Abstract C⁎-Algebras

In C*-Algebras and their Automorphism Groups (Second Edition), 2018

1.3.7

We say that a continuous real function f on an interval in R is operator monotone (increasing) if x y implies f ( x ) f ( y ) whenever the spectra of x and y belong to the interval of definition for f.

For each α > 0 , define f α on ] 1 / α , [ by

f α ( t ) = ( 1 + α t ) 1 t = [ 1 ( 1 + α t ) 1 ] / α .

Since the process of taking inverses is operator monotone decreasing by 1.3.6, it is easy to see that f α is operator monotone increasing on ] 1 / α , [ . The family of functions { f α } is further used repeatedly. Note that f α ( t ) < Min { t , 1 / α } and that Lim f α ( t ) = t uniformly on compact subsets of R as α 0 . Moreover, f α f β when α β , and f α f β = f α + β on ] ( α + β ) 1 , [ . Finally, if t > 0 , then Lim α f α ( t ) = 1 uniformly on compact subsets of ] 0 , [ as α .

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Integration—Methods and Applications

Colin McGregor , ... Wilson Stothers , in Fundamentals of University Mathematics (Third Edition), 2010

16.5 Volumes of Revolution

Let f be a real function, continuous on [a, b], and consider the solid of revolution formed by revolving, once about the x-axis, the curve C given by

y = f x a x b .

See Figure 16.5.1. We define the volume of such a solid by first approximating with cylinders. Recall the formula πR 2 H for the volume of a cylinder with radius R and height H.

Figure 16.5.1.

Let P = (x 0, x 1, …, xn ) be a partition of [a, b] and, for each r = l, …, n, let cr ∈ [x r–1, xr ]. Consider the cylinder shown in Figure 16.5.2. It is the solid of revolution formed by revolving, once about the x-axis, the corresponding rectangle, also shown in Figure 16.5.2. There are n such cylinders. The sum of their volumes is

Figure 16.5.2.

S = r = 1 n πf C r 2 x r x r 1 ,

which is a Riemann sum for the function g: [a, b] → ℝ defined by g(x) = π f (x)2. Since g is continuous on [a, b], S    a b g(x)dx as |P | → 0.

Definition 16.5.1 Let f be a real function, continuous on [a, b]. Then the volume of revolution V generated by the curve y = f (x) (axb) is defined by

(16.5.1) V = π a b f x 2 dx = π a b y 2 dx .

Theorem 16.5.2 Let C be the curve given by the parametric equations

x = x t , y = y t α t β ,

where x is injective with a continuous derivative, and y is continuous, on [α, β].Then the volume of revolution generated by C is

(16.5.2) V = π α β y y 2 x ˙ t dt .

Proof Suppose, without loss, that x: [α, β] → I where I = x([α, β]). Then x is a bijection and, by Example 7.3.16, I = [a, b] (say). Define f: [a, b] → ℝ by f (s) = y(x −1(s)). By Theorems 7.3.18 and 7.3.9 7.3.18 7.3.9 (3), f is continuous. Since C is the curve y = f (x) (axb), Definition 16.5.1 and Lemma 16.1.4 give

V = π a b f x 2 dx = π α β f x t 2 x ˙ t dt = π α β y t 2 x ˙ t dt

as required.

Example 16.5.3 Find the volume of a sphere, radius R.

Solution We can think of the required volume V as the volume of revolution generated by the upper half of the circle x 2  + y 2 = R 2. Then, using (16.5.1),

V = π R R R 2 x 2 dx = R 2 x x 3 3 R R = 4 3 π R 3 .

Alternatively, expressing the semi-circle in parametric form as

x = R cos t , y = R sin t 0 t π ,

equation(16.5.2) gives

V = π 0 π R sin t 2 R sin t dt = π R 3 0 π sin t dt = 4 3 π R 3 .

Verification of the final evaluation is left to the reader.

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Differentiation—Applications

Colin McGregor , ... Wilson Stothers , in Fundamentals of University Mathematics (Third Edition), 2010

9.4 More on Monotonic Functions

Recall part of Definitions 2.6.1. Let f be a real function whose domain contains the set A. Then f is strictly increasing on A if, for s,tA,

s < t f s < f t

and f is strictly decreasing on A if, for s,tA,

s < t f s > f t .

