this post was submitted on 04 Jun 2026
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I would really like to be able to solve it analitically (further analysis of the results is required so the more mathematical the solution, the better.) I will have x(0) and x'(0). This question comes from trying to solve for magnet acceleration inside a coil. This means that f is drawn from the space of solution ms of a second order homogenous liner differential equation (from an RLC circuit with no generator).
The fact that this comes from a physicall model means that there is a solution, no ?
Do you have a link to those equivalent system, i can't find it online and anything i can search up on is nice.
Thanks so much for such a complete answer. :D
Good instinct, but nope! There might be some phenomenon your model is neglecting where the neglected phenomenon allows the solution to exist but the model fails to have a unique solution. It also could have a non-unique solution (i.e. many solutions).
So the first move is to simply solve the equation for X'':
X'' = X^-4 f
Unfortunately, for the next technique, it's typically buried in any nonlinear control book. I'll give the technique below. I'm gonna do it for a fourth order system to make the technique clear, but it applies for ODE systems of any size.
Say you have the following fourth order scalar differential equation:
X''''(t) = F(t,X(t),X'(t),X''(t),X'''(t)) Make the following substitutions:
x1 = X
x2 = X'
x3 = X''
x4 = X'''
If you have initial conditions at some time t0, then they must be:
x1(t0) = X(t0)
x2(t0) = X'(t0)
x3(t0) = X''(t0)
x4(t0) = X'''(t0)
(In your problem, t0 = 0, plus your problem is time-invariant so it doesn't matter, but I'm describing the technique here for time-varying problems for generality, where it does matter.)
Differentiate the four equations as follows:
x1' = X'' = x2
x2' = X'' = x3
x3' = X''' = x4
x4' = X'''' = F(t,X(t),X'(t),X''(t),X'''(t)) = F(t,x1,x2,x3,x4)
Stack these into a vector as follows:
[x1,x2,x3,x4]^T^ = [x2,x3,x4,F(t,x1,x2,x3,x4)]^T^
(The ^T^ just means "make it a column vector".) You now have a system of four coupled first-order ODEs that any ODE system solver will solve. By applying the technique described above to the second-order ODE X'' = X^-4 f, you get [x1,x2]^T^ = [x2,x1(t)^-4 f(t)]^T^.
Got it, so the forcing is a sum of exponentials. So your input space is a subspace of the infinitely differentiable functions (except possibly at 0 if you choose to model the switching-on of the circuit as a step function). But I'm a little suspicious of the X^4^ term? Like I'm not a "magnets guy" although I've taken electromagnetic fields, but I don't think that's the type of nonlinearity I would expect to see (in particular , the fact that it's fourth power). I.e. if you derived this model yourself, double-check the derivation.
Hi again ! In your reply, you told me :
However. For the case wher f is of the form f=A•t•exp(b•t), i don't know how to convert that as a sum of exponential... I'm asking because i feel like i could just take x to be an indefinite sum of exponentials which might end up giving me some quite nice solutions... If only i could get a sum of exponentials on the other side of the equation ;)
Sorry, I was being a bit sloppy in the first answer. Your f=A•t•exp(b•t) term is, in general, part of the expected function space. (I just kinda forgot because in control systems, we often assume distinct real system poles for simplicity in teaching the material.)
Thanks you so much !! As for the 4th power, it comes from the fact that the magnetic field on the central axis of a coil is proportionnal to 1 over the cubed distance to the center of the coil. Since the force applied to the magnet is proportional to the spacial gradient of the magnetic fueld at that point, you have to differentiate the previous expression which give some x^-4.
Also, what specific further analysis do you want to do for the differential equation? Because you might not need the solution for it.
The limit of x' as t goes to infty would be good enough, but if i could get x'(t), it would allow me to take into account partial capacitor discharge which would be a much better fit.
Hmm...this feels like a Lyapunov-type problem, but there's some issues. The vector function [x2,x1^-4 f] has no equilibrium, except when f = 0. So assume that f = 0. Then, the system is [x1',x2'] = [x2,0] = [0 1;0 0][x1,x2]. I believe that the zero solution satisfies your original differential equation. You would then need to find a Lyapunov function that guarantees the system is globally asymptotically stable around the zero solution. (Make sure to check for "globally asymptotically" in the stability theorem, e.g. in Theorem 3.2 of Haddad and Chellaboina (LibGen). The "asymptotic" part is what gives you the steady-state result you're looking for.) By taking the second component, I would then conclude that the limit of X' is the limit of x2, which would be 0.
But that's assuming zero input and initial state, and Lyapunov functions can be hard to find. Linearizing fails; you would have to guess a Lyapunov function.
Honestly, from an engineering perspective, I think you should just simulate the system. In particular, instead of leaving f(t) as a function, I suggest plugging in the second-order ODE and numerically solving the resulting composite fourth-order ODE system for your RLC values and some deviations from those values. Simulating it might give you an ansatz (educated guess) on if or where the system has equilibria.
Damn, well i have my work cut our for me :D once again, thank you for your helpfulness.