Yeah, sure, I forget the x'(0) initial cond. However I don't see how it is possible to write the system with onlye 2 state variables, but I think I figured out how to do it with 3 state variables. Thanks
Yeah, you are right, my fault, but the notations are x1 = x, x2 = x' = x1'.
The ODE follows from the Lagrangian
L=x^(-2/a)*x'^2-b*x^(-2/a).
Therefore you have the Hamiltonian constrain
h=x^(-2/a)* (x'^2+b) where h=x(0)^(-2/a)*(x'(0)^2+b)
Hence x'=sqrt(h*x^(2/a)-b).
You can solve this problem analytically.
If you let dx/dt = y, then dy/dt = (y^2 + b)/(a*x)
then x(0) = x0 and x'(0) = y0
You can write: (dy/dt) / (dx/dt) = dy/dx = (y^2 + b)/(a*x*y)
This is separable: y/(y^2 + b) * dy = dx/(a*x)
Integrating: 1/2*ln( (y^2 + b) / (y0^2 + b) ) = 1/a*ln(x/x0)
Removing the "ln": ( (y^2 + b) / (y0^2 + b) )^(1/2) = (x/x0)^(1/a)
Solve for "x": x = x0*( (y^2 + b) / (y0^2 + b) )^(a/2)
and substitute into the original equation:
dy/dt = (y^2+b)^(1-a/2) * (y0^2 + b) )^(a/2) / (a*x0)
= c*(y^2+b)^(1-a/2)
Where c = (y0^2 + b) )^(a/2) / (a*x0)
The ODE is again separable:
[(y^2+b)^(a/2-1)] * dy = [c] * dt
The LHS is integrable, but the form of the function depends on the value of "a".
You will end up with f(y) = f(y0) + ct and x = (y^2 + b) / (a*c)
Example 1: a = 3, b = 0
y = sqrt(2*c*t + y0^2)
x = x0*( y/y0 )^3
Example 2: a = 2
y = c*t + y0
x = x0* (y^2 + b) / (y0^2 + b)
For random values of "a", it may get tricky.
Your equation is (quasi-linearly) implicit and therefore you must expext the appearance of geometric singularities. I am assuming in the sequel that a0, as otherwise you obtain a fully implicit first order equation requiring a different analysis. In your case the singularities are given by the vanishing of the separant (the derivative with respect to x''), i.e. all points where x=0 are singularities and it is also easy to determine that they are irregular singularities. Typically this means that the classical uniqueness results for ODEs do not hold at these points and solutions may intersect. In your equation one must distinguish whether b=0 or b0. If b=0, then the zero function x(t)=0 is a singular integral of the equation which can be considered as a kind of asymptote to the general solution x(t)= +/- sqrt(2*c*t+d) (obviously these square root solutions indeed intersect the singular integral at their zero). If b0 (actually one should have b
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