When the function involved is differentiable, strict increase and decrease can be related to the sign of the derivative.

Theorem 9.4.1 (The Monotonicity Theorem) Let f be a real function, continuous on [a, b] and differentiable on (a,b). Then

(1)

f ′(x)   =   0 for all x ∈ (a, b) ⇒ f is constant on [a, b],

(2)

f ′(x)   >   0 for all x ∈ (a, b) ⇒ f is strictly increasing on [a, b],

(3)

f ′(x)   <   0 for all x ∈ (a, b) ⇒ f is strictly decreasing on [a, b].

The variations given in Note 9.4.2 are included in the theorem.

Proof Let s, t ∈ [a, b] with s  < t. Then f is continuous on [s, t] and differentiable on (s, t). So, by the mean value theorem, there exists c ∈ (s, t)     (a, b) such that

f t f s = t s f c .

Since (ts)   >   0, we have

f x = 0 > 0 < 0 for all x a b f ( c ) = 0 > 0 < 0 f t f s = 0 > 0 < 0 f s = f t > f t < f t

and (1), (2) and (3) follow immediately.     □

Note 9.4.2 The proof of Theorem 9.4.1 allows for variations in the statement of the theorem. Specifically, we can replace [a, b] with I and (a, b) with J where I and J are given by a row of the following table.

I J
[a, b), (a, b] or (a, b) (a, b)
[a, ∞) or (a, ∞) (a, ∞)
(–∞, a] or (–∞, a) (–∞, a)

Example 9.4.3 Determine the intervals on which the real function f defined by

f x = x x + 1 2

is (a) strictly increasing, (b) strictly decreasing.

Solution Here,

f x = 1 x x + 1 3 .

So a table of signs for f ′(x) is as follows.

x –1 1
1 – x + + + 0
(x  +   1)3 0 + + +
f ′(x) ? + 0

From this we see that f ′(x)   >   0 when –1   < x  <   1 and f ′(x)   <   0 when x  <   –1 or x  >   1. Hence, by the monotonicity theorem, f is strictly increasing on (–1, 1] and strictly decreasing on (–∞,–1) and on [1, ∞).     □

Example 9.4.4 Prove that, for all x 0 π 2 ,

(9.4.1) sin x < x .

Deduce that (9.4.1) holds for all x  >   0.

Solution Define f : 0 π 2 → ℝ by

f x = x sin x .

Then f is continuous on 0 π 2 and differentiable on 0 π 2 with

f x = 1 cos x > 0 0 < x < π 2 .

So, by the monotonicity theorem, f is strictly increasing on 0 π 2 . Hence, for all x 0 π 2 , we have 0, x 0 π 2 and 0   < x so that

f 0 < f x , i . e .0 < sin x x , i . e . sin x < x .

Hence equation(9.4.1) holds for all x > 0.     □

Example 9.4.5 Show that, for all x ∈ [–1, 1],

Sin 1 x + cos 1 x = π 2 .

[See Problem 5.5.6.]

Solution Define f : [–1,1] → ℝ by

f x = sin 1 x + cos 1 x .

Then f is continuous on [–1, 1] and differentiable on (–1, 1) with

f x = 1 1 x 2 1 1 x 2 = 0 1 < x < 1 .

So, by the monotonicity theorem, f is constant on [–1, 1]. Hence, for all x ∈ [–1, 1],

f x = f 0 = sin 1 0 + cos 1 0 = 0 + π 2 = π 2

as required.     □

Hazard A real function whose domain D is not an interval may have zero derivative on the whole of D and yet not be constant on D. The next example illustrates this.

Example 9.4.6 Let f be the real function defined by

f x = tan 1 x 1 x + 1 + tan 1 x + 1 x 1

Find the maximal domain D of f. Show that f ′(x)   =   0 for all xD. Sketch the graph of f.

Solution Since tan–1 is defined on ℝ, the only x for which f ′(x) is not defined are –1 and 1. Hence

D = 1 , 1 = , 1 1 , 1 1 .

