On norm equivalence between the displacement and velocity vectors for free linear dynamical systems

Abstract: As the main new result, under certain hypotheses, for free vibration problems, the norm equivalence of the displacement vector y(t) and the velocity vector ẏ(t) is proven. The pertinent inequalities are applied to derive some two-sided bounds on y(t) and ẏ(t) that are known so far only for the state vector x(t) = [y(t), ẏ(t)]. Sufficient algebraic conditions are given such that norm equivalence between y(t) and ẏ(t) holds, respectively, does not hold, as the case may be. Numerical examples illustrate the results for vibration problems of n degrees of freedom with n ∈ {1, 2, 3, 4, 5} by computing the mentioned algebraic conditions and by plotting the graphs of y(t) and ẏ(t). Some notations and definitions of References Kohaupt (2008b, 2011) are necessary and are therefore recapitulated. The paper is of interest to Mathematicians and Engineers.

ABOUT THE AUTHOR Ludwig Kohaupt received the equivalent to the master's degree (Diplom-Mathematiker) in Mathematics in 1971 and the equivalent to the PhD (Dr.phil.nat.) in 1973 from the University of Frankfurt/Main. From 1974 until 1979, Kohaupt was a teacher in Mathematics and Physics at a Secondary School. During that time (from 1977 until 1979), he was also an auditor at the Technical University of Darmstadt in Engineering Subjects, such as Mechanics, and especially Dynamics. From 1979until 1990, he joined the Mercedes-Benz car company in Stuttgart as a computational engineer, where he worked in areas such as Dynamics (vibration of car models), Cam Design, Gearing, and Engine Design. Then, in 1990, Kohaupt combined his preceding experiences by taking over a professorship at the Beuth University of Technology Berlin (formerly known as TFH Berlin). He retired on 01 April 2014.

PUBLIC INTEREST STATEMENT
Under certain conditions, the norm equivalence of the displacement vector y(t) and the velocity vector ẏ(t) for free linear dynamical systems is derived. Hereby, a relation of the form c 0 ‖y(t)‖ ≤ ‖ẏ(t)‖ ≤ c 1 ‖y(t)‖, t ≥ t 1 , is understood with positive constants c 0 and c 1 . As a consequence, under norm equivalence, one has that lim t→0 y(t) = 0 is equivalent to lim t→0ẏ (t) = 0 and that boundedness of y(t) is equivalent to boundedness of ẏ(t).

Introduction
In free vibration problems with one degree of freedom and mild damping, the displacement as well as the velocity have zeros in any sufficiently large interval. But, with increasing dimension, it is likely that not all components of the displacement or of the velocity are zero simultaneously, in other words, it is increasingly likely with increasing dimension n that y(t) ≠ 0, t ≥ t 1 , and ẏ(t) ≠ 0, t ≥ t 1 , for sufficiently large t 1 . Then, the question arises as to whether an inequality of the form c 0 ‖y(t)‖ ≤ ‖ẏ(t)‖ ≤ c 1 ‖y(t)‖, t ≥ t 1 , can be proven for sufficiently large t 1 ; this property, if it is valid, will be called norm equivalence between y(t) and ẏ(t). Now, the contents of this paper will be outlined.
In Section 2, the state-space description ẋ = Ax, x(t 0 ) = x 0 , of the dynamical problem Mÿ + Bẏ + Ky = 0, y(t 0 ) = y 0 ,ẏ(t 0 ) =ẏ 0 is given where M, B, and K are the mass, damping, and stiffness matrices, as the case may be, and where y(t) is the displacement vector, y 0 is the initial displacement, and ẏ 0 is the initial velocity.
In Section 3, under certain hypotheses, the above-mentioned norm equivalence between y(t) and ẏ is proven for diagonalizable system matrix A, and in Section 4, the same is done for general system matrix A.
The pertinent norm inequalities between y(t) and ẏ are applied in Section 5 to improve Kohaupt (2011, Theorems 12 and 16). In both sections, also sufficient algebraic conditions are established that guarantee the validity, respectively, invalidity of norm equivalence of y(t) and ẏ. In Section 6, numerical examples illustrate the results for vibration problems of n degrees of freedom for n ∈ {1, 2, 3, 4, 5}. More precisely, in Section 6.1, the mentioned algebraic conditions are computed and the graphs of ‖y(t)‖ and ‖ẏ(t)‖ are plotted showing that, in the examples, for n ∈ {1, 2} there is no norm equivalence and that for n ∈ {3, 4, 5} there is norm equivalence; in Section 6.2, for n = 4, a case with x 0 [A] < [A] is illustrated by graphs, where x 0 [A] is the spectral abscissa of matrix A with respect to x 0 and [A] is the spectral abscissa of A; in Section 6.3, for n = 2, a model with nondiagonalizable matrix A is constructed and analyzed in detail. Section 7 contains computational aspects, and in Section 8, conclusions are drawn. The non-cited references [1], [2], [6], and [7] are given because they may be useful to the reader.
2. The state-space description of Mÿ + Bẏ + Ky = 0, y(t 0 ) = y 0 ,ẏ(t 0 ) =ẏ 0 Let M, B, K ∈ IR n×n and y 0 ,ẏ 0 ∈ IR n . Further, let M be regular. The matrices M, B, and K are the mass, damping, and stiffness matrices, as the case may be; y 0 is the initial displacement and ẏ 0 is the initial velocity. We study the initial value problem where y(t) is the sought displacement and z(t) =ẏ(t) is the associated velocity.

