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For any \(A(\cdot)\), the fundamental solution \(\Psi(t, s)\) associated to the adjoint equation $$ \dot{p}(t)=-A(t)^{*} p(t) $$ is \(\Psi(t, s):=\Phi(s, t)^{*}\).

Short Answer

Expert verified
The fundamental solution \(\Psi(t, s)\) associated with the adjoint equation \(\dot{p}(t) = -A(t)^{*} p(t)\) is given by the adjoint of the state-transition matrix \(\Phi(s, t)^{*}\).

Step by step solution

01

Concept of the Adjoint Equation

An adjoint equation is a linear ordinary differential equation (ODE) that is connected to another ODE through the adjoint operator, usually denoted by ^{*}. The adjoint operator associated with a linear operator \(A(t)\) applied to a scalar function \(p(t)\) is written as \(A(t)^{*}\). The adjoint equation is given by: \[ \dot{p}(t) = -A(t)^{*} p(t), \] where \(\dot{p}(t)\) denotes the time derivative of the scalar function \(p(t)\).
02

Defining the Fundamental Solution

The fundamental solution, also known as the Green's function, is a central concept in solving linear ODEs associated with a given adjoint equation. The fundamental solution is denoted by \(\Psi(t, s)\). The task is to find the expression for \(\Psi(t, s)\) using the provided formula.
03

Finding the Fundamental Solution

We are given the expression for the fundamental solution \(\Psi(t, s)\) associated with the adjoint equation: \[ \Psi(t, s) = \Phi(s, t)^{*}. \] Here, \(\Phi(s, t)^{*}\) denotes the adjoint of the state-transition matrix \(\Phi(s, t)\). Thus, the fundamental solution \(\Psi(t, s)\) for the given adjoint equation is the adjoint of the state-transition matrix, which is \(\Phi(s, t)^{*}\).

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Key Concepts

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

Fundamental Solution
In the context of linear ordinary differential equations and their adjoint equations, the concept of a fundamental solution plays a crucial role. When dealing with an adjoint equation, such as \( \dot{p}(t) = -A(t)^{*} p(t) \), our main objective is to find the fundamental solution \( \Psi(t, s) \). This function effectively describes how solutions to the differential equation evolve over time.

The fundamental solution, often referred to as the Green's function, acts as a powerful tool for harnessing the underlying dynamics of a system. It is defined in connection with the state-transition matrix \( \Phi(s, t) \) and is expressed as its adjoint \( \Phi(s, t)^{*} \).

For any given time points \(t\) and \(s\), the fundamental solution provides a mapping from the conditions at time \(s\) to those at a later time \(t\). This mapping encapsulates the entire influence of the system's internal changes across the specified interval. By analyzing \( \Psi(t, s) \), one gains insights into how the system's state shifts in response to internal operators over time.
Linear Ordinary Differential Equations (ODEs)
Linear ordinary differential equations, or ODEs, serve as the backbone of many mathematical models used to describe physical, biological, and engineering systems. These equations are categorized by their linearity, meaning that the function and its derivatives appear in a linear manner. In a standard form, a linear ODE can be expressed as:
  • \( \dot{x}(t) = A(t) x(t) + f(t) \)
where \( A(t) \) is a linear operator, \( x(t) \) represents the state of the system, and \( f(t) \) is a driving term or forcing function.

The power of linear ODEs lies in their simplicity and the ability to find explicit solutions known as state-transition matrices. These solutions describe how the system transitions from one state to another over time by using known conditions.

When applying these concepts to an adjoint equation like \( \dot{p}(t) = -A(t)^{*} p(t) \), the solutions must adhere to the adjoint operator's constraints. Here, understanding the linearity helps us to efficiently compute the fundamental and state-transition solutions that govern the system's behavior.
State-transition Matrix
The state-transition matrix, \( \Phi(t, s) \), is a mathematical object that captures the dynamics of linear systems over time. It acts as a bridge connecting the system's state at an initial time \(s\) to its state at a later time \(t\). In the realm of linear ordinary differential equations and their adjoints, this matrix serves as a cornerstone for solution strategies.

Defined for linear ODEs of the form \( \dot{x}(t) = A(t) x(t) \), the state-transition matrix allows us to express the solution to the ODE in terms of the initial state's evolution. When considering adjoint equations like \( \dot{p}(t) = -A(t)^{*} p(t) \), the state-transition matrix provides a way to understand how transformations reverse as they evolve backward in time.
  • It simplifies solving linear ODEs by giving direct transition information from \(x(s)\) to \(x(t)\).
  • Its adjoint, \( \Phi(s, t)^{*} \), offers insights into the dual problem or adjoint equation, wide-reaching in control theory and quantum mechanics.
Through the manipulation and understanding of \( \Phi(t, s) \), one can unlock deeper narratives about stability, control, and behavior of complex systems. The state-transition matrix is more than just a tool—it's an essential component of the mathematical landscape governing system dynamics.

