Solving Differential Eqs with Critical Point: Initial Conditions & Solutions

[Vrbic]

Hello,
I have two different discrepancies to this system:
a) How and when is possible to have more solution of differential eq. or their system for same initial problem? For example this is happening in following system. It is written about this system:
"Different value of constant $\dot{M}=4\pi r^2u\rho=const.$ lead to physically distinct classes of solution for the same boundary condition at infinity". Can anybody explain it to me?

$\frac{\rho'}{\rho}+\frac{u'}{u}+\frac{2}{r}=0$
$uu'+a^2(r)\frac{\rho'}{\rho}+\frac{\alpha}{r^2}=0$, where $a(r)=a_0\big(\frac{\rho(r)}{\rho_0}\big)^{(\Gamma-1)/2}$, $\rho(r)$ and $u(r)$ are function of r and $\Gamma, a_0, \rho_0, \alpha$ are constant.[itex][/itex]

b) How to solve this system? I am quit sure that is possible just numerically and I tried. I expressed this equation in this way:
$u'=\frac{D_1}{D}$
$u'=\frac{D_2}{D}$, where $D_1=\frac{2a^2/r-\alpha/r^2}{\rho}$, $D_2=\frac{2u^2/r-\alpha/r^2}{u}$ and $D=\frac{u^2-a^2}{\rho u}$.
No I see if I want smooth solution I need $D_1=D_2=D=0$ in same critical point $r_c$. These are two boundary condition (Im not sure what difference is between boundary and initial condition?). I find out $r_c$ and $u_c$, value of $u(r_c)=u_c$. So I have to condition $a(r_c)=u(r_c)$ and $u(r_c)=u_c$ Am I right till this time?
Now is coming my question:
Im not much familiar with numerical methods or numerical solving so I used Mathematica soft, concretely function NDSolve with conditions mentioned above. And Mathematica said "no". By "no" I mean it diverges in $r_c$. What could be wrong?

 

[Vrbic]

Hello,
I have two different discrepancies to this system:
a) How and when is possible to have more solution of differential eq. or their system for same initial problem? For example this is happening in following system. It is written about this system:
"Different value of constant $\dot{M}=4\pi r^2u\rho=const.$ lead to physically distinct classes of solution for the same boundary condition at infinity". Can anybody explain it to me?

$\frac{\rho'}{\rho}+\frac{u'}{u}+\frac{2}{r}=0$
$uu'+a^2(r)\frac{\rho'}{\rho}+\frac{\alpha}{r^2}=0$, where $a(r)=a_0\big(\frac{\rho(r)}{\rho_0}\big)^{(\Gamma-1)/2}$, $\rho(r)$ and $u(r)$ are function of r and $\Gamma, a_0, \rho_0, \alpha$ are constant.[itex][/itex]

b) How to solve this system? I am quit sure that is possible just numerically and I tried. I expressed this equation in this way:
$u'=\frac{D_1}{D}$
$u'=\frac{D_2}{D}$, where $D_1=\frac{2a^2/r-\alpha/r^2}{\rho}$, $D_2=\frac{2u^2/r-\alpha/r^2}{u}$ and $D=\frac{u^2-a^2}{\rho u}$.
No I see if I want smooth solution I need $D_1=D_2=D=0$ in same critical point $r_c$. These are two boundary condition (Im not sure what difference is between boundary and initial condition?). I find out $r_c$ and $u_c$, value of $u(r_c)=u_c$. So I have to condition $a(r_c)=u(r_c)$ and $u(r_c)=u_c$ Am I right till this time?
Now is coming my question:
Im not much familiar with numerical methods or numerical solving so I used Mathematica soft, concretely function NDSolve with conditions mentioned above. And Mathematica said "no". By "no" I mean it diverges in $r_c$. What could be wrong?

Mistake in second equation in section b) so again both:
$u'=\frac{D_1}{D}$
$\rho'=\frac{D_2}{D}$,