In these chapter 2A notes write vectors in boldface to reduce the ambiguity of the notation.

Transcription

1 1 2 Linear Systems In these chapter 2A notes write vectors in boldface to reduce the ambiguity of the notation 21 Matrix ODEs Let and is a scalar A linear function satisfies Linear superposition ) Linear scaling: Example: is not a linear function instead it is affine General form of a linear function: In matrix notation: Assume is a real matrix and General form for a linear ODE Example: (Linearization about an equilibrium point) Let Recall Taylor series expansion about Linear approximation at corresponds to throwing away the higher order terms Now suppose The same equation applies if the derivative is interpreted as a matrix Now suppose has an equilibrium point Expand the right hand side of the ODE in Taylor series about and use to obtain: Finally substitute let and make a linear approximation by throwing away the higher order terms to get This process of linearization about an equilibrium point is an important technique in the analysis of ODEs

2 2 Eigenvalues and Eigenvectors Let A be a real matrix and An eigenvector v of A is a nonzero solution of is the corresponding eigenvalue From linear algebra we know that the eigenvalues of a matrix A are given by the roots of the characteristic polynomial Look for solutions of [1] of form Substitution into [1] leads to Since is nonzero this implies with solution for an arbitrary constant The corresponding solution of [1] is Example: Consider the matrix Eigenvalues are solutions of They are and The first eigenvector satisfies or Letting we have This has the solution Eigenvectors are determined up to an arbitrary multiplicative factor Similarly find Two solutions of [1] for this matrix A are and Sketch examples of and in the ( plane Meiss calls these straight line solutions

3 3 Diagonalization Since [1] is linear linear combinations of solutions are themselves solutions Ex: If has linearly independent eigenvectors then these span In this case we also say that has a complete set of eigenvectors The matrix is nonsingular [Vertical bars separate column vectors] Then is the matrix with eigenvalues on the main diagonal and zeros in all of the other slots Multiplying by on the left to obtain In this case we say is diagonalizable or semisimple Example (continued) The inverse of any matrix is where Then use from the example above to obtain and Then we find If has linearly independent eigenvectors then any can be written as a linear combination of these where can be regarded as the coordinates of the point in eigenvector coordinates The matrix transforms from eigenvector coordinates to standard coordinates Correspondingly the matrix transforms back from standard coordinates to eigenvector coordinates With this in mind consider [1] Multiply on the left by to get:

4 4 [2] where are the coordinates of the point in the basis of eigenvectors Since is a diagonal matrix the equations [2] are not coupled The i^th component is just with solution for some constant In matrix notation this solution is written where Summary Our results in this subsection allow us to solve the following initial value problem in the case that the real matrix has a complete set of eigenvectors : The matrix initial condition by is nonsingular Multiply both the differential equation and the on the left to get Solve this uncoupled system to get and then multiply on the left by to get [3] where 22 Two-Dimensional Linear Systems Classify the two-dimensional linear systems according to the qualitative nature of their solutions These are systems of the form where is a matrix Eigenvalues are the roots of the characteristic polynomial Here is the trace of and is the determinant of The roots are

5 5 [6] where the is the discriminant of Qualitatively different cases of eigenvalues divide plane into 5 regions: Sketch of plane and the parabola Include in each of the 5 regions a sketch of the complex plane This is Meiss Fig 21 Note that the eigenvalues are distinct off of the parabola According to a theorem of linear algebra when the eigenvalues are distinct the corresponding eigenvectors will be too Then is diagonalizable and the general solution of the initial value problem has the form of [3] For the 2D case this may be written [4] The 5 regions of the plane correspond to 5 geometrically distinct phase portraits: [A] Unstable node: There are straight line solutions moving away from the origin that correspond to initial conditions on the eigenspaces and Other solutions are asymptotically parallel to as and asymptotically parallel to as Sketch -plane and several trajectories Like Meiss Fig 22 [B] Stable node: This is like [A] with arrows reversed and with and interchanged Straight line solutions move toward the origin on and Other solutions are asymptotically parallel to as and asymptotically parallel to as Sketch -plane and several trajectories [C] Saddle: Straight line solutions move away from the origin on and towards the origin on Other solution are asymptotic to as and asymptotic to as Sketch -plane and several trajectories Like Meiss Fig 23 [D] Unstable focus: Both the eigenvalues and eigenvectors are complex conjugate (To see this just take the complex conjugate of A ) Let where and are real vectors The solution is still given by [4] where and must also be conjugate in order that the solution be real Let these expressions into the right hand side of [4] one obtains [5] Substituting

6 6 We have written The solution is an exponentially expanding spiral Example The following phase portrait was constructed for the linear system with For this case and so that [E] Stable focus: The analysis is similar to that for an unstable focus The solution is an exponentially contracting spiral There are also several cases corresponding to boundaries between the 5 regions on the -plane The first two cases correspond to non-isolated equilibria which occur when there is a zero eigenvalue giving

