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The following question originates from a Physics problem, so I apologize if I am not using a suitable mathematical jargon.

The original system involves $N$ massless electric charges at position $\boldsymbol{r}_1$, $\boldsymbol{r}_2$, ..., $\boldsymbol{r}_N$ which can move on a plane pierced by a uniform and constant transverse magnetic field $\boldsymbol{B}$. The motion equation governing the dynamics of the $j$-th charge is: \begin{equation} \label{eq:Motion_eq} \boldsymbol{0}=q_j\,\dot{\boldsymbol{r}}_j\times\boldsymbol{B}\, +\, q_j\boldsymbol{E}(\boldsymbol{r}_j) \end{equation} where $q_j$ is the charge of the $j$-th electric charge and $\boldsymbol{E}(\boldsymbol{r}_j)$ is the electric field generated at position $\boldsymbol{r}_j$ by all the remaining charges. Moreover, at any time, the electric field at position $\boldsymbol{E}(\boldsymbol{r}_j)$ can be written as $$ \boldsymbol{E}(\boldsymbol{r}_j) = \sum_{i=1\\i\neq j}^N q_i\frac{\boldsymbol{r}_j-\boldsymbol{r}_i}{|\boldsymbol{r}_j-\boldsymbol{r}_i|^3}. $$ The motion equation above is manifestly of first order in time and the only initial conditions one needs are the initial positions of the massless charges, i.e. $\boldsymbol{r}_1(t=0)$, $\boldsymbol{r}_2(t=0)$, ..., $\boldsymbol{r}_N(t=0)$.

Now let us switch from massless to massive charges. This means that the aforementioned motion equation transforms as: \begin{equation} M_j \ddot{\boldsymbol{r}}_j=q_j\,\dot{\boldsymbol{r}}_j\times\boldsymbol{B}\, +\, q_j\boldsymbol{E}(\boldsymbol{r}_j) \end{equation} We assume that all the masses $M_j$ are very small. As far as I understand, the introduction of a non-zero mass constitutes a singular perturbation as it manifestly alters the order of the differential equations. Accordingly, the number of initial conditions which one should set doubles.

My question is: if one knows the solution $\{\boldsymbol{r}_1(t),\,\boldsymbol{r}_2(t),\,\dots,\,\boldsymbol{r}_N(t)\}$ of the massless problem (i.e. of the system of first-order differential equations), what can be said about the solution of the massive problem (i.e. of the system of second-order differential equations)?

For simplicity we can assume $M_j=M\,\forall j$ and we can also make additional assumptions, if needed. E.g. that the energy $$H_\mathrm{massive}=T+V=\sum_{j=1}^N\frac{1}{2}M_j\dot{\boldsymbol{r}}_j^2 + V$$ of the massive system is limited and does not significantly depart from that of the massless system $$H_\mathrm{massless}=V=\frac{1}{2}\sum_{j=1}^N\sum_{i\neq j}^N \frac{q_iq_j}{|\boldsymbol{r}_i-\boldsymbol{r}_j|}\,.$$ Another possible and reasonable assumption would be that the initial velocities of the 2nd-order problem deviate only little from the fixed initial velocities of the 1st order problem.

I guess that a term $\mathcal{O}(M^{-1})$ will show up in the solutions of the perturbed system. Is this correct? Apart from this term, is it possible to write a sort of Taylor expansion involving powers of $M$? Is there a general framework to approach this singular-perturbation problem?

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    $\begingroup$ As you note, the 1st order problem fixes the initial velocities, so for the 1st order problem to be a viable approximation for the 2nd order problem, you at the very least would also have to stipulate that the initial velocities of the 2nd order problem deviate only little from the fixed initial velocities of the 1st order problem, in addition to stipulating that $M$ be small. Perhaps looking at the $N=1$ problem would give some initial conceptual insight? There, the 1st order problem only has the trivial solution $v=0$, which also is a particular solution to the 2nd order problem ... $\endgroup$ Commented Feb 14, 2023 at 4:55
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    $\begingroup$ Since you invoke physics to motivate the question, I'd just remark that the massless equation you write isn't actually the correct equation of motion of a massless charge in a magnetic field. $\endgroup$ Commented Feb 15, 2023 at 3:40
  • $\begingroup$ @MichaelEngelhardt, thank you for your comment. Why do you say that the massless equation which I wrote isn't actually correct? In any case, I am not particularly interested to the case of massless/massive particles, but on the general way (if any) to approach and solve this kind of problems. $\endgroup$
    – AndreaPaco
    Commented Feb 15, 2023 at 19:33
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    $\begingroup$ Certainly, the physics comment is just a sidebar, the main question is how to set up a perturbation scheme of the described type. Physically, a massless particle is inherently relativistic, but the l.h.s. $M \ddot{r} $ in your equation of motion, in which, I guess, you simply set $M=0$, is nonrelativistic. The ostensible solution $v=0$ for $M=0$ is spurious - only solutions with $v=c$ are physical for $M=0$. $\endgroup$ Commented Feb 16, 2023 at 2:00
  • $\begingroup$ Yes, indeed the physics comment is just a sidebar. I see your point about the need of switching to a relativistic description for $M=0$. In any case, this issue goes beyond the purpose of my original question. $\endgroup$
    – AndreaPaco
    Commented Feb 16, 2023 at 9:25

1 Answer 1

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Let me change the notations to fit the mathematical literature. I will denote by $x(t) \in \mathbb{R}^{3N}$ the positions of the particles, by $y(t) \in \mathbb{R}^{3N}$ their velocities and by $0 < \varepsilon \ll 1$ their common small mass. Your massive problem can be written as a first order ODE as: $$ \begin{cases} \dot{x} = f(x,y), \\ \varepsilon \dot{y} = g(x,y), \end{cases} $$ with $f(x,y) = y$ and $g(x,y) = q y \times B + q E(x)$. The question now becomes whether solutions to the degenerate system $\dot{x} = f(x,y)$ and $g(x,y) = 0$ correctly approximate solutions with small $\varepsilon > 0$.

