Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision Next revision Both sides next revision | ||
|
linear_first-order_ode_system [2014/03/06 22:33] nikolaj |
linear_first-order_ode_system [2014/03/07 00:20] nikolaj |
||
|---|---|---|---|
| Line 22: | Line 22: | ||
| == Two special cases == | == Two special cases == | ||
| - | - For constant $A$, one has | + | * For constant $A$, one has |
| ^ $S(t,s)=\mathrm{exp}\left((t-s)\ A\right)$ ^ | ^ $S(t,s)=\mathrm{exp}\left((t-s)\ A\right)$ ^ | ||
| Line 30: | Line 30: | ||
| ^ $y(t)=\mathrm{exp}\left(t A\right)\cdot\left(y_0+\int_0^t\ \mathrm{exp}\left(-s A\right)\ b(s)\ \mathrm ds\right)$ ^ | ^ $y(t)=\mathrm{exp}\left(t A\right)\cdot\left(y_0+\int_0^t\ \mathrm{exp}\left(-s A\right)\ b(s)\ \mathrm ds\right)$ ^ | ||
| - | - 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 $y(t)=\mathcal T\exp(int_{t_0}^tA(t))y(0)$ | + | * The equation $y'(t)=A(t)y(t)$ is solved by $y(t)=\mathrm{e}^{\int A(t)\,\mathrm dt}y(0)$. |
| - | - For $A(t),b(t)$ one-dimensional one has | + | 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 | ||
| ^ $S(t,s)=\mathrm{e}^{I(t)-I(s)}\,\,\mathrm{with}\,\, I(s):=\int_0^s A(\tau)\ \mathrm d\tau$ ^ | ^ $S(t,s)=\mathrm{e}^{I(t)-I(s)}\,\,\mathrm{with}\,\, I(s):=\int_0^s A(\tau)\ \mathrm d\tau$ ^ | ||
