nonlinear 2nd-order diff


by dm164
Tags: 2ndorder, diff, nonlinear
dm164
dm164 is offline
#1
Dec26-11, 10:58 PM
P: 21
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.
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dm164
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#2
Dec26-11, 11:46 PM
P: 21
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
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#3
Dec27-11, 01:18 AM
P: 745
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.
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dm164
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#4
Dec27-11, 02:03 PM
P: 21

nonlinear 2nd-order diff


Very Interesting, and the method makes a lot of sense wish I thought of it. Thanks.
mbp
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#5
Dec28-11, 06:44 AM
P: 7
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
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#6
Dec28-11, 03:44 PM
P: 21
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
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#7
Dec29-11, 01:37 PM
P: 7
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.


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