Re: Matrix differential equation

*To*: mathgroup at smc.vnet.net*Subject*: [mg51035] Re: Matrix differential equation*From*: p-valko at tamu.edu (Peter Valko)*Date*: Sat, 2 Oct 2004 03:18:15 -0400 (EDT)*References*: <ciua6v$90v$1@smc.vnet.net> <cj0n8p$ldd$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

After exchanging several mails in the background I see what the propblem is: A = -{{1, 2, 3}, {4, 5 , 6}, {7, 8 , 9}}; K = -{{0, 10, 0}, {0, 0 , 0}, {0, 0 , 0}}; KT = Transpose[K]; and matrixExpA = First[X /. NDSolve[X'[t] == 0.5*IdentityMatrix[3]+ X + K.X + X.KT && X[0] == IdentityMatrix[3], X, {t, 0, 10}]] does not work. The reason is, that Mathematica rearranges 0.5*IdentityMatrix[3]+ X + K.X + X.KT before it realizes that X is a 3 by 3 matrix. To avoid the problem, first I write A = -{{1, 2, 3}, {4, 5 , 6}, {7, 8 , 9}}; K = -{{0, 10, 0}, {0, 0 , 0}, {0, 0 , 0}}; KT = Transpose[K]; then I write: f[X_]:= 0.5*IdentityMatrix[3]+ X + K.X + X.KT /; Dimensions[X]=={3,3} This way I have a right-hand-side that will evaluate only if Mathematica already knows that X is a 3 by 3 matrix. Now matrixExpA = First[X /. NDSolve[X'[t] == f[X[t]] && X[0] == IdentityMatrix[3], X, {t, 0, 10}]] works and provides an interpolating function. Finally, matrixExpA[1.] evaluates to {{311.319, -32.1828, 0.}, {-32.1828, 3.57742, 0.}, {0., 0., 3.57742}} Regards Peter p-valko at tamu.edu (Peter Valko) wrote in message news:<cj0n8p$ldd$1 at smc.vnet.net>... > I tried to reproduce the error message, but surprisingly, there was no > error and the sample problem worked. > > The result is an > InterpolatingFunction[{{0., 10.}}, <>] > and I can calculate it at time zero or at time = 1. > > For instance > matrixExpA[1] > gives > {{166.52, -16.4872, 0.}, {-16.4872, 1.64872, 0.}, {0., 0., 1.64872}} > > > Therefore, I have to conclude, that you do not input exactly what you > wish to input, or that something is not Cleared before you use it. > > I would suggest to print out > k > A > K > KT > before using the command > > > matrixExpA = > > X /. First[ > > NDSolve[X'[t] == 0.5*IdentityMatrix[3] +(X[t]) + K.X[t] + X[t].KT && > > X[0] == IdentityMatrix[3], X, {t, 0, 10}]] > > > to make sure that they really have the value I think they have. > > Regards > Peter > > > > "Marco Malvaldi" <M.Malvaldi at chem.rug.nl> wrote in message news:<ciua6v$90v$1 at smc.vnet.net>... > > It is possible to numerically solve a general matrix differential equation > > with Mathematica? From the example in Mathematica 5 tutorial, I found that > > this problem can be solved: > > > > k = 10 > > A = -{{1, 2, 3}, {4, 5 , 6}, {7, 8 , 9}}; > > K = -{{0, k, 0}, {0, 0 , 0}, {0, 0 , 0}}; > > X0 = {{1, 0, 0}, {0, 1 , 0}, {0, 0 , 1}}; > > KT = Transpose[K] > > > > matrixExpA = > > X /. First[ > > NDSolve[X'[t] == 0.5*IdentityMatrix[3] .(X[t]) + K.X[t] + X[t].KT && > > X[0] == IdentityMatrix[3], X, {t, 0, 10}]] > > > > Anyway, when I try to submit this problem: > > > > matrixExpA = > > X /. First[ > > NDSolve[X'[t] == 0.5*IdentityMatrix[3] +(X[t]) + K.X[t] + X[t].KT && > > X[0] == IdentityMatrix[3], X, {t, 0, 10}]] > > > > the answer is: > > > > NDSolve::ndfdmc: Computed derivatives do not have dimensionality consistent > > with the initial conditions. > > > > even if what I'm doing is to sum a 3x3 matrix to a system of 3X3 matrix. > > > > Regards > > Marco Malvaldi