nonlinear 2nd-order diff

[dm164]

Hi, I want to solve a differential equation but I have never dealt with nonlinear equations before.

Here it is : A - B/x = x" for x(t) A,B are constants you can absorb the negative if it matters.
I would like an analogical answer, but I'm not sure if it's possible. And any explanation or link to explain on how to solve would also be appreciated.

[dm164]

I tried some manipulation using rules of differentiation. And used the quadratic to get a new relation, then I assume there is only 1 solution and set the discriminate to 0. I don't know if I'm allowed to all of it. I then set d2x/dt2 = y And then got the equation C(y")2 = y another nonlinear I can't solve. If the discriminate is 0 then y-A = -By"/y/2

[JJacquelin]

It is not possible to express the function x(t) in terms of standard functions.
The function t(x) is obtained on the form of an integral, which is the closest form as a formal solution.
From the integral, x(t) can be computed thanks to numerical means.

[dm164]

Very Interesting, and the method makes a lot of sense wish I thought of it. Thanks.

[mbp]

The solution given by JJacquelin is correct. In fact your system is conservative, it can be written as a system of first order ODE
\begin{eqnarray}
x' & = & y \\
y' & = & \frac{-b}{x} + a
\end{eqnarray}
Divide one equation by the other to get
\begin{equation}
\frac{dy}{dx} = \frac{-b/x + a}{y}
\end{equation}
Separate the variables and integrate to obtain the Hamiltonian
\begin{equation}
H(x,y) = \frac{y^2}{2} - a x + b \ln x
\end{equation}
The level sets H(x,y) = E are the trajectories of your system in the phase space. If you need the time parametrization from this equation you can get
\begin{equation}
y = \pm \sqrt{2(a x - b \ln x + E)}
\end{equation}
Introduce this in the first equation, separate the variables and integrate
\begin{equation}
\int \frac{dx}{ \sqrt{2(a x - b \ln x + E)}} = t - t_0
\end{equation}
and you find the solution given by JJacquelin. This is a standard procedure to solve second order ODE stemming from conservative systems.

[dm164]

Well it better be conservative, otherwise my physics is wrong. If anyone is interested. This equation comes from the movement of a flat surface with two pressures that are reaching equilibrium. The motion I restrained to be linear. x is the length of a closed volume. I'm a little confused by the Hamiltonian. I've only seen it operate on a wave function. It makes sense to make it the total energy as A and B define parameters of the pressure and if there is a greater difference in pressure there should be more energy in the system. But, it doesn't make sense to me that the constant C1 would be the Energy, since there is a dependence. Maybe it is net energy and be 0 in the equation? And would be a value if it was in a theoretical potential but for mass? Could you help figure what C1 should be?

This is my work.
With a enclosed surface like box or cylinder like. Letting only one surface move. x is the length so x = 0 would have 0 volume.
Po*A - Pi*A = Fnet = m*d2x/dt2 ; Po - outside pressure ; Pi - inside pressure; A - area of surface.
Pi = nRT / (xA) ; m = σ*A ; σ being a surface mass density

to get to my equation above
a - b/x = d2x/dt2 ; where a = Po/σ and b = nRT/(σA)

I think from the integral is should be from x0 to xeq (equilibrium). Which xeq can be solved with a - b/x = 0 -> xeq = b/a
And then the integral can go from x0 to b/a to give a solution for Δt. But, JJacquelin is the more general solution. I think the ± comes from the x able to go either direction, but is dependent on a and b. Still not sure about C1



I don't really need an answer, but I like practicing problems I come up with. And, Thanks

[mbp]

In classical mechanics the Hamiltonian is a function of the
(generalized) coordinates and momenta. The energy is fixed by the
initial conditions
\begin{equation}
E = H(x(0),y(0))
\end{equation}
Starting from x(0), y(0) the system evolves along the curve
H(x(t),y(t)) = E for all t.

The constant C_1 actually is the energy. You can understand it if
in the derivation given by JJacquelin, you consider definite
integrals between x(0) and x(t) instead of indefinite integrals.