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linear_first-order_ode_system [2014/03/06 22:33] 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 | ||
| Line 22: | Line 19: | ||
| == 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 27: | ||
| ^ $y(t)=\mathrm{exp}\left(t A\right)\cdot\left(y_0+\int_0^t\ \mathrm{exp}\left(-s A\right)\ b(s)\ \mathrm ds\right)$ ^ | ^ $y(t)=\mathrm{exp}\left(t A\right)\cdot\left(y_0+\int_0^t\ \mathrm{exp}\left(-s A\right)\ b(s)\ \mathrm ds\right)$ ^ | ||
| - | - 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 | ||
| ^ $S(t,s)=\mathrm{e}^{I(t)-I(s)}\,\,\mathrm{with}\,\, I(s):=\int_0^s A(\tau)\ \mathrm d\tau$ ^ | ^ $S(t,s)=\mathrm{e}^{I(t)-I(s)}\,\,\mathrm{with}\,\, I(s):=\int_0^s A(\tau)\ \mathrm d\tau$ ^ | ||
| Line 38: | Line 37: | ||
| === 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]] | ||
