ENERGY DECAY ESTIMATES FOR LIENARD S EQUATION WITH QUADRATIC VISCOUS FEEDBACK

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1 Electronic Journal of Differential Equations, Vol. 00(00, No. 70, pp ISSN: URL: or ftp ejde.math.swt.edu (login: ftp ENERGY DECAY ESTIMATES FOR LIENARD S EQUATION WITH QUADRATIC VISCOUS FEEDBACK ALEXANDER Y. KHAPALOV & PARTHASARATHI NAG Abstract. This article concerns the stabilization for a well-known Lienard s system of ordinary differential equations modelling oscillatory phenomena. It is known that such a system is asymptotically stable when a linear viscous (motion-activated damping with constant gain is engaged. However, in many applications it seems more realistic that the aforementioned gain is not constant and does depend on the deviation from equilibrium. In this article, we consider a (nonlinear gain, introduced in [], which is proportional to the square of such deviation and derive an explicit energy decay estimate for solutions of the corresponding damped Lienard s system. We also discuss the place of our result in the framework of stabilization of so-called critical bilinear systems. 1. Introduction Motivation and main results. This article concerns the stabilization of a single oscillatory motion (or, more generally, of an oscillatory phenomenon by means of suitable damping. Without loss of generality (that is, one might need to perform a routine change of variables first such motion is described by a two dimensional system of ordinary differential equations of Lienard s type like ẋ = ( x + ubx, (1.1 where x = (x 1, x and ubx models a damping device of structure described by matrix B with gain u = u(x 1, x. A system like (1.1 is widely used to model various kinds of periodic oscillatory phenomena, for example, in electrical or civil engineering (see, e.g., [4, 7] and the references therein. In particular, if we assume that (1.1 models a motion of a particle, then x 1 (t would describe the deviation (position of this particle with respect to equilibrium x 1 = 0 at time t, while x (t would describe its velocity. One of the most typical damping devices used in applications is so-called the viscous (or motion-activated damping which, in the framework of model (1.1, 000 Mathematics Subject Classification. 9D05, 9D15, 9D0. Key words and phrases. Bilinear systems, stabilization, quadratic feedback, energy decay estimate. c 00 Southwest Texas State University. Submitted April 9, 00. Published June 1, 00. 1

2 A. Y. KHAPALOV & P. NAG EJDE 00/70 is associated with the matrix B = ( and a negative gain u, in which case system (1.1 looks as follows: ẋ 1 = x, t > 0, ẋ = x 1 + ux, (1. x 1 (0 = x 10, x (0 = x 0. (1. Note that the damping term in (1. is active only when the velocity x (t is not zero and it acts in the direction which is opposite of the motion of the particle at any given time t. It is classical result that system (1. (1. is asymptotically stable when a linear viscous damping with a negative constant gain is engaged, that is, when u in (1. is a negative number. (Note that in this case one can obtain the explicit formula for solutions. However, it seems more realistic in many applications that the aforementioned gain u should depend on deviation from equilibrium. In this respect we refer the reader to Jurdjevic and Quinn [4] who considered the quadratic gain u(x 1, x = x 1, (1.4 in which case the gain is proportional to the square of the magnitude of deviation. It was shown in [4] that system (1. (1.4 is asymptotically stable. On the other hand, the method of [4] does not provide any explicit estimates for the energy decay of the corresponding solutions, which evaluate the effectiveness of nonlinear feedback (1.4 and thus are of principal importance in applications. The main goal of this paper is to derive such an explicit energy decay estimate for system (1. (1.4. It is clear from the start that it cannot be of exponential type, because system (1. (1.4 is critical (see the discussion in the next section for details. We have the following result. Theorem 1.1. There exist positive constants β, γ, and t = t( (x 10, x 0 R such that the energy E(t = x 1(t + x (t of system (1. (1.4 satisfies E(t E(0 β + γe(0t, t > t. (1.5 The values of the constants β and γ are given in the proof of this theorem (see (.69 below. Problem background: Critical bilinear systems. System (1. (1.4 (or (1.1 is an important principal case of planar, so-called critical bilinear systems (BLS, which are of traditional interest in the context of asymptotic stabilization (see, e.g., [] for more detail. Let us remind the reader that a planar BLS of the form ẋ = Ax + u(x 1, x Bx, x(0 = x 0, (1.6 is called critical if the given matrices A and B satisfy the following two conditions: For each u R at least one eigenvalue of A + u B has nonnegative real part.

