19 Eigenvalues, Eigenvectors, Ordinary Differential Equations, and Control

Transcription

1 19 Eigenvalues, Eigenvectors, Orinary Differential Equations, an Control This section introuces eigenvalues an eigenvectors of a matrix, an iscusses the role of the eigenvalues in etermining the behavior of solutions of systems of orinary ifferential equations. An application to linear control theory is escribe Linear control theory: feeback Consier the initial value problem t x(t) = ax(t), t 0, x(0) = x 0, (1) where x(t) is a real-value function, a is a real constant, an the initial value x 0 is positive. We will refer to the variable t as time. The solution x(t) = exp(at)x 0 remains boune as t grows only when the constant a is zero or negative. When a is positive, the solution will grow without boun. In applications to control problems, the solution x(t) is often referre to as the system state. Moels of control problems often inclue an aitional term in equation (1), known as the feeback. The following equations illustrate a typical linear control problem with the feeback term fx(t): t x(t) = ax(t) bfx(t), t 0, x(0) = x 0, b 0, (2) y(t) = cx(t), c 0. The constants a, b, an c are etermine by the moel. The function y(t) is calle the system output; in this example it is a multiple of the system state x(t). In real-worl applications, the system output represents an observable or measurable property of the system. The parameter f is calle the feeback gain. The purpose of the feeback term is to prevent unboune growth of the solution x(t) as time increases. The solution of equation (2) is x(t) = exp ((a bf)t) x 0. (3) 1

2 Whether the system state x(t) is boune epens on the quantity a bf. The scalar-value linear control problem amounts to choosing a value of f, such that a bf 0. In most applications one requires a bf < 0. (4) In many control problems of interest, x(t) an f are column vectors, a is a matrix, an b an c are row vectors, e.g., x(t), f, b T, c T R n an a R n n. Then a bf is an n n matrix, an the conition (4) translates into the requirement that the matrix a bf only have negative eigenvalues. The following section reviews results on eigenvalue an eigenvector. Thereafter, we will return to control problems Matrices, eigenvalues, an eigenvectors Let A be a square n n matrix. A scalar λ an a nonzero vector v that satisfy the equation Av = λv (5) are calle an eigenvalue an eigenvector of A, respectively. The eigenvalue may be a real or complex number, an the eigenvector may have real or complex entries. The eigenvectors are not unique; see Exercises 19.5 an 19.7 below. Equation (5) may be rewritten as (λi A)v = 0, (6) which shows that the nonzero eigenvector v lies in the null space of the matrix λi A. Matrices with nontrivial null spaces are, by efinition, singular, an therefore, et(λi A) = 0. The function p A (z) := et(zi A) is a polynomial of egree n. This can be verifie by expaning the eterminant along rows or columns. The polynomial p A (z) is calle the characteristic polynomial of the matrix A. Example Consier the matrix A = Its characteristic polynomial is given by p A (z) = (z 1)(z 1)(z 2). 2

3 The eigenvalue λ = 1 is sai to be of algebraic multiplicity 2, because it is a zero of of p A (z) of multiplicity 2. The eigenvalue λ = 2 is of algebraic multiplicity 1. Example Expaning the characteristic polynomial for the matrix A = (7) along the last row yiels p A (z) = et ([ z z 3 ]) (z 1) = (z 2 4z + 1)(z 1). This shows that the eigenvalues of the matrix are λ 1,2 = 2 ± 3, λ 3 = 1. If we replace the entry 2 in position (1, 2) of the matrix (7) by any real number strictly smaller than 1, then the matrix has one pair of complex conjugate eigenvalues. For instance, setting the (1, 2)-entry to 2 yiels the eigenvalues λ 1 = 2 + i, λ 2 = 2 i, λ 3 = 1, where i = 1 is the imaginary unit, i.e., i 2 = 1. In particular, λ 1 an λ 2 are complexvalue eigenvalues. Both λ 1 an λ 2 are sai to have real part 2; λ 1 has imaginary part 1 an λ 2 has imaginary part 1. Thus, the imaginary part is the coefficient of i. Since λ 1 an λ 2 have the same real parts an have imaginary parts of opposite sign, these eigenvalues are sai to be complex conjugate; see below for further comments on complex numbers. Proposition 1 The roots of p A (z) are the eigenvalues of A. Proof. We alreay have seen that p A (z) vanishes at the eigenvalues of A. Conversely, assume that p A (λ) = 0 for some scalar λ. Then the matrix λi A is singular. Let v be a nontrivial solution of the homogeneous linear system of equations (6). Then v 0 satisfies (5). Thus, v is an eigenvector an λ an eigenvalue of A. By the Funamental Theorem of Algebra, a polynomial of egree n has precisely n zeros, counting multiplicities. Therefore, every n n matrix has n eigenvalues. Some of the zeros of p A (z) may be complex numbers. It follows that eigenvalues may be complex. However, the important class of symmetric matrices only have real eigenvalues; see Section 20. 3

