I want to solve a differential equation of the following form $$ \frac{d}{dt}x=A(t)x\, , $$ where $A(t)$ does not commute at different times. This equation holds on the interval $(a,b)$. Hence, the solution cannot be written as $e^{\int_a^td\tau A(\tau)}$. I have been searching now for several hours, and I only found (source: http://www.macs.hw.ac.uk/~simonm/linalg.pdf pg. 92) the so called "Neumann series" $$ x(t)=\left(I+\int_0^{t}A(\tau)d\tau + \int_0^{t}A(\tau_1)\int_0^{\tau_1}A(\tau_2)d\tau_2 d\tau_1+...\right) x_0\, . $$ (The series is very reminiscent of the Dyson series with the evolution operator in quantum mechanics, here $A(t)$ is not hermitian.) Here are my doubts:
- In the formula the lower bound is $0$. Is this part of the formula or can it be any number? in my case $a$.
- Is this an infinite sum? if yes how can I stop at a certain summand to have an approximation?
- How does this series continue? is it similar to the time ordering in quantum mechanics, i.e., one takes every possible commutation or is the third term just: $$ \int_0^{t}A(\tau_1)\int_0^{\tau_1}A(\tau_2)\int_0^{\tau_2}A(\tau_3)d\tau_2d\tau_3 d\tau_1\, . $$
- In the link above, it says, one can check that this is the solution by taking the derivative with respect to $t$. How can one take the derivative, when it depends on the upper bound of the integral and how is the product rule for non commutative matrices?
Thank you very much.
I think you probably already found the answers to your questions, but I'll still provide one for the others who like you and me would be interested by that:
Like for any Cauchy problem, $\,\,\dot{\vec{y}}(t)=f(t,\vec{y}(t))\,\,$, you must specify an initial value $(t_0,\vec{y}_0)$ that the solution $\vec{y}(t)$ should satisfy to. In other words, this should hold true: $\vec{y}(t_0)=\vec{y}_0$.
So as you might see, the author chooses the initial value to be $(0,\vec{y}_0)$ and with the Ansatz that their propose this makes totally sense since: $$ \vec{y}(0)=\left(I+\int_0^0A(\tau)\,d\tau+\int_0^0 A(\tau_1)\int_0^{\tau_1}A(\tau_2)\,d\tau_2\,d\tau_1\,+\,...\right)\vec{y}_0 = I\,\vec{y}_0=\vec{y}_0$$ So if now your initial value is $(t_0,\vec{y}_0)$, then the Ansatz for your solution would be: $$ \vec{y}(t)=\left(I+\int_{t_0}^t A(\tau)\,d\tau+\int_{t_0}^t A(\tau_1)\int_{t_0}^{\tau_1}A(\tau_2)\,d\tau_2\,d\tau_1\,+\,...\right)\vec{y}_0$$ such that when you substitute $t=t_0$, the integrals vanishe and you get your desired equality $\vec{y}(t_0)=\vec{y}_0$.
Replace $t_0$ by $a$ and you will obtain the formula you want.
As the author mentions it, this Ansatz for the solution of the initial value problem is written as a Neumann series which is a generalization of geometric series to operators, here the operator is: $$ T[O]=\int_{t_0}^{t} A(\tau)\,O(\tau)\,d\tau $$ with $O$ being a matrix of the same size as A. Then the Ansatz can be rewritten as: $$ \vec{y}(t)= \sum_{k=0}^\infty T^k[\,I\,] \,\, \vec{y}_0 \qquad \mathrm{where} \qquad T^k[\,I\,] := T^{k-1} \circ T[\,I\,] \equiv T^{k-1}[\,T[\,I\,]\,] $$ and $T^0[\,I\,] \equiv I\,$. This particular geometric series has also a convergence criteria which is that $|T|<1$ in the operator norm. We must typically prove that $T$ is a bounded linear operator on a Banach space. Then you will probably see that $|T|<1$ is equivalent to $\int_{t_0}^t ||A(\tau)||\,d\tau\,< \infty\,$. $\quad$ [to be done]
So to answer your question: If your matrix $A$ satisfies the convergence criteria, then yes stopping at a particular term in the sum above will give a more or less good approximation of your solution. Typically higher terms will contribute less than the first ones.
Yes, this is the third term of the series. Just take the operator $T$ defined above and repeatedly apply it to the last term to obtain the next one.
As the author mentions this is straightforward to prove. I use in the following lines the Leibniz integral rule since a matrix integration is simply a component wise integration. $$ \begin{align} \frac{d}{dt}\vec{y}(t) &= \frac{d}{dt}\left[ \left(I+\int_{t_0}^t A(\tau)\,d\tau+\int_{t_0}^t A(\tau_1)\int_{t_0}^{\tau_1}A(\tau_2)\,d\tau_2\,d\tau_1\,+\,...\right)\vec{y}_0 \right] = \\ &= \left(0+\frac{d}{dt}\left[\int_{t_0}^t A(\tau)\,d\tau \right] + \frac{d}{dt}\left[\int_{t_0}^t A(\tau_1)\int_{t_0}^{\tau_1}A(\tau_2)\,d\tau_2\,d\tau_1\,\right]+\,...\right)\vec{y}_0 = \\ &=\left(A(t)\,\frac{d}{dt} \bigg[\,t\,\bigg] + A(t)\,\int_{t_0}^t A(\tau_2)\,d\tau_2 \,\frac{d}{dt} \bigg[\,t\,\bigg] + \,...\right)\vec{y}_0 = \\ &= A(t) \,\, \left(I+\int_{t_0}^t A(\tau)\,d\tau+\int_{t_0}^t A(\tau_1)\int_{t_0}^{\tau_1}A(\tau_2)\,d\tau_2\,d\tau_1\,+\,...\right)\vec{y}_0 =\\ &= A(t)\,\vec{y}(t) \end{align} $$ So with that we have proven that the Ansatz is indeed a solution to the differential ordinary matrix equation. However, I must emphasize once again, the problem must satisfy the convergence criteria of the Neumann series.
(I did not get the question about the product rule, sorry)