# ODELS8

For the differential equation $u''+a_{1}(t)u'+a_{2}(t)u=0\,$, where $a_{i}(t+T)=a_{i}(t)\,$ for all $t(i=1,2)\,$, show that the characteristic multipliers $\rho _{1}\,$ and $\rho _{2}\,$ satisfy the relation $\rho _{1}\rho _{2}=\exp \left\{-\int _{0}^{T}a_{1}(t)dt\right\}\,$.

Let $x_{1}=u,x_{2}=u'\,$. Now the equation can be written as a first order system:

$x_{1}'=x_{2}\,$

$x_{2}'=-a_{1}(t)x_{2}-a_{2}(t)x_{1}\,$

$A={\begin{bmatrix}0&1\\-a_{2}(t)&-a_{1}(t)\\\end{bmatrix}}$

Assume we have two linearly independent solutions $u^{1}\,$ and $u^{2}\,$. Now a fundamental matrix is

$X={\begin{bmatrix}u^{1}&u^{2}\\u'^{1}&u'^{2}\\\end{bmatrix}}$

Also assume that $u^{1}(0)=1,u^{2}(0)=0,u'^{1}(0)=0,u'^{2}(0)=1\,$.

Now the monodromy matrix is

$B={\begin{bmatrix}u^{1}(T)&u^{2}(T)\\u'^{1}(T)&u'^{2}(T)\\\end{bmatrix}}$

The determinant of $B\,$ is $\exp \left\{\int _{0}^{T}{\mathrm {tr}}A\right\}=\exp \left\{-\int _{0}^{T}a_{1}(t)dt\right\}\,$.

The characteristic multipliers $\rho \,$ are the eigenvalues of $B\,$.

$\rho ^{2}-2\phi \rho +exp\left\{-\int _{0}^{T}a_{1}(t)dt\right\}\,$

Therefore $\rho _{1}\rho _{2}=\exp \left\{-\int _{0}^{T}a_{1}(t)dt\right\}\,$