Continuity is Uniform on Compact Sets

Sets, Functions and Metric Spaces

Alexander S. Poznyak , in Advanced Mathematical Tools for Automatic Control Engineers: Deterministic Techniques, Volume 1, 2008

Remark 14.4

The difference between the concepts of continuity and uniform continuity concerns two aspects:

(a)

uniform continuity is a property of a function on a set, whereas continuity is defined for a function in a single point;

(b)

$\delta$ participating in the definition (14.50) of continuity, is a function of $\epsilon$ and a point p, that is, $\delta =\delta \left(\epsilon ,p\right)$ , whereas $\delta$ , participating in the definition (14.17) of the uniform continuity, is a function of $\epsilon$ only serving for all points of a set (space) $\mathcal{X}$ , that is $\delta =\delta \left(\epsilon \right)$ .

Evidently, any uniformly continued function is continuous but not inverse. The next theorem shows when both concepts coincide.

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Elements of Mathematical Methods

Riccardo Sacco , ... Aurelio Giancarlo Mauri , in A Comprehensive Physically Based Approach to Modeling in Bioengineering and Life Sciences, 2019

2.2.2 Uniform Continuity, Absolute Continuity, Lipschitz Continuity, and Integrability

Definition 2.18 Uniform continuity

A function $f:I\to \mathbb{R}$ is said to be uniformly continuous in I if the property in Definition 2.13 holds for δ depending only on ε and not on $x_0$ . This means that for every $\epsilon >0$ there exists a ${\delta }_{\epsilon }>0$ such that for every $x_0\in I$ the following holds:

(2.60) $\forall x\in I\text{with}|x-x_0|<{\delta }_{\epsilon }\Rightarrow |f(x)-f(x_0)|<\epsilon .$

Definition 2.19

We denote by $UC(I)$ the set of real-valued functions $f:I\to \mathbb{R}$ satisfying Definition 2.18.

It is important to emphasize that the properties of the domain I on which the function is defined play a very important role in determining the continuity properties of the function f. For example, if I is an open set, then the notions of continuity and uniform continuity are not equivalent, as illustrated in Example 2.10 below. Conversely, if I is a compact set, namely, a closed and bounded set, then the Heine–Cantor theorem guarantees the equivalence between continuity and uniform continuity.

Theorem 2.7 Heine–Cantor

Let $I\subset \mathbb{R}$ be a compact set, namely, closed and bounded. If $f:I\to \mathbb{R}$ is continuous in I, then it is also uniformly continuous.

Thus, in general, the set of uniformly continuous functions is just a subset of all continuous functions. However, if I is compact, the two sets coincide, as stated by the following lemma.

Lemma 2.1

Given $I\subseteq \mathbb{R}$ , $UC(I)\subseteq C^0(I)$ . If I is compact, then $UC(I)=C^0(I)$ .

Definition 2.20 Absolute continuity

The function $f:I\subset \mathbb{R}\to \mathbb{R}$ is said to be absolutely continuous in I if for every $\epsilon >0$ there exists ${\delta }_{\epsilon }>0$ such that for any countable collection of nonoverlapping subintervals $(a_j,b_j)$ of I the following implication holds:

(2.61) $\sum _j(b_j-a_j)<{\delta }_{\epsilon }\Rightarrow \sum _j(f(b_j)-f(a_j))<\epsilon .$

Definition 2.21

We denote by $AC(I)$ the set of real-valued functions $f:I\to \mathbb{R}$ satisfying Definition 2.20.

The following proposition establishes an equivalence between absolute continuity and Lebesgue integrability for a function f defined on a compact subset of $\mathbb{R}$ . This property will be extremely useful when setting the foundation of continuum-based modeling approaches in Chapter 5.

Theorem 2.8 Fundamental theorem of Lebesgue integral calculus

The following conditions on a real-valued function f on a closed and bounded interval $[a,b]\subset \mathbb{R}$ are equivalent:

1.

f is absolutely continuous;

2.

f has a derivative $f^{\prime }$ defined almost everywhere on $[a,b]$ , the derivative is Lebesgue integrable, and

$f(x)=f(a)+{\int }_a^x{f}^{\prime }(s)ds\mathit{\text{for all}}x\in [a,b];$

3.

there exists a Lebesgue integrable function g defined on $[a,b]\subset \mathbb{R}$ such that

$f(x)=f(a)+{\int }_a^{x}g(s)ds\mathit{\text{for all}}x\in [a,b].$

These equivalences imply that $g=f^{\prime }$ almost everywhere on $[a,b]$ , namely, they may differ only on sets having zero measure.

The following property regarding the integral of continuous functions will be very useful when deriving the balance laws in local form starting from their integral formulations; see Chapter 7.

Lemma 2.2

Let $I\subset \mathbb{R}$ and let f be a continuous function in $\mathbb{R}$ . Then

(2.62) ${\int }_{I}f(s)ds=0\forall I\subset \mathbb{R}\Rightarrow f(s)=0\forall s\in \mathbb{R}.$

Note that the converse is obvious.