3 EJDE 00/70 LIENARD S EQUATION WITH QUADRATIC VISCOUS FEEDBACK There exists a u 0 R for which all the real parts of the eigenvalues of A + u 0 B are non positive. Note that the latter condition implies that system (1.6 is stable when the constant feedback u(x 1, x = u 0 is engaged, while the former condition implies that (1.6 is not asymptotically stabilizable by any constant feedback. It is well-known (e.g., [] that, by the standard linear change of coordinates and time-rescaling, all the matrices A satisfying the above-cited two conditions can be transformed either into the matrix ( as our matrix A in (1.1 or in (1. (1.4, associated with a periodic oscillatory motion, or into one of the following three simpler matrices: ( ( ( ,, We refer the reader to [1,,, 5] for various positive and negative results on stabilization of planar critical BLS by means of constant, linear, quadratic and piecewise constant feedback laws. However, it seems that all the available results do not provide any explicit energy decay rates. A q1 x A I r1 p1 II I A4 A5 p II q O qo IV A1 A8 x1 III p q IV III A6 A7 Figure 1. Trajectory for system (1. (1.4. Proof of Theorem 1.1 Our plan for this section as follows: First of all, as we are concerned with the large time behavior of (1. (1.4, without loss of generality (see Step 8 for the

4 4 A. Y. KHAPALOV & P. NAG EJDE 00/70 general case, we can assume that {(x 10, x 0 } B = {(ξ, η ξ + η 1 }. (.1 Select any angle θ 0 (0, π 4 and split the ball B into eight sectors I, I, II, II, III, III and IV,IV as shown on Figure 1, so that the central angles A 1 OA 8, A OA, A 4 OA 5 and A 6 OA 7 are all equal to θ 0. Again, without loss of generality, we may assume that the initial point (x 10, x 0 is located at point A 1, also denoted by q 0 on Figure 1 (see Step 8 otherwise, where respectively the solid line starting from q 0 describes the trajectory (x 1 (t, x (t of system (1. (1.4 as it crosses all the eight aforementioned sectors. Our plan is to derive the energy decay estimate for the solution to (1. (1.4 starting from q 0 by evaluating its energy subsequently in each of the above mentioned sectors I IV. Then we extend this result to the general case. Remark.1. Below we use the notation x 1 = x 1 (t, x = x (t, keeping in mind that (x 1, x = (x 1 (t, x (t represent the solution to (1. (1.4 at hand. Step 1: Denote by t p1 the time required for the trajectory of (1. (1.4 starting from q 0 = (x 1 (0, x (0 = (x 10, x 0 to reach point p 1. We have (see Figure 1: x 1 (t E(t p1 sin(θ 0, x (t E(0 sin(θ 0 (. and hence x 1(tx (t 4E(0E(t p1 sin 4 (θ 0. (. Multiplying equations (1. (1.4 respectively by x 1 and x and adding them, we obtain: x 1 (t 1 (t + x (t (t = de(t = 1 d x(t R dt dt Integrating this equation over (0, t p1 yields: E(t p1 = x 1(t p1 + x (t p1 Applying (. (.5, we obtain which yields: = tp1 0 = x 1(tx (t. (.4 x 1x dt + E(0. (.5 E(t p1 E(0 4E(0E(t p1 sin 4 (θ 0 t p1, (.6 E(t p1 E( E(0δ t p1, (.7 where δ = sin (θ 0. Thus, to estimate E(t p1, we need to evaluate t p1. This evaluation is accomplished in steps -4 below, based on the visualization technique, [5] which makes use of the phase-portrait of system (1. (1.4. Namely, to evaluate t p1, we intend to derive first explicit estimates both (a for the length of arc of the trajectory of (1. (1.4 lying in sector I and (b for the speed at which the point describing the position of system (1. (1.4 at time t moves along this arc. These estimates, obtained in terms of the energy function, will provide us with the bounds for the duration of the time-interval required by the system to traverse sector I, that is, for t p1. It turns out that the derived bounds for t p1, do not depend on the energy of the system in sector I.

