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I have such a system of ODEs, where A and B are constants.

$y_{1}^{''}(r)=-\frac{A}{R}r+B(y_{1}(r)R+y_{2}(r))\chi_{|r|<R}$
$y_{2}^{''}(r)=-\frac{A}{R}r+B(y_{2}(r)R+y_{1}(r))\chi_{|r|<R}$

Could you please give me some clue, how to deal with such nonautonomous systems of ODE?

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It's convenient to define $u_1=y_1+y_2$ and $u_2=y_1-y_2$. Then we find (removing the explicit use of the indicator function) $$u_1''=\left\{ \begin{array}{ll} -2\frac{A}{R}r, & |r|>R;\\ B(1+R)u_1 - 2\frac{A}{R}r, & |r|\leq R, \end{array} \right. $$ and $$u_2''=\left\{ \begin{array}{ll} 0, & |r|>R;\\ B(R-1)u_2, & |r|\leq R. \end{array} \right. $$

For $|r|>R$, we can write down general solutions by integrating twice.

For $|r|\leq R$, we obtain first general solutions of the homogeneous equations $u_1''=B(1+R)u_1$ and $u_2''=B(R-1)u_2$. Assuming that $R$ is also a constant, these are second order linear constant coefficient equations, so we can write down the required general solutions. For $u_2$, we're done, and for $u_1$, we look for a particular solution using the method of undetermined coeffients (that is, making an educated guess).

Assuming that the solutions should be twice-differentiable for all $r$, we then need to match the interior and exterior solutions (and their first and second derivatives) at $|r|=R$ to ensure this. This would involve choosing the constants of integration appropriately.


For a system with $n$ equations of the form $$y_i''=\alpha(r)+\beta(\rho y_i+\sum_{i\neq j}y_j),$$ we can write $$y''=a(r) + \beta(\rho I+J)y,$$ where $y\in\mathbb{R}^n$, $a$ is the column vector with every entry equal to $\alpha$, $I$ is the $n\times n$ identity matrix and $J$ is the matrix with ones everywhere except the diagonal, which has zeroes. $\beta$ and $\rho$ are constants corresponding to $B,R$ respectively. Let $M=\rho I +J$. This has eigenvalues $\rho+n-1$ (with eigenvector $(1,1,\dots,1)^T$) and $\rho-1$. The eigenvectors with eigenvalue $\rho-1$ are of the form $(y_1,y_2,\dots,y_n)^T$ with $$y_1+y_2+\cdots+y_n=0.$$ Thus $\rho-1$ has an $(n-1)$-dimensional eigenspace. Hence $M$ has a complete set of eigenvectors, and so can be diagonalized: there exists a non-singular matrix $S$ such that $$S^{-1}MS=\Lambda=\rm{diag}(\rho+n-1,\rho-1,\dots,\rho-1).$$

Now define $u\in\mathbb{R}^n$ by $y=Su$. Since $S$ is a constant matrix, we can follow the definitions above to get $$u''=S^{-1}a+\beta\Lambda u.$$ Since $\Lambda$ is diagonal, this provides decoupled second order linear equations for the components of $u$.

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many thanks for your kindly help. –  pcepkin Oct 24 '12 at 11:21
    
May I ask you an additional quetion? How can I reduce a system of n equations?e.g. $y_{i}^{''}(r)=-\frac{A}{R}r+B(y_{i}(r)R+\sum_{i\neq j} y_{j}(r))$ –  pcepkin Oct 24 '12 at 13:23
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See my updated answer. Nice question! –  user12477 Oct 24 '12 at 15:02
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