The concept of bounded variation will be very useful in the following examples to assess the absolute continuity of a given function. To this end, given $I=[a,b]\subset \mathbb{R}$ , let us consider finite collections of closed intervals $I_j=[l_j,r_j]$ , with $j=1,\dots ,N$ , and $a⩽r_{j-1}⩽l_j⩽r_j⩽b$ for $j=2,\dots ,N$ . We denote by $\mathcal{I}([a,b])$ the set of all such collections, so that

$\mathcal{I}([a,b])=\{\{I_1,I_2,\dots ,I_N\},n\in \mathbb{N}\},$

with $\mathbb{N}$ denoting the set of integers. A partition is an element $P\in \mathcal{I}([a,b])$ such that ${\bigcup }_{j=1}^N{I}_j=[a,b]$ .

Definition 2.22 Bounded variation

The function $f:[a,b]\to \mathbb{R}$ is said to be of bounded variation if

(2.63) $V_a^b(f)=\underset{P\in \mathcal{I}([a,b])}{\sup }\sum _{j=1}^N|f(r_j)-f(l_j)|$

is finite; $V_a^b(f)$ is called the total variation of f on $[a,b]$ .

Definition 2.23

We denote by $BV(I)$ the set of real-valued functions $f:I\to \mathbb{R}$ satisfying Definition 2.22.

The following proposition states that, on compact sets, the property of bounded variation is a necessary condition for absolute continuity. Thus, if a function defined on a compact set does not have bounded variation it cannot be absolutely continuous. This property will come handy in Example 2.11.

Proposition 2.5

Given $[a,b]\subset \mathbb{R}$ , the following property holds:

$AC([a,b])\subset C^0([a,b])\cap BV([a,b]).$

Another important notion of continuity is provided by the so-called Lipschitz continuity, which plays a crucial role in the well-posedness of initial value problems for differential equations (see Chapter 3).

Definition 2.24 Lipschitz continuity

The function $f:I\to \mathbb{R}$ is said to be uniformly Lipschitz continuous (or simply Lipschitz continuous) in I if there exists a positive constant $L_f$ (called Lipschitz constant of f) such that

(2.64) $|f(x_2)-f(x_1)|⩽L_f|x_2-x_1|\forall x_1,x_2\in I.$

A function is said to be locally Lipschitz continuous if the condition (2.64) holds only in the neighborhood of a given point $x_0\in I$ .

Definition 2.25

We denote by $LC(I)$ the set of real-valued functions $f:I\to \mathbb{R}$ satisfying Definition 2.24.

The various notions of continuity listed above define functions with increasing levels of smoothness, as stated by the proposition below.

Proposition 2.6

Given $I\subseteq \mathbb{R}$ , the following property holds:

$LC(I)\subset AC(I)\subset UC(I)\subseteq C^0(I).$

If $I=[a,b]$ , then the following property holds:

$LC([a,b])\subset AC([a,b])\subset UC([a,b])=C^0([a,b]).$

Since the whole theory of mathematical modeling based on the continuum approach presented in this book stems from the pivotal assumption that mass and charge can be represented mathematically as absolutely continuous functions, we will dwell a little more on these different notions of continuity. In the following, we utilize four examples to illustrate the concrete implications of each continuity requirement. As schematized in Fig. 2.8, we are going to consider:

Figure 2.8. The notions of continuity, uniform continuity, absolute continuity, and Lipschitz continuity define functions with increasing levels of smoothness. The functions g _i , with i = 1,…,4, are specific examples of functions that are discussed in the text.

•

a function $g_1$ that is not continuous (see Example 2.9);

•

a function $g_2$ that is continuous but not uniformly continuous (see Example 2.10);

•

a function $g_3$ that is uniformly continuous but not absolutely continuous (see Example 2.11);

•

a function $g_4$ that is absolutely continuous but not Lipschitz continuous (see Example 2.12).

The four functions $g_i$ , $i=1,\dots ,4$ , are graphed in Fig. 2.9.

Figure 2.9. This figure illustrates the four functions g _i , i = 1,…,4, that are used in the examples about the property of continuity. Top, left panel: g ₁(x). Top, right panel: g ₂(x). Bottom, left panel: g ₃(x). Bottom, right panel: g ₄(x).

Example 2.9

The function $g_1:I=[-2,2]\to \mathbb{R}$ , also known as step function, graphed in Fig. 2.9 (top, left) and defined as

(2.65) $g_1(x)=\{\begin{array}{ccc}0,\hfill & \hfill & \text{for}x\in [-2,0],\hfill \\ 1,\hfill & \hfill & \text{for}x\in (0,2],\hfill \end{array}$

is not continuous in I. Intuitively, a function is not continuous if we need to lift our pencil off the paper when drawing it.