5 EJDE 00/70 LIENARD S EQUATION WITH QUADRATIC VISCOUS FEEDBACK 5 Remark.. From (.4 we conclude that the distance x(t R monotonically decreases as t. This in turn implies that the trajectory initiated at q 0 remains in B for all t > 0. In other words, system (1. (1.4 is stable (see also [4]. Step : Estimate for 1 (t + (t in sector I. In this step we obtain an estimate for the value of the speed 1 (t + (t at which the point (x 1 (t, x (t moves in sector I. Since (x 1, x B, in sector I we have 0 < x 1 1, 0 < x 1 and x 1 x 1. (.8 Hence, using (1. and applying (.8, 1 = x, Therefore, from (.9 we derive: = x 1 x 1x = x 1 (1 x 1 x x 1. ( x x x 1 + x = E(t 4. (.10 Since Ė(t 0, in view of (.10, we have the following estimates: E(t 1 (t + E(tp1 (t, t [0, t p1 ]. (.11 Applying the triangular inequality to the second equation in (1. and using (.8, we obtain: (t = x 1 (t x 1(tx (t x 1 (t + x 1(t x (t x 1 (t + x 1(t = x 1(t. (.1 Remark.. Note that inequalities (.10 (.1 hold everywhere in B. Using (.1 and the first equation in (1., we further derive that for t [0, t p1 ], 9x 1 (t + (t 1 (t 9(x + x 4 (t 1 (t + x (t E(0. (.1 4 Combining equations (.11 and (.1 yields E(tp1 1 (t + (t E(0, t [0, t p1 ]. (.14 These estimates will be used to evaluate t p1 in step 4 below. Step : Estimate for q 0 p 1. In this step we obtain an estimate for the length of the curve q 0 p 1 connecting the points q 0 and p 1 (see Figure 1. It follows from (.8 and (.9 that the curve q 0 p 1 is rising as t increases in [0, t p1 ]. Hence, we can describe it geometrically as a graph of some monotone decreasing function x = x (x 1, x 1 [ E(t p1 sin(θ 0, E(0 cos(θ 0 ] and E(0 cos(θ 0 q 0 p 1 = 1 + ( dx dx 1. (.15 dx 1 E(tp1 sin(θ 0 Using the first equation of (1. and the fact that (t is strictly positive on the interval t [0, t p1 ], we have from (.1: 1 = x (t, x 1(t. (.16

6 6 A. Y. KHAPALOV & P. NAG EJDE 00/70 Note that in sector I, x 1 (t and x (t are both positive, while 1 (t is strictly negative. Hence, x 1(t 1 x (t, t [0, t p 1 ]. (.17 Therefore, [8, pp.105], we conclude that Thus, dx dx 1 < x 1(t x (t, t [0, t p 1 ]. ( ( dx 9(x 1(t + x (t dx 1 4x (t = 9E(t x (t, t [0, t p 1 ]. (.19 Using (.19 and the fact that E(t is a decreasing function of t, we further obtain that 1 + ( dx 9E(0 dx 1 x (t or 1 + ( dx dx E(0 1 x (t, t [0, t p 1 ]. (.0 Since (t is strictly positive, E(0 sin(θ0 x (t E(t p1 cos(θ 0 E(t p1, t [0, t p1 ]. (.1 Applying (.1 to (.0 gives 1 + ( dx dx 1 sin(θ 0, t [0, t p 1 ]. (. Remark.4. Note that, in view of Remark.1, inequality (. also holds in sector I, namely in the portion that lies in the first quadrant. Combining (.15 with (. yields: E(0 cos(θ 0 q 0 p 1 = 1 + ( dx dx dx 1 1 E(tp1 sin(θ 0 ( E(0 cos(θ 0 E(t p1 sin(θ 0. sin(θ 0 (. Moreover, since E(0 > E(t p1 and cos(θ 0 sin(θ 0 for any θ 0 [0, π 4 ], further from (., we obtain E(0 cos(θ 0 q 0 p 1 = 1 + ( dx dx dx 1 E(0 1 sin(θ0. (.4 E(tp1 sin(θ 0 We shall now evaluate the length of the curve q 0 p 1 from below. Once again, (.9 yields = dx x 1(t 1 dx 1 x (t, t [0, t p 1 ]. (.5 Therefore, 1 + ( dx 4x 1(t + x (t dx 1 4x (t x 1(t + x (t 4x (t, t [0, t p1 ]. Applying (.4 to this inequality, we obtain 1 + ( dx dx E(tp1 1 x (t, t [0, t p 1 ]. (.6

