7
$\begingroup$

Given non-commuting matrices $A$ and $B$ of order $n$, is there a closed-form solution to the differential equation $$\frac{dX}{dt} = AX + tBX$$ with $X(0) = I$?

I know that for the reals, $x = a \exp{\int f(t)}$ is the general solution to $\dot{x} = f(t)x$, 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 $$\frac{dX}{dt} = f(t)X,\;\;X(0)=I$$ but in circumstances where $f'(t)$ may be large and a $1$st order piecewise approximation would be far more accurate than $0$th order for any given $\Delta t$. Ultimately my concern is computing $X(T)$ as quickly as possible.

Are there better techniques for accomplishing this?

$\endgroup$

1 Answer 1

5
$\begingroup$

You may use the Dyson series to solve the equation like $$\frac{d}{dt}X(t)=A(t)X(t)$$ Where $X:\mathbb{R}^{+}\rightarrow\mathbb{C}^{n}$ and $A:\mathbb{R}^{+}\rightarrow\mathbb{M}_{n\times{n}}(\mathbb{C})$. First let $$X(t)=U(t)X(0)=U(t)I$$ Where $U:\mathbb{R}^{+}\rightarrow\mathbb{M}_{n\times{n}}(\mathbb{C})$, such that $U(0)=I$ (to match the initial conditions), thus $$\frac{d}{dt}U(t)=A(t)U(t)$$ This may be converted to an integral equation $$U(t)=I+\int_{0}^{t}dt_{1}A(t_{1})U(t_{1})$$ The von-Neumann expansion than may be written $$U(t)=I+\int_{0}^{t}dt_{1}A(t_{1})+\int_{0}^{t}dt_{1}\int_{0}^{t_{1}}dt_{2}A(t_{1})A(t_{2})+\int_{0}^{t}dt_{1}\int_{0}^{t_{1}}dt_{2}\int_{0}^{t_{2}}dt_{3}A(t_{1})A(t_{2})A(t_{3})+..$$ $$=I+\sum_{k=1}^{\infty}\int_{0}^{t}dt_{1}...\int_{0}^{t_{k-1}}dt_{k}\mathcal{T}[A(t_{1})...A(t_{k})]$$ This series is known as the dyson series. There $\mathcal{T}[.]$ is the time ordering operator, i.e. $\mathcal{T}[A(t_{3})A(t_{1})A(t_{2})]=A(t_{1})A(t_{2})A(t_{3})$, it is nessesary as the matricies taken at different times do not commute, e.g. in your example $[A+tB, A+t'B]=(t-t')[B, A]$. The convergence condition for $t\in[0, \tau]$ may be shown to be $$\int_{0}^{\tau}||A(t)||_{2}dt<\pi$$ One may additionally show that this series is equaivalent to the so called time ordered exponential $$U(t)=\mathcal{T}\Big[\exp\Big(\int_{0}^{t}A(s)ds\Big)\Big]=\exp\Big(\sum_{k=1}^{\infty}\Omega(t)\Big)$$ With $$\Omega_{1}(t)=\int_{0}^{t}dt_{1}A(t_{1})$$ $$\Omega_{2}(t)=\frac{1}{2}\int_{0}^{t}dt_{1}\int_{0}^{t_{1}}dt_{2}[A(t_{1}), A(t_{2})]$$ $$\Omega_{3}(t)=\frac{1}{6}\int_{0}^{t}dt_{1}\int_{0}^{t_{1}}dt_{2}\int_{0}^{t_{2}}dt_{3}([A(t_{1}), [A(t_{2}), A(t_{3})]]-[A(t_{3}), [A(t_{2}), A(t_{1})]])$$ $$...$$ This series is known as Magnus series, has the same convergence conditions and may be shown to be equivalent to the Dyson one.

$\endgroup$
1
  • 1
    $\begingroup$ Many thanks. If I could upvote you 100 times, I would. $\endgroup$
    – COTO
    Aug 30, 2017 at 15:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .