Notes | Mechatronic Systems and Laboratory

3.2 Class of functions and optimality criteria

Having defined an objective functional $F [x]$ and the set of admissible trajectories $x (t) \in 𝒟 \subset 𝒳$ , one must then decide about the class of functions with respect to which the optimization shall be performed. The traditional choice in the calculus of variations is to consider the class of continuously differentiable functions, e.g., $𝒞^{1} [t_{1}, t_{2}]$ . Yet, as shall be seen later on, an optimal solution may not exist in this class. In response to this, a more general class of functions shall be considered, such as the class ${\hat{𝒞}}^{1} [t_{1}, t_{2}]$ of piecewise continuously differentiable functions.

At this point, we need to define what is meant by a minimum (or a maximum) of $F [x]$ on $𝒟$ . Similar to finite-dimensional optimization problems we shall say that $F$ assumes its minimum value at $x ⋆$ provided that

F [x ⋆] \leq F [x] \forall x \in 𝒟 .

This assignment is global in nature and may be made without consideration of a norm (or, more generally, a distance). Yet, the specification of a norm permits an analogous description of the local behavior of $F [x]$ at a point $x ⋆ \in 𝒟$ . In particular, $x *$ is said to be a local minimum for $F [x]$ in $𝒟$ , relative to the norm $‖ \cdot ‖$ , if

\exists δ > 0 such that F [x ⋆] \leq F [x], \forall x \in ℬ_{δ} (x ⋆) \cap 𝒟

with $ℬ_{δ} (x ⋆) : = {x \in 𝒳 : ‖ x - x ⋆ ‖ < δ}$ . Unlike finite-dimensional linear spaces, different norms are not necessarily equivalent in infinite-dimensional linear spaces, in the sense that $x ⋆$ may be a local minimum with respect to one norm but not with respect to another.

Having chosen the class of functions of interest as $𝒞^{1} [t_{1}, t_{2}],$ several norms can be used. Maybe the most natural choice for a norm on $𝒞^{1} [t_{1}, t_{2}]$ is

{‖ x ‖}_{1, \infty} : = \max_{a \leq t \leq b} | x (t) | + \max_{a \leq t \leq b} | ẋ (t) |

since $𝒞^{1} [t_{1}, t_{2}]$ endowed with ${‖ \cdot ‖}_{1, \infty}$ is a Banach space. Another choice is to endow $𝒞^{1} [t_{1}, t_{2}]$ with the maximum norm of continuous functions:

{‖ x ‖}_{\infty} : = \max_{a \leq t \leq b} | x (t) | .

The maximum norms, ${‖ \cdot ‖}_{\infty}$ and ${‖ \cdot ‖}_{1, \infty}$ are called the strong norm and the weak norm, respectively. Similarly, we shall endow $𝒞^{1} {[t_{1}, t_{2}]}^{n_{x}}$ with the strong norm ${‖ \cdot ‖}_{\infty}$ and the weak norm ${‖ \cdot ‖}_{1, \infty}$ :

\begin{aligned} {‖ x ‖}_{\infty} & : = \max_{a \leq t \leq b} ‖ x (t) ‖ \\ {‖ x ‖}_{1, \infty} & : = \max_{a \leq t \leq b} ‖ x (t) ‖ + \max_{a \leq t \leq b} ∥ \dot{x} (t) ∥ \end{aligned}

where $‖ x (t) ‖$ stands for any norm in $ℝ^{n_{x}}$ . The strong and weak norms lead to the following definitions for a local minimum:

Definition 3.3: Strong Local Minimum, Weak Local Minimum

$x ⋆ \in 𝒟$ is said to be a strong local minimum for $F [x]$ in $𝒟$ if

\exists δ > 0 such that F [x ⋆] \leq F [x], \forall x \in ℬ_{δ}^{\infty} (x ⋆) \cap 𝒟 .

Likewise, $x ⋆ \in 𝒟$ is said to be a weak local minimum for $F [x]$ in $𝒟$ if

\exists δ > 0 such that F [x ⋆] \leq F [x], \forall x \in ℬ_{δ}^{1, \infty} (x ⋆) \cap 𝒟 .

Remark 3.2

Note that for strong (local) minima, we completely disregard the values of the derivatives of the comparison elements $x \in 𝒟$ . That is, a neighborhood associated to the topology induced by ${‖ \cdot ‖}_{\infty}$ has many more curves than in the topology induced by ${‖ \cdot ‖}_{1, \infty}$ . In other words, a weak minimum may not necessarily be a strong minimum. Consider the three curves $y_{0}, y_{1}, y_{2}$ illustrated in Figure 3.3. $y_{0}$ and $y_{2}$ are close in the strong sense that is according to ${‖ \cdot ‖}_{\infty}$ but not in a weak sense. While $y_{0}$ and $y_{1}$ are close both according to ${‖ \cdot ‖}_{\infty}$ and ${‖ \cdot ‖}_{1, \infty}$ . ^a ^a Note that actually to take the ${‖ \cdot ‖}_{\infty}$ of $y_{2}$ we should make it $𝒞^{1}$ hence smoothing the corners

Figure 3.3:: Closeness in weak and strong sense

These important considerations are illustrated in the following example.