7 7 [a] Unstable degenerate equilibrium: The eigenspace is a line of unstable equilibria Solutions that originate off this line move away exponentially as along lines parallel to [b] Stable degenerate equilibrium: The eigenspace is a line of stable equilibria Solutions that originate off this line move away exponentially as along lines parallel to [c] Center: are obtained for and The analysis is given by [5] with Trajectories are ellipses This case is particularly important both because it occurs in bifurcations (chapter 8) and because Hamiltonian systems always have To analyze the boundary states corresponding to the parabola we need to recall some concepts from linear algebra beginning with two concepts of the multiplicity of an eigenvalue: An eigenvalue has algebraic multiplicity if the characteristic polynomial can be written where ie is a -fold root of the characteristic polynomial An eigenvalue has geometric multiplicity if it has linearly independent eigenvectors ie and A theorem of linear algebra is that the geometric multiplicity is at most the algebraic multiplicity When the geometric multiplicity is less than the algebraic multiplicity the matrix is defective The column space or range of a matrix is defined to be the span of its column vectors If then The rank of B is the dimension of the range: Then the geometric multiplicity of is The null space or kernel of a matrix B is the set of vectors that the matrix send to zero: For example consider an eigenvalue and eigenvector of matrix These satisfy or equivalently Then The dimension of the kernel is called the nullity of a matrix:

8 8 If matrix has columns a theorem of linear algebra implies that Continue the example If the geometric multiplicity of is then has linearly independent eigenvectors associated with and Return to the analysis of the boundary states in the -plane on the parabola where is the trace of and is the determinant of Then [6] implies that the matrix has an eigenvalue of algebraic multiplicity 2 When the geometric multiplicity of is 2 That is for every vector This implies Every nonzero vector is an eigenvector and the solution of is uniform expansion away from the origin for or uniform contraction toward the origin for When the geometric multiplicity of is 1 the matrix has only a single eigenvector This means [4] cannot be the general solution of the initial value problem The solution will be obtained in section Exponentials of Operators Let be a vector space We may have or but results in this section also apply to infinite dimensional vector spaces An operator maps a vector to some other vector A linear operator T satisfies linear superposition and scaling If and and are scalars then If and is a basis for then T will be represented by a matrix in that basis In this case may replaced by in the equations below We write the space of linear operators on is itself an dimensional vector space Notice that a matrix representing has components If then represents the Euclidean norm: We will also define an operator norm The operator norm satisfies the usual properties for norms In particular the triangle inequality is satisfied If S and T are operators on then [The matrix norm is a continuous function of its argument See Hirsch and Smale p 78]

9 9 A linear operator also satisfies since for For any integer we have by the same logic Since this is true for all we conclude [6] An operator for which is said to be bounded Example Suppose On a finite dimensional vector space the sphere is compact and the Euclidean norm is continuous Therefore attains a maximum value on the sphere and is bounded A series in a normed vector space is absolutely convergent if the series converges This condition implies that converges (Hirsch and Smale 1974) Recall the Taylor series for about from calculus: We formally define the exponential of an operator in the same way: [7] Lemma 21 If T is a bounded linear operator then is as well Proof By [6] and the series of real numbers converges to Therefore the series [7] converges absolutely by the comparison test It also follows that Properties of the exponential The first 6 of these are from Meiss His Roman numerals are given but in my order I have added a property (vii) [i] This is a direct consequence of the definition of the exponential [v] If then

10 10 Notice and insert in the definition of the exponential [vi] If is an eigenvector of with eigenvalue then Notice and let [iv] If B is nonsingular then Using the definition of the exponential [iii] If A and B commute then (the law of exponents!) See exercise #6 The idea of part (a) is that if and are real numbers then the series for their exponentials can be manipulated to show that Since matrices and commute their series can be manipulated in the same way However the manipulation is complicated Parts (b) and (c) provide a much shorter proof [ii] Follows directly from the property [iii] [vii] The matrices and commute Let be a positive integer then and let The three examples given at the end of this section are important The first anticipates results in section 26 Example (Nilpotent matrices) A matrix N has nilpotency k if but For example consider where has nilpotency 2 Every matrix commutes with the identity matrix; in particular the matrices and N commute Then from property (iii) where only the first two terms in the series for are nonzero Example (Roots of Identity) This result will be used in the next example The matrix

11 11 has powers Insert into the definition of the matrix exponential and recall the MacLaurin series To obtain Example (Rotation Matrix) If is real and has a pair of complex conjugate eigenvalues then the associated eigenvectors will also be complex conjugate Let the eigenvectors be denoted and We will see in section 25 that when A is transformed using the matrix we get the form From the example above The matrix in the last expression rotates a vector in the plane clockwise by an angle 24 Fundamental Solution Theorem Theorem 23 Let be an matrix Then the initial value problem [1] has the unique solution Proof First demonstrate that the solution works The differential equation is satisfied:

12 12 The initial condition is also satisfied: To show that the solution is unique suppose Then use the product rule to obtain is another solution of the initial value problem where we have also used the fact that and commute Therefore is a constant and by property (ii) in section 23 Moreover since the two solutions must satisfy the same initial condition Example Mass-spring system Consider a spring of natural length L mass m and Hooke s law constant k on a frictionless table: The figure shows the spring extended a distance of motion gives beyond its natural length Newton s law This ODE is affine but we can make it linear by substituting Then get This can be further simplified by introducing the dimensionless time where Denote to get Write as a first order system

13 13 Consider the initial value problem using : Note that and According to the classification of linear systems the solution is a center By the Fundamental Theorem the solution of the initial value problem is Finally express the answer in terms of : where and Consider the result of Theorem 23 for a set of initial condition vectors Put the initial condition vectors together to form a matrix and the solutions together to form We then have Theorem 24 The matrix initial value problem Has the unique solution A matrix initial value problem is used in the beautiful proof of the law of exponents for the matrix exponential given by problem 6 parts (b) and (c)! The fundamental matrix solution is the solution of the initial value problem Its name arises from the fact that solve the initial value problem [1] for any Check by calculating (t) and using this formula When is a constant matrix We will return to fundamental matrix solution in section 28 when we will consider a function of time