Mathematica Solve complex equation analytically

[djymndl07]

I was going through an article. I have the following functions,

$$f(\text{r})\text{=}-\frac{2 M r^2}{g^3+r^3}+\frac{8}{3} \pi P r^2+1$$
$$\text{v}(\text{r})\text{=}\frac{f(r)}{r^2}$$
Now I want to solve v'[r]=0 for r. Clearly This cannot be solved analytically. But there they solved it somehow and got the following result.

$$\text{r}=\sqrt[3]{-g^3+\sqrt{g^6-2 g^3 M^3}+M^3}+\frac{M^2}{\sqrt[3]{-g^3+\sqrt{g^6-2 g^3 M^3}+M^3}}+M$$

Can anyone tell me how can it be solved in mathematica or what method did they use?
Thanks in advance.

 

[renormalize]

I was going through an article. I have the following functions,

f[r_] := 1 - (2*M*r^2)/(r^3 + g^3) + (8/3)*Pi*P*r^2
v[r_] := f[r]/r^2
Now I want to solve `v'[r]==0` for r. Clearly This cannot be solved analytically. But there they solved it somehow and got the following result.

r = g^3/(2*r^2) + r/2 + (4/3)*(g^3 + r^3)*P*Pi + ((3 + 8*Pi*r^2)^2*(g^3 + r^3)^2)/(36*r^4*\[Xi]^(1/3)) + \[Xi]^(1/3)
Where,

\[Xi] = ((3 + 8*Pi*r^2)^3*(g^3 + r^3)^3)/(216*r^6) + Sqrt[g^6 - (g^3*(3 + 8*Pi*r^2)^3*(g^3 + r^3)^3)/(108*r^6)] - g^3

Can anyone tell me how can it be solved in mathematica or what method did they use?
Thanks in advance.

It's tedious to read and understand your formulas as written. Can you repost them using Latex?

 

[djymndl07]

It's tedious to read and understand your formulas as written. Can you repost them using Latex?

Done

 

[pasmith]

The result looks like the solution of a cubic. These can be solved analytically, as can quartics.

 

[Frabjous]

What’s the article?

 

[DaveE]

I was going through an article. I have the following functions,

$$f(\text{r})\text{=}-\frac{2 M r^2}{g^3+r^3}+\frac{8}{3} \pi P r^2+1$$
$$\text{v}(\text{r})\text{=}\frac{f(r)}{r^2}$$
Now I want to solve v'[r]=0 for r. Clearly This cannot be solved analytically. But there they solved it somehow and got the following result.

$$\text{r}=\sqrt[3]{-g^3+\sqrt{g^6-2 g^3 M^3}+M^3}+\frac{M^2}{\sqrt[3]{-g^3+\sqrt{g^6-2 g^3 M^3}+M^3}}+M$$

Can anyone tell me how can it be solved in mathematica or what method did they use?
Thanks in advance.

Are you sure Mathematica was right? I* tried to verify it by setting M=g=1 and it didn't work, but I could have made mistakes.

I found that the problem reduces to $r^6 + 3Mr^5 + 2g^3r^3 + g^6 =0$. Again, no guarantee I didn't screw it up.

* plus a bit of ancient MathCad

edit: should be $r^6 - 3Mr^5 + 2g^3r^3 + g^6 =0$

 

[Frabjous]

I found that the problem reduces to r6+3Mr5+2g3r3+g6=0. Again, no guarantee I didn't screw it up.

I got a minus sign in front of the second term.

 

[DaveE]

I got a minus sign in front of the second term.

Yes, you're right.
$r^6-3Mr^5+2g^3r^3+g^6=0$

 

[djymndl07]

What’s the article?

Can I post arxiv link of the article here?

 

[djymndl07]

Are you sure Mathematica was right? I* tried to verify it by setting M=g=1 and it didn't work, but I could have made mistakes.

I found that the problem reduces to $r^6 + 3Mr^5 + 2g^3r^3 + g^6 =0$. Again, no guarantee I didn't screw it up.

* plus a bit of ancient MathCad

edit: should be $r^6 - 3Mr^5 + 2g^3r^3 + g^6 =0$

Well This is not an exact solution. I think they made some approximation. But still I cant figure out the way they did.

 

[renormalize]

Can I post arxiv link of the article here?

Yes, that's OK here.

 

[renormalize]

Well This is not an exact solution. I think they made some approximation. But still I cant figure out the way they did.

Since the solution of $v^{\prime}\left(r\right)=0$ in the particular case $g=0$ is $r=3M$, I think the author(s) are approximating the general solution for the specific case $g/\left(3M\right)\ll1$ and $1-r/\left(3M\right)\ll1$, thereby arriving to lowest order at a cubic equation to be solved for $r$.