Example 3.4 (A Weak Minimum that is Not a Strong One). Consider the problem P to minimize the functional

F [x] : = \int_{0}^{1} ẋ^{2} (1 - ẋ^{2}) d t

on $𝒟 : = {x \in {\hat{𝒞}}^{1} [0, 1] : x (0) = x (1) = 0}$ . We first show that the function $\bar{x} (t) = 0, 0 \leq t \leq 1,$ is a weak local minimum for P in the topology induced by ${‖ \cdot ‖}_{1, \infty}$ . Consider the open ball of radius 1 centered at $\bar{x},$ i.e. $ℬ_{1}^{1, \infty} (\bar{x}) .$ For every $x \in ℬ_{1}^{1, \infty} (\bar{x}),$ we have

ẋ (t) \leq 1, \forall t \in [0, 1]

hence $F [x] \geq 0 .$ This proves that $\bar{x}$ is a local minimum for P since $F [\bar{x}] = 0$ . In the topology induced by $∥ \cdot ∥_{\infty}$ , on the other hand, the admissible trajectories $x \in$ $ℬ_{δ}^{\infty} (\bar{x})$ are allowed to take arbitrarily large values $ẋ (t), 0 \leq t \leq 1$ . Consider the sequence of functions defined by:

x_{k} (t) : = {\begin{matrix} \frac{1}{k} + 2 t - 1 & if \frac{1}{2} - \frac{1}{2 k} \leq t \leq \frac{1}{2} \\ \frac{1}{k} - 2 t + 1 & if \frac{1}{2} \leq t \leq \frac{1}{2} + \frac{1}{2 k} \\ 0 & otherwise \end{matrix}

clearly, $x_{k} \in {\hat{𝒞}}^{1} [0, 1]$ and $x_{k} (0) = x_{k} (1) = 0$ for each $k \geq 1,$ i.e., $x_{k} \in 𝒟$ . Moreover

∥ x_{k} ∥ = \max_{0 \leq t \leq 1} | x_{k} (t) | = \frac{1}{k}

meaning that for every $δ > 0,$ there is a $k \geq 1$ such that $x_{k} \in ℬ_{δ}^{\infty} (\bar{x})$ . Finally

F [x_{k}] = \int_{0}^{1} ẋ_{k} {(t)}^{2} (1 - ẋ_{k} {(t)}^{2}) d t = {[- 12 t]}_{\frac{1}{2} - \frac{1}{2 k}}^{\frac{1}{2} \frac{1}{1}} = - \frac{12}{k} < 0

for each $k \geq 1$ . Therefore, the trajectory $\bar{x}$ cannot be a strong local minimum for P.

3.2.1 Existence of an Optimal Solution

Prior to deriving conditions that must be satisfied for a function to be a minimum (or a maximum) of a problem of the calculus of variations, one must ask the question whether such solutions actually exist for that problem. In the case of optimization in finite-dimensional Euclidean spaces, it has been shown that a continuous function on a nonempty compact set assumes its minimum (and its maximum) value in that set (see Weierstrass Theorem 2.1). It is possible to extend this result to continuous functionals on a nonempty compact subset of a normed function space. But as attractive as this solution to the problem of establishing the existence of maxima and minima may appear, it is of little help because most of the sets of interest are too ”large” to be compact.

The principal reason is that most of the sets of interest are not bounded with respect to the norms of interest. As just an example, the set
$𝒟 : = {x \in 𝒞^{1} [a, b] : x (a) = x_{a}, x (b) = x_{b}}$ is clearly not compact with respect to the strong norm $∥ \cdot ∥_{\infty}$ as well as the weak norm $∥ \cdot ∥_{1, \infty},$ for we can construct a sequence of curves in $𝒟$ (e.g., parabolic functions satisfying the boundary conditions) which have maximum values as large as desired. That is, the problem of minimizing, say, $F [x] = - ∥ x ∥$ or $F [x] = \int_{a}^{b} x (t) d t,$ on $𝒟,$ does not have a solution.

A problem of the calculus of variations that does not have a minimum is addressed in the following:

Example 3.5 ( A Problem with No Minimum ). Consider the problem P to minimize the functional:

\begin{aligned} F [x] : = \int_{0}^{1} \sqrt{x {(t)}^{2} + ẋ {(t)}^{2}} d t \\ on 𝒟 : = {x \in 𝒞^{1} [0, 1] : x (0) = 0, x (1) = 1} . \end{aligned}

Observe first that for any admissible trajectory $x (t)$ joining the two end-points, we have

F [x] = \int_{0}^{1} \sqrt{x^{2} + ẋ^{2}} d t > \int_{0}^{1} | ẋ | d t \geq \int_{0}^{1} ẋ d t = x (1) - x (0) = 1

That is,

F [x] > 1, \forall x \in 𝒟, and \inf {F [x] : x \in 𝒟} \geq 1 .

Now, consider the sequence of functions ${x_{k}}$ in $𝒞^{1} [0, 1]$ defined by $x_{k} (t) : = t^{k} .$ Then

\begin{aligned} F [x_{k}] & = \int_{0}^{1} \sqrt{t^{2 k} + k^{2} t^{2 k - 2}} d t = \int_{0}^{1} t^{k - 1} \sqrt{t^{2} + k^{2}} d t \\ \leq \int_{0}^{1} t^{k - 1} (t + k) d t = 1 + \frac{1}{k + 1} \end{aligned}

so we have $F [x_{k}] \overset{k \to \infty}{\to} 1 .$ Overall, we have thus shown that $\inf F [x] = 1 .$ But since $F [x] > 1$ for any $x$ in $𝒟$ we know with certainty that $F$ has no global minimizer on $𝒟$ .