For xD,

f x = 1 1 + x 1 x + 1 2 x + 1 x 1 x + 1 2 + 1 1 + x + 1 x 1 2 x 1 x + 1 x 1 2 = 2 x + 1 2 + x 1 2 2 x 1 2 + x + 1 2 = 0 .

It follows from the monotonicity theorem that f is constant on each of the intervals (–∞, –l), (–1, 1) and (l, ∞). Since

f 0 = tan 1 1 + tan 1 1 = π 4 + π 4 = π 2

and

f 2 + 3 = tan 1 1 3 + tan 1 3 = π 6 + π 3 = π 2

it follows that f(x) =  –π/2 for all x ∈ (–1, 1) f(x)   =   π /2 for all x ∈ (l, ∞). Since f is an even function, f(x) =  π/2 for all x ∈ (–∞, –l,). Alternatively, evaluate f(–2 – 3 ).

The graph of f is sketched in Figure 9.4.1.     □

Figure 9.4.1.

Problem 9.4.7 For the function f of Example 9.4.6, establish the constant value of f on (l, ∞) by considering lim x f x .

Corollary 9.4.8 (to the Monotonicity Theorem) Let g and h be real functions, continuous on [a, b] and differentiable on (a, b) with

g x = h x a < x < b .

Then there exists K ∈ ℝ such that

g x = h x + K a x b ,

i.e. g and h differ by a constant on [a, b].

Proof Define f : [a, b]     ℝ by

f x = g x h x .

Then f is continuous on [a, b] and differentiable on (a, b) with

f x g x h x = 0 a x b .

So, by the monotonicity theorem, f is constant on [a, b] and the result follows immediately. □

Remark Corollary 9.4.8 will be used in Section 14.3 to prove the fundamental theorem of calculus.

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Integral and Finite Difference Inequalities and Applications

B.G. Pachpatte , in North-Holland Mathematics Studies, 2006

Theorem 2.3.3

Let u (x,y) ,a (x,y) ,b(x,y) ,g (x,y), h (x, y) ∈ C ( R + 3 , R + ) and p > 1 be a real constant.

(c1 ) If

(2.3.27) u p ( x , y ) a ( x , y ) + b ( x , y ) 0 x 0 y [ g ( s , t ) u p ( s , t ) + h ( s , t ) u ( s , t ) ] d t d s ,

for x,yR+ , then

(2.3.28) u ( x , y ) { a ( x , y ) + b ( x , y ) e ( x , y ) × exp ( 0 x 0 y [ g ( s , t ) + h ( s , t ) p ] b ( s , t ) d t d s ) } 1 p ,

for x,yR+ , where

(2.3.29) e ( x , y ) = 0 x 0 y [ g ( s , t ) a ( s , t ) + ( p 1 p + a ( s , t ) p ) h ( s , t ) ] d t d s ,

for x,y ∊ R+.

(c 2) Let c(x, y) be a real-valued, continuous, positive and nondecreasing function defined for x, yR +. Ifup (x,y) ≤ cp (x,y) + b(x,y)

(2.3.30) u p ( x , y ) c p ( x , y ) + b ( x , y ) + 0 x 0 y [ g ( s , t ) u p ( u , t ) + h ( s , t ) u ( s , t ) ] d t d s ,

for x,yR+ , then

(2.3.31) u ( x , y ) c ( x , y ) { 1 + b ( x , y ) e 0 ( x , y ) × exp ( 0 x 0 y [ g ( s , t ) + h ( s , t ) p c 1 p ( s , t ) ] b ( s , t ) d t d s ) } 1 p ,

for x, yR +, where

(2.3.32) e 0 ( x , y ) = 0 x 0 y [ g ( s , t ) + h ( s , t ) c 1 p ( s , t ) ] d t d s ,

for x, y ∈ R+.

Proof. (c 1) Define a function z(x,y) by

(2.3.33) z ( x , y ) = 0 x 0 y [ g ( s , t ) u p ( s , t ) + h ( s , t ) u ( s , t ) ] d t d s ,

then z(0,y) = z(x,0) = 0 and (2.3.27) can be written as

(2.3.34) u p ( x , y ) a ( x , y ) + b ( x , y ) z ( x , y ) .

From (2.3.34) and using the elementary inequality (1.3.11) (see [30, p. 30]) we observe that

(2.3.35) u ( x , y ) ( a ( a , y ) + b ( x , y ) z ( x , y ) ) 1 p ( 1 ) 1 ( p / p 1 ) p 1 p + a ( x , y ) p + b ( x , y ) p z ( x , y ) .

From (2.3.33)-(2.3.35) it is easy to observe that

(2.3.36) z ( x , y ) e ( x , y ) + 0 x 0 y [ g ( s , t ) + h ( s , t ) p ] b ( s , t ) z ( s , t ) d t d s ,

where e(x,y) is defined by (2.3.29). Clearly e(x,y) is nonnegative, continuous and nondecreasing for x, yR+ . A suitable application of Theorem 4.2.2 given in [34, p. 325] to (2.3.36) yields

(2.3.37) z ( x , y ) e ( x , y ) exp ( 0 x 0 y [ g ( s , t ) + h ( s , t ) p ] b ( s , t ) d t d s ) ,

for x,y ∈R+ The required inequality (2.3.28) follows from (2.3.34) and (2.3.37).

(c 2) Since c(x, y) is positive, continuous and nondecreasing function for x, yR+ , from (2.3.30) we observe that

(2.3.38) ( u ( x , y ) c ( x , y ) ) p 1 + 0 x 0 y [ g ( s , t ) ( u ( s , t ) c ( s , t ) ) p + h ( s , t ) c 1 p ( s , t ) u ( s , t ) c ( s , t ) ] d t d s .

Now a suitable application of the inequality given in part (c1 ) to (2.3.38) yields the desired inequality in (2.3.31).

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Probabilities and Potential C

In North-Holland Mathematics Studies, 1988

Ray Resolvents

80.

We denote by F a compact metrisable space, by C the space of continuous real functions on F, by (Vp) a submarkovian resolvent on F, having the following property

(80.1) T h e K e r n e l s ¯ V p ( p > 0 ) l e a v e c i n v a r i a n t . ¯

In this paragraph, we denote by G p the cone of continuous p-supermedian functions on F; for p < q we have G p G q , but the relation f ▭ [f + (q-p)Vpf]-(q-p)Vpf shows that every element of G p is the differenceof two elements of G p (if f is q-supermedian, f + (q-p)Vpf is p-supermedian: this is clear when f = Vqg, g ≥ 0, then by passing to the limit for f q-excessive; finally a general q-supermedian function is the sum of an excessive function and a positive function vanishing a.e.). In other words, the vector space S p S p does not depend on p; we denote it by R in the sequel. As S p is ∧-stable and contains the constant 1, R is a lattice and contains the constants. By the Stone-Weierstrass theorem, R is dense in e if and only if separate points of F. This amounts to saying that S p separates F for some p (for all p). Note well that the elements of S p are continuous p-supermedian functions, but not necessarily p-excessive.

DEFINITION. If (Vp) satisfies (80.1) and if R separates points, we say that (Vp) is a Ray resolvent.

81.

We will say that a submarkovian semigroup (Pt) on F is a Feller semigroup if

(81.1) For all t , p t maps C to C ( p t is a Feller Kernel in the sense of IX .8 ) .

(81.2) For f C , we have lim t 0 p t f = f ( pointwise convergence ) .

Recall that F is compact. On an LCC space, Feller semigroups are defined by replacing e by e 0, the space of continuous functions tending to zero at infinity. See XIII.16.

It is then clear that the resolvent U p = e -ps p s ds is a Ray resolvent: for, if × ≠ y and f is a continuous function such that f(x) ≠ f(y), we have Upf(x) ≠ Upf(y) for large p, because pUpf → f (p → ∞). The notion of a Ray resolvent was created as a more flexible version of the notion of a Feller semigroup but in fact, thanks to the compactification methods, a tremendous gain in generality was achieved.

In the situation of Feller semigroups, we have limp pUpf = f for f ∈ e . Conversely, consider a Ray resolvent (Vp) with this property; for f G q , the functions pVp+qf increase to f and Dini's lemma implies that convergence is in fact uniform. In other words, not only R , but the image of the resolvent is dense in e . By the Hille-Yosida theorem, which we will see in chapter XIII, (Vp) is thus the resolvent of a strongly continuous submarkovian semigroup on e , hence of a Feller semigroup.