4 Let λ 1, λ 2,...,λ n be the eigenvalues of the n n matrix A, an let v 1, v 2,...,v n enote the corresponing eigenvectors, i.e., Av j = λ j v j, j = 1, 2,..., n. (8) A matrix is sai to be iagonalizable if the n eigenvectors v 1, v 2,..., v n can be chosen to be linearly inepenent. Assume this is the case an introuce the eigenvector matrix V = [v 1, v 2,...,v n ]. This matrix is nonsingular since its columns are linearly inepenent. Define the iagonal matrix etermine by the eigenvalues of A, Then the equations (8) can be expresse compactly as D = iag[λ 1, λ 2,...,λ n ]. (9) AV = V D. Since the matrix V is nonsingular, the above equation yiels A = V DV 1. (10) This formula shows that the matrix A is iagonal when expresse in terms of the eigenvector basis {v 1, v 2,..., v n }. This is the reason for the importance of eigenvalues an eigenvectors. The set of eigenvalues of a matrix is sometimes referre to as the spectrum of the matrix, an the factorization (10) as the spectral factorization. Most, but not all, square matrices are iagonalizable. Example The matrix A = [ has the eigenvalues λ 1 = 1 an λ 2 = 1, but only one linearly inepenent eigenvector. This follows from equation (6), which can be expresse as [ ] 0 2 v = The above equation shows that all solutions are of the form v = [α, 0] T, where α is a nonvanishing scalar. Thus, all eigenvectors of A are a multiple of the axis vector e 1 = [1, 0] T. Perturbing any one of the iagonal entries of A slightly gives a matrix with istinct eigenvalues. Matrices with pairwise istinct eigenvalues have linearly inepenent eigenvectors. We conclue that there are matrices arbitrarily close to A with linearly inepenent eigenvectors; see Exercise 19.8 for an illustration. 4 ].

5 Exercise 19.1 Let A = Determine the characteristic polynomial p A (z) of A. Exercise 19.2 Let A = Determine the characteristic polynomial p A (z) of A. Exercise 19.3 Give a simple expression for A 2 in terms of the matrices V an D by using the spectral factorization (10). What is the corresponing expression for A k when k is a positive integer? Assume that A is nonsingular. What is the corresponing expression for A k when k is a negative integer? Exercise 19.4 Let A be an n n matrix with nonnegative eigenvalues. Give an expression for A 1/2 by using the spectral factorization (10). Exercise 19.5 Eigenvectors are not unique. Let v be an eigenvector of A. Show that any nonzero multiple of v is also an eigenvector of A. Exercise 19.6 Consier the matrix. A = iag[1, 2, 3]. What are the eigenvalues? Describe all eigenvectors, e.g., by investigating the solution set of (6) for λ = 1, 2, 3. 5.

