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While studying the derivation of the normal modes of oscillation of a liquid sphere in the paper "Nonradial oscillations of stars" by Pekeris (1938), which can be found here, on page 193 and 194 two coupled second order differential equations in two variables are merged into one fourth order differential equation in one variable. I really can't get my head around the way you eliminate one of the two variables.
The two second order equations are:
[itex]c^2\ddot{X}+\dot{X}\left[\dot{c}^2-\left(\gamma-1\right)g+\frac{2c^2}{r}\right]+X\left\{\sigma^2+\left(2-\gamma\right)\left(\dot{g}+\frac{2g}{r}\right)-n\left(n+1\right)\frac{c^2}{r^2}\right\}=g\ddot{w}+\dot{w}\left(2\dot{g}+\frac{2g}{r}\right)+\left[2-n\left(n+1\right)\right]\frac{wg}{r^2}[/itex]
[itex]\dot{X}r^2+X\left[2r+\left(g-\gamma g-\dot{c}^2\right)\left(n+1\right)\frac{n}{\sigma^2}\right]=r^2\ddot{w}+4r\dot{w}+w\left[2-n\left(n+1\right)\right][/itex]
In these equations, all variables depend on [itex]r[/itex] except [itex]\gamma[/itex], [itex]n[/itex] and [itex]\sigma[/itex], which are constants.
Apparently, according to the paper, this can be written as a single, fourth order differential equation in [itex]X[/itex]:
[itex]\ddot{G}+\dot{G}\left(\frac{6}{r}-2\frac{\dot{A}}{A}\right)+G\left\{-\frac{\ddot{A}}{A}+\left(\frac{6-n-n^2}{r^2}\right)-\frac{6 \dot{A}}{Ar}+\frac{2\ddot{A}^2}{A^2}\right\}-AH=0[/itex]
Where
[itex]A=2\left(\frac{\dot{g}}{g}-\frac{1}{r}\right)[/itex]
[itex]gG=c^2\ddot{X}+\dot{X}\left(\dot{c}^2-\gamma g+\frac{2c^2}{r}\right)+X\left[\sigma^2+\left(2-\gamma\right)\dot{g}+\left(1-\gamma\right)\frac{2}{r}-n\left(n+1\right)\frac{c^2}{gr^2}+\frac{n}{\sigma^2r^2}\left(n+1\right)\left(\dot{c}^2-g+\gamma g\right)\right][/itex]
[itex]H=\ddot{X}+\dot{X}\left[\frac{4}{r}-\frac{n}{\sigma^2 r^2}\left(n+1\right)\left(\dot{c}^2-g+\gamma g\right)\right]+X\left[\frac{2}{r^2}-\frac{n}{\sigma^2r^2}\left(n+1\right)\left(\ddot{c}^2-\dot{g}+\gamma\dot{g}\right)\right][/itex]
Does someone know a general way to transform two second order coupled differential equations into one fourth order equation? Thanks for any hints or help!
The two second order equations are:
[itex]c^2\ddot{X}+\dot{X}\left[\dot{c}^2-\left(\gamma-1\right)g+\frac{2c^2}{r}\right]+X\left\{\sigma^2+\left(2-\gamma\right)\left(\dot{g}+\frac{2g}{r}\right)-n\left(n+1\right)\frac{c^2}{r^2}\right\}=g\ddot{w}+\dot{w}\left(2\dot{g}+\frac{2g}{r}\right)+\left[2-n\left(n+1\right)\right]\frac{wg}{r^2}[/itex]
[itex]\dot{X}r^2+X\left[2r+\left(g-\gamma g-\dot{c}^2\right)\left(n+1\right)\frac{n}{\sigma^2}\right]=r^2\ddot{w}+4r\dot{w}+w\left[2-n\left(n+1\right)\right][/itex]
In these equations, all variables depend on [itex]r[/itex] except [itex]\gamma[/itex], [itex]n[/itex] and [itex]\sigma[/itex], which are constants.
Apparently, according to the paper, this can be written as a single, fourth order differential equation in [itex]X[/itex]:
[itex]\ddot{G}+\dot{G}\left(\frac{6}{r}-2\frac{\dot{A}}{A}\right)+G\left\{-\frac{\ddot{A}}{A}+\left(\frac{6-n-n^2}{r^2}\right)-\frac{6 \dot{A}}{Ar}+\frac{2\ddot{A}^2}{A^2}\right\}-AH=0[/itex]
Where
[itex]A=2\left(\frac{\dot{g}}{g}-\frac{1}{r}\right)[/itex]
[itex]gG=c^2\ddot{X}+\dot{X}\left(\dot{c}^2-\gamma g+\frac{2c^2}{r}\right)+X\left[\sigma^2+\left(2-\gamma\right)\dot{g}+\left(1-\gamma\right)\frac{2}{r}-n\left(n+1\right)\frac{c^2}{gr^2}+\frac{n}{\sigma^2r^2}\left(n+1\right)\left(\dot{c}^2-g+\gamma g\right)\right][/itex]
[itex]H=\ddot{X}+\dot{X}\left[\frac{4}{r}-\frac{n}{\sigma^2 r^2}\left(n+1\right)\left(\dot{c}^2-g+\gamma g\right)\right]+X\left[\frac{2}{r^2}-\frac{n}{\sigma^2r^2}\left(n+1\right)\left(\ddot{c}^2-\dot{g}+\gamma\dot{g}\right)\right][/itex]
Does someone know a general way to transform two second order coupled differential equations into one fourth order equation? Thanks for any hints or help!