82.

Thus the essential difference between Feller semigroups and Ray resolvents comes from the existence of points × where it is not true that, for every continuous function f on F, limp pVpf(x) = f(x). These pointsare called branch points. We will denote by B the set of branch points, by N its complement (N is the initial of normal: the word normal Markov process was sometimes used by the Russian school to denote a property close to non-branching).

This may be expressed differently: every function f G q admits a q-excessive regularisation f ˆ = lim p p V p + q f , which depends only on f, not on q. This map f f ˆ ( x ) is a linear form on R , which extends by continuity, to e as a measure (subprobability) which we will denote provisionally ε ˆ X . (Later this will be the measure εxP0, Pt denoting the Ray semigroup.) We have for every continuous function f

(82.1) ε ˆ X ( f ) = f ˆ ( x ) = lim p p V p f ( x ) .

Moreover

(82.2) x ε N ε ˆ X = ε X .

For every f ε S p , we have f ˆ f . In the language of chapter X, no. 28, this means that ε ˆ X is a sweeping of εx relative to the cone S p . The language of the theory of sweeping may be put to use, N being interpreted as a Choquet boundary.

THEOREM. Suppose that 1 ˆ = 1 . Then × belongs to N if and only if εx admits, relative to S p , no sweeping of mass 1 distinct from εx.

Proof: Suppose that x ε N ( ε ˆ X ε X ) , and let μ be a sweeping of εx, of mass 1. Let g be a continuous function lying between 0 and 1; we have

l = p V p g + p V p ( 1 g ) + 1 p V p l = u + v + w ,

where the three terms belong to S p - for the third write

q V p + q ( 1 p V p l ) = q V p + q 1 p q V p + q v p l = q V p + q p ( v p 1 V p + q l ) = ( p + q ) V p + q l p V p l 1 p V p l .

More generally, if h is 0-supermedian, (I-pVp)h is p-supermedian (the proof is indicated above in 80).

As μ(u) ≤ u(x), and similarly for v, w, and as the sum of these three inequalities is the equality μ(1) = 1, we have in fact μ(u) = u(x). In other words, μVpg = εxVpg; multiplying by p and letting p tend to ∞, we obtain μ ( g ˆ ) = g ˆ ( x ) for every continuous function g.

Take g ε G p and use the hypothesis × ∈ N: we have g ( x ) = g ˆ ( x ) = μ ( g ˆ ) , and g ˆ g , so g(x) ≤ μ(g). As μ(g) ≤ g(x) by hypothesis, μ and εx are equalon S p , hence on R , and finally on C by density.

The converse assertion is clear: the hypothesis 1 ˆ = 1 implies that ε ˆ X is a sweeping of mass 1 of εx, hence uniqueness implies that ε ˆ X = ε X , in other words that × belongs to N.

The result means that N is the Choquet boundary of F, relative to the cone S p . It follows for example that N is a G δ of F. This can be seen directly in a very simple way (and without supposing that 1 ˆ = 1 ): the set N is the intersection of the sets { f ˆ = f } , f running over a dense sequence in S p ; it suffices thus to show that each one of these is a G δ . But this follows from the inequality f ˆ f , and from the fact that f ˆ = sup q q V p + q f is l.s.c.

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Homeomorphisms and Diffeomorphisms of the Circle

A. Zumpano , A. Sarmiento , in Encyclopedia of Mathematical Physics, 2006

Further Results

In this section we shall state some additional results about homeomorphisms of the circle in the area of Fourier analysis.

The first result is a theorem of Pál (1914) and Bohr (1935): let f : T R be a real continuous function; then, there exists a homeomorphism of the circle h such that f h U ( T ) . The best proof of this theorem is due to Salem (1945). In 1978, Kahane and Katznelson showed that the result is still valid for f : T C continuous.

A similar question was posed by Lusin: given a continuous function f : T R , is there a homeomorphism of the circle h such that f h A ( T ) ? The problem remained open until 1981, when Olevskii, Kahane, and Katznelson answered negatively the question: there exists a real (or complex) continuous function f on the circle, such that, for all homeomorphism of the circle h, f h A ( T ) .

