Fakultäten für Informatik & Mathematik Technische Universität München Dr. Tobias Neckel Summer Term Dr. Florian Rupp Exercise Sheet 3 Dynamical Systems & Scientic Computing: Homework Assignments 3. [ ] Application of the Routh-Hurwitz Method) Apply the Routh-Hurwitz method to determine the location of all roots of the following polynomials: a) px) = 3x + 5, b) px) = x 5x, c) px) = 53x 57x + 89, d) px) = x + x )x + x + ), and e) px) = x 3 + 5x + x 3. Solution: In the following the components of the polynomial are denoted as px) = a n x n + a n x n +... + a a) The rst polynomial has degree and hence we only have to determine, which is given by = a n = 5 and a n = 3. Now we calculate the number of sign changes k according to the Routh-Hurwitz theorem as k = V a n, ) = V 3, 5) =. In this case we know, that the only root is in the left half-plane, i.e. it has negative real part and the stability criterion is satised. b) The degree of the second polynomial is. Therefore we have to calculate and : = a n = 5 an = det a n =. a n Thus k calculates as k = V a n,, ) = V a n )) )) 5 = det = 5, 5, 5 ) = V, 5, ) =. 5 This means that all roots are in the left half-plane. For the Routh-Hurwitz theorem we have to consider a n = 5 = and = 5. Both values are positive, which means that all roots have negative real part. c) The degree of this polynomial is as well and the determinants calculate as = 57 )) 57 = det = 57 89 = 773 53 89 a n = 53.
By this k is given by k = V 53, 57, 773 ) = V 53, 57, 89) =. 57 Since there are two sign changes k =, we know that both roots are in the right half-plane and hence the Routh-Hurwitz stability criterion is not satised in this case. d) This polynomial is given as the product of two factors. In order to make sure that all roots have negative real part the Routh-Hurwitz stability criterion must be satised for both factors. We denote the factors as follows q x) = x + x q x) = x + x + and our polynomial is then px) = q x) q x). At rst we analyse q. Since the degree of q is, we calculate and as and for k follows k = V = = det a n = )) =,, ) = V,, ) =. This means that there is a root of q in the right half-plane. Because all roots of q are also roots of p, there is at least one root of p in the right half-plane. Thus the Routh-Hurwitz stability criterion cannot be satised for p and we are done here. e) The last polynomial has degree 3 and we need to calculate, and 3 as follows = 5 )) )) an a = det n 3 5 3 = det = 5 + 3 = 53 a n a n a n a n 3 5 3 3 = det a n a n = det = 5 9 = 59 a n a n 3 5 3 a n =. For k we have k = V a n,,, ) 3 = V, 5, 53 5, 59 ) = 53 and hence this polynomial is not stable either, since there is one root with positive real part. 3. [ ] plementation of the Routh-Hurwitz Algorithm) plement the Routh-Hurwitz method for determining the location of the roots of a polynomial of degree 4 within the complex plane in Matlab. Use the Matlab root nding methods to plot the roots in the complex plane. a) Test your code against the analytic results derived in exercise 3.. b) Apply your code to determine the stability of the zero solution of the linear oscillator dierential equation ẍt) + cdẋt) + d xt) =, x C R, R), with constants c =.64 and d = 5.56, cf. exercise.4.
c) Finally, consider the parameters c and d from part b) to be arbitrary, i.e., c, d) R. Dene a function depending on c and d that is positive if the origin is asymptotically stable in exercise b), zero if the origin is stable and negative if the origin is unstable. Plot the graph of this function against the c-d-plane. Solution: We implement the Routh-Hurwitz algortihm with the help of MATLABs determinant function det). The notation for the polynomial of degree 4 is fz) = α z 4 + α z 3 + α 3 z + α 4 z + α 5, by this the vector alpha contains the coecients in a descent order. coecient Matrix as α α 4 A = α α 3 α 5 α α 4. α 3 α 5 Then the determinants easily calculate in MATLAB notation as: Thus we initialize the Delta)=alpha_; Delta)=detA:,:)); Delta3)=detA:3,:3)); Delta4)=detA); And the sign change can be checked by a simple if condition. If the polynomial has a lower degree, then α takes the place of the highest order coecient, e.g. for a polynomial of degree we have fz) = α z + α z + α 3 and α 4 = α 5 =. By this the input for the algorithm does not change and we can use it for any degree smaller or equal 4. Now we are able to test the obtained algorithm for the polynomials from 3.. a) We notice that our results from 3. coincide with the results obtained by the algorithm, as we see in Figure -3. The number of roots with positive real part are precisely derived. 8.8 6.6 4.4...4.6.8 4 6 3.5.5.5.5 8.5.5.5 Figure : Roots of the polynomial from 3. a.) left) and 3. b.) right).
