3
$\begingroup$

I would like to calculate the Peano-Baker series with mathematica, namely an integral expansion to approximate the state transition matrix:

Say I have a matrix $A(t)$ and I want to calculate:

$\Phi(t,t_0) = e^{\int_{t_0}^t A(\sigma)d \sigma}$

To approximately calculate this I can expand the Peano-Baker Series as follows:

$\Phi(t,t_0) = I + \int_{t_0}^t A(\sigma_1)d\sigma_1+ \int_{t_0}^{t} A(\sigma_1) \int_{t_0}^t A(\sigma_2) d \sigma_2 \sigma_1+ \ldots \int_{t_0}^{t} A(\sigma_1) \int_{t_0}^t A(\sigma_2) \dots \int_{t_0}^{\sigma_{j-1}}A(\sigma_j)d\sigma_j \dots d\sigma_2 d \sigma_1 + \ldots$

What would be a smart way to do this with Mathematica? I would like a function of this sort:

PeanoBakerSeries[A_?MatrixQ,t_,t0_,j_?IntegerQ]:=Module[{n, dummies, PB}, 
n = Length@A;
dummies = Table[Symbol["\[Sigma]" <> ToString[i]], {i, 1, k}];
....]

I was also wondering if there is an undocumented function for this (I bet there is!).

$\endgroup$
  • 1
    $\begingroup$ MatrixExp[Integrate[A[s],{s, t0, t}]]? $\endgroup$ – Henrik Schumacher Aug 1 '19 at 12:05
  • 1
    $\begingroup$ Your form for the PB series looks different from the one I am accustomed to (e.g. the one here). You can use @Henrik's expression if the appropriate commutator is zero, but in general, you will want to exploit Cauchy's iterated integral formula. $\endgroup$ – J. M. will be back soon Aug 1 '19 at 12:15
1
$\begingroup$

I actually feel like I figured out a possible solution:

     PeanoBakerSeries[A_,t_,t0_,k_?IntegerQ]:=Module[{n,dummies,PB},n=Length@A[t];
dummies=Table[Symbol["\[Sigma]"<>ToString[i]],{i,1,k}];
PB=Integrate[A,{dummies[[1]],t0,t}];
PB=IdentityMatrix[n];
Do[With[{intervals=FlattenAt[{{dummies[[1]],t0,t},Sequence@@@Table[{{dummies[[i]],t0,dummies[[i-1]]}},{i,2,j}]},2]},
PB=PB+Integrate[Product[testA[dummies[[i]]],{i,1,j}],Sequence@@intervals]],{j,1,k}];
PB
]

Nevertheless I will keep the question open, in case someone feels to suggest something smarter or an undocumented function that does this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.