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.
D[...,r]
orD[...,{r,2}]
$\endgroup$