.8.8.6.6.4.4.....4.4.6.6.8.8 3.5.5.5.5 3.5.5.5.5 Figure : Roots of the polynomial from 3. c.) left) and 3. d.) right)..5.5.5.5.5.5 3.5.5.5.5 Figure 3: Roots of the polynomial from 3. e.). b) Since the dierential equation is linear and homogeneous) we simply obtain the characteristic polynomial as fz) = z + cdz + d Applying our algorithm we obtain that there are no roots with real part greater than. In fact we have x = 9.9584 +.9559i, x = 9.9584.9559i. c) The idea is to nd a suitable function, that changes signs with the stability. Hence it should be negative, if there are any roots of the characteristic polynomial with positive real part. We take a look at the Hurwitz determinants, which are in this case = cd, = cd 3 and hence for the function of sign changes we have k = V, cd, d ). We easily recognize, that the number of sign changes k is either or and depends only on the second argument cd. Thus we dene the stability indicating function as hc, d) = cd. This function is positive if the zero solution is asymptotically stable and negative if not. For
c = we have a stable solution, but if d = the solution is already unstable, since the two roots are both exactly and the dierential equation reduces to ẍ =. In order to satisfy the condition that hc, d) is only if the zero solution is stable we need to make a minor adjustment to { cd, for c, d) R \{d = } hc, d) =. ), if d = See Figure 4 for the contour plot. 8 8 6 6 4 4 6 4 4 6 8 8.8.6.4...4.6.8 Figure 4: Contour plot of the stability indicating function hc, d) as given in ). The jump at d = can be seen if one looks closely. 3.3 [ ] Application of the Lozinskii-Measure) In an article by Rupp and Scheurle [3]) a reduced model for sh-jellysh interactions is proposed, where sh being assumed to represent the dominant predatory species feeding on jellysh. This model is given by the following set of coupled non-linear ordinary dierential equations ẋ = c + y + y ) x x, and ẏ = β d x + y where c R and β, d > are parameters, and x denotes the sh population whereas y stands for the jellysh population, respectively. a) Show that the origin is an equilibrium and give the linearization of the equations at this point. ) y, b) Apply the method of Lozinskii-Measures to determine linearized) stability of the origin. Solution: ) ) a) We have that c + y +y x x = for x = and β d x +y y = for y =. Thus, the origin x, y) =, ) is an equilibrium. At any evaluation point x, y) = ξ, η ) the right
hand side's matrix Jξ, η ) of linearization evaluated at this point reads as Jξ, η ) = c + η +η ξ ξ +η ). βd η ξ +η β dβ +η ) In particular, the matrix of linearizion evaluated at the origin reads as J, ) = diagc, β). Its eigenvalues are thus given by λ = c corresponding to the eigenvector e =, ) T and λ = β corresponding to the eigenvector e =, ) T. I.e., from the principle of linearized stability we immediately have asymptotic stability if both c and β are negative. Due to our restriction β > the origin can hence be at best of saddle-type. b) To apply the Lozinskii-measure method, we require the second compound matrix A [] of our matrix Jξ, η ) = a i,j ) i,j=,, and ) detjξ, η )). Here, A [] is just a scalar and reads as and A [] = a + a = trjξ, η )) = c + η + η ξ + β dβ + η ) = c + β =: K for ξ, η ) =, ) ) detj, )) = cβ =: K. In the lecture, we had the following theorem: Let A R d d. For sa) < it is sucient and necessary that sa [] ) < and ) d deta) >. Here, K > if and only if c, β > or c, β <. Moreover, K = sa [] ) < holds, together with K > only in the case c, β <. Thus, for c, β < we would have asymptotic stability. 3.4 [ ] RDE Properties) In order to describe in some qualitative manner the nature of an earthquake disturbance, Bogdano et al. [], [], considered the following model for ground accelerations yt) = n ta j exp α j t) cosω j t + Θ j ), for t, and yt) = for t <, ) j= where a j, α j and ω j are given real positive numbers and the parameters Θ j are independent random variables uniformly distributed over an interval of length π. Let us assume a one-story building that is at rest at t = and let Xt), t, denote the relative horizontal displacement of its roof with respect to the ground. Then, based upon an idealized linear model, the relative displacement Xt) subject to ground accelerations is governed by a) Show that 3) has a path-wise solution. b) Compute this solution analytically. Solution: ẍt) + ζω ẋt) + ω xt) = yt), for t. 3) a) We apply the following theorem: Theorem Existence & Uniqueness of Path-Wise Solutions): Let Ω, A, P) be a probability space and the following three prerequisites be satised: a) The functions fx, t, ω) are A-measurable for all x, t) R d I. b) fx, t, ω) is continuous on R d I for almost all ω Ω. c) For almost all ω Ω there is a real continuous function Lt, ω) on I such that where t I and x, x R d. fx, t, ω) fx, t, ω) Lt, ω) x x, ξ
