Given non-commuting matrices A and B of order n, is there a closed-form solution to the differential equation d X d t = A X + t B Xwith X ( 0 ) = I?I know that for the reals, x = a exp ⁡ ∫ f ( t ) is the general solution to x ˙ = f ( t ) x, but I'm also 99% certain this relies on the commutivity of the reals.I'm more specifically looking to numerically compute X ( T ) given the more general differential equation d X d t = f ( t ) X , X ( 0 ) = Ibut in circumstances where f ′ ( t ) may be large and a 1st order piecewise approximation would be far more accurate than 0th order for any given Δ t. Ultimately my concern is computing X ( T ) as quickly as possible.Are there better techniques for accomplishing this?

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Given non-commuting matrices   A and   B of order   n, is there a closed-form solution to the differential equation            d      X              d      t        =  A  X  +  t  B  Xwith   X  (  0  )  =  I?I know that for the reals,   x  =  a  exp  ⁡      ∫    f    (    t    )   is the general solution to             x      ˙        =  f  (  t  )  x, but I&#039;m also 99% certain this relies on the commutivity of the reals.I&#039;m more specifically looking to numerically compute   X  (  T  ) given the more general differential equation            d      X              d      t        =  f  (  t  )  X  ,      X  (  0  )  =  Ibut in circumstances where       f    ′    (  t  ) may be large and a 1st order piecewise approximation would be far more accurate than 0th order for any given   Δ  t. Ultimately my concern is computing   X  (  T  ) as quickly as possible.Are there better techniques for accomplishing this?

Sydnee Villegas · Accepted Answer

You may use the Dyson series to solve the equation like      d          d      t        X  (  t  )  =  A  (  t  )  X  (  t  )Where   X  :            R              +        →            C              n       and   A  :            R              +        →            M              n      ×              n              (      C    ). First let  X  (  t  )  =  U  (  t  )  X  (  0  )  =  U  (  t  )  IWhere   U  :            R              +        →            M              n      ×              n              (      C    ), such that   U  (  0  )  =  I (to match the initial conditions), thus      d          d      t        U  (  t  )  =  A  (  t  )  U  (  t  )This may be converted to an integral equationThe von-Neumann expansion than may be written  U  (  t  )  =  I  +      ∫          0              t        d      t          1        A  (      t          1        )  +      ∫          0              t        d      t          1            ∫          0                      t                  1                      d      t          2        A  (      t          1        )  A  (      t          2        )  +      ∫          0              t        d      t          1            ∫          0                      t                  1                      d      t          2            ∫          0                      t                  2                      d      t          3        A  (      t          1        )  A  (      t          2        )  A  (      t          3        )  +  .  .  =  I  +      ∑          k      =      1              ∞            ∫          0              t        d      t          1        .  .  .      ∫          0                      t                  k          −          1                      d      t          k            T    [  A  (      t          1        )  .  .  .  A  (      t          k        )  ]This series is known as the dyson series. There       T    [  .  ] is the time ordering operator, i.e.      T    [  A  (      t          3        )  A  (      t          1        )  A  (      t          2        )  ]  =  A  (      t          1        )  A  (      t          2        )  A  (      t          3        ), it is nessesary as the matricies taken at different times do not commute, e.g. in your example   [  A  +  t  B  ,  A  +      t    ′    B  ]  =  (  t  −      t    ′    )  [  B  ,  A  ]. The convergence condition for   t  ∈  [  0  ,  τ  ] may be shown to be      ∫          0              τ            |        |    A  (  t  )      |              |              2        d  t  &amp;lt;  πOne may additionally show that this series is equaivalent to the so called time ordered exponential  U  (  t  )  =      T              [        exp  ⁡            (            ∫          0              t        A  (  s  )  d  s            )                  ]        =  exp  ⁡            (            ∑          k      =      1              ∞        Ω  (  t  )            )      With      Ω          1        (  t  )  =      ∫          0              t        d      t          1        A  (      t          1        )      Ω          2        (  t  )  =      1    2        ∫          0              t        d      t          1            ∫          0                      t                  1                      d      t          2        [  A  (      t          1        )  ,  A  (      t          2        )  ]      Ω          3        (  t  )  =      1    6        ∫          0              t        d      t          1            ∫          0                      t                  1                      d      t          2            ∫          0                      t                  2                      d      t          3        (  [  A  (      t          1        )  ,  [  A  (      t          2        )  ,  A  (      t          3        )  ]  ]  −  [  A  (      t          3        )  ,  [  A  (      t          2        )  ,  A  (      t          1        )  ]  ]  )  .  .  .This series is known as Magnus series, has the same convergence conditions and may be shown to be equivalent to the Dyson one.

Given non-commuting matrices A and B of order n , is there a closed-form solution t

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