Proof

The function $g_1$ is not continuous in I because it is not continuous at $x_0=0$ . This is due to the fact that the left and right limits do not coincide, namely,

$\underset{x\to 0^{-}}{\lim }{g}_1(x)=0\text{whereas}\underset{x\to 0^{+}}{\lim }{g}_1(x)=1.$

  □

Example 2.10

The function $g_2:I=(0,1)\to \mathbb{R}$ graphed in Fig. 2.9 (top, right) and defined as

(2.66) $g_2(x)=\frac{1}{x}\text{for}x\in I$

is continuous but not uniformly continuous on I . Intuitively, uniform continuity guarantees that $g_2(x_1)$ and $g_2(x_2)$ are as close to each other as we please by requiring only that $x_1$ and $x_2$ are sufficiently close to each other, regardless of where $x_1$ and $x_2$ are located on I. Note that the interval $I=(0,1)$ on which the function $g_2$ is defined is not compact.

Proof

The continuity of $g_2$ follows from the fact that it is differentiable in I. Indeed, its derivative $g_2^{\prime }$ is the function $g_2^{\prime }(x)=-1/x^2$ , which is defined at each point in I. Next, we show that $g_2$ is not uniformly continuous by means of a counterexample, namely, by showing that

$\exists {\epsilon }_0\text{such that}\forall \delta >0\exists x_1,x_2\in I\text{with}|x_1-x_2|<\delta \text{and}|g_2(x_1)-g_2(x_2)|⩾{\epsilon }_0.$

For example, let ${\epsilon }_0=1$ and consider $x_1=\delta /6$ and $x_2=\delta /3$ for any $\delta >0$ so that

$|x_1-x_2|=|\frac{\delta }{6}-\frac{\delta }{3}|=\frac{\delta }{6}<\delta .$

Since the interval I has unitary width, we can safely assume that $0<\delta <1$ ; this assumption, combined with the definition of $g_2$ , implies that

$|g_2(x_1)-g_2(x_2)|=|\frac{6}{\delta }-\frac{3}{\delta }|=\frac{3}{\delta }⩾1={\epsilon }_0,$

thereby proving that $g_2$ is not uniformly continuous.   □

Example 2.11

The function $g_3:I=[0,2/\pi ]\to \mathbb{R}$ , graphed in Fig. 2.9 (bottom, left) and defined as

(2.67) $g_3(x)=\{\begin{array}{ccc}0,\hfill & \hfill & \text{for}x=0,\hfill \\ x\sin \left(\frac{1}{x}\right),\hfill & \hfill & \text{otherwise,}\hfill \end{array}$

is uniformly continuous but it is not absolutely continuous. Intuitively, a function that is not absolutely continuous is not regular enough to be written as the integral of some function.

Proof

The uniform continuity of $g_3$ follows from Theorem 2.7, since $g_3$ is continuous on a compact interval. Next, we show that $g_3$ is not absolutely continuous by means of Proposition 2.5, namely, showing that $g_3$ does not have bounded variation. To this end, let us consider the partition $P_N=\{0,a_N,a_{N-1},\dots ,a_1\}$ with

$a_n=\frac{2}{n\pi },n=1,\dots ,N,$

with N odd. Then the variation of $g_3$ on the partition $P_N$ is given by

(2.68) $V_0^{2/\pi }(g_3)=\frac{2}{\pi }+2\sum _{n=1}^N\frac{2}{(2n+1)\pi }.$

Since the above series diverges as $N\to +\infty$ , the total variation of f over $[0,2/\pi ]$ is not finite. Thus, Proposition 2.5 implies that $g_3$ is not absolutely continuous.   □

Example 2.12

The function $g_4:I=[0,1]\to \mathbb{R}$ , graphed in Fig. 2.9 (bottom, right) and defined as

(2.69) $g_4(x)=\sqrt{x}\text{for}x\in I,$

is absolutely continuous but not Lipschitz continuous in I. Intuitively, a Lipschitz continuous function has an upper bound for the slope of its tangent at any point that prevents the function from changing too rapidly.

Proof

The absolute continuity of $g_4$ follows from Proposition 2.8, since (i) it is defined on a compact interval; (ii) its derivative $g_4^{\prime }$ defined as $g_4^{\prime }(x)=1/(2\sqrt{x})$ is defined for every $x\ne 0$ ; and (iii) the following relation holds:

$\sqrt{x}={\int }_0^x\frac{1}{2\sqrt{s}}ds\text{for all}x\in [0,1].$

Next, we prove that $g_4$ is not Lipschitz continuous by showing that it is not possible to find a positive constant $L_f$ independent of $x_1$ and $x_2$ satisfying condition (2.64). To this end, let us consider the sequence

$x_n=\frac{1}{n},n\in \mathbb{N},n⩾1.$

Note that $x_n\in I=[0,1]$ for all positive integers. Considering condition (2.64) with $f=g_4$ , $x_2=x_n$ , and $x_1=0$ , we evaluate the ratio

$\frac{g_4(x_n)-g_4(0)}{x_n-0}=\frac{\sqrt{1/n}-\sqrt{0}}{1/n-0}=\sqrt{n}\underset{n\to +\infty }{\to }+\infty$

to find that it becomes unbounded as $x\to 0$ , thereby preventing the function from being Lipschitz continuous.   □

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Elements of Stability Theory

Alexander S. Poznyak , in Advanced Mathematical Tools for Automatic Control Engineers: Deterministic Techniques, Volume 1, 2008

