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linear_first-order_ode_system [2014/03/06 22:34] nikolaj |
linear_first-order_ode_system [2014/03/07 00:20] nikolaj |
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| * The equation $y'(t)=A(t)y(t)$ is solved by $y(t)=\mathrm{e}^{\int A(t)\,\mathrm dt}y(0)$. | * 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 $y(t)=\mathcal T\exp(int_{t_0}^tA(t))y(0)$ | + | 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 | * For $A(t),b(t)$ one-dimensional one has | ||
