NONLOCAL LONGITUDINAL VIBRATION IN A NANOROD, A SYSTEM THEORETIC ANALYSIS

. Analysis of longitudinal vibration in a nanorod is an important subject in science and engineering due to its vast application in nanotechnology. This paper introduces a port-Hamiltonian formulation for the longitudinal vibrations in a nanorod, which shows that this model is essentially hyperbolic. Furthermore, it investigates the spectral properties of the associated system operator. Standard distributed control and feedback are shown not to be controllable nor stabilizing.


Introduction
The physics of micro-and nano-scale are fundamentally different from macro-scale.For example, carbon nanotubes have hollow structure [31], low density defect [25], high electrical and thermal conductivity [20], ZnO nanowires have wide band gap of 3.37 eV and an exciton binding energy of 60 meV [22] and boron nitride (BN) nanotubes are light in weight, stable at high temperatures, resistant to oxidation, and have outstanding thermal and electrical conductivity [16].Nitrogen-doped carbon nanotube sponge can load capacity 100 times larger than its weight [24].These distinguished properties of nanomaterials have led to its practical usages in NEMS.
According to the number of dimensions less than 100 nm, nanostructures can be classified into twodimensional (2-D, nanofilm), 1-D (nanorod, nanotube or nanowire), and 0-D (nanoparticle) structures.In particular, 1-D nanostructures have attracted much attention due to the confinement of the other two dimensions perpendicular to the longitudinal direction.Due to the combination of quantum confinement in the nanoscaled dimensions and the bulk properties in another dimension, a host of interesting properties and applications can be expected based on a wide variety of 1-D nanostructures.
Therefore, dynamical analysis and control of nanorods are essential tools for industrial applications.There are many interdisciplinary research works related to vibrations of nanoroads.Hadi et al. investigate axial vibration analysis of nonlocal Rayleigh nanorod [4].Nonlocal longitudinal vibration of viscoelastic coupled double-nanorod systems is studied by Karlicic et al. [13].Eren and Aydogdu introduced a PDE model for describing nonlinear free vibration of nanotube with finite strain [7].Akbas studied the vibration dynamic of cracked nanotube under an axial force [3].Numanoglu et al. studied vibration of nanorod under different boundary conditions using Eringen's nonlocal theory [19].Wang and Wand analyzed the effect of surface energy on dynamics of vibrating cracked nanorod [30].
The dynamics of many physical systems can be suitably modeled using the port-Hamiltonian (pH) formulation.This formulation gives valuable information on the energy function, the interconnection structure and the passivity and dissipation of the system [12,28].The ports model the connection of the system to its environment, and this makes these models very suitable for control design.We refer to the work of e.g.[8,27,29] for applications on macro-scale models.For more details we refer to [28].
Investigating the dynamics of interconnected nanorods is essential due to nanotechnology rapid development and its wide applications in different areas such as cable-payload system [14] (space elevator [21], drug delivery [24]), nanomedicine [23], NEMS circuits [26], energy harvesting [19] optical [11] and chemical sensors [5].Since the interconnection of pH systems via the ports is again a pH system, pH-models will form a powerful tool for analyzing the dynamics and control of interconnected nanorods.
To the best of our knowledge, in spite of the large amount of research on vibration of nanorod and pH system, there is little research on pH modelling of vibration of nanorods.As for Hamiltonian systems, there can be one than one pH-models.In [10], we introduced two, but they are less suitable for system analysis and control design, therefor we introduce a new pH model.We show how its ports are related to the boundary conditions of the original PDE, and show that its eigenfunctions form a Riesz basis.This latter property implies that simple tests can be used to check system properties like controllability and stabilizability, but is also very desirable when designing and analysing controllers, see e.g.[6].Our controllability and stability analysis complements that of [9].
The paper is organized as follows.In the next section, A short review on nonlocal theory and governing equation are given.The port-Hamiltonian formulation, its semigroup, spectral properties of the port-Hamiltonian model, controllability and feedback stabilization are investigated in next sections.Conclusions and some future works are given in the last section.