State-space description
Let and x is called state vector and A is called system matrix. Herewith, the above second-order initial value problem is equivalent to the first-order initial value problem of double size,

The spectral abscissa of
A with respect to the initial vector x 0 ∈ IR n Let u * k , k = 1, … , m = 2n be the eigenvectors of A * corresponding to the eigenvalues k , k = 1, … , m = 2n. Under (H1), (H2), and (HS), the solution x(t) of (1) has the form with uniquely determined coefficients c 1k , k = 1, … , m = 2n. Using the relations Kohaupt, 2008b, Section 3.1 for the last relation), then according to Kohaupt (2008a), the spectral abscissa of A with respect to the initial vector x 0 ∈ IR n is given by

Index sets
In the sequel, we need the following index sets: and

Appropriate representation of bdy(t) and ẏ(t)
Let with q k , r k ∈ C I m×m , q (r) k , r (r) k , q (i) k , r (i) k ∈ IR n , k = 1, … , m = 2n. Then, from (8), (9), Page 5 of 33 Kohaupt, Cogent Mathematics (2015) with k = 1, … , n as well as with k = 1, … , n After these preparations, for the quantities J 0 , g k (t), h k (t), we formulate the following conditions: There exists a t 1 ≥ t 0 such that For these conditions, there are sufficient algebraic conditions, as the case may be: In the examples of Section 6, also the case occurs that the above conditions are not fulfilled. For this, we formulate the following conditions: For every t 1 ≥ t 0 there exists a t ≥ t 1 such that For these conditions, there are sufficient algebraic conditions, as the case may be (see Kohaupt, 2011): Further, here and in the sequel, we denote by ‖ ⋅ ‖ any vector norm.

Norm equivalence inequalities for general matrix A
In this section, we prove the same statement for a general square matrix A as in Theorem 1 for a diagonalizable matrix A. This cannot be deduced in a similar way as for Theorem 1, that is, it cannot be done by Kohaupt (2011, Theorems 9 and 15) since they contain the factor e (t−t 0 ) on the righthand side. Neither can it be done by Kohaupt (2011, Theorems 12 and 16) since they contain the factor e (t−t 0 ) on the right-hand side and the factor e − (t−t 0 ) on the left-hand side. Nevertheless, the same equivalence inequalities as in Theorem 1 hold true in the general case. The proof, however, is much more involved.
Again, first we formulate some hypotheses and conditions. We mention that for the special hypothesis (HS � ) similar remarks hold as for (HS) in the case of diagonalizable matrices A.

Hypotheses and conditions for general square matrix
Let (H1 � ), (H2 � ), and (HS � ) be fulfilled and Ap (l) where the indices are chosen such that +l = l , l = 1, … , and p ( +l) k are the principal vectors of stage k corresponding to the eigenvalue l of A.
In the case of a general square matrix A, we also have to collect some definitions, respectively, notations and representations of x(t) from Kohaupt (2008bKohaupt ( , 2011.