 

[djymndl07]

Since the solution of $v^{\prime}\left(r\right)=0$ in the particular case $g=0$ is $r=3M$, I think the author(s) are approximating the general solution for the specific case $g/\left(3M\right)\ll1$ and $1-r/\left(3M\right)\ll1$, thereby arriving to lowest order at a cubic equation to be solved for $r$.

Thanks, I will try this.

 

[djymndl07]

Yes, that's OK here.

Lookk at equations 7,8,16,17 in Shadow thermodynamics

 

[renormalize]

Lookk at equations 7,8,16,17 in Shadow thermodynamics

I'm not seeing how eqs.(7,8,16) in this reference lead to the first 2 equations in your post #1. Can you post your derivation?

 

[djymndl07]

I'm not seeing how eqs.(7,8,16) in this reference lead to the first 2 equations in your post #1. Can you post your derivation?

There is a typo in equation 16. The correct expression will be, $f(r)=h(r)$. Now using this in equation 7 and taking E=0, L=1 leads to Second equation of my post.

 

[renormalize]

There is a typo in equation 16. The correct expression will be, $f(r)=h(r)$. Now using this in equation 7 and taking E=0, L=1 leads to Second equation of my post.

Can you motivate why you set the photon energy $E=0$? Does eq.(17) in your reference require this condition?

 

[djymndl07]

Yes there are plenty of papers where they did so.... E and L are arbitrary constants. The physical importance does not change from such choices. Its for simplification purpose.

 

[renormalize]

OK, it turns out that eq.(17) in your reference is the immediate result of a simple approximation.
Your original equations:$$f\left(r\right)\text{=}-\frac{2Mr^{2}}{g^{3}+r^{3}}+\frac{8}{3}\pi Pr^{2}+1,\quad v\left(r\right)=\frac{f(r)}{r^{2}},\quad\left.v^{\prime}\left(r\right)\right|_{r=r_{p}}=0\tag{1a,b,c}$$
Calculating (1c) gives:$$0=\frac{6Mr_{p}^{2}}{\left(g^{3}+r_{p}^{3}\right)^{2}}-\frac{2}{r_{p}^{3}}\;\Rightarrow\;0=r_{p}^{3}-3Mr_{p}^{2}+g^{3}\left(2+\frac{g^{3}}{r_{p}^{3}}\right)\approx r_{p}^{3}-3Mr_{p}^{2}+2g^{3}\;\text{for}\;\frac{g^{3}}{r_{p}^{3}}\ll1$$and, per Mathematica, the only real solution of the cubic $r_{p}^{3}-3Mr_{p}^{2}+2g^{3}=0$ is:
$$r_p=\sqrt[3]{-g^3+\sqrt{g^6-2 g^3 M^3}+M^3}+\frac{M^2}{\sqrt[3]{-g^3+\sqrt{g^6-2 g^3 M^3}+M^3}}+M$$i.e., eq. (17).
(Edited to correct exponent from 3 to 2 in $6Mr_{p}^{2}$ term.)

 

[djymndl07]

OK, it turns out that eq.(17) in your reference is the immediate result of a simple approximation.
Your original equations:$$f\left(r\right)\text{=}-\frac{2Mr^{2}}{g^{3}+r^{3}}+\frac{8}{3}\pi Pr^{2}+1,\quad v\left(r\right)=\frac{f(r)}{r^{2}},\quad\left.v^{\prime}\left(r\right)\right|_{r=r_{p}}=0\tag{1a,b,c}$$
Calculating (1c) gives:$$0=\frac{6Mr_{p}^{3}}{\left(g^{3}+r_{p}^{3}\right)^{2}}-\frac{2}{r_{p}^{3}}\;\Rightarrow\;0=r_{p}^{3}-3Mr_{p}^{2}+g^{3}\left(2+\frac{g^{3}}{r_{p}^{3}}\right)\approx r_{p}^{3}-3Mr_{p}^{2}+2g^{3}\;\text{for}\;\frac{g^{3}}{r_{p}^{3}}\ll1$$and, per Mathematica, the only real solution of the cubic $r_{p}^{3}-3Mr_{p}^{2}+2g^{3}=0$ is:
$$r_p=\sqrt[3]{-g^3+\sqrt{g^6-2 g^3 M^3}+M^3}+\frac{M^2}{\sqrt[3]{-g^3+\sqrt{g^6-2 g^3 M^3}+M^3}}+M$$i.e., eq. (17).

Thanks a lot