7 EJDE 00/70 LIENARD S EQUATION WITH QUADRATIC VISCOUS FEEDBACK 7 Combining (.1 with (.6 yields 1 + ( dx dx 1 1, t [0, t p 1 ]. (.7 Applying (.7 to (.15, we obtain the following estimate for the length of the curve q 0 p 1 from below: E(0 cos(θ 0 q 0 p 1 = 1 + ( dx E(tp1 sin(θ 0 dx dx 1 1 ( E(0 cos(θ 0 (.8 E(t p1 sin(θ 0 K E(0, where K = cos(θ 0 sin(θ 0. Combining (.4 and (.8, we finally obtain: E(0 sin(θ0 q 0p 1 K E(0. (.9 Step 4: Estimate for t p1. In this step we shall derive an estimate for t p1, i.e., the time required for the trajectory of (1. (1.4, initiated at q 0, to reach the point p 1. We begin by computing the lower bound estimate for t p1. Making use of (.14and (.9, we obtain that K E(0 q0 p 1 = which in turn yields: tp1 Analogously, it follows from (.14 and (.9 that E(tp1 tp1 t p1 q 0 p 1 = 1 (t + (tdt which implies 0 0 t p1 Combining (.1 and (. yields 1 (t + (tdt E(0 t p1, (.0 K t p 1. (.1 E(0 sin(θ0, (. E(0 sin(θ 0 E(t p1. (. K t E(0 p 1 sin(θ 0 E(t p1. (.4 Since (t is strictly positive in sector I, we have x (t p1 x (0. Hence, E(t p1 = x 1(t p1 + x (t p1 (see Figure 1, which implies that x 1(t p1 Applying inequality (.5 to (.4 yields: x (0 = E(0 sin (θ 0 E(0 E(t p1 1 sin(θ 0. (.5 K t p 1 sin (θ 0. (.6

8 8 A. Y. KHAPALOV & P. NAG EJDE 00/70 Step 5: Estimate for the time taken by the trajectory of (1. (1.4 to pass sectors I and I. In this step we shall derive an estimate for t q1, i.e., for the time required by the trajectory of (1. (1.4, initiated at q 0, to reach point q 1 (see Figure 1. Split the trajectory in sector I into two arcs p 1 r 1 and r 1 q 1 (see Figure 1. Denote the time required by the trajectory of (1. (1.4 to connect the points p 1 and r 1 by t p1r 1 and the time required to connect the points r 1 and q 1 by t r1q 1. Step 5.1: Estimate for time t p1r 1. The arc p 1 r 1 can be described geometrically as a graph of function x = x (x 1, x 1 [0, E(t p1 sin(θ 0 ]. Then, applying the inequality (. (see Remarks.1 and.4 to (.7 in sector I, we obtain E(t p1 sin(θ 0 p 1 r 1 = ( dx dx 1 dx 1 E(tp1 sin(θ 0 sin(θ 0 E(t p1. (.7 Since Ė(t < 0 for any time t [t p 1 ], we have from (.10, x 1 (t + (t 1 (t + x (t E(t E(tp1 + t p1r = 1. (.8 4 Hence, in view of (.7, E(tp1 + t p1r 1 tp1 +t p1 r 1 t p1r 1 p 1 r 1 = t p1 which yields t p1r 1 1 (t + (tdt E(tp1, (.9 E(t p1 E(tp1 + t p1r 1. (.40 Since (t is positive, we have x (t p1 + t p1r 1 x (t p1. Hence, E(t p1 + t p1r 1 = x (t p1 + t p1r 1 which implies x (t p1 = E(t p1 cos (θ 0, E(t p1 E(t p1 + t p1r 1 1 cos(θ 0. (.41 Finally, combining (.40 and (.41, we obtain the following estimate for t p1r 1, t p1r 1 cos(θ 0. (.4 Step 5.: Estimate for the time t r1q 1. Once again, the arc r 1 q 1 can be described geometrically as the graph of a function x = x (x 1, with x 1 in the interval [ E(t p1 + t p1r 1 sin(θ 0, 0]. Then 0 r 1 q 1 = 1 + ( dx dx 1. (.4 dx 1 E(t p1 +t p1 r 1 +t r1 q 1 sin(θ 0 Note that x 1 (t, 1 (t, and (t are strictly negative on the interval t [t p1 + t p1r 1 ], where x (t is positive. Since it follows from (1. (1.4 and (.1 (see Remark. that x 1 and 1 = x,