6 Exercise 19.7 Consier the matrix A = iag[1, 1, 3]. Describe all eigenvectors, e.g., by investigating the solution set solutions of (6) for λ = 1, 3. Exercise 19.8 Compute eigenvectors associate with the istinct eigenvalues of the matrix [ ] A =. 0 1 Are they linearly inepenent? Are they almost parallel? Cf. the iscussion of Example Exercise 19.9 Use the MATLAB/Octave comman magic to etermine a 4 4 matrix, whose entries form a magic square. Use the MATLAB/Octave comman eig to compute the spectral factorization of the matrix. How are the eigenvectors normalize? One of the eigenvectors has all entries equal. Can this be expecte? Hint: What is the corresponing eigenvalue? 19.3 Systems of linear orinary ifferential equations Consier the system of linear orinary ifferential equations (ODEs) in time t 0: t x 1(t) = a 11 x 1 (t) + a 12 x 2 (t) + a 13 x 3 (t), t x 2(t) = a 21 x 1 (t) + a 22 x 2 (t) + a 23 x 3 (t), (11) t x 3(t) = a 31 x 1 (t) + a 32 x 2 (t) + a 33 x 3 (t). This system has a unique solution when the initial values x 1 (0) = x 1, x 2 (0) = x 2, x 3 (0) = x 3 (12) are prescribe. We can write the system of ODEs (11) as x(t) = Ax(t), (13) t 6

7 where A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33, x(t) = x 1 (t) x 2 (t) x 3 (t). The solution of (13) for t 0 is given by x(t) = exp(at)x(0), where exp(at) is the matrix exponential function. Let A be an n n matrix with spectral factorization (10). We efine exp(a) := V exp(d) V 1 (14) an exp(d) := iag[exp(λ 1 ), exp(λ 2 ),..., exp(λ n )], where D is given by (9). The solution of the system of ifferential equations (13), where now A is this n n matrix, can be expresse as x(t) = exp(at)x(0) = V exp(dt) V 1 x(0) = V iag[exp(λ 1 t), exp(λ 2 t),...,exp(λ n t)] V 1 x(0), where x(0) is the n-vector of initial values. We often are intereste in whether solution x(t) ecreases to zero or grows in magnitue as t increases. Therefore the norm x(t) = V exp(dt)v 1 x(0) κ(v ) x(0) max 1 j n { exp(λ jt) } (15) is of interest. Here κ(v ) := V V 1 is the conition number of the eigenvector matrix V. Note that neither κ(v ) nor x(0) epen on t. Therefore, the growth or ecay of the norm of the solution epens entirely on the factor max 1 j n { exp(λ j t) }; see Exercises Exercise Let A be an n n matrix. MATLAB an Octave allow the commans exp(a) an expm(a), Which comman yiels the matrix exponential? What oes the other comman compute? Hint: Compare with (14). 7

8 Exercise Show the boun (15). Exercise Assume that all the eigenvalues λ j are strictly negative. Does the norm of the solution x(t) of (13) increase or ecrease as t becomes large? What oes this imply for each component x j (t), j = 1, 2, 3, of x(t)? Exercise Assume that all the eigenvalues λ j are strictly positive. Does the norm of the solution x(t) of (13) increase or ecrease as t becomes large? Exercise Assume that the 3 3 matrix A has the eigenvalues λ 1 = λ 2 = 1 an λ 3 = 0. How can we expect the solution x(t) of (13) to behave as t increases? Exercise Let A be the matrix from Exercise 19.9 an let x(0) = [1, 1, 1, 1] T. Plot the solution x(t) of (13) for 0 t 0.2. How oes the solution behave? Exercise Let the matrix A be obtaine by subtracting 34.5I from the matrix use in Exercise 19.15, where I enotes the ientity matrix. Let x(0) = [1, 1, 1, 1] T an plot the solution x(t) of (13) for 0 t 5. How oes the solution behave? Explain! 19.4 Linear control theory The subject of linear control theory is concerne with choosing a feeback gain vector f R n, so that the solution x(t) of the control system t x(t) = Ax(t) bft x(t), A R n n, x(t), b R n, t 0, (16) y(t) = c T x(t), c R n, 8

