Linear first-order ODE system


context $ A:\mathbb R\to\mathrm{Matrix}(n,\mathbb R) $
context $ b:\mathbb R\to\mathbb R^n $
definiendum $ y \in \mathrm{it} $
postulate $ y:C^k(\mathbb R,\mathbb R^n) $
postulate $ y'(t)=A(t)\ y(t)+b(t) $


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


Wikipedia: Dyson series


Subset of

Link to graph
Log In
Improvements of the human condition