Model formulation
In this section, we recall the reference [18] to obtain the mathematical modelling of vibration of nanorods.We consider a nanorod with length and cross-sectional area A. In our case, the cross-sectional area is constant along the x-coordinate, but in general it could have arbitrary shapes along this x coordinate.We assume the material of the nanorod is elastic and homogeneous.Also, we consider the free longitudinal vibration of nanorod in x-direction.An infinitesimal element of length dx is taken at a typical coordinate location x.Further, we take that a force N is the resultant of an axial stress σ xx acting internally on A, where σ xx is assumed to be uniform over the cross-section.The stress resultant N may vary along the length, and is also a function of time N = N (x, t).Using our assumptions, we find that (2.1) In addition, an axially distributed force F is assumed, having dimensions of force per unit length, which results from external sources, either internally or externally applied.
The equilibrium of forces in the x-direction is where dm = ρAdx is the mass of an infinitesimal element and w is the displacement in the x-direction.Substituting dm = ρAdx and simplifying (2.2) gives Next we must model the stress-strain relation.Based on nonlocal Eringen's theory [1,2], it is assumed that the stress at a point is related to the strain at all other points in the domain.The nonlocal constitutive equations for an elastic medium are as follows where E is the elastic modulus and µ = e 0 a 2 is the nonlocal parameter (length scales).By substituting equation (2.1) into equation (2.4), the stress resultant for the nonlocal theory is obtained as The equation of motion can be expressed in terms of the displacement w for nonlocal constitutive relation.By differentiating both sides of equation (2.5) with respect to x and using equation (2.3), we get the following equation of motion We consider F = a 2 w as the distributional axial force.For simplicity, without loss of generality, we assume that A = 1.Therefore, Next, we analyze equation (2.7) for the development of a port-Hamiltonian model.

Port-Hamiltonian formulation
To find a port-Hamiltonian formulation of equation (2.7), we proceed formally first.To simplify notation we write w tt for ∂ 2 w ∂t 2 and w xx for ∂ 2 w ∂x 2 .With this notation and the formal operator J = 1 − µ d 2 dx 2 , we can write the PDE (2.7) as This equation can equivalently be written as To design a port-Hamiltonian system, we begin by finding an operator Q such that where Q is the (formal) adjoint of Q.Since the (formal) adjoint of d dx equal − d dx , we have that if we choose Q as then and so Using this, our PDE (3.2) becomes Let us introduce the state then The last equation can be rewritten as with We see that this format is very similar to the general port-Hamiltonian format as given in [15].The difference is that the first derivative in that paper is replaced by Q and Q .It is easy to see that P 1,Q and P 0 are (formally) skew-adjoint, and hence z T Hzdx is a conserved quantity.
To prove that this is a conserved quantity, we must, like in [15], define the state space and the proper domain of the operators.

