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Linear ODE’s in NonCommutative Associative Algebras Gordon Erlebacher Florida State University Garret Sobczyk UDLA 1/27/2004 UDLAP, Department of Mathematics Linear Scalar ODE Constant coefficient a dx ax, dt x 0 x 0 x eat x0 a Coefficient a(t) dx a t x, dt x 0 x 0 t a ( s ) ds 0 xe x 0 Linear Matrix ODE dX AX , X 0 X 0 dt A Gl n, dX XA, X 0 X 0 dt A Gl n, X e At X 0 X X 0 e At The exponential matrix function is defined by n A e At t n n 0 n ! Note: AX XA Linear Matrix ODE(t) dX At X , X 0 X 0 dt A Gl n, t X e 0 A s ds dX XA, X 0 X 0 dt A Gl n, t A s ds 0 X X 0e X0 The exponential matrix function is defined by n A s ds 0 n At e t n! n 0 t If and only if A s A(t ) A( s ) A t s, t Matrix ODE More generally, if the commutator A s , A t 0 , use Picard iteration: dX At X dt X (t ) X 0 A s X ( s )ds t 0 s X 0 A s X 0 A s ' X ( s ')ds ' 0 0 t X 0 A s X 0 ds A s A s ' X 0 ds ' ds t t s 0 0 0 A s A s ' A s "X 0 ds '' ds ' ds H .O.T . t s s' 0 0 0 H.O.T. : Higher Order Terms More general Linear Matrix ODEs dX A t X XB (t ) dt dX A(t ) XB (t ) dt n dX Ai t XBi t dt i 0 Simple Matrix Equation dX AXB dt d x11 dt x21 x12 41 x11 21 x21 x22 31 x11 11 x21 1 0 4 2 A , B 0 3 1 2 42 x21 22 x22 32 x21 12 x22 Note that it is not possible to solve the equation in this form. Furthermore, the structure of A and B have been destroyed. Matrix Equation (cont.) Vec X x11 The solution is to define x12 x21 x22 T d Vec X CVec( X ) dt 41 0 C 31 0 0 42 0 32 21 0 1 0 0 22 0 2 BT A Disadvantage: the natural structure of X is destroyed. Often, X has physical significance, e.g., through its invariants. These invariances are lost when using Vec(X). and solve Further Considerations • In many problems, a scalar matrix represents a choice of coordinate system • It is often desirable to solve problems in a coordinate-free manner! • A coordinate free solution is expressed using invariants of the individual elements of the equation. Matrix Invariants Consider the matrix A. Its invariants are constant when A is subjected to a similarity transformation: A SAS 1 Invariants of A (aij) include its eigenvalues, its determinant, and i i j i j i j k ... i j k Quaternion invariants Not clear what these are, since the concept of eigenvalues is not defined in the same manner as for matrices. Example: Quaternion equation It is well-known that a quaternion can be encoded as either a 2x2 complex matrix, a 4x4 real matrix. Other representations are also possible. They are widely used in computer graphics to express 3 rotations in q q0 xi yj zk Where the basis functions have the properties: i 2 j 2 k 2 ijk 1 Based on these, it is easy to show that if q1 , q2 H where H is the space of quaternions, then q q1q2 H , q1 , q2 0 Quaternion ODE dq aqb, dt a, q, b H Further Generalization The space H of quaternions and the space G(n,C) of invertible matrices form an associative, non-commutative Algebra. Given an arbitrary associative, non-commutative algebra Alg, we wish to solve analytically dx n ai t xbi t f t dt i0 where ai t , bi t , f t , x t Alg An analytical solution of this general problem is not possible without further restrictions. We assume: bi , b j 0, i, j Spectral Basis Given an algebra Alg, any element a in the algebra has the spectral decomposition n a pi i 0 i1qi , pi , qi Alg, ij i 0 The basis functions pi , qi have remarkable properties: pi p j ij pi q j q j ij , no implicit summation qi q j 0, i j qiri 0, and qiri 1 0 Spectral Basis (cont.) pi qi are idempotents; are nilpotents of index ri i 0 are the eigenvalues of the element a in Alg Consequence of this decomposition: n 1 ri 1 f (a ) f i 0 j 0 where n i 0 pi qij n d f x 1 n f i 0 n ! dx n x i0 Minimal Polynomial Given an element a in Alg, and the spectral basis S S0 Sn i ,r 1 Si i 0 i The minimal polynomial is defined as n a a ai mi i 1 n m mi i 1 Note: pi pi a , qi qi a , polynomials of degree < m in the element a Assumptions Consider the equation dx n ai t xbi t f t dt i1 x 0 x0 with initial condition Also consider a general basis S= Si , Si pi , qi , with idempotents pi , qiri 1 , i 1, and nilpotents qi ,r Assumptions (cont.) We assume that all bi can be expanded in the elements of S: n bi t bij t j 1 rj bij t ijk t pij qijk k 1 By the properties of the spectral basis, bi t , b j s 0, when i j , t , s Linear Superposition We would like to express the solution to dx n ai t xbi t f t dt i1 We would like to express the solution as a linear superposition of simpler solutions. Right-multiply the equation by p( n ) p1 j1 p2 j2 dxp n dt Note that bi , ji t pnjn n ai t xbi t p n f t p n i1 n ai t xp n bi, j t f n t i i1 has lives in the space spanned by Si , ji pi , ji r 1 q i ,jij i Linear Superposition (cont.) Given the solution to the reduced equation, the full solution is simply n1 x i1 1 nn xp in 1 i1 pi2 pin where we have used the idempotent property: n1 1 i1 1 nn p in 1 i1 pi2 pin New Problem