It was proved by the author that there are C homeomorphisms of the circle, not necessarily of finite type, that transport A ( T ) into U ( T ) . It is a very technical work, published in 1998, and it gives a necessary and sufficient condition for a homeomorphism of the circle with a flat point to transport A ( T ) into U ( T ) .

Finally, the Denjoy theorem (Theorem 9) is rather close to being optimal. The example constructed here can be improved by obtaining a circle diffeomorphism whose first derivatives have Hölder exponent arbitrarily close to 1 (see Katok and Hasselblatt (1995)). Recent work has dealt with the existence of a differentiable conjugacy between a diffeomorphism f with irrational rotation number λ and R λ . Arnol, Moser, and Herman have obtained results (see Melo and Strien (1993) for a discussion of this results and references).

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Convex Functions, Partial Orderings, and Statistical Applications

In Mathematics in Science and Engineering, 1992

5.59 Remark

The following result similar to (5.48) is also valid (Delange, 1947): Suppose that f(x) and g(x) are real-valued and continuous functions on [a, b]. Let f″ be continuous on [a, b] with f″ ≤ 0 and let g′ be continuous on the same interval. Furthermore, let g(a) = f(a), g(b) = f(b), and g(x) ≥ f(x) for a < x < b. If h(t) is a real-valued function having continuous derivatives h′ and h″ with h″(t) > 0 on [a, b], then

a b h ( g ( x ) ) d x a b h ( f ( x ) ) d x ,

and equality holds iff f(x) is identical to g(x) on [a, b].

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Preliminaries

Feng-Yu Wang , in Functional Inequalities, Markov Semigroups and Spectral Theory, 2005

Remark 0.2.1

Assume that μ is finite andPt -supermedian. LetC(E) (resp.Cb (E)) be the set of all continuous (resp. bounded continuous) real functions onE. Then for anyfCb (E), the right-continuity of the process impliesPtffand hence P t f f L 2 ( μ ) 0 ast→ 0. If moreover σ(C(E)) = ℱ, i.e. the Borel σ-field is induced by the class of continuous functions, thenCb (E) is dense inL 2(μ) and hence {Pt } t≥0is a sub-MarkovC 0-contraction semigroup onL 2(μ). For general μ, {Pt } t≥0is strongly continuous if

E : = { f C b ( E ) L 2 ( μ ) : { P t f } t [ 0 , 1 ) is uniformly integrable in L 2 ( μ ) }

is dense inL 2(μ).

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Differential Equations, Ordinary

Anthony N. Michel , in Encyclopedia of Physical Science and Technology (Third Edition), 2003

II.C n th-Order Ordinary Differential Equations

Thus far we have concerned ourselves with systems of first-order ordinary differential equations. It is also possible to characterize initial value problems by means of nth-order ordinary differential equations. To this end, we let h be a real function which is defined and continuous on a domain D of the real (t, y 1,   …, y n ) space. Then

(E n ) y n = h t , y 1 , , y n 1

is an nth-order ordinary differential equation. A solution of (E n ) is a real function ϕ which is defined on a t interval J  =   (a, b) which has n continuous derivatives on J and satisfies (t, ϕ(t),   ⋯,ϕ(n−1)(t))   D for all t  J and

ϕ ( n ) ( t ) = h ( t , ϕ ( t ) , , ϕ ( n 1 ) ( t ) )

for all t  J

Now for a given (τ ξ1,   ⋯,ξ n )   D, the initial value problem for (E n ) is

(I n ) y ( n ) = h ( t , y , y ( 1 ) , , y ( n 1 ) ) y ( τ ) = ξ 1 , , y ( n 1 ) ( τ ) = ξ n

A function ϕ is a solution of (I n ) if ϕ is a solution of Eq. (E n ) on some interval containing τ and if ϕ(τ)   =   ξ1,   …,ϕ(n−1)(τ)   =   ξ n .

As in the case of systems of first-order equations, we single out several special cases.

1.

Consider equations of the form

(1) y n + a n 1 t y n 1 + + a 1 t y 1 + a 0 t y = g t

where a n−1(t),   …,a 0(t ) are real continuous functions defined on the interval J. We refer to Eq. (1) as a linear homogeneous ordinary differential equation of order n.
2.

