# dropping a particle into a vector field, part 2

Okay, so earlier I posted this question "dropping a particle into a vector field " as sort of a feeler question as i study line integrals in order to go into surface integrals and eventually differential forms and differential geometry and I got to thinking about my question's example and how to solve it. I've taken an advanced linear algebra course and the required ODE course, but the ODE course never made it to solving systems of differential equations. So my posited vector field in which I'm dropping a particle into is $$\mathbf{F}(x,y)=y\mathbf{i}-x\mathbf{j}$$ An answer came in the form that in order to solve the question I posited, I needed to find an $\mathbf{r}$ such that $\mathbf{r}'(t)=\mathbf{F}(\mathbf{r}(t))$. So my arbitrary $\mathbf{r}(t)=(\mathbf{x}(t),\mathbf{y}(t))$, if i take the derivative of this since it's just a vector valued function i get $$\mathbf{r}'(t)=(\mathbf{x}'(t),\mathbf{y}'(t))$$ Now, $$\mathbf{F}(\mathbf{r}(t))=\mathbf{F}(\mathbf{x}(t),\mathbf{y}(t))=\mathbf{y}(t)\mathbf{i}-\mathbf{x}(t)\mathbf{j}=(\mathbf{y}(t),-\mathbf{x}(t))$$ So my equation has become $$\mathbf{r}'(t)=(\mathbf{x}'(t),\mathbf{y}'(t))=(\mathbf{y}(t),-\mathbf{x}(t))$$ This gives me a system of differential equations to solve: $$\mathbf{x}'(t)=\mathbf{y}(t)$$ $$\mathbf{y}'(t)=-\mathbf{x}(t)$$ And here I'm not sure how to procede, or if this is even the correct intution. The person that wrote the answer on the last post gave me numeric Euler method as a general case, but this seems pretty solvable and straight forward, but I'm not seeing where to move.

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Notice that $x''(t) = y'(t) = -x(t)$. What functions do you know of which satisfy $x''(t) = -x(t)$? – Adam Saltz Sep 19 '13 at 1:52
It's funny because I had an inkling because the particular vector field that intuitively the cosine and sine functions would play a role, but seeing that the answer lies in simply taking the derivative of $\mathbf{x}'(t)$ is enlightening, and a little embarassingly easy... – Eleven-Eleven Sep 19 '13 at 1:55
and plus, this little particle would just be going in circles....I should have picked up on that!' – Eleven-Eleven Sep 19 '13 at 1:57
It's often helpful to sketch the vector field and a few trajectories before doing any heavy-duty work. – Adam Saltz Sep 19 '13 at 2:02
Actually this helps because again, I've never studied systems of differential equations. I'm sure there are much more difficult cases, but It's always good to have methods to solving things in your pocket. Thanks! – Eleven-Eleven Sep 19 '13 at 2:05

It's pretty easy to see the "little particle" is going in circles; we don't need to solve the differential equation or directly introduce sines and cosines; just look at

$\mathbf r^2(t) = \mathbf r(t) \cdot \mathbf r(t) = \mathbf x^2(t) + \mathbf y^2(t); \tag{1}$

if we differentiate this with respect to $t$ we obtain

$\frac{d}{dt}(\mathbf r^2(t)) = 2\mathbf x(t) \dot{\mathbf x}(t) + 2\mathbf y(t) \dot{\mathbf y}(t), \tag{2}$

and if we now use our differential equation

$\dot{\mathbf x}(t) = \mathbf y(t), \tag{3}$

$\dot{\mathbf y}(t) = -\mathbf x(t), \tag{4}$

substituting (3) and (4) into (2) we find that

$\frac{d}{dt}(\mathbf r^2(t)) = 2\mathbf x(t) \mathbf y(t) - 2\mathbf y(t) \mathbf x(t) = 0. \tag{5}$

(5) shows that $\Vert \mathbf r(t) \Vert^2 = \mathbf r^2(t)$ is constant, hence $\Vert \mathbf r(t) \Vert$ is constant; the little particle moves on a circle of radius $\Vert \mathbf r(t_0) \Vert$, where $t_0$ is some initial moment in time. We can do better, still without solving a differential equation: note the vector $\dot{\mathbf r}(t) = (\mathbf y(t), -\mathbf x(t))^T$ is in fact orthogonal to $\mathbf r(t) = (\mathbf x(t), \mathbf y(t))$, so it is tangent to the circle on which the particle is constrained to move, which is as we have seen is of constant radius $\Vert \mathbf r(t_0) \Vert$ about the origin $(0, 0)^T$. In fact it is easy to see that the vector $\dot{\mathbf r}(t) = (\mathbf y(t), -\mathbf x(t))^T$ points in a clockwise direction; furthermore the speed of the particle, that is, the rate at which it traverses distance along the circle on which it moves, is clearly given by

$\Vert \dot{\mathbf r}(t) \Vert = \sqrt{\mathbf y^2(t) + \mathbf x^2(t)} = \Vert \mathbf r(t) \Vert = \Vert \mathbf r(t_0) \Vert; \tag{6}$

notice that the speed is of constant magnitude; in fact, this magnitude is just the size of the radius of the circle! And since the circumference of a circle of radius $\Vert \mathbf r(t_0) \Vert$ is given by $2\pi \Vert \mathbf r(t_0) \Vert$, it follows that the entire circle is traversed exactly once every $2\pi$ seconds (and here I'm assuming we are measuring time in seconds); and since a circle, in terms of angular measure, is precisely $2\pi$ radians, we see that the angular velocity of the particle about the point $(0, 0)^T$ is a constant one radian per second in magnitude. So if $\theta$ is the central or polar angle in our coordinate system, then we must have

$\dot \theta = -1, \tag{7}$

or

$\theta (t) =\int_{t_0}^t (-1)ds = t_0 -t + \theta_0, \tag{8}$

where we choose $\theta$ increasing counter-clockwise, and $\theta_0 = \theta(t_0)$. At this point it is convenient to introduce sines and cosines. Since $(\mathbf x(t), \mathbf y(t))^T$ lies on the cirle of radius $\mathbf r(t_0)$, we may write

$\mathbf x(t) = \Vert \mathbf r(t_0) \Vert \cos \theta(t) = \Vert \mathbf r(t_0) \Vert \cos (t_0 -t + \theta_0), \tag{9}$

$\mathbf y(t) = \Vert \mathbf r(t_0) \Vert \sin \theta(t) = \Vert \mathbf r(t_0) \Vert \sin (t_0 -t + \theta_0), \tag{10}$

which apparently gives a complete solution to the equation(s) (3)-(4) with initial conditions

$\mathbf x(t_0) = \Vert \mathbf r(t_0) \Vert \cos \theta_0, \tag{11}$

$\mathbf y(t_0) = \Vert \mathbf r(t_0) \Vert \sin \theta_0. \tag{12}$

The preceding analysis provides an example of how a differential equation, or system of differential equations, may be analyzed and sometimes even solved without actually "solving" it, if you take my meaning: sometimes there are ways to "get at" the solution without resorting to formal quadrature or related, systematic techniques. For example, here we used a geometrical analysis of the vector field $(-\mathbf y, \mathbf x)^T$. Methods related to that exploited here were used in my answers to this question and this one.

Of course using (3), (4) to derive $\ddot {\mathbf x} = \dot{\mathbf y}$, $\ddot {\mathbf y} = - \dot {\mathbf x}$ and from there move to $\ddot{\mathbf x} + \mathbf x = 0$, $\ddot{\mathbf y} + \mathbf y =0$, as suggested by Adam Salz in his comment, and from there to solutions which are of the form $C_1\sin (t + \theta_0)$, $C_2\cos(t + \theta_0)$, or linear combinations thereof, with $C_1, C_2$ constants, is part of a generally accepted and systematic method which forms an essential part of any ODE solver's toolkit. And on a related note, the theory of matrix exponentials, addressed by automaton in his/her comment, is another facet of the jewel which itself forms an indispensable tewel ($\equiv \text {tool}$ ;)!); in the present case this approach is particularly simple; we simply write (3), (4) as