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Most popular questions from this chapter

Assume that \((A, B)\) is controllable and \(\mathcal{U} \subseteq \mathbb{R}^{m}\) is a neighborhood of 0 . Then \(J_{k}^{\mathrm{R}} \subseteq \mathcal{R}_{u}(0)\) for all \(k\).

Consider, as an example, the system \(\Sigma\) corresponding to the linearized pendulum \((2.31)\), which was proved earlier to be controllable. In Appendix C.4 we compute $$ e^{t A}=\left(\begin{array}{cc} \cos t & \sin t \\ -\sin t & \cos t \end{array}\right) $$ Thus, for any \(\delta>0\), $$ \mathbf{R}\left(e^{\delta A}, B\right)=\left(\begin{array}{ll} 0 & \sin \delta \\ 1 & \cos \delta \end{array}\right) $$ which has determinant \((-\sin \delta)\). By Lemma \(3.4 .1, \Sigma\) is \(\delta\)-sampled controllable iff $$ \sin \delta \neq 0 \text { and } 2 k \pi i \neq \pm i \delta \text {, } $$ i.e., if and only if \(\delta\) is not a multiple of \(\pi\). Take, for instance, the sampling time \(\delta=2 \pi\). From the explicit form of \(e^{t A}\), we know that \(e^{\delta A}=I\). Thus, $$ A^{(\delta)}=A^{-1}\left(e^{\delta A}-I\right)=0 $$ so \(G=0\). This means that the discrete-time system \(\Sigma_{[\delta]}\) has the evolution equation $$ x(t+1)=x(t) $$ No matter what (constant) control is applied during the sampling interval \([0, \delta]\), the state (position and velocity) is the same at the end of the interval as it was at the start of the period. (Intuitively, say for the linearized pendulum, we are acting against the natural motion for half the interval duration, and with the natural motion during the other half.) Consider now the case when \(\delta=\pi\), which according to the above Lemma should also result in noncontrollability of \(\Sigma_{[\delta]} .\) Here $$ F=e^{\delta A}=-I $$ and $$ A^{(\delta)}=A^{-1}\left(e^{\delta A}-I\right)=-2 A^{-1}=2 A $$ so $$ G=2 A B=\left(\begin{array}{l} 2 \\ 0 \end{array}\right) . $$ Thus, the discrete-time system \(\Sigma_{[6]}\) has the evolution equations: $$ \begin{aligned} &x_{1}(t+1)=-x_{1}(t)+2 u(t) \\ &x_{2}(t+1)=-x_{2}(t) \end{aligned} $$ This means that we now can partially control the system, since the first coordinate (position) can be modified arbitrarily by applying suitable controls \(u\). On the other hand, the value of the second coordinate (velocity) cannot be modified in any way, and in fact at times \(\delta, 2 \delta, \ldots\) it will oscillate between the values \(\pm x_{2}(0)\), independently of the (constant) control applied during the interval.

One could also define a class of systems as in (3.28) with other choices of \(\theta\). Theorem 8 may not be true for such other choices. For instance, the theorem fails for \(\theta=\) identity (why?). It also fails for \(\theta=\arctan\) : Show that the 4-dimensional, single-input system $$ \begin{aligned} \dot{x}_{1} &=\arctan \left(x_{1}+x_{2}+x_{3}+x_{4}+2 u\right) \\ \dot{x}_{2} &=\arctan \left(x_{1}+x_{2}+x_{3}+x_{4}+12 u\right) \\ \dot{x}_{3} &=\arctan (-3 u) \\ \dot{x}_{4} &=\arctan (-4 u) \end{aligned} $$ satisfies that \(B \in \mathbf{B}_{n, m}\) but is not controllable. Explain exactly where the argument given for \(\theta=\tanh\) breaks down.

Let \(\Sigma\) be a continuous-time system as in Definition \(2.6 .7\) and let \(\sigma<\tau\). With the present terminology, Lemma \(2.6 .8\) says that \((x, \sigma) \sim\) \((z, \tau)\) for the system \(\Sigma\) iff \((z, \sigma) \sim(x, \tau)\) for the system \(\Sigma_{\sigma+\tau}^{-}\). This remark is sometimes useful in reducing many questions of control to a given state to analogous questions (for the time-reversed system) of control from that same state, and vice versa. Recall that a linear system is one that is either as in Definition 2.4.1 or as in Definition 2.7.2.

Let \(\mathcal{U} \subseteq \mathbb{R}^{m}\) and pick any two \(S, T \geq 0\). Then $$ \mathcal{R}_{u}^{T}(0)+e^{T A} \mathcal{R}_{u}^{S}(0)=\mathcal{R}_{u}^{S+T}(0) . $$

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