Contraction semigroup
In this section we show that by imposing proper boundary conditions, the abstract differential equation given in (3.9)-(3.11)generates a (contraction) semigroup on a state space.As the state space X, we choose the Hilbert space X = L 2 ((0, ); R 3 ) with inner product with H ginven by (3.11).By Lemma 7.2.3 of [12] we know that A H := P 1,Q H + P 0 H generates a contraction semigroup on X if and only if A := P 1,Q + P 0 generates a contraction semigroup on L 2 ((0, ); R 3 ) with the standard inner product.Hence without loss of generality, we may assume that H is the identity.Before we formulate and prove the contraction property, we need some more notation.By R and R we denote the (formal) operators Thus the operator R is the formal adjoint of R, and We can write Again formally we see that R(f ) = g if and only if f = R −1 (g).The last expression is nothing else than the ordinary differential equation The solution of this equation is given by The above expression defines a mapping, when we fix c.This can be done by imposing boundary conditions on g.By R c (f ), we denote the right-hand side of the above equation, i.e., Similarly, we define It is not hard to see that for f , f d smooth With the mappings R c and R c d we define the A as, see (4.4), The following result is almost immediate.
) is a bounded linear operator.Hence it is the infinitesimal generator of a group on L 2 ((0, ); R 3 ) and thus on X.
Now we consider the question when this semigroup on X will be a contraction.Therefor we assume that c and c d are such that where W is a full-rank 2 × 4 matrix and f ∂ , e ∂ ∈ R 2 are defined as ) and (4.8), respectively.In order to link the constants c and c d to the boundary conditions, we introduce the matrices Proof.We begin by showing that A is well-defined.Since R c and R c d are defined using the constants c and c d , respectively, and A is defined via boundary values of g 2 and g d 3 , respectively, this is not immediately clear.Using the definitions of g 2 and g d 3 , we see that We see that this is uniquely solvable for all f 2 and f 3 if and only if implies that ( c c d ) = 0.By our assumption, this holds, and so (4.14) uniquely determines c and c d .Furthermore, we see that c and c d are linear bounded mappings of f 2 and f 3 and so by Lemma 4.1, it follows that A is a bounded linear operator from X to X. Hence A generates a contraction semigroup if and only if for all f ∈ X the following inequality holds: and since the first operator is skew-adjoint, we find that To study the sign of this expression, we use that, see (4.13), With this notation we have that By construction, we have that Using this we find Using this in (4.16) have the following Applying the definition of f ∂ and e ∂ the last expression equals √ µf T ∂ e ∂ .By Theorem 7.2.4 of [12] or Theorem 4.1 of [15] we have that f T ∂ e ∂ ≤ 0 for all f ∂ e ∂ ∈ ker W if and only if W ΣW T ≥ 0. This proves the assertion for the contraction semigroup.
Since A is bounded, we have that it generates a unitary group if and only if we have equality in (4.15).Or equivalently, when f T ∂ e ∂ = 0 for all f ∂ e ∂ ∈ ker W . Again by using Theorem 4.1 of [15] we see that this holds if and only if W ΣW T = 0.
In the above theorem we see that W has to satisfy two conditions.However, in a special case we have that whenever the second one is satisfied, so is the first.
E2 is invertible if and only if where W = (W 1 W 2 ) and Proof.By Lemma 7.3.1 of [12] we see that any W satisfying W ΣW T ≥ 0 can be written as where S is invertible, and V V * ≤ I. Furthermore, W ΣW T > 0 if and only if V V * < I.
Assume now that W E1 E2 is not invertible.Then there exists a non-zero vector v ∈ R 2 such that A direct calculation gives that which is an unitary matrix.We see that this unitary matrix equals U * 0 .Defining ṽ := (E 2 − E 1 )v we write equation (4.20) as Hence if V V * < I, then there does not exists a non-zero v such that (4.21) holds, which proves the assertion for and since U 0 is unitary, we see that there exists a non-zero ṽ such that (4.21) holds if and only if (4.19) holds.

More on the boundary conditions
In (4.11), Lemma 4.1, and Theorem 4.2, we see that we have defined A via c and c d .However, it is not a-priori clear how c and c d are related to "normal" boundary conditions like (5.2) Using (4.9) and (4.10) we find or equivalently This can equivalently be written as So we have an expression for the left-hand side of (5.2).Letting (I + √ µ d dx ) operate on (5.2) and using (4.10) we find Using (5.6) we find Since sinh(y) + cosh(y) = e y and γ = 1 √ µ , we can simplify this as Thus Now we repeat the procedure and let (I − √ µ d dx ) operate on (5.2) and using (4.9) we find Using (5.6) we find Since cosh(y) − sinh(y) = e −y , we can simplify this as Thus (5.8) From (5.7) and (5.8) we see that the pairs (c, c d ) and (c 1 , c2 ) are one-to-one correspondence with each other.Since f 2 = w t , we find from (5.1) that and similarly, Thus with (5.5) and using (5.6) we derive from The two equations for the pair (c 1 , c 2 ) can be written as . (5.9) If the matrix on the left is invertible, then we can uniquely derived c 1 and c 2 and thus c and c d .This will happen in most cases, for instance, when (α 1 , α 2 ) = (β 1 , β 2 ) = (1, 0).We summarise the above in a theorem Theorem 5.1.Consider the PDE (2.7) with boundary conditions (5.1) and let From the above we see that the condition for the one-to-one relation between the PDE and the abstract formulation depends on many parameters.Thus if it does not hold, then for a slightly different value of µ and/or it will hold.However, when α 1 = γα 2 and β 1 = γβ 2 then there is no value of for which (5.10) will hold.