Representation of the basis x (l,r)
k (t), x (l,i) k (t), k = 1, … , m l , l = 1, … , Under the hypotheses (H1 � ), (H2 � ), and (HS � ), from Kohaupt (2008b) we obtain the following real basis functions for the ODE ẋ = A x: is the decomposition of p (l) k into its real and imaginary part.

Index sets
For the sequel, we need the following index sets: and (15) x (l,r)

Appropriate representation of y(t) and ẏ(t)
Then, from (24), (25) with After these preparations, for the quantities J we formulate the following conditions: There exists a t 1 ≥ t 0 such that For these conditions, there are sufficient algebraic conditions, as the case may be (see Kohaupt, 2011). However, the following sufficient conditions are not so stringent in that only conditions on components of eigenvectors are used and not on the set of components of all principal vectors. The sufficient algebraic conditions read: The above conditions are not always fulfilled. For this, we formulate the following conditions: , are linearly independent.
For these conditions, there are sufficient algebraic conditions, as the case may be, (see Kohaupt, 2011). However, the following sufficient conditions are not so stringent in that we do not suppose on the algebraic multiplicity that m l 0 = 1. The sufficient algebraic conditions read: For the next lemma, we set: The definition of the spectral abscissa 0 = x 0 [A] of matrix A with respect to the initial vector x 0 can also be found in Kohaupt (2011).
Proof We prove only the first relation. The second one is proven in a similar way. One has This implies for sufficiently large t 1 ≥ t 0 since the fraction in the bracket tends to zero. Further, for sufficiently large t 1 ≥ t 0 , again since the fraction tends to zero. □ For the formulation of the next lemma, we introduce some abbreviations. So, we define where the quantities c (l,r) k , c (l,i) k , and c (l) 1k are contained in each of the quantities g (l)

This clearly implies
Further, define and as well as After these preparations, we are now in a position to state the following lemma.
Proof We prove only the first relation. The second one is proven in a similar way. One has

This delivers
We note that the terms containing the vectors q (l,r) 1 and q (l,r) 1 with c (l,r) give us the function u(t), and we mention that it has the factor p(t), both defined above. The rest of the sum is denoted by R(t); it can be estimated from above by polynomials of degree less than m l ′ . So, we obtain This entails, taking into account the definition of the function (t), for sufficiently large t 1 ≥ t 0 since the last fraction tends to zero as t tends to infinity. Similarly, for sufficiently large t 1 ≥ t 0 . □ The next lemma is also important for results in the sequel.
On the other hand, if additionally the sufficient algebraic condition (A ′ g,1 ), respectively, (A ′ h,1 ) is satisfied, then for every t 1 ≥ t 0 there exists a t ≥ t 1 such that as the case may be, meaning correspondingly, Proof We prove only the first relation. The second one is proven in a similar way. Assume that for all t 1 ≥ t 0 there exists a t ≥ t 1 such that (t) = 0. Then, u(t) = 0 so that Now, due to (A ′ g,1 ), the vectors q (l,r) 1 , q (l,i) 1 , l ∈ J � 0 are linearly independent. Therefore, . From this, we conclude that or, This delivers a contradiction since we have seen above that c (l) . □ Now, we state the following corollary.
(ii) If, instead, the sufficient algebraic conditions (A ′ g,1 ) and (A ′ h,1 ) are fulfilled, then, Proof This follows from Lemmas 2 to 4. □ The last corollary is the basis for the derivation of the following theorem that is the main theoretical result of this paper.

Proof
(i) Due to Corollary 5, we have for sufficiently large t 2 ≥ t 1 ≥ t 0 . Now, Y(t) and Z(t) are positive for t ≥ t 2 . Thus, due to the periodicity and continuity of Y(t) and Z(t) the extreme values exist and are positive. Therefore, so that Theorem 6 follows with c 0 = 1 4 Z min ∕Y max and c 1 = 4 Z max ∕Y min . (ii) Due to Corollary 5, we have Now, (t) and (t) are periodic and continuous as well as positive for sufficiently large t 1 ≥ t 0 . Thus, the extreme values 1 4 exist and are positive. Therefore, so that Theorem 6 follows with c 0 = 1 16 min ∕ max and c 1 = 16 max ∕ min . □
Proof The proof is similar to that of Theorem 7 and is therefore omitted. □

Numerical examples
In this section, we illustrate the obtained results by examples.
The associated initial value problem is given by where y = [y 1 , … , y n ] T and or, in the state-space description Remark We mention that, in all examples, condition (H3) resp. (H3 � ) is fulfilled, i.e., that all eigenvalues are different from zero. Therefore, the sufficient algebraic conditions (A g,1 ) and (A h,1 ) are equivalent (since then r (k) Kohaupt, 2011)). The same holds true for (A ′ g,1 ) and (A ′ h,1 ), for (A g,1 ) and (A h,1 ), and for (A ′ g,1 ) and (A ′ h,1 ). So, we need only the first sufficient algebraic condition with index g, in each case. The stepsize in all figures is Δt = 0.01.