This question has been studied by many authors, probably starting with Tikhonov's paper:

Andrei Nikolaevich Tikhonov. "Systems of differential equations containing small parameters in the derivatives." Mat. Sb.(NS) 31.73 (1952): 575-586.

As a starting point, I would suggest the book:

Wolfgang Wasow, Asymptotic expansions for ordinary differential equations. Reprint of the 1976 edition. Dover Publications, Inc., New York, 1987. x+374 pp. ISBN: 0-486-65456-7

In particular, Section 39 contains an introduction to the topic (and surveys Tikhonov's results), and Section 40 discusses series expansions (with respect to $\varepsilon$), which correspond to the framework you are looking for.

Eventually, let me mention two difficulties:

  1. As already mentioned in the comments, a key difficulty in such singularly perturbed problems lies in the discrepancy between the boundary conditions, which can lead to the presence of boundary layers in the solutions with $\varepsilon > 0$ (in your case, boundary layers in time, located near $t = 0$). The study of boundary layers is a whole field by itself. If you are not familiar with it, Wikipedia's page on singular perturbations already features a nice example with the ODE $\varepsilon \ddot{x}(t) + \dot{x}(t) = - e^{-t}$.
  2. In your physical problem, there is another source of singularity, since the electric field becomes singular as two particles become very close to each other. This difficulty is typically not included in the above mentioned literature, which always requires some kind of regularity on $f$ and $g$ (at least continuous for instance). Note that this difficulty is somehow independent, since it is already present with $\varepsilon = 0$ or $\varepsilon = 1$. In order to extend the known results for the general setting explained above to your case, you would first typically have to derive a priori estimates proving that the particles won't be collapsing together on the considered time interval.

Following your request, here are two (hopefully more "practical") examples:

a) Concerning boundary layers, let us look at $\dot{x} = 1$ on $\mathbb{R}$, with initial condition $x(0) = 0$. Then the solution is $x(t) = t$. Note that, in this massless case, $\dot{x}(0) = 1$.

Now the massive problem is $\varepsilon \ddot{x}_\varepsilon + \dot{x}_\varepsilon = 1$, with boundary conditions $x_\varepsilon(0) = 0$ and $\dot{x}_\varepsilon(0) = y^0$ given. The explicit solution is $$ x_\varepsilon(t) = t + \varepsilon (y^0 - 1) (1-e^{-t/\varepsilon}). $$ In particular, when $y^0 = 1$ (the initial speed according to the massless problem), one has $x_\varepsilon(t) = t$, so the massive solution is (in this particular case) exactly the same.

On the contrary, when $y^0 \neq 1$, there is a short period of time, say $t \in [0,5\varepsilon]$ where the solution $x_\varepsilon(t)$ does not resemble the massless solution $t$. The particle quickly transitions from its initial speed to the one predicted by the massless model. After this initial period called "boundary layer", one has $x_\varepsilon(t) \approx t + \varepsilon (y^0-1)$ so resembles the massless solution $t$, with a slight correction due to the initial phase.

b) Concerning expansions, let us consider the massless model $\dot{x} + x = 0$, with initial condition $x(0) = 1$, so that $x(t) = e^{-t}$ is the solution, with $\dot{x}(0) = -1$ the initial speed in this model. Now consider the massive version $\varepsilon \ddot{x}_\varepsilon + \dot{x}_\varepsilon + x_\varepsilon = 0$ with $x_\varepsilon(0) = 1$ and let us use $\dot{x}_\varepsilon(0) = -1$ as the initial condition matching the one of the massless problem, to avoid the boundary layers mentioned above.

You can then indeed look (at least formally) for a solution under the form $$ x_\varepsilon(t) = \sum_{k=0}^{+\infty} \varepsilon^k x_k(t) $$ with $\dot{x}_0 + x_0 = 0$, $x_0(0) = 1$ and, for $k \geq 0$, $$ \dot{x}_{k+1} + x_{k+1} = - \ddot{x}_k. $$ Here $x_0$ is the solution to the massless problem so $x_0(t) = e^{-t}$, and you can solve iteratively using the variation of constant formula. For example, $x_1(t) = - t e^{-t}$, and so on.

In this case, you can prove that if you truncate the series, you have a solution which approximates $x_\varepsilon$ at any precision.

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  • $\begingroup$ Thank you for setting my question into the right framework. Indeed, since I am not familiar with the mathematical background and jargon which this topic is linked to, I would kindly ask you to provide a more "practical" answer, i.e. to show me, if possible, how this boundary layer is and how the solution of the perturbed problem can be approximated. $\endgroup$
    – AndreaPaco
    Commented Feb 23, 2023 at 22:58
  • $\begingroup$ Here you go. I tried to add 2 examples. Unfortunately, handling thoroughly your initial question from a fully mathematical point of view would probably require a whole research paper. I hope the examples allow you to grasp some of the aspects of this subject. $\endgroup$
    – cs89
    Commented Feb 26, 2023 at 18:23

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