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Problem 5

For each of the following problems, solve the Euler-Lagrange equation (with appropriate boundary conditions). 1\. \(\int_{0}^{\pi} x^{2}-\dot{x}^{2} d t, x(0)=x(\pi)=0\). 2\. \(\int_{0}^{\pi} 4 \dot{x}^{2}+2 x \dot{x}-x^{2} d t, x(0)=2, x(\pi)=0\). 3\. \(\int_{0}^{1} \cos \dot{x} d t, x(0)=0\). (Find only solutions \(x(\cdot)\) that are continuously differentiable.) 4\. \(\int_{0}^{1} \sqrt{1+\dot{x}^{2}} d t, x(0)=a, x(1)=b\), where \(a\) and \(b\) are any two constants. In the last problem, after you have solved it, explain why the solution should have been obvious.

Expert verified

1\. For the functional \(\int_{0}^{\pi} x^{2}-\dot{x}^{2} dt\), the solution to the boundary-value problem \(x(0)=x(\pi)=0\) is found using the Euler-Lagrange equation, and we obtain \(x(t) = 0\).
2\. For the functional \(\int_{0}^{\pi} 4 \dot{x}^{2}+2 x \dot{x}-x^{2} d t\), apply the given boundary conditions, and solve the Euler-Lagrange equation to find the specific solution \(x(t)\).
3\. For the functional \(\int_{0}^{1} \cos \dot{x} d t\), find only continuously differentiable solutions, and apply the boundary condition \(x(0)=0\) to find the specific solution \(x(t)\).
4\. For the functional \(\int_{0}^{1} \sqrt{1+\dot{x}^{2}} d t\), find the solution for any two constants \(a\) and \(b\) as boundary values, \(x(0)=a, x(1)=b\), and explain why the solution should have been obvious.

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Chapter 9

Consider the system with \(X=\mathbb{R}^{3}, \mathcal{U}=\mathbb{R}^{2}\), discussed in Example 9.2.14. Take the control $\tilde{\omega} \in \mathcal{L}_{u}(0,2 \pi)$ whose coordinates are $$ u_{1}(t) \equiv 0, \quad u_{2}(t)=\sin (t) $$ Since \(\tilde{\omega} \not \equiv 0\), it is nonsingular for any initial state \(x^{0}\). (a) Compute a formula for the pseudoinverse operator \(L^{\\#}\), when the control \(\tilde{\omega}\) is used, and \(x^{0}\) is arbitrary. (b) Provide a formula for the basic recursive "Newton" step in (9.33), when \(p(x)=\|x\|^{2}\), and compute two iterates, starting from any nonzero state of your choice. (You may use \(h=1\).)

Chapter 9

A problem for which \(q \equiv 0\) is often called a Meyer problem of optimal control (this follows similar terminology used in the calculus of variations). A Lagrange problem, in contrast, is one for which \(p \equiv 0\). (The general problem which we considered, incidentally, is sometimes called a "Bolza" problem.) The main idea which was used in deriving Theorem 43 from Lemma $9.2 .3$ was the observation that every Lagrange problem can be recast as a Meyer problem. Show that, conversely, every Meyer problem can be reformulated as a Lagrange problem, in such a manner that optimal controls are the same, for every possible initial state. (You may assume that \(p \in \mathcal{C}^{2}\).)

Chapter 9

The matrix $W(\sigma, \tau)=\int_{\sigma}^{\tau} \Phi(\tau, s) B(s) B(s)^{\prime} \Phi(\tau, s)^{\prime} d s$ appears in the computation of the perturbation given by formula (9.33). It would seem that in order to compute \(W(\sigma, \tau)\), one must first solve the differential equation \(\dot{x}=f(x, \widetilde{\omega})\), then solve for the fundamental solution \(\Phi\) of the variational equation, and, finally, integrate numerically using the formula for \(W(\sigma, \tau)\). Such an approach involves large amounts of storage (values of \(\widetilde{x}\) to be used in the fundamental equation, as well as in \(B(t)\), values of \(\Phi\), etc). There is an alternative way of computing \(W(\sigma, \tau)\), in "one pass", however, as follows. Let $$ A(x, u)=f_{x}(x, u) \text { and } B(x, u)=f_{u}(x, u), $$ seen as functions from \(X \times \mathcal{U}\) into \(\mathbb{R}^{n \times n}\) and \(\mathbb{R}^{n \times m}\) respectively. Now consider the following system of \(n+n^{2}\) differential equations for \(x:[\sigma, \tau] \rightarrow X\) and \(Q:[\sigma, \tau] \rightarrow \mathbb{R}^{n \times n}:\) $$ \begin{aligned} \dot{x} &=f(x, \widetilde{\omega}) \\ \dot{Q} &=A(x, \widetilde{\omega}) Q+Q A(x, \widetilde{\omega})^{\prime}+B(x, \widetilde{\omega}) B(x, \widetilde{\omega})^{\prime} \end{aligned} $$ Show that the solution of this equation with initial conditions \(x(\sigma)=x^{0}\) and \(Q(\sigma)=0\) is so that \(Q(\tau)=W(\sigma, \tau)\).

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