# BVP - shooting method

I really don't know how to realize this method. The examples in internet and the reference aren't helpful for me.

1) I have the values for today:

 x[0] == 0, y[0] == 0.84, r1[0] == 0.39, r2[0] == 0.39, z[0] == 0


Now I need the initial conditions at t=-20 that reproduces the values today.

alpha = 0.5
beta = 1

sol = First@NDSolve[{

x'[t] == -x[t]*a[t] - 3*x[t] + alpha*y[t]^2 +
1/2*beta*(r1[t]^2 - r2[t]^2),
y'[t] == -y[t]*a[t] - alpha*x[t]*y[t],
r1'[t] == -r1[t]*a[t] - 3/2*r1[t] + beta*x[t]*r1[t],
r2'[t] == -r2[t]*a[t] - 3/2*r2[t] - beta*x[t]*r2[t],
z'[t] == -z[t]*a[t] - 2*z[t],
x[0] == 0, y[0] == 0.84, r1[0] == 0.39, r2[0] == 0.39, z[0] == 0},
{x, y, r1, r2, z},
{t, 0, -20}];
a[t_] = -1/2*(3 - 3*y[t]^2 + 3*x[t]^2 + z[t]^2)


Most of the examples are integrated in a plot. First I don't want that.

2)Is it maybe also possible to say that I have the conditions at the present as follows:

  x[t]^2 + y[t]^2==0.5
r1[t]^2 + r2[t]^2==0.5


PS: This topic can be deleted: boundary problem

• In your previous question, you ask for $x^2+y^2=\text{const}$ only at $t=0$, but in this question you ask for conservation of $x^2+y^2$ respecting to $t$, am I right? – Silvia Jun 25 '13 at 18:52
• I just want to mention, that you can delete your question all by yourself, by click the delete button at the bottom of your question. – xzczd Jul 27 '13 at 8:18

I think you're almost finished on 1) :

a[x_, y_, z_] = -1/2*(3 - 3*y^2 + 3*x^2 + z^2);

sol[t_] = {x[t], y[t], r1[t], r2[t], z[t]} /.
First@NDSolve[
{x'[t] == -x[t]*a[x[t],y[t],z[t]] - 3 x[t]+alpha*y[t]^2 + 1/2*beta*(r1[t]^2 - r2[t]^2),
y'[t] == -y[t]*a[x[t], y[t], z[t]] - alpha*x[t]*y[t],
r1'[t] == -r1[t]*a[x[t], y[t], z[t]] - 3/2*r1[t] + beta*x[t]*r1[t],
r2'[t] == -r2[t]*a[x[t], y[t], z[t]] - 3/2*r2[t] -
beta*x[t]*r2[t],
z'[t] == -z[t]*a[x[t], y[t], z[t]] - 2*z[t],
x[0] == 0, y[0] == 0.84, r1[0] == 0.39, r2[0] == 0.39,  z[0] == 0},
{x[t], y[t], r1[t], r2[t], z[t]}, {t, -20, 0}];


Check :

sol[0]
(* {0., 0.84, 0.39, 0.39, 0.} *)


sol[-20]