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linear_first-order_ode_system [2014/03/06 22:34]
nikolaj
linear_first-order_ode_system [2017/01/17 01:06]
nikolaj
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 ===== Linear first-order ODE system ===== ===== Linear first-order ODE system =====
 ==== Set ==== ==== Set ====
-| @#88DDEE: $ A:\mathbb R\to\mathrm{Matrix}(n,​\mathbb R) $ | +| @#55CCEE: context ​    | @#55CCEE: $ A:\mathbb R\to\mathrm{Matrix}(n,​\mathbb R) $ | 
-| @#88DDEE: $ b:\mathbb R\to\mathbb R^n $ |+| @#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) ​$ |
  
-| @#FFBB00: $ y \in \mathrm{it} $ | +-----
- +
-| @#55EE55: $ y:​C^k(\mathbb R,\mathbb R^n) $  | +
- +
-| @#55EE55: $ y'​(t)=A(t)\ y(t)+b(t) $ | +
- +
-==== Discussion ====+
 === Theorems === === Theorems ===
 There exists a matrix $S(t,s)$ such that the solution of the equation above is of the form  There exists a matrix $S(t,s)$ such that the solution of the equation above is of the form 
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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
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 === Reference === === Reference ===
 Wikipedia: [[http://​en.wikipedia.org/​wiki/​Dyson_series|Dyson series]] Wikipedia: [[http://​en.wikipedia.org/​wiki/​Dyson_series|Dyson series]]
-==== Parents ====+ 
 +-----
 === Context === === Context ===
 [[Square matrix]] [[Square matrix]]
 === Subset of === === Subset of ===
 [[ODE system]] [[ODE system]]
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