How to do Magnus expansion for time-dependent companion matrix?

[adf89812]

TL;DR

Magnus expansion of companion matrix time-dependent or solve by substitution how?

For example, consider the following system of 2 first order ODEs:
$$
\left\{\begin{array}{l}
x_1^{\prime}=2 t x_1+t^2 x_2 \\
x_2^{\prime}=t^3 x_1+4 t x_2
\end{array}\right.
$$

This is a linear homogeneous system of 2 first order ODEs with $$A(t)=\left[\begin{array}{ll}2 t & t^2 \\ t^3 & 4 t\end{array}\right]$$.


"Secondly, the substitution method works in the same manner as usual. Indeed, the first line of the system leads to $$y = t^{-2}\dot{x} + 2t^{-1}x$$, which can be differentiated in order to find $\dot{y}$ in terms of $$x$$ and $$t$$. Next, plugging these expressions into the second line, you will end up with a second-order linear ODE with non-constant coefficients for $$x$$, which itself might not be easy to solve in the present case."

"
Firstly, you mentioned diagonalization; however, in that case, the eigenvalues and the eigenvectors will be themselves time-dependent. If $$S$$ denotes the change of basis allowing the diagonalization of $$A$$ as $$D$$, i.e. $$ D= SAS^{-1}$$, then the system of equations $$\dot{u} = Au$$, where $$u = (x,y)$$, becomes
$$
\dot{v} = \partial_t(Su) = S\dot{u} + \dot{S}u = \left(SA + \dot{S}\right)u = \left(SAS^{-1} + \dot{S}S^{-1}\right)Su = \left(D + \dot{S}S^{-1}\right)v
$$
for $$v = Su$$. This new system might be even harder to solve yet, because of the extra (non-diagonal) term $$\dot{S}S^{-1}$$."

 

[anuttarasammyak]

Please let me know
x’=\dot{x}=\frac{dx}{dt}?

 

[billtodd]

So is it basically $\dot{x}(t)=A(t)x(t)$?
One sort of solution is by $x(t)=e^{\int^t A(s)ds}x(0)$.

I think so, am I wrong?

 

[renormalize]

I think so, am I wrong?

You are correct only if $x,A$ are scalars (1-dimensional). Otherwise, if $x(t)$ is an n-dimensional vector and $A(t)$ is a $n\times n$ matrix then the correct solution involves the ordered-exponential involving nested integrals over products of $A(t)$. This will reduce to your usual exponential for special matrices that commute at different times, like a constant matrix.

 

[billtodd]

You are correct only if $x,A$ are scalars (1-dimensional). Otherwise, if $x(t)$ is an n-dimensional vector and $A(t)$ is a $n\times n$ matrix then the correct solution involves the ordered-exponential involving nested integrals over products of $A(t)$. This will reduce to your usual exponential for special matrices that commute at different times, like a constant matrix.

I was refering to the vector case.
Obviously using $e^A=\sum_{n=0}^\infty A^n/n!$.
And here (in the OP) you first integrate $A$, and then plug it to the sum. Though I am not sure if there's a closed form solution here.

 

[renormalize]

I was refering to the vector case.
Obviously using $e^A=\sum_{n=0}^\infty A^n/n!$.
And here (in the OP) you first integrate $A$, and then plug it to the sum. Though I am not sure if there's a closed form solution here.

Sorry, that doesn't work to solve your vector ODE for a general matrix $A(t)$. Just try it out: start from your proposed "solution" $x\left(t\right)=\exp\left[\intop_{0}^{t}A\left(t^{\prime}\right)dt^{\prime}\right]x\left(0\right)$ with the exponential expanded as a series. Now differentiate the series term-by-term to see if $x(t)$ satisfies $\dot{x}\left(t\right)=A\left(t\right)x\left(t\right)$. (Hint: it won't unless $A(t_1)$ commutes with $A(t_2)$ for all pairs $t_1,t_2$ (i.e., it's a constant matrix) so that $A$ can be moved completely to the left of the exponential.

 

[julian]

Hi @billtodd. If you write $x(t) = M(t) x(0)$ then $\dot{x} (t) = \dot{M} (t) x(0)= A(t) x(t) = A(t) M(t)x(0)$. Implying you need to solve:

\begin{align*}
\frac{d}{dt} M(t) = A(t) M(t)
\end{align*}

subject to $M(0)= \mathbb{1}$. I derived the formal solution to this for the general case of a time dependent matrix $A(t)$ here:

https://www.physicsforums.com/threa...unction-valued-matrices.1046714/#post-6814958

and it involves the time-ordered product of matrices $A(t_1),A(t_2), \dots$. This solution reduces to $M(t) = \exp (\int_0^t A (t') dt')$ in the case where $A(t)$ is a constant matrix.

 

[billtodd]

Sorry, that doesn't work to solve your vector ODE for a general matrix $A(t)$. Just try it out: start from your proposed "solution" $x\left(t\right)=\exp\left[\intop_{0}^{t}A\left(t^{\prime}\right)dt^{\prime}\right]x\left(0\right)$ with the exponential expanded as a series. Now differentiate the series term-by-term to see if $x(t)$ satisfies $\dot{x}\left(t\right)=A\left(t\right)x\left(t\right)$. (Hint: it won't unless $A(t_1)$ commutes with $A(t_2)$ for all pairs $t_1,t_2$ (i.e., it's a constant matrix) so that $A$ can be moved completely to the left of the exponential.

So how would you solve it?
One can try power series for both $x_1(t),x_2(t)$. But then you translate the differential equations to recurrence equations, which doesn't necessarily make the problem any easier.
OK, now I understand what is that ordered exponential.
First time I had seen it was in QM2.
https://en.wikipedia.org/wiki/Dyson_series
But one can only solve this numerically with the ordered exponential.