Illustration of the sufficient algebraic conditions
In this subsection, we illustrate the sufficient algebraic conditions that guarantee the validity, respectively, invalidity of the equivalence inequalities, as the case may be.
Remark In the following Examples 1-5, we have to consider the quantities ũ j = (Ux 0 ) j = (x 0 , u * j ), j = 1, … , m = 2 n because they play a role in the definition of 0 = x 0 [A] (see Kohaupt, 2011). Due to the numbering j+n (A) = j (A) = j (A * ), j = 1, … , n, we have to study only the quantities ũ j for j = 1, … , n (and not for j = 1, … , m = 2 n), see also the definition of 0 on this.
Example 1: n = 1. We choose Here, ũ j = (Ux 0 ) j ≠ 0, j = 1 and Thus, Sufficient algebraic condition (A g,1 ): We have are linearly dependent, the equivalence inequalities between y(t) and ẏ(t) do not hold. This is consistent with the fact that |y(t)| and |ẏ(t)|have zeros (see Figures 2 and 3).
Remark Here, we have a nontrivial example of a case with x 0 Example 5: n = 5. This model was often used before by the author. We choose Here, ũ j = (Ux 0 ) j ≠ 0, j = 1, … , 5 and Re j (A) = −0.06080170524095.  We have Since q (5,r) 1 , q (5,i) 1 are linearly independent, the equivalence inequalities between y(t) and ẏ(t) hold. This is consistent with the fact that ‖y(t)‖ 2 and ‖ẏ(t)‖ 2 do not have zeros for t > 0 (see Figures 10 and 11).
We have  In what follows, we give the point of contact t s,u,2 between curve and upper bound as well as the optimal constants Y 1,2 and Z 1,2 computed by the differential calculus of norms.
For the upper bound y = Y 1,2 e [A](t−t 0 ) , we obtain The curve y = ‖y(t)‖ 2 and its upper bound can be seen in Figure 12.
For the upper bound y = Z 1,2 e [A](t−t 0 ) , we obtain The curve y = ‖z(t)‖ 2 and its upper bound can be seen in Figure 13.
For the upper bound y = Y 1,2 e 0 (t−t 0 ) , we obtain The curve y = ‖y(t)‖ 2 and its upper bound can be seen in Figure 14.
For the upper bound y = Z 1,2 e 0 (t−t 0 ) , we obtain The curve y = ‖z(t)‖ 2 and its upper bound can be seen in Figure 15.  Comparing the corresponding figures, it is evident that the spectral abscissa with respect to the initial vector x 0 , i.e. 0 = x 0 [A], has not only theoretical meaning, but sometimes also practical significance.

Illustration of a case with non-diagonalizable matrix A
In this subsection, we first construct an example with n = 2 degrees of freedom so that A ∈ IR 4×4 is not diagonalizable. The aim is then to apply Theorems 7 and 8, where we restrict ourselves to the upper bounds ‖y(t)‖ 2 ≤ 1,2 ‖ (t)‖ 2 and ‖z(t)‖ 2 ≤ 1,2 ‖ (t)‖ 2 .
(i) Construction of a non-diagonalizable matrix Remark Since 2 (t) = 1 (t), we could also use the vector (t) without component 2 (t). Then merely the constants in the upper bounds would change. (

v) Optimal upper bounds on y(t)and z(t) =ẏ(t)
The displacement y(t) and the velocity ẏ(t) can be computed from x(t) = [y T (t),ẏ T (t)] T . For comparison reasons, we have determined x(t) also by x(t) = e A (t−t 0 ) x 0 and obtained numerically identical results. The advantage of the representation of x(t) by the real basis functions is that we get more insight into its vibration behavior than without it.