary $x(-1)=0$, $x(1)=e^2 - e^{1 + 1/e}$.

What if we try to work backward from the final state? This is sometimes useful for fixed final states.

If $u(t)=1$ on $[t_0, 1]$, $x(t) = -1 + e^{t-t_0}$. $x(1) = -1 + e^{1-t_0}$. If $u(t)=0$ on $[-1, t_0]$, $x(t) = 0$. This gives $x(t_0)=0$. Then $x(1) = -1 + e^{1-t_0}$. Setting this equal to the target: $-1 + e^{1-t_0} = e^2 - e^{1 + \frac{1}{e}}$. This requires $e^{1-t_0} = e^2 - e^{1 + \frac{1}{e}} + 1$. Let's evaluate the right side numerically: $e^2 - e^{1 + 1/e} + 1 \approx 7.389 - 3.922 + 1 = 4.467$. So $e^{1-t_0} = 4.467$. $1-t_0 = \ln(4.467) \approx 1.5$. $t_0 = 1 - 1.5 = -0.5$. This value $t_0 = -0.5$ is within $(-1, 1)$.

So the proposed optimal control is $u^*(t) = \begin{cases} 0 & t < -0.5 \\ 1 & t > -0.5 \end{cases}$.

Let's check the costate $\lambda(t)$. We need $\lambda(t_0) = -1$ for the switch to occur at $t_0$. We found $\lambda(t) = -e^{t_0-t}$. With $t_0=-0.5$, $\lambda(t) = -e^{-0.5-t}$. At $t = t_0 = -0.5$, $\lambda(-0.5) = -e^{-0.5 - (-0.5)} = -e^0 = -1$. This confirms the switch condition.

Now, we need to verify that this control maximizes the integral $\int_{-1}^1 (tx - u^2) dt$ subject to $x(-1)=0, x(1)=e^2 - e^{1 + \frac{1}{e}}$, and $\dot{x} = x + u^2$, $u extrm{ in } [0,1]$.

The PMP guarantees that if the Hamiltonian is maximized, we have an optimal control candidate. The issue with fixed final states is ensuring the transversality conditions are met, which determines the constant $C$ for $\lambda$. In our case, we found $C$ implicitly by setting $\lambda(t_0)=-1$ and then solving for $t_0$ using the state equation and boundary conditions. This implicitly satisfies a form of transversality.

The value of the integral can be computed. With $u^*(t) = \begin{cases} 0 & t < -0.5 \\ 1 & t > -0.5 \end{cases}$:

For $t \in [-1, -0.5]$, $x(t)=0$. For $t \in [-0.5, 1]$, $x(t) = -1 + e^{t+0.5}$.

Integral $= \int_{-1}^{-0.5} (t imes 0 - 0^2) dt + \int_{-0.5}^1 (t(-1+e^{t+0.5}) - 1^2) dt$

$= 0 + \int_{-0.5}^1 (-t + t e^{t+0.5} - 1) dt$

We need to evaluate $\int t e^{t+0.5} dt$. Let $s = t+0.5$, $ds=dt$. $t = s-0.5$. $\int (s-0.5) e^s ds = \int s e^s ds - 0.5 \int e^s ds = (s-1)e^s - 0.5 e^s = (s - 1.5) e^s = (t+0.5-1.5)e^{t+0.5} = (t-1)e^{t+0.5}$.

So, $\int_{-0.5}^1 (-t + t e^{t+0.5} - 1) dt = [-\frac{t^2}{2} + (t-1)e^{t+0.5} - t]_{-0.5}^1$

$= [-\frac{t^2}{2} - 2t + (t-1)e^{t+0.5}]_{-0.5}^1$

At $t=1$: $-\frac{1}{2} - 2 + (1-1)e^{1.5} = -2.5$. At $t=-0.5$: $-\frac{(-0.5)^2}{2} - 2(-0.5) + (-0.5-1)e^{-0.5+0.5} = -\frac{0.25}{2} + 1 + (-1.5)e^0 = -0.125 + 1 - 1.5 = -0.625$.

Value $= -2.5 - (-0.625) = -2.5 + 0.625 = -1.875$.

This is the value of the integral. The problem asked for the maximum value of the integral. We found the optimal control. The value itself is $-1.875$. The problem phrasing is