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omyojj
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The problem is about mathematics but it originates from the self-gravitational instability of incompressible fluid, so let me explain the situation first.
I have an incompressible uniform fluid disk that is infinite in the x-y direction.
The disk has a finite thickness [tex] 2a [/tex] along the z-direction. (-a<z<a)
The space exterior to the disk is assumed to be filled with a rarefied medium that has constant pressure equal to the fluid's, which prevents the disk from dispersing.
Thus, the initial density distribution has a step discontinuity and can be written as
[tex] \rho(x,y,z) = \rho_0 ( \theta(z-a) - \theta(z+a) ) [/tex]
where [tex] \theta(z) [/tex] is a step function.
Now I want to apply small Lagrangian perturbation to the fluid of the form
[tex] \xi_{x,z}(x,z) = \xi_{x,z}(z) e^{ikx + i\omega t} [/tex]
where [tex] \xi_{x,z} [/tex] is the x,z-component of Lagrangian displacement vector.
Perturbation has its wavenumber k along the x-direction, and I assumed time dependence.
Also I consider only the perturbation that has even reflection symmetry for displacement, that is , [tex] \xi_{x,z}(z) = - \xi_{x,z}(-z) [/tex] ()
(Sausage type: The rectangular shape of the slab changed slightly (though infinitesimally) so that it looks more like a cylinder now)
Deep inside the disk, there would be no change in density because the fluid itself is incompressible.
([tex] \nabla \cdot {\mathbf{\xi}} = 0 [/tex])
But near the boundary surfaces, discrete density changes in Eulerian density variable [tex] \delta\rho(x,z) [/tex] could occur if the difference b/w a and the height from the midplane(z) is smaller than the Lagrangian displacement vector at z=a.
[tex] \delta\rho(x,z) = \rho_0 [ \theta(z - \xi_z(z=a)e^{ikx} - \theta(z - a) ] + \rho_0 [ \theta(z+a) - \theta(z - \xi_z(z=-a)e^{ikx}) ] [/tex]
(Of course, Lagrangian density perturbation is everywhere zero, i.e., [tex] \Delta \rho = 0 [/tex] .)
Now I want to introduce self-gravity at this point because I want to examine the strength of perturbed gravity that makes the system unstable to this small disturbances.
[tex] \nabla^2 \delta\psi = 4\pi G \delta\rho [/tex]
Can I solve the above equation for [tex]\delta\psi(x,z) [/tex] with right-hand side involving step functions of sinusoidal behavior in x-direction?
Any hint or help would be much appreciated.
Thank you.
BTW, excuse my English..
I have an incompressible uniform fluid disk that is infinite in the x-y direction.
The disk has a finite thickness [tex] 2a [/tex] along the z-direction. (-a<z<a)
The space exterior to the disk is assumed to be filled with a rarefied medium that has constant pressure equal to the fluid's, which prevents the disk from dispersing.
Thus, the initial density distribution has a step discontinuity and can be written as
[tex] \rho(x,y,z) = \rho_0 ( \theta(z-a) - \theta(z+a) ) [/tex]
where [tex] \theta(z) [/tex] is a step function.
Now I want to apply small Lagrangian perturbation to the fluid of the form
[tex] \xi_{x,z}(x,z) = \xi_{x,z}(z) e^{ikx + i\omega t} [/tex]
where [tex] \xi_{x,z} [/tex] is the x,z-component of Lagrangian displacement vector.
Perturbation has its wavenumber k along the x-direction, and I assumed time dependence.
Also I consider only the perturbation that has even reflection symmetry for displacement, that is , [tex] \xi_{x,z}(z) = - \xi_{x,z}(-z) [/tex] ()
(Sausage type: The rectangular shape of the slab changed slightly (though infinitesimally) so that it looks more like a cylinder now)
Deep inside the disk, there would be no change in density because the fluid itself is incompressible.
([tex] \nabla \cdot {\mathbf{\xi}} = 0 [/tex])
But near the boundary surfaces, discrete density changes in Eulerian density variable [tex] \delta\rho(x,z) [/tex] could occur if the difference b/w a and the height from the midplane(z) is smaller than the Lagrangian displacement vector at z=a.
[tex] \delta\rho(x,z) = \rho_0 [ \theta(z - \xi_z(z=a)e^{ikx} - \theta(z - a) ] + \rho_0 [ \theta(z+a) - \theta(z - \xi_z(z=-a)e^{ikx}) ] [/tex]
(Of course, Lagrangian density perturbation is everywhere zero, i.e., [tex] \Delta \rho = 0 [/tex] .)
Now I want to introduce self-gravity at this point because I want to examine the strength of perturbed gravity that makes the system unstable to this small disturbances.
[tex] \nabla^2 \delta\psi = 4\pi G \delta\rho [/tex]
Can I solve the above equation for [tex]\delta\psi(x,z) [/tex] with right-hand side involving step functions of sinusoidal behavior in x-direction?
Any hint or help would be much appreciated.
Thank you.
BTW, excuse my English..
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