$\dot {\mathbf r}(t) = \begin{pmatrix} \dot{\mathbf x}(t) \\ \dot{\mathbf y}(t) \end{pmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{pmatrix} \mathbf x(t) \\ \mathbf y(t) \end{pmatrix} = J \mathbf r(t), \tag{13}$

where

$J = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \tag{14}$

which satisfies

$J^2 = - I. \tag{15}$

Now it is a fact which our OP Christopher Ernst will soon encounter, if he has not done so already, that for any constant matrix $A$ we can define the matrix $e^{At}$, just as we can define $e^{\alpha t}$ for scalars $\alpha$, and that we have

$\frac{d}{dt}e^{A(t - t_0)} = Ae^{A(t - t_0)}, \tag{16}$

just as

$\frac{d}{dt}e^{\alpha(t - t_0)} = \alpha e^{\alpha (t - t_0)}, \tag{17}$

so that just as

$\mathbf x(t) = e^{\alpha (t - t_0)} \mathbf x(t_0) \tag{18}$

solves

$\dot {\mathbf x}(t) = \alpha \mathbf x(t) \tag{19}$

with initial condition $x(t_0)$ at $t = t_0$, so we have

$\mathbf r (t) = e^{A(t - t_0)} \mathbf r(t_0) \tag{20}$

solves

$\dot {\mathbf r}(t) = A\mathbf r(t) \tag{21}$

with initial condition $\mathbf r(t_0)$ at $t = t_0$. Applying these ideas to our equation (13), we find that

$\mathbf r(t) = e^{J(t - t_0)} \mathbf r(t_0), \tag{22}$

and in the present case the matrix $e^{J(t - t_0)}$ is particularly easy to evaluate. From (15), it follows that the algebra involved in computing $e^{J(t - t_0)}$ precisely parallels that of calculating $e^{i(t - t_0)}$ where $i^2 = -1$ is the standard complex scalar $\sqrt{-1}$. Thus in fact just as

$e^{i(t - t_0)} = \cos(t - t_0) + i \sin(t - t_0), \tag{23}$

so

$e^{J(t - t_0)} = \cos(t - t_0) + J \sin(t - t_0), \tag{24}$

which can easily be seen via a term-by-term comparison of the power series for $e^{i(t - t_0)}$ and $e^{J(t - t_0)}$; I'll leave the relatively simple issues of convergence of these series to my readers; they are not difficult. From (24) combined with (14):

$e^{J(t - t_0)} = \begin{bmatrix} \cos(t - t_0) & \sin(t - t_0) \\ -\sin(t - t_0) & \cos(t - t_0) \end{bmatrix}, \tag{25}$

and so from (22) and (11), (12) we obtain

$\mathbf x(t) = \Vert \mathbf r(t_0) \Vert (\cos \theta_0 \cos (t - t_0) + \sin \theta_0 \sin(t - t_0)) \tag{26}$

and

$\mathbf y(t) = \Vert \mathbf r(t_0) \Vert (-\cos \theta_0 \sin (t - t_0) + \sin \theta_0 \cos(t - t_0)), \tag{27}$

which after a little trigono-algebraic manipulation are seen to agree with (9) and (10). So be some of the more conventional, systematic procedures for solving (3), (4), (13) and their kindred equations.

There are of course numerical methods, such as Euler's, which can be used to calculate solutions to these equations and much more complex, general systems; but for problems like the one put forth in this question, which have relatively easily derived, closed-form solutions, use of such numerical approximation schemes is of questionable value. But such schemes can be used to find the solutions to systems which are otherwise intractable; their study is a vast and complex subject in its own right. So with these words I say, adieu, 'till we meet again!

Hope this helps. Cheers, and as always

Fiat Lux!!!

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