Trace Properties of Certain Damped Linear Elastic Systems

We study the spectrum of a damped linear elastic system with discrete eigenvalues, showing the relationship between the sum of the real parts of the eigenvalues of the (generally unbounded) generator and the trace of the damping operator, assuming the latter to be a trace type operator. Some relationships between the sequence of eigenvectors and a corresponding orthonormal sequence, constructed by means of a variant of the Gram-Schmidt method, are also explored. A simple hybrid system is presented as an example of application.


Form of the Abstract System
• Z, Y separable real, resp.complex, Hilbert spaces; • A positive self-adjoint, generally unbounded; • B non-negative, self-adjoint, typically compact (trace class); The "energy", E (U (t), V (t)) ≡ 1 2 Then (cf., e.g., Dunford Schwartz) the operator A, generates a strongly continuous semigroup on H.All eigenvalues λ, λ = µ ± i ν of A have µ ≤ 0. Typical eigenvalue plots in complex plane: For Y = E n A is represented by a 2n × 2n matrix and we have Our objective is to show that this relationship continues to hold for many unbounded self-adjoint operators A with trace class B important in the study of distributed systems.
Assumption σA consists of discrete conjugate eigenvalues to conjugate eigenvectors X k , X k forming a complete and strongly independent set in H.
Theorem Let the eigenvalues of A be conditionally summed as (In general this sum could be infinite.)If B is a trace class operator the sequence {µ k } is summable and Remark If A 1/2 is also a trace class operator, this result is trivial.Otherwise the proof is not immediate because we do not have a definition of Tr A nor a result on the relationship between such a trace, were it to be defined, and the sum of the eigenvectors.That requires the following auxiliary lemma.
Lemma There is a complete orthonormal set of vectors

Proof of the Theorem (assuming the Lemma)
With the Y k , Y k as above, since Tr B = Tr B, the last result following from a standard theorem on trace class operators.Since A 0 is real and antihermitian so that the first term in the last line is zero.If B, hence B, is a trace class operator Tr B is finite and thus {µ k } is summable.
Remark If B = γ I for some positive γ, a common model for viscous damping, then since Tr I = ∞ when Y is infinite dimensional, the foregoing discussion implies µ k is infinite.
We begin with a lemma preliminary to the one stated above.
Operator A with Basis of Eigenvectors Let A be an operator on H with eigenvalues λ k , k = 1, 2, 3, ... and corresponding eigenvectors X k , k = 1, 2, 3, ..., assumed complete and strongly independent in H. Then we have (cf.Dunford Schwartz): Theorem Given the above, there is an orthonormal basis Y k , k = 1, 2, 3, ..., for H such that, for all k, Proof is essentially the Gram -Schmidt process; we show only the first step.For Y The general inductive step proceeds in essentially the same way, obtaining orthonormality and A Y k , Y k = λ k at each step.
Why preliminary?Gram -Schmidt is not independent of the order in which λ k , λ k , X k , X k are introduced; as a result one cannot conclude that the vectors Y k , Ŷk are conjugates.
To simplify the presentation of the result needed, in the sequel we replace the inner product notation Y, Ŷ H with Ŷ * Y .
Theorem Suppose H includes a complete and strongly independent sequence of paired complex conjugate vectors X k , X k , k = 1, 2, 3, ... .From these we can construct, via a modified Gram-Schmidt process, a conjugated orthonormal sequence

Outline of Proof
Let us recall, for a finite, linearly independent set X k , k = 1, 2, ..., K, one can obtain an orthonormal set with the same span via the sequential Gram -Schmidt process, which involves an ordering of the X k as indicated, or we can do it all at once in a manner independent of any ordering.One defines the positive K × K Gram matrix X with Then, with X 1/2 the positive square root of X, we have we immediately see that the columns Y k of Y = Ξ X −1/2 are the desired orthonormal vectors.
The main idea of the proof is to combine sequential Gram -Schmidt with this "simultaneous" method carried out two steps at a time with each successive pair X k , X k .For this approach the sequential formula becomes and conclude that Ỹk , Ỹk , hence also their normalizations, for which we will continue to use the same symbols, are orthogonal to the Y j , Y j , j = 1, 2, ... k − 1.A further elementary step produces Y k , Y k mutually orthonormal and we conclude that Y j , Y j , j = 1, 2, ... k form an orthonormal basis for the space spanned by X j , X j , j = 1, 2, ... k.Continue by induction.
A long and algebraically complicated argument then shows that with Y k , Y k so constructed we have, for each k, as noted earlier, completing the proof.

Flexible String with Boundary Mass
A flexible string with frictionally damped boundary mass can be modeled as follows: • y(x, t), 0 ≤ x ≤ L, string displacement; • y(L, t) ≡ η(t), displacement of a discrete mass at x = L; • mass subject to restoring force −K 2 η(L, t), friction −γ dη dt (L, t) (natural, or via linear feedback on the mass velocity).Thus Thus B has trace γ and we have an instance of A as described earlier: