1
$\begingroup$

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: enter image description here enter image description here

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: enter image description here

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.

$\endgroup$
4
  • $\begingroup$ Can you please define more precisely how $r^\star$ is defined? Which midpoint exactly are you referring to? $x=0$? $\endgroup$
    – Domen
    Jul 18 at 11:12
  • $\begingroup$ @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 $x=0$ $\endgroup$ Jul 18 at 12:32
  • $\begingroup$ 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? $\endgroup$
    – Domen
    Jul 19 at 16:59
  • $\begingroup$ @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. $\endgroup$ Jul 20 at 7:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.