Spectral properties of A
In this section we study the spectrum of the operator associated to the PDE (2.7).That is we study the (bounded) operator, see (3.8) and (4.11) with boundary conditions given by (5.1).To uniquely link these boundary conditions to the c and c d in (4.11) we assume that det = 0, see Theorem 5.1.We begin by determining the eigenfunctions and eigenvalues of A H . Using (4.11) we see that which concludes Thus we find the following equation of f 2 and since (6.3) implies that f 2 is smooth, we obtain by using (4.9) and (4.10) that Using the expression for J, this becomes the differential equation The boundary conditions (5.1) give that, see also Section 5, Equation (6.6) and (6.7) form a special case of the Sturm-Liouville eigenvalue problem, and so we have that there exists infinite infinitely many solutions, φ n , n ∈ N and when normalised they form an orthonormal basis of L 2 (0, ).However since we are studying the eigenfunctions of A H , we cannot directly exclude the case that f 2 = 0. From (6.3) we see that f 2 = 0 implies that λ = 0 or f 1 = 0, f 3 = 0.The latter we can ignore, since we are looking for eigenfunctions of A H .When λ = 0, then (6.3) implies that Since f 2 = 0, we find from Theorem 5.1 that c d = 0, and so the above equation becomes Hence the eigenspace corresponding to λ = 0 is infinite-dimensional.We summarise and extend the above results in a theorem.For that we need the following notation.Let {φ n , n ∈ N} be the orthonormal basis of eigenfunctions corresponding to the eigenvalues problem We denote the corresponding eigenvalues by κ n , n ∈ N.
Theorem 6.1.Consider the operator A H of equation (6.1) with boundary conditions given by (5.1).Assume that det as defined in Theorem 5.1 is unequal to zero, κ n = µ −1 , and a 2 E+µa 2 = κ n for all n, where κ n are the eigenvalues/solutions of (6.9).
The eigenvalues are given by λ 0 = 0, and For n = 0 the corresponding eigenvectors are where φ −n := φ n , n ∈ N and λ n is given in (6.10).
For n = 0 the eigenspace is given by The only accumulation points of λ n are ± E+µa 2 µρ i.The eigenfunctions f n , n ∈ Z \ {0} form a Riesz basis of the linear subspace If {q n , n ∈ N} are chosen such that they form a Riesz basis of V 0 , then the union of {f n , n ∈ Z \ {0}} and {q n , n ∈ N} form a Riesz basis of X or L 2 ((0, ); R 3 ).
Proof.a. Since (6.6) and (6.7) are the same as (6.9), we see that that λ is an eigenvalue of A H only if for some n ∈ N.This gives as solutions (6.10).In that case f 2 can be chosen as φ n .By (6.4) we have that f 1,n = 1 ρλn f 2,n , and Note that by our assumption a 2 E+µa 2 = κ n , λ n = 0 for n = 0.By the assumption κ n = µ −1 equation (6.12) is solvable for λ.
b.The eigenspace expression for the eigenvalue zero follows directly from (6.8).
c. Since κ n lie on the real axis and |κ n | → ∞, we have that ± E+µa 2 µρ i are the only accumulation points of the λ n 's.
d.We first show that V 1 is a closed subspace of L 2 ((0, ); R 3 ).It is easy to see that this holds if and only if (−I + R c ) has closed range.Since R c is a compact operator, this holds.
On V 1 we have the following equivalent norm Using the fact that a set remains a Riesz basis when the norm is replaced by and equivalent one, we can check the Riesz basis property of f n , n ∈ Z \ {0} with respect to this norm.Note that where we used that φ −n = φ n , λ −n = −λ n , and that φ n , n ∈ N is an orthonormal basis of L 2 (0, ).Since λ n is bounded and bounded away from zero, we have that there are positive constants m, M independent of a n and N such that