9 remains boune as time t increases. However, only the output y(t) can be observe. The initial state x(0) an the vectors b an c are efine by the moel, as is the matrix A, which we assume to be iagonalizable. The solution to the initial value problem (16) is given by x(t) = exp ( (A bf T )t ) x(0); compare this formula with the analogous scalar expression in equation (3). The problem (16) is sai to be controllable if a feeback gain vector f can be foun, such that x(t) remains boune as time t increases. This entails choosing f so that the eigenvalues of the matrix A bf T are non-positive (or, more generally, have non-positive real part). There are several algorithms available for etermining such a vector f. Some of these are calle eigenvalue assignment methos, because they seek to etermine a feeback gain vector f, such that the matrix A bf T has specifie eigenvalues. For many matrix-vector pairs {A, b}, a gain vector, such that A bf T has a prescribe spectrum can be foun. However, the computation of this vector is for some pairs {A, b} too ill-conitione to yiel useful results in finite precision arithmetic. The gain vector f inicates to engineers how a structure, such as the space station, shoul be reinforce to avoi harmful unampe oscillations. Another example is provie below A control example The electro-magnetic suspension of a ferrous ball is a classic example in the control theory literature. The physical goal of the example is to maintain the position of the suspene ball. Our mathematical goal is to express the physical moel as a linear control problem of the form (16), an to solve for a feeback gain vector to stabilize the system an keep the ball suspene in a fixe position. We present a slightly moifie version of the stanar example in orer to simplify the computations. The example is illustrate in Figure 1. A voltage v is applie to the coil at the top of the illustration. The current i(t) flowing through the coil at time t generates a magnetic force F that pulls the ball up. At the same time, the force of gravity G is pulling the ball back own towars the groun. We enote the istance between the en of the coil an the top of the ball at time t as h(t). The ball will remain suspene in miair whenever the forces F an G balance out. The otte line at h(0) inicates the set position at which we esire to suspen the ball (as illustrate, the ball is well below the set point). The eviation from this position is enote by h(t) := h(t) h(0). Similarly, the eviation from the current i(0) is written ĩ(t) := i(t) i(0). A stanar moel of the motion of the ball is given by 2 t 2h(t) = 1 (G F), (17) m 9

10 Figure 1: Magnetic suspension of a ball above the groun. t i(t) = v l r i(t), (18) l where m is the mass of the ball, r is the resistance of the wire, an l is the impeance of the coil. The gravitational force G is constant. The magnetic force F = ki(t) 2 /h(t) is a nonlinear function of the istance h(t) an current i(t) (k is a constant). The nonlinearity of F prevents us from irectly setting up a linear control problem of the form (16). We can, however, assume that the moel is approximately linear near the set point h(0), an use the Taylor series approximation: F F(h(0), i(0)) + F h = F(h(0), i(0)) + F h (h(t) h(0)) + F i(0) i(0) h(t) + F i i ĩ(t). h(0) h(0) (i(t) i(0)) Noting that 2 h(t) = 2 2 h(t), we can substitute the linear approximation of F into equation t 2 t (17), 2 t 2 h(t) = G/m k 1 + k 2 h(t) + k3 ĩ(t), (19) where we have simplifie the notation by introucing the constants k 1 := F(h(0), i(0))/m, k 2 := F, an, k 3 := F. m h i(0) m i h(0) 10

11 Define the state vector x(t) := [ h(t), t h(t), ] T ĩ(t) an let the system output be the eviation from the set point, Let the constants have the values y(t) := h(t). k 2 = 1, k 3 = 1, r/l = 10, G/m k 1 = 1, v/l = 1. The efinition of x(t), y(t), an the constants, together with (19) an (18), yiel the control system t x(t) = k 2 0 k 3 x(t) + G/m k r/l v/l = x(t) = Ax(t) + b, (20) y(t) = c T x(t) of the form (16), where c := [1, 0, 0] T. The system output y(t) measures eviation from the esire set position; this value shoul be riven to zero in orer to suspen the ball at the set position h(0). Exercise What are the eigenvalues of the matrix A in (20)? Solve the system of ifferential equations (20). Graph the system output y(t) is a function of time t. You will fin from the above exercise that the system as escribe is not stable; the system output iverges over time an the ball will not remain suspene at the set point. We can control the system output by aing a feeback term f T x(t) to equation (20), transforming it into a linear control problem of the form (16). Specifically, we woul like to etermine a feeback gain vector f, such that the solution x(t) converges to 0 as t increases. The matrix A bf T may have two complex-value eigenvalues, say, λ 1 an λ 2, for certain vectors f. We therefore have to iscuss how exp(λ j t) behaves as t increases for complexvalue λ j. 11

12 Let α R an i = 1. We efine exp(iα) = cos(α) + i sin(α). (21) We refer to cos(α) as the real part an sin(α) as the imaginary part of the complex number exp(iα). Complex numbers can be thought of as vectors in R 2. They iffer from orinary vectors in R 2 only in that complex numbers can be multiplie, using a special rule, while vectors in R 2 cannot. Thus, we may ientify the complex number cos(α) + i sin(α) with the point (cos(α), sin(α)) in R 2. This point lives on the unit circle in R 2. We efine the magnitue of the complex number cos(α) + i sin(α) to be the length of the corresponing vector (cos(α), sin(α)) in R 2. We have cos(α) + i sin(α) = cos 2 (α) + sin 2 (α) = 1. We therefore say that cos(α) + i sin(α) lives on the unit circle in the complex plane. Returning to the eigenvalues of A bf T, express the complex eigenvalue λ 1 in the form λ 1 = λ 1,1 + iλ 1,2, λ 1,1, λ 1,2 R, i = 1, where λ 1,1 is the real part an λ 1,2 the imaginary part of λ 1. Using that the exponential of a sum is the prouct of the exponential of each term yiels From (21) we now obtain It follows that exp(λ 1 ) = exp(λ 1,1 + iλ 1,2 ) = exp(λ 1,1 ) exp(iλ 1,2 ). exp(λ 1 ) = exp(λ 1,1 )(cos(λ 1,2 ) + i sin(λ 1,2 )). exp(λ 1 ) = exp(λ 1,1 ) (cos(λ 1,2 ) + i sin(λ 1,2 ) = exp(λ 1,1 ), where we have use that exp(λ 1,1 ) > 0 an (cos(λ 1,2 ) + i sin(λ 1,2 ) = 1. When we stuy the stability of solutions of systems of ODEs, we are intereste in whether expressions of the form exp(λ j t) increase or ecrease as t increases; cf. (15). Our iscussion above shows that exp(λ j t) = exp(λ j,1 t), where λ j,1 enotes the real part of the eigenvalue λ j. We therefore only are concerne with the sign of the real part of the eigenvalues when etermining whether the solution x(t) converges to zero as t increases. 12

13 Let the matrix A an vector b be efine by (20), an let V be the eigenvector matrix an D the eigenvalue matrix of the A; cf. (10). Introuce the vectors b = [ b1, b 2, b 3 ] T := V 1 b, f := [α, 0, 0] T, f := V T f, (22) where α is a scalar to be etermine. Then A bf T = V (D b f T ) V 1. (23) Exercise Let the vectors b an f, as well as the matrix D be the same as in (22). What are the eigenvalues of the matrix D b f T? Exercise Equation (23) an Exercise provie us with an algorithm for solving the linear control problem associate with this example: 1. Compute [V,D]=eig(A) 2. Let b=[0,1,1] an compute btile=v\b 3. Let ftile=[1,0,0] 4. Fin a value of alpha so that the eigenvalues of D - alpha*btile*ftile are all have negative real part. 5. Compute f = V \ftile Coe the above algorithm in MATLAB or Octave, solve the example control problem, an make a new plot of the system output y(t) over time for your solution. 13