Solve dx n ai t xbi t f n t dt i1 where the b’s form a commuting set and live in the space spanned by a simple basis Si 1 qi Given the solution, right-multiply by permutations. qiri p n and sum over all index We concentrate on the homogeneous equation. The forcing term is handled via the standard technique of the method of variation of constants (see paper). First Step: Multiply to the right by dx a (t ) xb t dt S 1 q T q r 1 T dxS T a(t ) xb t S T a t xBS T dt where Ir N 0 1 0 0 B 0 0 0 0 0 r Nr 0 is the identity matrix of order is a nilpotent matrix, r 1 r 2 0 0 t Ir N First Step (cont.) Note: B is matrix of scalars, so commutes with elements of Alg. Therefore, dxS T a(t ) xBS T a t BxS T dt which is a matrix equation for the (1xr) matrix variable y xS Note that Bx=xB since matrices of complex numbers always commute with elements of Alg. T Solve this equation making use of the fundamental solution of the scalar equation dx t x, dt x 0 1 We denote the solution symbolically by t Fundamental Solution The solution to is dx t x, dt x 0 x0 x t t t x0 Similarly, the solution to the time-dependent linear matrix equation dX At X , X 0 X 0 dt is X t At t X 0 Note that this is a symbolic representation of the solution. One must still determine how to compute it in practice. Back to our problem … We are trying to solve dy a t B t y dt a t 0 t I r y a t N t y It is easy to solve this equation symbolically using the fundamental solution: T 0aI r 1aI N aI 0 0 r 0 r y t y xS The subscript to the fundamental solution are elements of Gl(r,Alg), the algebra of rxr matrices whose elements lie in Alg. The solution x(t) is found by left-multiplying y(t) by the row matrix Er 1 0 0 x t Er 0aIr 1 0 aI r N 0 aI r x0 S T New Notation Let a 0 I r , The ODE becomes K aN , 1K dy K y dt The solution becomes: xS T Em t t x0 S T The solution to the inhomogeneous problem can be shown to be: x t Em t t x0 S T t Em t t 1 s 1 s f s ds S T 0 General Equation: dx n ai (t ) xbi t dt i1 Rewrite the equation in the matrix form: dx A t xBT t dt where A t a1 an B t b1 bn Notation To each bi associate a matrix Bi i 0 t I ri Ni t where 0 i1 t 0 0 Ni t 0 0 0 0 Is a nilpotent of multiplicity ri i ,r 1 t i ,ri 2 t 0 i 0 Notation (cont.) Define the tensor product of spectral bases: S S1 Sn S T S1T S nT and the overbar symbol (same definition for all subscripted matrices): Clearly, Bi I r1 Bi B B1 Bn Bi i 0 I ri Ni I rn Solution Multiply to the right by ST d T xS AxBT S T AB T xS T dt which is an equation linear in y xS T Note that x, B 0 but A, B 0 The above equation is symbolically equivalent to the earlier differential equation with a single term on the right-hand side. Therefore, the solution is simply Solution (cont.) n AB T ai t Bi t i 1 K where n i i1 n K Ki i 1 The equation to solve reduces to the previous one (symbolically) dy K y dt Solution (cont.) xS y EM t t x0 S T where M m1m2 T mn The non-homogeneous solution also has a structure similar to that of the simpler ODE (with one term on the right-hand side): x t EM t t x0 S T t EM t t 1 s 1 s f s ds S T 0 Explicit Computations (advantages of the decomposition) By definition, t is the solution to d t , dt 0 I M which can be solved by Picart iteration: I M s s ds t 0 n t i 1 0 I M i s s ds Explicit Computation (cont.) One more iteration will reveal the general structure of the solution: n t I M ds1i1 s1 1 ds2 i1 s2 s2 0 i11 i2 1 n t n t n n I M ds1i1 s1 ds1 ds2 i1 s1 i2 s2 s2 i11 j 0 i1 1 i2 1 t s1 0 0 t j 1 where k n t 0 ds1 0 i 1 i 1 1 I M 0 n k t sk 1 dsk ij s j k j 1 However … n n i 1 i 1 i 0i aia I mi I M 0i aia I M t where t Alg It follows that i 1 Ki where Ki I m1 ai Ni I mn is a tensor project of nilpotent matrices: a t N t 0 mi j 1 j i j The series for therefore has a finite number of terms!!! Exact solutions may be computed. Example: Quaternion equation Consider the equation: dx a t xb t f t dt where a,b,x,f lie in the space H of quaternions. We consider 3 cases: I II III no nilpotent, single subspace b t b0 t no nilpotent, two subspaces b t b1 t p1 b2 t p2 Nilpotent of index 1, single subspace b t b0 t b1 t q bi t Case I: b t b0 t t t x t b0 a t x0 b0a f s ds 0 Case II: b t b1 t p1 b2 t p2 Solution to homogeneous equation: xH t b1a t x0 p1 b2a t x0 p2 Solution to non-homogeneous equation: 2 t xNH t bi a x0 bi a1 s f s ds pi 0 i 1 Case III: b t b0 t b1 t q Solution to homogeneous equation: xH t b0 a t x0 ds s x0 q t 0 Solution to non-homogeneous equation: xNH t xH t b0 a ds bi a1 s f s t 0 b0 a t ds1 t 0 1 bi a s1 f s1 0 where t b1a t ab a t b1 t H 0 0 s1 ds2 s2 q Possible extensions • • • • Remove the constraint that bi , b j 0 Solve higher order linear ODEs in Alg Solve systems of linear equations in Alg Asymptotic analysis of equations defined in Alg • Develop efficient numerical algorithms to compute the spectral basis of an arbitrary element of Alg FULL PAPER http://www.csit.fsu.edu/~erlebach/publications/sobczyk_erlebacher_2004.pdf