Then for any initial condition X, t ) S d there exists an unique path-wise solution on I of the random dierential equation dx t dt = fx t ), t, ω), X t ) R d. The required conditions hold in our example: a) With z := x, x ) T we have fz, t, ϖ) = ) } ω ζω {{ } =: A + yt, ϖ) with a constant matrix A. The process yt, ) : Ω R is measurable, as the pre-image of every measurable set in R is measurable w.r.t. A. Thus, f is measurable, too, for all x, t) R R. b) For any xed ϖ Ω the stochastic process y, ϖ) is continuous as a composition products, nite sum) of continuous functions. Hence, f,, ϖ) is continuous on R R. c) Finally, ), fz, t, ϖ) fz, t, ϖ) = Az z ) A z z. b) As shown in the lecture, we can compute the general solution X t of 3) explicitly analogous to the deterministic case. I.e., let Qt) be the fundamental matrix of the homogeneous system ż = Az, then for all initial conditions X S it holds that X t t = Qt)X + Qt) Q s)z s ds, t Z t =, yt, ϖ)) T, is the unique path-wise solution of 3). Compare with exercise. to obtain the fundamental solutions of ż = Az the techniques are essentially the same). 3.5 [ ] RDE Simulation) For equation 3) let the parameters be given according to [], p.67, as ω = [rad/ sec] and ζ =.5. a) Let yt) be the stochastic process dened by ) the coecients of which are given according to [], p.67, as ω = 6 [rad/ sec], ω = 8 [rad/ sec], ω 3 = [rad/ sec], ω 4 =.5 [rad/ sec], ω 5 =, 75 [rad/ sec], ω 6 =..., ω = [rad/ sec], ω =.5 [rad/ sec], ω 3 = 3.5 [rad/ sec], ω 4 = 5 [rad/ sec], ω 5 = 7 [rad/ sec], ω 6 = 9 [rad/ sec], ω 7 = 3.5 [rad/ sec], ω 8 = 3 [rad/ sec], ω 9 = 34 [rad/ sec], ω = 36 [rad/ sec], as well as α = = α = /3, a = = a =.5. Plot yt) against the time t [, ]. b) Apply the usual deterministic) Euler scheme to solve 3) for the given parameter values. c) Apply the usual deterministic) Heun scheme to solve 3) for the given parameter values. d) Compare the solution from b) and c) with the analytical solution obtained in 3.4 b). Solution: a) For the uniformly distributed random variables Θ j we are using the MATLAB expression theta=*pi*unifrnd,*pi,,); by this the resulting function yt) is plotted in Figure 5. We note that yt) is highly oscillating, as we would suspect for a suitable model for the ground acceleration of an earthquake.
4 3 yt) 3 3 4 5 6 7 8 9 time t Figure 5: The ground acceleration yt) plotted in the interval [, ]. b+c) At rst we take a look at the solution in the x ẋ-phase plane in Figure 6. We used the initial condition x) = and ẋ =, i.e. the building is at rest at t =. We see that the acceleration ẋ increases quite fast and alternates between positive and negative acceleration. Linked to that the dislocation xt) alternates as well. The solutions produced by the Euler and Heun scheme behave similar, in Figure 7 the convergence of the known methods is illustrated. Like in problem.4 the convergence rate of the implicit and explicit Euler is O h), but the implicit Euler produces an useful solution even for small step sizes. The Heun scheme has a convergence rate of O h ), which is clearly illustrated in the convergence plot. Like the explicit Euler, the Heun scheme needs a minimal step size to produce an accurate solution. d) Classication: easy, easy with longer calculations, a little bit dicult, challenging. ferences [] J.L. Bogdanoff, J.E. Goldberg and M.C. Bernard 96): sponse of a Simple Structure to a Random Earthquake-Type Disturbance, Bull. Seism. Soc. Amer., 5, pp. 93-3. [] J.E. Goldberg, J.L. Bogdanoff, and D.R. Sharpe 964): The sponse of Simple Nonlinear Systems to Random Disturbance of the Earthquake Type, Bull. Seism. Soc. Amer., 54, pp. 63-74.
.5.4 plicit Euler Starting point Endpoint Heun.3.....3.4.5.5..5..5.5..5..5 Figure 6: Solution in the x ẋ-phase plane. Explicit Euler plicit Euler Heun Accuracy 3 4 5 6 7 3 4 5 Number of steps Figure 7: Convergence of the dierent methods compared to a high order solution computed with MATLAB. [3] F. Rupp and J. Scheurle ): On the Jellysh Joyride: Mathematical Analysis of Catastrophes in Maritime Ecosystems, to appear at the proceedings of the 9th AIMS Conference on Dynamical Systems.