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Solve matrix differential system with mathématica

  1. Jul 6, 2011 #1
    hello ,

    I need to solve a differential system with mathematica that has the form :


    knowing that B and A are (12,12) Matrix and with initial conditions B(0)=I

    Can you help me please

  2. jcsd
  3. Jul 6, 2011 #2
    Try something like

    Code (Text):
    n = 12;
    B[t_] = Array[Subscript[b, ##][t] &, {n, n}];
    Bt = Array[Subscript[b, ##] &, {n, n}];
    id = IdentityMatrix[n];
    A = SparseArray[{Band[{1, 1}] -> 2., Band[{1, 2}] -> -1, Band[{2, 1}] -> -1}, {n, n}];
    The key command for getting a matrix equation into something that DSolve and friends like is the command LogicalExpand. (You can also use combinations of Flatten and Thread, but that's not as neat).

    Code (Text):

    soln = DSolve[LogicalExpand[B'[t] == A.B[t] && B[0] == id], Flatten@Bt, t];
    Where we could have used B[t] instead of Bt, but then the resultant replacement rules (soln) would be less useful. For the purely numerical example above, NDSolve would be quicker to get a solution for any particular range of t.

    We can then plot the result using ArrayPlot:
    Code (Text):

    Animate[ArrayPlot[Evaluate[B[t] /. soln[[1]]]], {t, 0, 5 n}]
  4. Jul 6, 2011 #3
    Thank you very much for your help....but the problem is that since I'm not used to Mathematica I don't understand so functions that you have used.

    For example I supposed in this part you defined the Matrix A
    A = SparseArray[{Band[{1, 1}] -> 2., Band[{1, 2}] -> -1, Band[{2, 1}] -> -1}, {n, n}];

    But for me the Matrix A has been calcultaed before.
    Can you show me please how to implemented in my programm solution that you proposed to me .

    Here is my programm for calcultaing the matrix A :

    A = Table[0, {12}, {12}];
    premier = -Inverse[M] . CM;
    second = -Inverse[M]. KM;
    For[k = 1, k <= 3, k++,
    For[l = 1, l <= 3, l++,
    A[[k]][[l]] = premier[[k]][[l]];
    A[[k]][[l + taille]] = second[[k]][[l]];
    A[[k + taille]][[l]] =
    ID[[k]][[l]]; (*Calcul de la matrice d'état*)

    A[[k + taille]][[l + taille]] = Nul[[k]][[l]]

    Where M, CMand KM are periodic matrix (6,6)

    Thank you very much I really appreciate
  5. Jul 6, 2011 #4
    I only defined A like that since I needed some coefficient matrix. You can see what I (almost) arbitrarily chose by running MatrixForm[A].
    By the way, the stuff in (* ... *) is a comment - an alternative choice of A that I should have deleted before posting.

    So to make the code I posted in your program work, you just need your own A.
    There are quite a few things not defined in your code, which makes it hard to tell what you want.
    It kind of looks like you're constructing the block matrix
    {{premier, second}, {ID, Nul}}
    but then your iterators should go up to 6....
    Anyway, I'll assume that's what you're doing.
    If it's not, can you describe in words how you construct A and what the result should be?

    Let's first construct some test cyclic matrices:
    Code (Text):

    M  = NestList[RotateRight, RandomReal[{0, 1}, 6], 5];
    CM = NestList[RotateRight, RandomReal[{0, 1}, 6], 5];
    KM = NestList[RotateRight, RandomReal[{0, 1}, 6], 5];
    and your premier and second matrices:
    Code (Text):

    premier = -Inverse[M].CM;
    second = -Inverse[M].KM;
    I'll assume that Nul is the zero matrix and ID the identity matrix
    Code (Text):

    Nul = Array[0 &, {6, 6}];
    ID = IdentityMatrix[6];
    Then your loop becomes
    Code (Text):

    A = Array[0 &, {12, 12}];
    taille = 6;
    For[k = 1, k <= taille, k++,
      For[l = 1, l <= taille, l++,
       A[[k]][[l]] = premier[[k]][[l]];
       A[[k]][[l + taille]] = second[[k]][[l]];
       A[[k + taille]][[l]] = ID[[k]][[l]];
       A[[k + taille]][[l + taille]] = Nul[[k]][[l]]
    Which is much more easily created using ArrayFlatten:
    Code (Text):
    ArrayFlatten[{{premier, second}, {ID, Nul}}]
    ArrayFlatten would also be faster since the internal code would be optimized.
  6. Jul 6, 2011 #5
    Oops... I confused periodic matrix (M^(k+1) = M) with cyclic matrix (M_{i,j}=M_{i+1,j+1}) - not that it makes any difference to the above example.

    Also, if Nul really is the zero matrix then you can replace Nul[[k]][[l]] with 0 in the loop and Nul with 0 in the ArrayFlatten.

    Also, when accessing elements of arrays, it is about twice as fast to use M[[k,l]] than M[[k]][[l]]. To check this, run

    Code (Text):
    M = RandomReal[{0,1},{6,6}]
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