Solving a complicated equation for approximate analytical Solution using Mathematica

  • #1
djymndl07
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Solving a complicated equation for approximate analytical Solution
Hello there, I am trying to solve the Following equation for r,
$$2 a Q^4+5 r^4 \left(3 c (\omega +1) r^{1-3 \omega }-2 r (r-3 M)-4 Q^2\right)=0$$
Clearly this is unsolvable. But if we substitute a=0 and c=0 we get one of the solution, ##r=\frac{1}{2} \left(\sqrt{9 M^2-8 Q^2}+3 M\right)##. Can I obtain approximate analytical solution of the above equation which gives the same value when substitutions a=0 and c=0 are applied. If yes, then how? I have tried AsymptoticSolve, but got no answer.
Thanks in advance.
 
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  • #2
Are a, c, Q, M, and ω all constants (i.e. not functions of r)? I take it you want r(a,c,Q,M,ω). If you put in values for a,c,Q,M,ω, you could get numerical solutions that might help guide you. What's the magnitude of ω compared to 1? Could you do an expansion if ω is much larger or smaller than 1?
 
  • #3
a,c,Q,M, ##\omega## Are arbitrary constants. ##\omega## lies between -1 and -3. Other constants may take any positive value.
 
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  • #4
Then I am afraid you're doomed. Techniques exist if you know something about these parameters (depending on what it is that you know) but if they can literally be anything, you need the general solution. Which does not exist.
 
  • #5
I assume this equation came from some physical problem. So maybe you know some possible values of the constants. Then you could put in those constants and then solve numerically for r as a function of ω, for example. Is that a possible approach?
 
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  • #6
djymndl07 said:
Can I obtain approximate analytical solution
phyzguy said:
solve numerically
I don't think that's what he wants.
 
  • #7
phyzguy said:
I assume this equation came from some physical problem. So maybe you know some possible values of the constants. Then you could put in those constants and then solve numerically for r as a function of ω, for example. Is that a possible approach?
Yes, numerically I can do that. but some analytical solution, even if it is an approximate one would be better.
 
  • #8
djymndl07 said:
Yes, numerically I can do that. but some analytical solution, even if it is an approximate one would be better.
As @Vanadium 50 said, I don't think that's possible.
 
  • #9
How do approximations work? You have a big piece plus a small piece, and you neglect the small piece. What is the small piece here? You tell us you cannot tell - it could be anything. OK, that's fair, but it also means you can't approximate.
 
  • #10
I got some way to do that in mathematica. Thank you everyone for the reply. One can see the link Here if interested.
 
  • #11
Note the continued questioning about the size of the parameters. Also note that the expression you got assumes a and c are small (which you told us was not something we can assume).

In short -if you ask different questions, you get different answers.
 

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