Proof

Suppose that such function V (t, x) exists. Show that $\overline{x}\left(t,x_0,t_0\right)\to 0$ as $t\to \infty$ whenever $\Vert x_0\Vert$ is small enough, that is, show that for any $\epsilon >0$ there exists $T=T\left(\epsilon \right)$ such that $\Vert \overline{x}\left(t,x_0,t_0\right)\Vert <\epsilon$ for all $t>T$ . Notice that by the uniform continuity V (t,x) and by Corollary 20.1 all trajectories of $\overline{x}\left(t,x_0,t_0\right)$ (20.1) remain within the region where $\Vert \overline{x}\left(t,x_0,t_0\right)\Vert <\epsilon$ if $\Vert x_0\Vert <\delta$ . So, the function $W\left(t,\overline{x}\left(t,x_0,t_0\right)\right)$ remains bounded too. Suppose that $\Vert \overline{x}\left(t,x_0,t_0\right)\Vert$ does not converge to zero. By monotonicity of $V\left(t,\overline{x}\left(t,x_0,t_0\right)\right)$ , this means that there exists $\in >0$ and a moment $T=T(\in )$ such that for all $t\ge T(\in )$ we have $\Vert \overline{x}\left(t,x_0,t_0\right)\Vert <\epsilon$ . Since W (t, x) is a positive-definite function, it follows that

for all $t\ge T(\in )$ and, hence, by (20.20) we have

$\begin{array}{ll}V\left(t,\overline{x}\left(t,x_0,t_0\right)\right)\hfill & =V\left(t,\overline{x}\left(t,x_0,t_0\right)\right)-{\displaystyle \underset{s=t_0}{\overset{t}{\int }}W\left(s,\overline{x}\left(s,x_0,t_0\right)\right)ds}\hfill \\ \hfill & \le V\left(t,\overline{x}\left(t,x_0,t_0\right)\right)-\alpha t\to -\infty \hfill \end{array}$

which contradicts the condition that V (t, x) is a positive-definite function. The fact that this result is uniform on t ₀ follows from Corollary 20.1. Theorem is proven.

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Selected Topics of Real Analysis

Alexander S. Poznyak , in Advanced Mathematical Tools for Automatic Control Engineers: Deterministic Techniques, Volume 1, 2008

Theorem 16.10. (First sufficient (Riemann's) condition)

Assume that α ↑ on [a, b]. If for any ε > 0 there exists a partition P_ε of [a, b] such that P_n is finer than P _ε implies

(16.75) $0\le U(P_n,f,\alpha )-L(P_n,f,\alpha )<\epsilon$

then f ∈ R _[a,b] (α).

Proof

Since by α ↑ on [a, b] we have

$U(P_n,f,\alpha )\le S(P_n,f,\alpha )\le L(P_n,f,\alpha )$

In view of (16.75) this means that S (P_n, f, α) has a limit when n → ∞ which, by the definition (15.11), is the Riemann–Stieltjes integral. Theorem is proven.

Theorem 16.11(Second sufficient condition)

If f is continuous on [a, b] and α is of bounded variation on [a, b], then f ∈ R _[a,b] (α).

Proof

Since by (15.55) any α of bounded variation can be represented as α (x) = α⁺ (x) − α⁻ (x) (where α⁺ ↑ on [a, b] and α⁻ ↑ on [a, b]), it suffices to prove the theorem when α ↑ on [a, b] with α (a) < α (b). Continuity of f on [a, b ] implies uniform continuity, so that if ε > 0 is given, we can find δ = δ (ε) > 0 such that |x − y| < δ implies |f (x) − f (y)| < ε/A where A = 2 [α (b) − α (a)]. If P_ε is a partition with the biggest interval less than δ, then any partition P_n finer than P_ε gives

since

$M_i-m_i=\text{sup}\left\{f(x)-f(y):x,y\in [x_{i-1},x_i]\right\}$

Multiplying (16.76) by Δ α _i and summing, we obtain

$U(P_n,f,\alpha )-L(P_n,f,\alpha )\le \epsilon /A{\displaystyle \sum }_{i=1}^n\Delta {\alpha }_i=\frac{\epsilon }{2}<\epsilon$

So, Riemann's condition (16.75) holds. Theorem is proven.

Corollary 16.4

For the special case of the Riemann integral when α (x) = x Theorem 16.11 together with (15.23) state that each of the following conditions is sufficient for the existence of the Riemann integral ${\displaystyle {\int }_{x=a}^{b}f(x)dx}$ :

1.

f is continuous on [a, b];

2.

f is of bounded variation on [a, b].

The following theorem represents the criterion (the necessary and sufficient condition) for the Riemann integrability.

Theorem 16.12. (Lebesgue's criterion for integrability)

Let f be defined and bounded on [a, b]. Then it is the Riemann integrable on [a, b], which is f ∈ R _[a,b] (x), if and only if f is continuous almost everywhere on [a, b].

Proof

Necessity can be proven by contradiction assuming that the set of discontinuity has a nonzero measure and demonstrating that in this case f is not integrable. Sufficiency can be proven by demonstrating that Riemann's condition (16.75) (when α (x) = x) is satisfied assuming that the discontinuity points have measure zero. The detailed proof can be found in Apostol (1974).

