Constrained Optimization in the Calculus of Variations and Optimal Control Theory

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Triangle inequality. In this manner, a normed linear space becomes a metric space. The elements of a normed linear space can be objects of any kind, for example, numbers, matrices, functions, etcetera. The following normed spaces are important for our subsequent purposes: 1. C[ti , tf ]. However, this is not true. For example, arc length is continuous if we use the norm in C 1 [ti , tf ], but not1 continuous if we use the norm in C[ti , tf ]. Since we want to be able to use ordinary analytic operations such as passage to the limit, then, given a functional, it is reasonable to choose a function space such that the functional is continuous.

In fact, the set of functions satisfying the constraints of a given variational problem, called the admissible functions is in general not a linear space. Nevertheless, the concept of a normed linear space and the related concepts of the distance between functions, continuity of functionals, etcetera, play an important role in the calculus of variations.

Let X be a normed linear space. Prove that the following are equivalent: a L is continuous. Theorem 3. Calculus in function spaces and beyond which contradicts 3. Note that this is a necessary condition for the existence of an extremum.

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We will apply the necessary condition for an extremum established xf xi t ti tf Figure 3. The simplest variational problem. Proof The proof is long and so we divide it into several steps. Step 1. So Theorem 3. Step 2. Calculus of variations Step 3.

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Next we show that if the map in 3. Finally we will complete the proof by proving the following. Lemma 3. Applying Lemma 3. This completes the proof of Theorem 3. Euler-Lagrange equation 27 given boundary conditions. Note that the Euler-Lagrange equation is only a necessary condition for the existence of an extremum. However, in many cases, the Euler-Lagrange equation by itself is enough to give a complete solution of the problem.

In fact, the existence of an extremum is sometimes clear from the context of the problem. From example, in the brachistochrone problem, it is clear from the physical meaning. If in such scenarios, there exists only one critical curve2 satisfying the boundary conditions of the problem, then this critical curve must a fortiori be the curve for which the extremum is achieved.

We indicate some special cases in Exercise 2 on page , where in each instance, F is independent of one of its arguments. Using Theorem 3. Is this a curve which maximizes or minimizes the functional I? Brachistochrone problem, continued. The constants are chosen so that the curve passes through the points 0, 0 and x0 , y0. This curve is known as a cycloid, and in fact it is the curve described by a point P in a circle that rolls without slipping on the x axis, in such a way that P passes through x0 , y0 ; see Figure 3. Euler-Lagrange equation 29 Example.

It can be shown that the following three cases are possible, depending on the positions of the points x0 , y0 and x1 , y1 : 1. If a single curve of the form 3. If two critical curves can be drawn through the points x0 , y0 and x1 , y1 , then one of the curves actually corresponds to the surface of revolution if minimum area, and the other does not.

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  • If there does not exist a curve of the form 3. This is intuitively expected: imagine a soap bubble between concentric rings which are being pulled apart. This example shows that a critical curve need not always exist in the class of curves under consideration.

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    Calculus of variations y1 y0 x1 x0 Figure 3. Find the curve which has minimum length between 0, 0 and 1, 1. See the Exercise on page 2. This pricing model allows the cost of mining to increase with the extent of the mined region and speed of production. Find an optimal mining operation. Free boundary conditions 3. Proceeding as in the proof of Theorem 3. But recall that in Step 3 and Step 4 of the proof of Theorem 3. Integration by parts in 3.

    This completes the proof. We can summarize the results by the following: critical curves for 3. Generalization xf xi tf ti t ti tf t Figure 3. Note that we eliminated high order derivatives at the price of converting the scalar function into a vector-valued function. Since we can always do this, this is one of the reasons in fact for considering functionals of the type 3.

    Chapter 4 Optimal control 4. We prove the following result. Theorem 4. Thus applying the necessity of the condition 36 Chapter 4. Optimal control that the derivative must vanish at extremal points now simply for a function from R to R! Then from Theorem 1. We now introduce an function p in order to rewrite 4. Multiplying 4. With this choice of p, 4.

    Thus 4. This completes the proof of the theorem.

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    This is known as the relative stationarity condition. It should be emphasized that Theorem 4. However, if we already know that an optimal solution exists and that there is a unique critical control, then this critical control is obviously optimal. This analogy with Hamiltonian mechanics was responsible for the original motivation of the Pontryagin minimum principle, which we state below without proof. Let F x, u, t and f x, u be continuously differentiable functions of each of their arguments. Equation 4.

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    • What is the value of k? Optimal control From Theorem 4. From 4. Generalization to vector inputs and states 41 Example.

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