If in (1) we let g(t)     0, then

(2) y n + a n 1 t y n 1 + + a 1 t y 1 + a 0 t y = 0

We call Eq. (2) a linear homogeneous ordinary differential equation of order n.
3.

If in (2) we have a i (t)   a i , i  =   0,1,   …,n−1, then

(3) y n + a n 1 y n 1 + + a 1 y 1 + a 0 y = 0

and we call Eq. (3) a linear, autonomous, homogeneous ordinary differential equation of order n.

We can, of course, also define periodic and linear periodic ordinary differential equations of order n in the obvious way.

It turns out that the theory of nth-order ordinary differential equations reduces to the theory of a system of n first-order ordinary differential equations. To this end, we let y  = x 1, y (1)  = x 2,   …, y (n−1)  = x n in Eq. (I n ). Then we have the system of first-order ordinary differential equations:

(4) x 1 = x 3 x 2 = x 3 x n = h ( t , x 1 , , x n )

which is defined for all (t, x 1,   …,x n )   D. Assume that the vector ϕ   =   1,   …,ϕ n )T is a solution of (4) on an interval J. Since ϕ2  =   ϕ1′, ϕ3  =   ϕ′2,   …,ϕ n   =   ϕ1 (n−1), and since

h ( t , ϕ 1 ( t ) , , ϕ n ( t ) ) = h ( t , ϕ 1 ( t ) , , ϕ n ( n 1 ) ( t ) ) = ϕ n ( n ) ( t )

it follows that the first component ϕ1 of the vector ϕ is a solution of Eq. (E n ) on the interval J. Conversely, if ϕ1 is a solution of (E n ) on J, then the vector (ϕ, ϕ(1),   …,ϕ(n−1))T is clearly a solution of Eq. (4). Moreover, if ϕ1(τ)   =   ξ1,   …, ϕ1 (n−1)(τ)   =   ξ n , then the vector ϕ satisfies ϕ(τ)   =   ξ   =   1,   …,ξ n )T. The converse is also true.

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Elements of Stability Theory

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

20.1.2 Positive definite functions

First, let us introduce the following definitions which will be intensively used hereafter.

Definition 20.3

A real function V = V (t, x), specified in the domain x h ( x n , h > 0 ) for all t t 0 , is called positive-definite if there exists a real continuous function W (x) defined for x 0 h such that

1.

(20.3) W ( 0 ) = 0

2.

for x > 0

(20.4) W ( x ) > 0

3.

for all t t 0

(20.5) V ( t , x ) W ( x )

If the properties (2)–(3) are replaced by W (x) < 0 and V (t, x) ≤ W (x), then the function V (t, x) will be negative-definite.

Example 20.1

is precisely such a function. Indeed,

V ( t , x ) = x 1 2 + x 2 2 + x 1 x 2 sin t x 1 2 + x 2 2 | x 1 | | x 2 | = ( | x 1 | | x 2 | ) 2 + | x 1 | | x 2 | | x 1 | | x 2 | : = W ( x )

So, all conditions of Definition 20.3 are fulfilled for the function W (x).

Definition 20.4

Denote by x ¯ ( t , x 0 , t 0 ) the dynamic motion (trajectory) which satisfies (20.1) when x (t 0) = x 0. Then, if there exists the time derivative of the function V ¯ ( t ) : = V ( t , x ¯ ( t , x 0 , t 0 ) ) , then the function V (t,x) is said to be differentiable along the integral curves (or the path) x ¯ ( t , x 0 , t 0 ) of the system (20.1).

Claim 20.1

The full time derivative of the differentiable function V ¯ (t) is calculated as follows

(20.6) d d t V ¯ ( t ) = d d t V ( t , x ¯ ( t , x 0 , t 0 ) ) = t V ( t , x ¯ ( t , x 0 , t 0 ) ) + i = 1 n V x i ( t , x ¯ ( t , x 0 , t 0 ) ) f i ( t , x ¯ ( t , x 0 , t 0 ) )

In short (20.6) is written as

(20.7) d d t V ( t , x ¯ ) = t V ( t , x ¯ ) + i = 1 n V x i ( t , x ¯ ) f i ( t , x ¯ )

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