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Computational Methods for Modelling of Nonlinear Systems

In Mathematics in Science and Engineering, 2007

Example 2

[26] Let X be the space of sequences

$x=\left({\xi }_1,{\xi }_2,\dots \right)$

with ξ → 0 and set

$\Vert x\Vert =max\left|{\xi }_k\right|$

Suppose that ω ⊂ X consists of all sequences of the form

$X_m=\left\{x={\xi }_1,\dots {\xi }_m,0,0,\dots )\right\}$

where σ_i is + 1 or −1. The set ω belongs to the unit ball of the space X and is closed.

We define a functional f on ω; by

$f\left(y\right)={\sigma }_1{\sigma }_2\dots {\sigma }_k$

Since we have

$\Vert y^{\prime }-y^{″}\Vert \ge 1$

for distinct points y′, y″ ∈ ω;, the functional f is uniformly continuous. It is easily extended to the unit ball in X with preservation of uniform continuity.

Let

$X_m=\left\{x={\xi }_1,\dots {\xi }_m,0,0,\dots )\right\}$

be a subspace of X. We make no distinction between this space and the space $l\frac{}{m}$ . The restriction of a polynomial functional p_n of degree n on X_m is a polynomial in m variables of degree not greater than n. Therefore if

${\Omega }_m\subset \Omega \cap X_m$

is the set of vertices of the M –dimensional cube [−1, 1] ^m and if f_m is the restriction of the functional f to ω_m, then for any polynomial p_n of degree n and for any m, the inequality

$sup\left\{\left|f\left(x\right)-p_n\left(x\right)\right|forx\in \Omega \right\}\ge E_n\left(f_m\right)$

holds, where E_n (f_m ) is the best uniform approximation of f_m by polynomials in m variables of degree n.

The function f_m is odd with respect to each argument, and in the set of all best-approximation polynomials of this function, there also exists an odd polynomial with respect to each argument. In particular, for m > n the zero polynomial is the best approximation polynomial, since this is the unique polynomial of degree n of m variables which is odd with respect to all arguments, and therefore

$E_n\left(f_m\right)=1$

for m > n.

Thus for any polynomial function p_n, we have

$sup\left\{\left|f\left(x\right)-p_n\left(x\right)\right|forx\in \Omega \right\}\ge 1.$

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Existence and Uniqueness Revisited

Morris W. Hirsch , ... Robert L. Devaney , in Differential Equations, Dynamical Systems, and an Introduction to Chaos (Third Edition), 2013

17.4 Extending Solutions

Suppose we have two solutions $Y(t),Z(t)$ of the differential equation $X^{\mathfrak{\prime }}=F(X)$ where F is C ¹. Suppose also that Y(t) and Z(t) satisfy $Y(t_0)=Z(t_0)$ and that both solutions are defined on an interval J about t ₀. Now the Existence and Uniqueness Theorem guarantees that Y(t) = Z(t) for all t in an interval about t ₀ that may a priori be smaller than J. However, this is not the case.

To see this, suppose that J* is the largest interval on which Y(t) = Z(t). If $J^{\ast }\ne J$ , there is an endpoint t ₁ of J* and $t_1\mathfrak{\in }J$ . By continuity, we have $Y(t_1)=Z(t_1)$ . Now the uniqueness part of the theorem guarantees that, in fact, Y(t) and Z(t) agree on an interval containing t ₁. This contradicts the assertion that J* is the largest interval on which the two solutions agree.

Thus we can always assume that we have a unique solution defined on a maximal time domain. There is, however, no guarantee that a solution X(t) can be defined for all time. For example, the differential equation

$\begin{array}{l}\hfill x^{\mathfrak{\prime }}=1+x^2\end{array}$

has as solutions the functions $x(t)=\tan (t-c)$ for any constant c. Such a function cannot be extended over an interval larger than

$\begin{array}{l}\hfill c-\frac{\pi }{2}<t<c+\frac{\pi }{2}\end{array}$

since $x(t)\to \pm \mathfrak{\infty }$ as $t\to c\pm \pi ∕2$ .

Next, we investigate what happens to a solution as the limits of its domain are approached. We state the result only for the right-hand limit; the other case is similar.

Theorem

Let $\mathcal{O}\subset {ℝ}^n$ be open, and let $F:\mathcal{O}\to {ℝ}^n$ be C ¹ . Let $Y(t)$ be a solution of $X^{\mathfrak{\prime }}=F(X)$ defined on a maximal open interval $J=(\alpha ,\beta )\subset ℝ$ with $\beta <\mathfrak{\infty }$ . Then, given any compact set $\mathcal{C}\subset \mathcal{O}$ , there is some $t_0\mathfrak{\in }(\alpha ,\beta )$ with $Y(t_0)\notin \mathcal{C}$ .

This theorem says that if a solution Y(t) cannot be extended to a larger time interval, then this solution leaves any compact set in $\mathcal{O}$ . This implies that, as $t\to \beta$ , either Y(t) accumulates on the boundary of $\mathcal{O}$ or else a subsequence $|Y(t_i)|$ tends to ∞ (or both).

