# Solving two coupled partial differential equations

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

• Can you provide the code you have used to try to solve the problem? Feb 20, 2021 at 18:25
• 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}]` Feb 20, 2021 at 19:50
• @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. Feb 20, 2021 at 21:15
• Are you sure that a general solution for arbitrary a,b,c,k exists? Feb 20, 2021 at 21:28
• @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, Feb 20, 2021 at 22:28  