this post was submitted on 04 Jun 2026
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So i fell upon the following differential equation : X''(t)X⁴(t)=f(t) I couldn't find anything similar on internet... Thanks in advance !

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[–] PM_ME_VINTAGE_30S@anarchist.nexus 2 points 1 week ago* (last edited 1 week ago) (2 children)

The fact that this comes from a physicall model means that there is a solution, no ?

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).

Do you have a link to those equivalent system, i can't find it online and anything i can search up on is nice.

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^.

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).

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.

[–] polotype@lemmy.ml 1 points 4 days ago (1 children)

Hi again ! In your reply, you told me :

this means f is a sum of exponetial

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 ;)

[–] PM_ME_VINTAGE_30S@anarchist.nexus 1 points 3 days ago* (last edited 3 days ago)

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.)

[–] polotype@lemmy.ml 2 points 1 week ago

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.