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    • CommentRowNumber1.
    • CommentAuthorzskoda
    • CommentTimeJul 22nd 2011
    • (edited Jul 22nd 2011)

    I want to solve the nonhomogeneous system of n 2n^2 ODEs with constant coefficients for n 2n^2 functions P ij=P ij(μ)P_{i j} = P_{i j}(\mu):

    dP ijdμt s=1 nk isP js=k ij \frac{d P_{i j}}{d\mu} - t \sum_{s=1}^n k_{i s}P_{j s} = k_{i j}

    where i,j=1,,ni,j = 1,\ldots,n for arbitrary nn where tt is a parameter and the initial condition is

    P ij(μ=0)=q ij P_{i j}(\mu = 0)= q_{i j}

    In matrix form one writes dP/dμ=ktkP Td P/d\mu = k - t k P^T, and P(0)=qP(0) = q, where P TP^T is the transpose of PP and kP Tk P^T assumes the matrix product.

    For small nn one can try to solve by brute force (solving the homogeneous first and then, say, variation of constants method), but I am not sure I can get a formula for general nn. Help from more skillful ones appreciated. Of course, if the parameter t=0t = 0 the solution is P ij=μk ij+q ijP_{i j} = \mu k_{i j} + q_{i j} but I want for nonzero tt.

    • CommentRowNumber2.
    • CommentAuthorzskoda
    • CommentTimeJul 22nd 2011
    • (edited Jul 22nd 2011)

    No! I am wrong. The differential equation for my purpose should be (if I am not mistaken again) without the transpose (cf.coproduct+for+Ugln+dual (zoranskoda)), i.e.

    dP ijdμt s=1 nk isP sj=k ij \frac{d P_{i j}}{d\mu} - t \sum_{s=1}^n k_{i s}P_{s j} = k_{i j} P ij(μ=0)=q ij P_{i j}(\mu = 0)= q_{i j}

    and in a matrix form dP/dμ=ktkPd P/d\mu = k - t k P, P(0)=qP(0)=q.

    Thus we can use, for t0t\neq 0, the matrix substitution R=tPIR = t P-I to get simply the homogeneous matrix equation dR/dμ=tkRd R/d\mu = -t k R (here tt is a scalar, kk and RR are n×nn\times n-matrices and kRk R denotes the matrix product) with the initial condition R(0)=tqIR(0) = t q -I i.e. R(0) ij=tq ijδ ijR(0)_{i j} = t q_{i j} - \delta_{i j}. This is now much easier than the one with the transpose I had before.

    To relate it to known examples, one sees that the equation in this case, without the transpose, do not mix different jj-s, so one has nn independent systems, one for each jj (what is not the case in 1 where we had a transpose).

    So for a fixed jj, call (for emphasis) R ij=:R˜ iR_{i j} =: \tilde{R}_i and we have simply

    dR˜ i/dμ=t sk isR˜ s d \tilde{R}_i/d\mu = -t \sum_s k_{i s}\tilde{R}_s

    what is in a rather standard form for homogeneous systems of first order ODEs with standard exponential matrix solution.

    Thus the formal solution is

    R=e μtkR(0) R = e^{-\mu t k} R(0)

    i.e. R ij= sexp(μtk) isR(0) sjR_{i j} = \sum_s exp(-\mu t k)_{i s} R(0)_{s j}. where exp(μtk)exp(-\mu t k) is the matrix exponential.

    Now this, with R(0) sj=tq sjδ sjR(0)_{s j} = t q_{s j}-\delta_{s j} gives, in matrix form,

    tPI=e μtk(tqI) t P - I = e^{-\mu t k} (t q- I)


    P(μ)=e μtkq+Ie μtkt P(\mu) = e^{-\mu t k} q + \frac{I - e^{-\mu t k}}{t}

    Note that qq and kk are constant matrices and tt,μ\mu are scalars. For μ=0\mu = 0 we have P(0)=qP(0) = q and for small tt we have P=μk+q+(μ 2k 2/2μkq)t+O(t 2)P = \mu k + q + (\mu^2 k^2/2 - \mu k q) t + O(t^2), agreeing with (i.e. with continuous limit to) t=0t = 0 result P=μk+qP = \mu k + q.

    • CommentRowNumber3.
    • CommentAuthorzskoda
    • CommentTimeJul 22nd 2011

    Now one needs (for my original purpose) also to compute P(K 1(k),q)P(K^{-1}(k),q) where K 1K^{-1} is inverse of the function kP(k,0)k\mapsto P(k,0), and P(k,q)P(k,q) denotes the function P(μ=1)P(\mu = 1) calculated above with kk and qq as above. In our case, we need to invert

    kIe tkt k \mapsto \frac{I - e^{-t k}}{t}

    what gives k1tln(Itk)k \mapsto -\frac{1}{t} ln(I-t k) where we symbolically use the matrix logarithm. Thus we obtain

    P(K 1(k),q)=k+qtkq P(K^{-1}(k),q) = k + q - t k q

    what gives a simple quadratic coproduct.