Using that
So if we can show that f n , n ∈ Z \ {0} span V 1 , then we have shown that it is a Riesz basis for this subspace.

Let
This concludes the assertion that e. Since V 0 is the eigenspace of A H corresponding to eigenvalue zero, and since zero is not in the spectrum of A H restricted to V 1 , the sets must have trivial intersection. Let where c = c(f 2,1 ).Hence f 2 = f 2,1 and f 1,1 , f 3,0 should satisfy with R 0 := R c=0 , R 0 := R c d =0 .Since c depends on f 2,1 we may consider that given, and we obtain the following set of equation for f 11 , f 3,0 ∈ L 2 (0, )

.16)
Using the second, the first equation gives which is equivalent to Since R 0 and R 0 are compact operators, we have that this equation is uniquely solvable for any right hand side if and only if The equation implies that f 1,1 (0) = 0 and it can be rewritten as Thus with (4.10) x−τ √ µ f 1,1 (τ )dτ = 0, which gives df1,1 dx (0) = 0. Differentiating this once more leads to a second order differential equation in f 1,1 .From the zero initial conditions, we conclude that f 1,1 is identically zero, and thus (6.18) holds.Hence (6.17) gives a unique f 1,1 , and substituting this in (6.16) gives f 3,0 .Concluding we see that any f1 f2 f3 ∈ L 2 ((0, ); R 3 ) lies in V 0 + V 1 as well.
f. Since V 0 + V 1 = L 2 ((0, ); R 3 ), and since {f n , n ∈ Z \ {0}} is a Riesz basis of V 1 , and {q n , n ∈ N} is a Riesz basis of V 0 , it follows almost directly the their union is a Riesz basis of L 2 ((0, ); R 3 ).Since the norm of X and that of L 2 ((0, ); R 3 ) are equivalent, this union is also a Riesz basis of X.

Controllability and stability
In this section, the control system is considered in the following form Using the notions and results from the previous section, we see that B maps into V 1 .Since V 1 is spanned by eigenfunctions of A H , we see that any solution of ż(t) = A H z(t) + Bu(t), z 0 = 0 will stay in V 1 for all t ≥ 0. Hence the system (7.1) is not controllable.
Next we take as input the velocity feedback, i.e., u(t) = α ∂w ∂t .With this choice, the operator A H becomes It is easy to see that A f b V 0 = 0, and so zero remains an eigenvalue of the closed loop system operator, and thus the system (7.1) has not become asymptotically stable.

Conclusion and future works
This paper investigates longitudinal vibration of a nanorod by proposing a new port-Hamiltonian model.This gives the energy (Hamiltonian) function, which is usually a good Lyapunov function, explicitly appears in the dynamics of the system [17].This model always generates a strongly continuous group, but for special boundary conditions it can generate a contraction semigroup.Moreover, we showed that the eigenfunctions from a Riesz basis, but that that the eigenspace associated to the eigenvalue zero was infinite-dimensional, and this prevented controllability and stabilizability.For the future work, we plan to extend this work for viscoelastic vibrating nanorod [13], and to study the system theoretic properties of the port-Hamiltonian formulations given in [10].

Lemma 4 . 1 .
If c and c d are bounded linear mappings from L 2 ((0, ); R 3 ) to R, then the operator A of(4.11)