===== Linear first-order ODE system =====
==== Set ====
| @#55CCEE: context     | @#55CCEE: $ A:\mathbb R\to\mathrm{Matrix}(n,\mathbb R) $ |
| @#55CCEE: context     | @#55CCEE: $ b:\mathbb R\to\mathbb R^n $ |
| @#FFBB00: definiendum | @#FFBB00: $ y \in \mathrm{it} $ |
| @#55EE55: postulate   | @#55EE55: $ y:C^k(\mathbb R,\mathbb R^n) $  |
| @#55EE55: postulate   | @#55EE55: $ y'(t)=A(t)\ y(t)+b(t) $ |

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=== Theorems ===
There exists a matrix $S(t,s)$ such that the solution of the equation above is of the form 

^ $y(t)=S(t,0)\ y_0+\int_0^t\ S(t,s)\ b(s)\ \mathrm ds$ ^

We don't know $S(t,s)$ in general, but 

^ $S(t,0)=\lim_{n\to\infty}\ \mathrm{exp}\left({\frac{t}{n}A(\frac{n-1}{n}t)}\right)\cdots \ \mathrm{exp}\left({\frac{t}{n}A(0)}\right)$ ^

== Two special cases ==

  * For constant $A$, one has 

^ $S(t,s)=\mathrm{exp}\left((t-s)\ A\right)$ ^

and so

^ $y(t)=\mathrm{exp}\left(t A\right)\cdot\left(y_0+\int_0^t\ \mathrm{exp}\left(-s A\right)\ b(s)\ \mathrm ds\right)$ ^

  * The equation $y'(t)=A(t)y(t)$ is solved by $y(t)=\mathrm{e}^{\int A(t)\,\mathrm dt}y(0)$. 

We can in fact sketch how to deal with this equation in cases where $A(t)$ is a more general operator. Dyson series: Say we at least know how to apply $A(t)$. The iterative solution technique for the equation is $y_{n+1}(t)=y(0)+\int_{0}^t A(t)\,y_n(t)\,\mathrm dt$. Note that "$f(x):=y(0)+\mathrm{int}x $" iterated with initial condition $y(0)$ gives $\left(\sum_{n=0}^\infty\mathrm{int}^n\right)y(0)$. Factors $\frac{1}{n!}$ are introduces when time-ordering the integrand and the resulting series is hence mnemonically written as $y(t)=\mathcal T\exp(\mathrm{int}_{t_0}^tA(t))y(0)$

  * For $A(t),b(t)$ one-dimensional one has

^ $S(t,s)=\mathrm{e}^{I(t)-I(s)}\,\,\mathrm{with}\,\, I(s):=\int_0^s A(\tau)\ \mathrm d\tau$ ^

=== Reference ===
Wikipedia: [[http://en.wikipedia.org/wiki/Dyson_series|Dyson series]]

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=== Context ===
[[Square matrix]]
=== Subset of ===
[[ODE system]]