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Find the particular solution which vanishes at t=0\, and identify the Green's matrix G_0(t,s)\, of the system \begin{cases}x_1'=x_2+g_1(t) \\ x_2'=-x_1+g_2(t)\end{cases}\,.

The homogeneous equations can be written \vec x' = A(x)\vec x\, where

A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\,


Find the eigenvalues \lambda = \pm i\,.

The eigenvector for \lambda_1=i\, is \vec{u_1} = \begin{bmatrix}1\\i\end{bmatrix}\,.

The real and imaginary parts are \vec v = \begin{bmatrix}1\\0\end{bmatrix}, \vec w=\begin{bmatrix}0\\1\end{bmatrix}\,.

So if \lambda = \alpha+i\beta\, then

e^{\lambda t} = e^{\alpha t} (\cos \beta t + i \sin \beta t)\,

\vec x = e^{\lambda t} \vec{u_1}\,

\vec x = e^0(\cos t + i\sin t)\begin{bmatrix}1\\i\end{bmatrix}\,

\vec x = \begin{bmatrix} \cos t + i \sin t \\ -\sin t + i\cos t \end{bmatrix}\,

So two linearly independent solutions are the real and imaginary parts of \vec x\,.

x_1 = \begin{bmatrix}\cos t\\-\sin t\end{bmatrix}, x_2 = \begin{bmatrix}\sin t\\\cos t\end{bmatrix}\,


So the fundamental matrix is

X(t) = \begin{bmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{bmatrix}\,


Now the particular solution is

x_p = X(t)\int_0^t X^{-1}(s)g(s)ds\,


Define Green's matrix

G_0(t,s) = \begin{cases} 0 & t\le s\le b \\ X(t)X^{-1}(s) & a\le s < t \end{cases} \,

 = \begin{bmatrix}\cos t\cos s + \sin t\sin s & -\cos t\sin s + \sin t \cos s \\ -\sin t\cos s + \cos t\sin s & \sin t\sin s + \cos t \cos s\end{bmatrix}\,


Main Page : Ordinary Differential Equations : Linear Systems

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