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I am trying to solve the following system of two coupled partial differential equations (both equations equal 0):

enter image description here

Here, $V$ and $Y$ are functions that only depend on $r$, i.e., $V=V(r)$ and $Y=Y(r)$. $\alpha$, $\beta$, $\gamma$ and $k^{-2}$ are just constants, $H^{L\mu}_\mu=0$ and $H^{Li}_i-H^{Lt}_t=0$. I've tried to use DSolve but it doesn't work:

DSolve[{2*(3*b - a)*1/r^4*(r^2*v'[r])'' - c*k*1/r^2*(r^2*v'[r])' - 
4*(3*b - a)*1/r^2*(r^2*Y'[r])' + 2*c*k*Y[r] == 0, 2*b*1/r^4*(r^2*v'[r])'' - c*k*1/r^2*(r^2*v'[r])' + 
2*(a - 2*b)*1/r^2*(r^2*Y'[r])' == 0}, {v, Y}, r]

Since the prime only works for the pure function, I've switched the initial code to this equivalent one:

`DSolve[{2*(3*b - a)*(2 v'[r] + 4 r v''[r] + r^2 v'''[r]) - 
c*k*1/r^2*(2 r v'[r] + r^2 v''[r]) - 
4*(3*b - a)*1/r^2*(2 r Y'[r] + r^2 Y''[r]) + 2*c*k*Y[r] == 0, 
2*b*1/r^4*(2 v'[r] + 4 r v''[r] + r^2 v'''[r]) - 
c*k*1/r^2*(2 r v'[r] + r^2 v''[r]) + 
2*(a - 2*b)*1/r^2*(2 r Y'[r] + r^2 Y''[r]) == 0}, {v, Y}, r] `

But still doesn't work. I don't know much about mathematica and this problem is untractable without numerical methods. Can someone send some help? Thank you so much.

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  • $\begingroup$ Can you provide the code you have used to try to solve the problem? $\endgroup$
    – infinitezero
    Feb 20, 2021 at 18:25
  • $\begingroup$ Writing ``(..)'` is wrong syntax. You can use this syntax only if the argument is a function, not an expression. You would have to write: D[...,r] or D[...,{r,2}] $\endgroup$
    – Daniel Huber
    Feb 20, 2021 at 19:50
  • $\begingroup$ @DanielHuber Thanks for your answer, if you see my edit I did exactly that writing explicitly the derivative, but mathematica doesn't give any answer either. $\endgroup$
    – Axionlike particles
    Feb 20, 2021 at 21:15
  • $\begingroup$ Are you sure that a general solution for arbitrary a,b,c,k exists? $\endgroup$
    – Daniel Huber
    Feb 20, 2021 at 21:28
  • $\begingroup$ @DanielHuber Yes, in some sense the only non arbitrary constants are $k$, which has to do with newton constant, and $\gamma$ which is approximately 2 for a good Newtonian limit of the theory. These equations arise up in studying higher derivatives theories of gravity, you can see for instance researchgate.net/publication/…, and it has a solution of raising and falling Yukawa's potential. However, I can't derive by myself those solutions because I'm doing something wrong in mathematica, $\endgroup$
    – Axionlike particles
    Feb 20, 2021 at 22:28

1 Answer 1

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If you solve the NOT coupled equations separately, you will get a result:

DSolve[{2b1/r^4*(2 v'[r] + 4 r v''[r] + r^2 v'''[r]) - ck1/r^2*(2 r v'[r] + r^2 v''[r]) + 2*(a - 2*b)1/r^2(2 r Y'[r] + r^2 Y''[r]) == 0}, {v}, r]

enter image description here

DSolve[{2b1/r^4*(2 v'[r] + 4 r v''[r] + r^2 v'''[r]) - ck1/r^2*(2 r v'[r] + r^2 v''[r]) + 2*(a - 2*b)1/r^2(2 r Y'[r] + r^2 Y''[r]) == 0}, {Y}, r]

enter image description here

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