Proof

Suppose $Y(t)\subset \mathcal{C}$ for all $t\mathfrak{\in }(\alpha ,\beta )$ . Since F is continuous and $\mathcal{C}$ is compact, there exists M > 0 such that $|F(X)|\le M$ for all $X\mathfrak{\in }\mathcal{C}$ .

Let $\gamma \mathfrak{\in }(\alpha ,\beta )$ . We claim that Y extends to a continuous function $Y:\left[\gamma ,\beta \right]\to \mathcal{C}$ . To see this, it suffices to prove that Y is uniformly continuous on J. For $t_0<t_1\mathfrak{\in }J$ we have

$\begin{array}{ll}\hfill |Y(t_0)-Y(t_1)|& =\left|\underset{t_0}{\overset{t_1}{\int }}{Y}^{\mathfrak{\prime }}(s)ds\right|\\ \hfill & \le \underset{t_0}{\overset{t_1}{\int }}|F(Y(s))|ds\\ \hfill & \le (t_1-t_0)M.\end{array}$

This proves uniform continuity on J. Thus we may define

$\begin{array}{l}\hfill Y(\beta )=\underset{t\to \beta }{lim}Y(t).\end{array}$

We next claim that the extended curve $Y:\left[\gamma ,\beta \right]\to {ℝ}^n$ is differentiable at $\beta$ and is a solution of the differential equation. We have

$\begin{array}{ll}\hfill Y(\beta )& =Y(\gamma )+\underset{t\to \beta }{lim}\underset{\gamma }{\overset{t}{\int }}{Y}^{\mathfrak{\prime }}(s)ds\\ \hfill & =Y(\gamma )+\underset{t\to \beta }{lim}\underset{\gamma }{\overset{t}{\int }}F(Y(s))ds\\ \hfill & =Y(\gamma )+\underset{\gamma }{\overset{\beta }{\int }}F(Y(s))ds,\end{array}$

where we have used uniform continuity of F(Y(s)). Therefore,

$\begin{array}{l}\hfill Y(t)=Y(\gamma )+\underset{\gamma }{\overset{t}{\int }}F(Y(s))ds\end{array}$

for all t between γ and β. Thus Y is differentiable at β, and, in fact, $Y^{\mathfrak{\prime }}(\beta )=F(Y(\beta ))$ . Therefore, Y is a solution on $\left[\gamma ,\beta \right]$ . Since there must then be a solution on an interval $\left[\beta ,\delta \right]$ for some $\delta >\beta$ , we can extend Y to the interval $\left[\alpha ,\delta \right]$ . Thus $(\alpha ,\beta )$ could not have been a maximal domain of a solution. This completes the proof of the theorem.

This important fact follows immediately from the preceding theorem.

Corollary

Let $\mathcal{C}$ be a compact subset of the open set $\mathcal{O}\subset {ℝ}^n$ and let $F:\mathcal{O}\to {ℝ}^n$ be C ¹ . Let $Y_0\mathfrak{\in }\mathcal{C}$ and suppose that every solution curve of the form $Y:\left[0,\beta \right]\to \mathcal{O}$ with $Y(0)=Y_0$ lies entirely in $\mathcal{C}$ . Then there is a solution $Y:\left[0,\mathfrak{\infty }\right]\to \mathcal{O}$ satisfying $Y(0)=Y_0$ , and $Y(t)\mathfrak{\in }\mathcal{C}$ for all t ≥ 0, so this solution is defined for all (forward) time.

Given these results, we can now give a slightly stronger theorem on the continuity of solutions in terms of initial conditions than the result discussed in Section 17.3. In that section we assumed that both solutions were defined on the same interval. In the next theorem we drop this requirement. The theorem shows that solutions starting at nearby points are defined on the same closed interval and also remain close to each other on this interval.

Theorem

Let $F:\mathcal{O}\to {ℝ}^n$ be C ¹ . Let Y(t) be a solution of $X^{\mathfrak{\prime }}=F(X)$ that is defined on the closed interval $\left[t_0,t_1\right]$ , with $Y(t_0)=Y_0$ . There is a neighborhood $U\subset {ℝ}^n$ of Y ₀ and a constant K such that, if $Z_0\mathfrak{\in }U$ , then there is a unique solution Z(t) also defined on $\left[t_0,t_1\right]$ with $Z(t_0)=Z_0$ . Moreover, Z satisfies

$\begin{array}{l}\hfill |Y(t)-Z(t)|\le |Y_0-Z_0|\exp (K(t-t_0))\end{array}$

for all $t\mathfrak{\in }\left[t_0,t_1\right]$ .

For the proof of the preceding theorem, will need the following lemma.

Lemma

If $F:\mathcal{O}\to {ℝ}^n$ is locally Lipschitz and $\mathcal{C}\subset \mathcal{O}$ is a compact set, then $F|\mathcal{C}$ is Lipschitz.

