# Solving 2nd order coupled differential equations using shooting method

I'm trying to solve these two coupled 2nd order differential equations:

with the following boundary conditions:

where $$r_{*}$$ is the value of the $$r$$ at midpoint. I was trying to solve these equations using shooting method and for $$l=3$$ and $$t=4$$, I succeeded by guessing the value of $$v'(\frac{-l}{2})$$ and $$r*$$:

m = 1/2 (Tanh[v[x]/(1/3)] + 1);

rv3 = ParametricNDSolve[{r[x]^4/(Guess)^2 - r[x]^2 -
2 r'[x] v'[x] + (r[x]^2 - m) v'[x]^2 == 0,
r[x]^2 - r[x]^2 v'[x]^2 - r[x] v''[x] + 2 r'[x] v'[x] == 0,
r[-1.5] == 10, v[-1.5] == 4, v'[-1.5] == guess}, {r, v}, {x, -1.5,
1.5}, {guess, Guess}, MaxSteps -> Infinity]

Manipulate[
Plot[Evaluate[r[guess, Guess][t] /. rv3], {t, -1.5, 1.5}], {{Guess,
1.1039}, 1.1, 1.2}, {{guess, -9.051}, -10, -9}]
Manipulate[
Plot[Evaluate[v[guess, Guess][t] /. rv3], {t, -1.5, 1.5}], {{Guess,
1.1039}, 1.1, 1.2}, {{guess, -9.051}, -10, -9}]


but for length intervals greater $$l$$ than 3, for example 6, I'm having problem finding the right value for guessing. Also we have symmetry along the $$x$$-axis at midpoint and the derivates of $$r$$ and $$v$$ with respect to $$x$$ are both zero $$r'=v'=0$$ but for lengths greater than 3, I keep getting solutions that doesn't respect the symmetry and doesn't have those derivatives zero at midpoint. I'm trying to get some results similar to these:  I could replicate the $$l=3$$ one but for the rest, especially $$l=6$$ and greater I'm having problems finding the right values, because I'm getting solutions that are not correct. For example something like this for the $$v-x$$ plot: where I've chose r[-6]==100.

Can anyone point me in the right direction so I can find the values for the equations? any help would be appreciated.

• Can you please define more precisely how $r^\star$ is defined? Which midpoint exactly are you referring to? $x=0$? Jul 18 at 11:12
• @Domen r* is the value of r at the midpoint of the length interval, for example in these examples that starts from $-l/2$ and ends on $l/2$, the midpoint is $x=0$ Jul 18 at 12:32
• I don't really know how to solve this kind of differential equations with singular BVP. I have tried rewriting the DEs in terms of $1/r(x)$, so that I could use boundary condition $r(-l/2)=0$, but this didn't really help. Perhaps this can be solved by series expansion: $r(x) = \sum_n a_n / (1- (l/2)^{2n}), v(x) = \sum_n b_n x^{2n}$. Did the authors of your reference plots provide any information about their method of solving? Jul 19 at 16:59
• @Domen yeah sadly changing the variable to $\frac{1}{r(x)}$ doesn't help. At first I tried to solve it using just the BVPs but Mathematica couldn't do it, so I started using shooting method and turning it into an IVP. well not this reference per se, but others with similar approach just mentioned shooting method. Jul 20 at 7:32