- #1

adf89812

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- TL;DR Summary
- Magnus expansion of companion matrix time-dependent or solve by substitution how?

For example, consider the following system of 2 first order ODEs:

$$

\left\{\begin{array}{l}

x_1^{\prime}=2 t x_1+t^2 x_2 \\

x_2^{\prime}=t^3 x_1+4 t x_2

\end{array}\right.

$$

This is a linear homogeneous system of 2 first order ODEs with $$A(t)=\left[\begin{array}{ll}2 t & t^2 \\ t^3 & 4 t\end{array}\right]$$.

"Secondly, the substitution method works in the same manner as usual. Indeed, the first line of the system leads to $$y = t^{-2}\dot{x} + 2t^{-1}x$$, which can be differentiated in order to find $\dot{y}$ in terms of $$x$$ and $$t$$. Next, plugging these expressions into the second line, you will end up with a second-order linear ODE with non-constant coefficients for $$x$$, which itself might not be easy to solve in the present case."

"

Firstly, you mentioned diagonalization; however, in that case, the eigenvalues and the eigenvectors will be themselves time-dependent. If $$S$$ denotes the change of basis allowing the diagonalization of $$A$$ as $$D$$, i.e. $$ D= SAS^{-1}$$, then the system of equations $$\dot{u} = Au$$, where $$u = (x,y)$$, becomes

$$

\dot{v} = \partial_t(Su) = S\dot{u} + \dot{S}u = \left(SA + \dot{S}\right)u = \left(SAS^{-1} + \dot{S}S^{-1}\right)Su = \left(D + \dot{S}S^{-1}\right)v

$$

for $$v = Su$$. This new system might be even harder to solve yet, because of the extra (non-diagonal) term $$\dot{S}S^{-1}$$."

$$

\left\{\begin{array}{l}

x_1^{\prime}=2 t x_1+t^2 x_2 \\

x_2^{\prime}=t^3 x_1+4 t x_2

\end{array}\right.

$$

This is a linear homogeneous system of 2 first order ODEs with $$A(t)=\left[\begin{array}{ll}2 t & t^2 \\ t^3 & 4 t\end{array}\right]$$.

"Secondly, the substitution method works in the same manner as usual. Indeed, the first line of the system leads to $$y = t^{-2}\dot{x} + 2t^{-1}x$$, which can be differentiated in order to find $\dot{y}$ in terms of $$x$$ and $$t$$. Next, plugging these expressions into the second line, you will end up with a second-order linear ODE with non-constant coefficients for $$x$$, which itself might not be easy to solve in the present case."

"

Firstly, you mentioned diagonalization; however, in that case, the eigenvalues and the eigenvectors will be themselves time-dependent. If $$S$$ denotes the change of basis allowing the diagonalization of $$A$$ as $$D$$, i.e. $$ D= SAS^{-1}$$, then the system of equations $$\dot{u} = Au$$, where $$u = (x,y)$$, becomes

$$

\dot{v} = \partial_t(Su) = S\dot{u} + \dot{S}u = \left(SA + \dot{S}\right)u = \left(SAS^{-1} + \dot{S}S^{-1}\right)Su = \left(D + \dot{S}S^{-1}\right)v

$$

for $$v = Su$$. This new system might be even harder to solve yet, because of the extra (non-diagonal) term $$\dot{S}S^{-1}$$."