9 EJDE 00/70 LIENARD S EQUATION WITH QUADRATIC VISCOUS FEEDBACK 9 we have Therefore, 0 x 1(t 1 x (t, t [t p 1 + t p1r 1 ]. (.44 1+( dx 9(x 1(t + x (t dx 1 4x (t 9E(t x (t, t [t p 1 +t p1r 1, t p1 +t p1r 1 +t r1q 1 ]. (.45 Using (.45 and the fact that E(t is a decreasing function of t, we obtain 1 + ( dx dx E(tp1 + t p1r 1. (.46 1 x (t Since (t is strictly negative, for t in [t p1 + t p1r 1 ], E(t p1 + t p1r 1 cos(θ 0 x (t E(t p1 + t p1r 1. (.47 Applying (.47 to (.46 gives 1 + ( dx E(t dx p1 + t p1r 1 1 E(tp1 + t p1r 1 cos(θ 0, (.48 for t [t p1 + t p1r 1 ]. Thus applying inequality (.48 to (.4, we obtain an upper estimate for the length of the arc r 1 q 1 : 0 r 1 q 1 = 1 + ( dx dx 1 E(t p1 +t p1 r 1 +t r1 q 1 sin(θ 0 dx 1 (.49 E(t p1 + t p1r 1 tan(θ 0. Since Ė(t < 0 for any t [t p 1 + t p1r 1 ], making use of (.10 (see Remark., we derive that x 1 (t + (t 1 (t + x (t E(tp1 + t p1r 1. (.50 4 Using inequalities (.49 and (.50, we obtain E(tp1 + t p1r 1 t r1q 1 r 1 q 1 = which, in turn, provides tp1 +t p1 r 1 +t r1 q 1 t p1 +t p1 r 1 E(t p1 + t p1r 1 tan(θ 0. 1 (t + (tdt (.51 t r1q 1 tan(θ 0 E(t p1 + t p1r 1 E(tp1 + t p1r 1. (.5 Once again, integrating (.4 over (t p1 with (.6 + t p1r 1 yields (compare E(t p1 + t p1r 1 = E(t p1 + t p1r 1 + tp1 +t p1 r 1 +t r1 q 1 t p1 +t p1 r 1 x 1x dt. (.5

10 10 A. Y. KHAPALOV & P. NAG EJDE 00/70 Observe that x (t 1 and 0 x 1(t E(t p1 + t p1r 1 sin (θ 0 for any t [t p1 Hence, + t p1r 1 ]. Hence, it follows from (.5 that E(t p1 + t p1r 1 E(t p1 + t p1r 1 (1 sin (θ 0. (.54 E(t p1 + t p1r 1 E(t p1 + t p1r sin (θ 0 t r1q 1. (.55 Applying inequality (.55 to (.5 yields: 0 < t r1q 1 tan(θ sin (θ 0 t r1q 1, (.56 which gives the following estimate for t r1q 1 : 0 t r1q 1 C = 9 tan (θ 0 sin (θ 0 + tan(θ 0 9 tan (θ 0 sin 4 (θ (.57 Step 5.: Estimate for time t q1. Combining (.6, (.4 and (.57 yields K t q 1 = t p1 + t p1r 1 sin (θ C. (.58 cos(θ 0 Step 6: Estimates for E(t p1 and E(t q1. Substituting the lower bound for t p1 from (.6 into (.7 and recalling that since E(t is decreasing in time, we obtain from (.7 that E(0 E(t q1 E(t p1. ( E(0δ K Step 7: Estimate for E(t for any time t > 0. It is not hard to see that in the next two sectors II and II one can apply the same strategy as described in the above steps 1-6, namely, leading to estimate (.59. Denote the time required by the trajectory of (1. (1.4 to reach points p and q from q 0 respectively by t p and t q. Then, similarly as in (.59 and (.6, (.58 (see Remark., we obtain: E(t q E(t p where, analogously to (.58, we have: E(t q E(t q1 δ (t p t q1 E(t q1, ( E(t q1 δ K K t p t q1 t q t q1 sin (θ 0 + cos(θ 0 + C. Remark.5. Consider the function f(x = x 1+κx, with κ a positive constant which is associated with the right hand sides in (.59, (.60, considered as functions of energy. Since f (x > 0 for all x > 0, f(x is a monotonically increasing function. Since E(t q1 E(t p1 and (.59 holds, Remark.5 implies E(t q E(0 1+4E(0δ K E(0 1+4E(0δ K δ K E(0 = 1 + K. (.61 4E(0δ

