I'm studying ordinary differential equations and learned some methods to solve linear ones. I'm definitely a novice. Well I think I can follow the procedures and math behind it. However, I'm not sure that differential equations actually model realistic phenomena. For example, a car reduces its speed continuously and finally stops in real world. Let's denote the velocity of the car as $v(t)$ which is a function of time $t$. And let $\tau$ be the time when the car stops. Then $v$ is not an analytic function. It is singular at $\tau$. We may assume that $v$ is infinitely differentiable at $\tau$ but anyway it is not analytic there. Can we model this movement of the car with a differential equation? All solutions of the differential equations I have learned were analytic. I think the equations themselves don't put such restrictions. But I don't get why the solutions are all analytic. Are there special methods to obtain non-analytic solutions? Or should the solutions of differential equations be analytic? If so, how can we model the movement of the car with differential equations? I have navigated my textbook but found no clue (The book may be too elementary).

  • $\begingroup$ If you have non-analytic coefficients or inputs you can have non-analytic solutions. Does your differential equations textbook have a chapter on step functions and impulse functions? $\endgroup$ – bof Apr 11 '17 at 4:59
  • $\begingroup$ @bof I think it does not have. Should we allow the coefficient or input to be singular to model realistic phenomena? $\endgroup$ – LuianChu Apr 11 '17 at 5:03
  • $\begingroup$ Actually, even things like step functions are idealizations: the point however is that you can model interesting phenomena and the solutions can be used to gain insight, even if they don't correspond to reality exactly (e.g. the simulations that car companies do to test car components). But within the limitations of the model, there are situations where the solutions are very accurate representations of reality (e.g. the calculation of the energy eigenstates of the hydrogen atom, including very small corrections from QED) $\endgroup$ – NickD Apr 11 '17 at 5:11
  • $\begingroup$ Yes, I think that for many applications you will have to have non-analytic (in fact discontinuous, or generalized) functions in the equation. If your book has problems with step functions and impulses, they are probably right after the section on Laplace transforms. $\endgroup$ – bof Apr 11 '17 at 5:12

First of all don't confuse the possibility of modeling something with a differential equation with whether a particular differential equation models something. Basically what you have is a differential equation that doesn't capture some of the aspects of reality very well, this cannot be taken as argument that these aspects can't be modeled using a(nother) differential equation.

Note that most models are meant to be mere approximations, they will therefore in some cases deviate significantly from reality in some cases or at some aspects.

Also there's no real requirement that the functions in a differential equation is analytic, obviously there's no reason to assume that the solution is either. However it's often so in physics that one assumes that one can differentiate all functions, but strictly that might not be required anyway in the normal sense (one can use so called distribution to get around some situations where the function is apparently non-differentiable).

As for the stopping car, there's no actual reason to believe that the movement of it isn't differentiable. Even if it looks like it comes to a sudden stop it could be that it's just that we don't look close enough, maybe the deceleration decreases quickly instead of just transiting directly to zero?

If we're to use the model that the braking force is constant whenever the car is in motion the solution for $v(t)$ would have a corner-point where it comes to a stop, but the model would as a differential equation be $v'(t)=F_0\operatorname{sgn}(v)/m$. We can handle this in various way, one is to rewrite it as an integral equation instead $v(t) = \int F_0\operatorname{sgn}(v) dt/m$ which won't be hindered by the corner-point. Another way is to use distributions where $v$ would be differentiable anyway. A third approach is to allow non-differentiability at isolated points where other requirements governs the solution (for example allowing $v(t)$ do be non-differentiable at isolated points given it's at least continuous there).

  • $\begingroup$ For the first approach, $v$ exponentially decreases, but never vanishes. The car never stops. I think the corner by $sgn$ does not matter in this case. Anyway accepting small enough velocity as zero may be a practical solution but are there really no better way? How can we measure how long does it take the car stop? For the second approach, I think I don't know what you mean. $\endgroup$ – LuianChu Apr 11 '17 at 5:55
  • $\begingroup$ For the third approach, I understand allowing non-analytic equations may allow non-analytic solutions. But is that the only way to obtain non-analytic solution? For example, can a second order linear ODE with continuous coefficients have non-analytic but second differentiable or smooth solution? $\endgroup$ – LuianChu Apr 11 '17 at 5:56
  • $\begingroup$ @LuianChu That the speed exponentially decreases is not a very realistic solution to a braking car, the model does not reflect normal reality well. Normally you would try to apply constant braking force to achieve constant retardation rate. With constant braking force the car actually gets to a full stop. $\endgroup$ – skyking Apr 11 '17 at 6:03
  • $\begingroup$ I would guess that you can't get a non-analytical solution out of an ODE with only analytic functions (but I don't have any evidence for it). The ODE given as a model obviously has a non-analytic function $\operatorname{sgn}$. OTOH if the functions in the ODE is only continuous I'd guess that one can find equations that has non-analytical solutions. Note that the solution has to be derivable somehow or one normally need to put other restrictions on the solution to handle these points. $\endgroup$ – skyking Apr 11 '17 at 6:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.