Proof

Suppose not. Then for every k > 0, no matter how large, we can find X and Y in $\mathcal{C}$ with

$\begin{array}{l}\hfill |F(X)-F(Y)|>k|X-Y|.\end{array}$

In particular, we can find $X_n,Y_n$ such that

$\begin{array}{l}\hfill |F(X_n)-F(Y_n)|>n|X_n-Y_n|\text{for}n=1,2,\dots \end{array}$

Since $\mathcal{C}$ is compact, we can choose convergent subsequences of the X _n and Y _n . Relabeling, we may assume $X_n\to X^{\ast }$ and $Y_n\to Y^{\ast }$ with X* and Y* in $\mathcal{C}$ . Note that we must have $X^{\ast }=Y^{\ast }$ , since, for all n,

$\begin{array}{l}\hfill |X^{\ast }-Y^{\ast }|=\underset{n\to \mathfrak{\infty }}{lim}|X_n-Y_n|\le \underset{n\to \mathfrak{\infty }}{lim}{n}^{-1}|F(X_n)-F(Y_n)|\le \underset{n\to \mathfrak{\infty }}{lim}{n}^{-1}2M,\end{array}$

where M is the maximum value of $|F(X)|$ on $\mathcal{C}$ . There is a neighborhood ${\mathcal{O}}_0$ of X* on which $F|{\mathcal{O}}_0$ has Lipschitz constant K. Also there is an n ₀ such that $X_n\mathfrak{\in }{\mathcal{O}}_0$ if $n\ge n_0$ . Therefore, for $n\ge n_0$ ,

$\begin{array}{l}\hfill |F(X_n)-F(Y_n)|\le K|X_n-Y_n|,\end{array}$

which contradicts the assertion just made for $n>n_0$ . This proves the lemma.

The proof of the theorem now goes as follows.

Proof

By compactness of $\left[t_0,t_1\right]$ , there exists $ϵ>0$ such that $X\mathfrak{\in }\mathcal{O}$ if $|X-Y(t)|\le ϵ$ for some $t\mathfrak{\in }\left[t_0,t_1\right]$ . The set of all such points is a compact subset $\mathcal{C}$ of $\mathcal{O}$ . The C ¹ map F is locally Lipschitz, as we saw in Section 17.2. By the lemma, it follows that $F|\mathcal{C}$ has a Lipschitz constant K.

Let $\delta >0$ be so small that $\delta \le ϵ$ and $\delta \exp (K|t_1-t_0|)\le ϵ$ . We claim that if $|Z_0-Y_0|<\delta$ , then there is a unique solution through Z ₀ defined on all of $\left[t_0,t_1\right]$ . First of all, $Z_0\mathfrak{\in }\mathcal{O}$ since $|Z_0-Y(t_0)|<ϵ$ , so there is a solution Z(t) through Z ₀ on a maximal interval $[t_0,\beta )$ . We claim that $\beta >t_1$ because, if we suppose $\beta \le t_1$ , then, by Gronwall's Inequality, for all $t\mathfrak{\in }\left[t_0,\beta \right]$ , we have

$\begin{array}{ll}\hfill |Z(t)-Y(t)|& \le |Z_0-Y_0|\exp (K|t-t_0|)\\ \hfill & \le \delta \exp (K|t-t_0|)\\ \hfill & \le ϵ.\end{array}$

Thus Z(t) lies in the compact set $\mathcal{C}$ . By the preceding results, $[t_0,\beta )$ could not be a maximal solution domain. Therefore, Z(t) is defined on $\left[t_0,t_1\right]$ . The uniqueness of Z(t) then follows immediately. This completes the proof.

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WEAK-INVARIANCE AND REST POINTS IN CONTROL SYSTEMS

M.L.J. Hautu , ... Ronald J. Stern , in Dynamical Systems, 1977

4.6 Theorem

Consider system 1.1 with n = 2 and assume the following hold:

(a)

f(x,u) is continuous in both arguments for all $\text{x}\in {\underset{\_}{\underset{\_}{\text{R}}}}^2$ and $\text{u}\in {\underset{\_}{\underset{\_}{\text{R}}}}^{\text{m}}.$

(b)

For all $\text{x}\in {\underset{\_}{\underset{\_}{\text{R}}}}^2$ the set f(x,Ω) is compact and convex.

Assume that for an admissible control u ε U_Ω there exists a solution ϕ_u (t) which for some T > 0 is a proper parametrization of a Jordan curve K on [0,T]. If K has no Ω-rest states of 1.1 then for each $\overline{\text{u}}\in \Omega$ there exists $\overline{\text{x}}\in \text{enc}\left(\text{K}\right)$ such that $\text{f}\left(\overline{\text{x}}\text{,}\overline{\text{u}}\right)=0$ .

Proof

Assume that K has no Ω-rest states of 1.1, i.e., the vector field V defined by V(x): = f(x,Ω) is regular on K.

