# Diff. eq. for the Einstein metric inside a body

1. Aug 8, 2008

### gijeqkeij

Solving the Einstein eq. i found the following differential eq. that would descrive the metric inside a body:
x2d2y(x)/dx2+2xD1(y)dy(x)/dx+D2(y)=0
where D1(y) and D2(y) are known function of y and 0<=x<=1.
I try to solve numerically but looks like there is a cusp; any suggestion how to approach that eq. or some links where that eq. is treated?
Ty

2. Aug 9, 2008

### gijeqkeij

If we change the x with x=exp(-t) we have the following diff. eq.:
d2y/dt2+(1-2D1)dy/dt+D2=0
where 0<=t<=$$\infty$$.

Any hint? Do you know where this kind of eq. has been analyzed?

Ty

3. Aug 11, 2008

### MrSnoopy

Hi.

Yes this is very difficult task you want to solve, but for my opinion is unsolvable in analytical way!
Why?
You don't know what kind of functions are $$D_{1}(y), D_{2}(y)$$ this is main problem and even not knowing whether you can solve analytical or numerical, but for solving numerical this is very difficult task. If you know what function are $$D_{1}(y), D_{2}(y)$$ than is trivial task as you already know

4. Aug 11, 2008

### gijeqkeij

Thanks Snoopy. Actualli I know the D1(y)=A1*(B1-ye)/ye+1/2 and D2(y)=A2(1-ye)(B2-ye)/y2e-1... (Ai, Bi, e are costants) but the problems is that when i try to solve that equation numerically y->0 for a finite value of t and that has no physical meaning (the metric degenerates at a finite (>0) distance from the center of the star). Why u said it's trivial if i would know D1 and D2?

5. Aug 11, 2008

### MrSnoopy

No problem gijeqkeij.
It is only expression we use to simplify our life

Trivial task is only when"thinking" in numerical solving problem without analytical solution.
On the other hand you can see, with your note, that $$D(y), y \neq y(x)$$ but you are want to solve DE $$y''(t)+C_{1}\; y'(t)= C_{2}$$
where $$C_{1}, C_{2}$$ are only constants.
If I understand you correctly $$D_{1,2}(y),\; D_{1,2} \neq D_{1,2}(y(t))$$, if so you just have to solve DE second degree analytical

MrSnoopy

6. Aug 12, 2008

### gijeqkeij

Sorry Snoopy I guess I was unclear. The eq. to solve is: $$y''(t)+C_{1}(y(t))\; y'(t)= C_{2}(y(t))$$. I thought that kind of eq. received some attention in literature and I wonder if somebody can help me in finding papers that analyze that type of DE. Thanks

7. Aug 12, 2008

### MrSnoopy

Hy

You wrote something like this:
$$D_{1}(y) = A_{1}\frac{B_{1}-y^{e}}{y^{e}}+\frac{1}{2}$$ and
$$D_{2}(y) = A_{2} (1-y^{e})\frac{B_{2}-y^{e}}{y^{2e-1}}\;\;\; ...$$

where are $$A_{1}, A_{2}, B_{1},B_{2}, e=exp(1) \in const$$. OK this DE cannot be solve analytical

Numerical approach

$$x^{2}y''(x)+\left[2xA_{1}(B_{1}y(x)^{-e}-1)+x \right]y'(x) +A_{2}(B_{2}-y^{e})(y^{1-e}-y^{1-2e}) = 0$$

For knowing $$y(x=?)= ..., y'(x= ?)= ...$$ than make DE something like this:
$$y'_{1} = y_{2}$$
$$y'_{2} = -\frac{1}{x^{2}}\left[\left(2A_{1}x(B_{1}y^{-e}_{1}-1)+x\right)y_{2}+A_{2}(B_{2}-y^{e}_{1})(y^{1-e}-y^{1-2e})\right]$$

This is best to solve with Runge-Kutta 5 degree or with ode45 in MatLab or Octave.
I attach file with Runge-Kutta method in file.m (MatLab, Octave).
Write like this in MatLab:
$$f=inline([\; 'y(1)' \; ;\; '\frac{1}{x^2}\left[..... \right]'\;], 'x' , 'y' )$$

In Runge-Kutta file will output vector dimension 2*n where n is steps make in calculation.
Where you want do solve this equation on interval $$I$$ second column in function RungeKutta5, $$y0(x=?)=[y(x=?) , y'(x=?)]$$ and for n put 100 or more.

I hope I helped you

MrSnoopy

#### Attached Files:

• ###### RungeKutta5pure.m
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8. Aug 13, 2008

### gijeqkeij

Thanks Snoopy... that help but the problem is that a numerical solution bring to a singularity somewhere for t=t* (finite value). I wonder if somebody has analyzed that kind of eq. to identify what are (if any) the conditions for it's integrability.