11 EJDE 00/70 LIENARD S EQUATION WITH QUADRATIC VISCOUS FEEDBACK 11 Denote the following subsequent intersecting points of the trajectory at hand with the lines OA 6, OA 7,... by p, q,..., p n, q n,.... Analogously, we have E(0 E(t qn E(t pn 1 + nk, n = 1,.... (.6 4E(0δ Consider any t > 0. Then there exists a positive integer n such that t qn 1 t t qn n sin (θ 0 + n + nc, (.6 cos(θ 0 where, thus, t sin (θ + n. (.64 0 cos(θ 0 + C Note that, since t > t qn 1 and E(t decreases in time, (.6 also implies that E(0 E(t E(t qn (n 1K 4E(0δ. (.65 Combined with (.6, this gives the following energy decay estimate for any t > 0 in the case when the initial state (x 10, x 0 lies in B on the line OA 1 as shown on Figure 1: E(0 E(t 1 4KE(0δ + 4KE(0δ ( sin (θ + 0 cos(θ + 0 C 1 t. (.66 Step 8: The general case. Consider any initial point (x 10, x 0. Since (1. (1.4 has an isolated equilibrium point (0, 0, by Poincare-Bendixson theorem (see [7], the trajectory to (1. (1.4 starting from (x 10, x 0 after some finite time will enter the ball B and intersect the line OA 1. Denote the time of this intersection by t 0. Then, by the scheme discussed in steps 1-7, we have where E(t 0 E(t 1 4KE(t0δ + αe(t 0 (t t 0, t t 0, (.67 α = 4K sin (θ 0 cos(θ 0 sin 4 (θ 0 C sin (θ 0 cos(θ sin (θ cos(θ 0. Using (.1 (i.e., E(t and the argument of Remark.5, we derive from (.67 that E(t 0 E(t 1 4E(t0Kδ + αe(t 0 (t t 0 E(t 0 (.68 1 Kδ + αe(t 0 (t t 0 E(0 1 Kδ + αe(0(1 t0 t t t > t 0. Choose t = t 0 and denote β = 1 K sin4 (θ 0, γ = α. (.69 Recall that δ = sin (θ 0 and θ 0 is any number in (0, π 4. Then (.68 implies (1.. This concludes the proof of Theorem 1.1.

12 1 A. Y. KHAPALOV & P. NAG EJDE 00/70 Acknowledgement. This work was supported in part by grant DMS from the NSF. References [1] A. A. Agrachev and D. Liberzon, Lie algebraic stability criteria for switched systems, SIAM J. Control Optim., 40 (001, pp [] A. Bacciotti, Local stabilizability of nonlinear control systems, Series on Advances in Mathematics and Applied Sciences, vol. 8, World Scientific, Singapore, 199. [] J. P. Hespanha, A. S. Morse, and D. Liberzon, Stability of switched systems:a Lie-algebraic condition, System Control Lett., 7 (1999, pp [4] V. Jurdjevic and J. P. Quinn, Controllability and Stability, J. Differential Equations, 8 (1978, pp [5] A. Y. Khapalov and R. R. Mohler, Asymptotic stabilization of the bilinear time-invariant system via piecewise-constant feedback, System Control Lett., (1998, pp [6] R. R. Mohler and A. Y. Khapalov, Bilinear Control and Application to Flexible a.c. Transmission System, Journal of Optimization Theory and Applications, 105 (000, pp [7] L. Perko, Differential Equations and Dynamical Systems, Texts in Applied Mathematics, vol. 7, Springer-Verlag, NY, [8] W. Rudin, Principles of Mathematical Analysis, McGraw Hill, Alexander Y. Khapalov Department of Mathematics, Washington State University, Pullman, WA , USA address: khapala@wsu.edu Parthasarathi Nag Department of Mathematics, Washington State University, Pullman, WA , USA address: pnag@math.wsu.edu