For ɛ > 0 and all $\text{x}\in {\underset{\_}{\underset{\_}{\text{R}}}}^2$ define

${\text{V}}_{\epsilon }\left(\text{x}\right):=\left\{\text{y+}\epsilon \text{z}|\text{y}\in \text{V}\left(\text{x}\right);\left|\left|\text{z}\right|\right|\le 1\right\}.$

Clearly ${\text{V}}_{\epsilon }\in \underset{\_}{\text{C}}\left({\underset{\_}{\underset{\_}{\text{R}}}}^2\to \Gamma \right)$ and by the continuity of V and the compactness of K, there exists an ɛ > 0 such that V_ɛ is also regular on K. Since V(x) ⊂ V_ɛ (x) for all $\text{x}\in {\underset{\_}{\underset{\_}{\text{R}}}}^2$ it is clear that ρ_K (V_ɛ) = ρ_K (v) where v is any continuous selection of V. By Lemma 4.4 and the fact that every constant control provides a continuous selection of f(.,Ω), the proof will be complete upon showing that ρ_K(V_ɛ) ≠ 0. By Lemma 4.5 this will be accomplished if we can show that for sufficiently small δ > 0, w_δ(t): = δ⁻¹ [ϕ_u (t+δ)-ϕ_u (t)] ∈ V_ɛ (ϕ_u (t)) for all 0 ≤ t ≤ T (with ϕ_u being extended periodically outside the interval [0,T]). In view of the uniform continuity of V (and of ϕ _u) on K there clearly exists a δ > 0 such that V(ϕ_u (τ)) ⊂ V_ɛ (ϕ(t)) for all t ε [0,T] and all τ such that |t-τ| < δ. Now suppose that ${\text{w}}_{\delta }\left({\text{t}}_0\right)\in {\text{V}}_{\epsilon }\left({\varphi }_{\text{u}}\left({\text{t}}_o\right)\right)$ for some t_o ε [0,T]. Then there exists a vector c ≠ 0 and a number α such that 〈c,w_δ(t_o)〉 > α and 〈c,y〉 ≤ α for all y ε V_ɛ (ϕ_u (t_o)).

Since V(ϕ_u (t)) ⊂ V_ɛ (ϕ_u (t_o)) for all t₀ ≤ t ≤ t₀ + δ and since ${\dot{\varphi }}_{\text{u}}\left(\text{t}\right)\in \text{V}\left({\varphi }_{\text{u}}\left(\text{t}\right)\right)$ a.e., it follows that

$〈{\text{c,w}}_{\delta }\left({\text{t}}_o\right)〉={\delta }^{-1}{\displaystyle \underset{{\text{t}}_o}{\overset{{\text{t}}_o+\delta }{\int }}〈\text{c,}{\dot{\varphi }}_{\text{u}}\left(\tau \right)〉\text{d}\tau \le \alpha \text{,}}$

a contradiction. This completes the proof.

As a consequence of Theorem 4.6 we also have

4.7 Theorem

Consider system 1.1 with n = 2 and assume 4.6a and 4.6b hold. Assume that for some admissible control u ∈ U_Ω and some T > 0 there exists a solution ϕ_u (t) of 1.1 defined on [0,T] such that ϕ_u (T) = ϕ_u (0). Then SCH(L) contains an Ω-rest state of 1.1 where L: = {ϕ_u (t) |0≤t≤T} is the trajectory of ϕ_u.

The proof of Theorem 4.7 as well as that of the following Theorem can be found in HAUTUS, HEYMANN and STERN [1976].

4.8 Theorem

Consider system 1.1 with n = 2, and assume that 4.6a and 4.6b hold. Let $\text{S}\subset {\underset{\_}{\underset{\_}{\text{R}}}}^2$ be a compact, simply connected and locally connected subset. If for some admissible control u ∈ U_Ω there exists a solution ϕ_u (t) of 1.1 which is contained in S for all t ∈ [0,∞), then S contains an Ω-rest state.

Theorem 4.5 implies the interesting fact that under conditions 4.6a and 4.6b, a planar control system (i.e., n = 2) has Ω-rest states whenever there are bounded trajectories. If 4.6b fails to hold then 1.1 may not have any Ω-rest states in the plane even when there exist bounded trajectories as the following example illustrates

$\begin{array}{l}{\dot{\text{x}}}_1=\text{sin}\text{u}\hfill \\ {\dot{\text{x}}}_2=\text{cos}\text{u}\hfill \end{array}\begin{array}{l}\Omega =\left[0,\pi \right]\hfill \end{array}$

When n ≥ 3, the existence of bounded trajectories no longer insures the existence of Ω-rest states even when 4.6a and 4.6b hold. Indeed, this can be verified to be the case in the system

$\begin{array}{l}\dot{\text{x}}\text{=}\text{ux}\text{−}\text{y}\hfill \\ \dot{\text{y}}\text{=}\text{x}\text{+}\text{uy}\hfill \\ \dot{\text{z}}\text{=}\text{1}-{\text{x}}^2-{\text{y}}^2\hfill \end{array}\begin{array}{l}\Omega =\left\{\text{u}|\left|\text{u}\right|\le 1\right\}\hfill \end{array}$

which has bounded trajectories (e.g. starting at ${\text{x}}_o={\text{y}}_o=\frac{\sqrt{2}}{2},$ , z_o = 0 under the constant control u(t) ≡ 0) but has no Ω-rest states at all.

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