# Solving 2nd order coupled differential equations using shooting method

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

$m(v)=\frac{1}{2}&space;(\tanh(3v)+1)&space;\\&space;\frac{r(x)^4}{r_{*}^{2}}=&space;r(x)^2&space;+&space;2r'(x)v'(x)-(r(x)^2-m(v))v'(x)^2&space;\\&space;r(x)^2&space;-&space;r(x)^2v'(x)^2&space;-&space;r(x)v''(x)&space;+&space;2&space;r'(x)v'(x)=0$

with the following boundary conditions:

$r(0)=r(l)=\infty&space;,\;\;\;&space;v(0)=v(l)=t$

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:

``````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.