windpipe33u

2022-04-22

Solution to a system of ODE:s

Consider a system of linear ODE:s:

${x}^{\prime}\left(t\right)=Ax\left(t\right)+C$

Here A and C are coefficient matrices. If$C=0$ , the solution is $x\left(t\right)=\mathrm{exp}\left(Ax\left(t\right)\right)x\left(0\right)$ , where exp represents the matrix exponential. Can we write a similar solution for a general C?

Consider a system of linear ODE:s:

Here A and C are coefficient matrices. If

Pedro Taylor

Beginner2022-04-23Added 19 answers

Step 1

Using the matrix exponential as integrating factor you get

${\left({e}^{-At}x\left(t\right)\right)}^{\prime}={e}^{-At}C$

which can be integrated to (at first only for regular A, but in the end for all A)

Step 2

${e}^{-At}x\left(t\right)-x\left(0\right)=-{A}^{-1}({e}^{-At}-I)C\Rightarrow x\left(t\right)={e}^{At}x\left(0\right)+t{\varphi}_{1}\left(At\right)C$

where$\varphi}_{1$ is the matrix version of the function $\varphi}_{1}\left(z\right)=\frac{{e}^{z}-1}{z$ , continued with ${\varphi}_{1}\left(0\right)=1$ as per its power series. Note that $\varphi}_{1$ , despite its singular definition, is an analytical or entire function, similar to the exponential.

These modified exponentials, matrix phi functions or whatever name they got in-between,$\varphi}_{2}\left(z\right)=\frac{{e}^{z}-1-z}{{z}^{2}$ etc. occur also in exponential Runge-Kutta methods, thus they are also implemented in good numerical linear algebra libraries along the matrix exponential.

Using the matrix exponential as integrating factor you get

which can be integrated to (at first only for regular A, but in the end for all A)

Step 2

where

These modified exponentials, matrix phi functions or whatever name they got in-between,

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