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I was brushing up on my complex arithmetic in preparation for a class in ODE's this semester and I found myself looking at Exercise 2.7.5 in Introduction to Complex Analysis for Engineers by Michael Alder, which reads

The exponential function is a procedure for turning vector fields into flows; if you take the vector field which is given by$$V\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$$ you call the matrix $A$ and then the flow is given as $e^{tA}$.
Draw a picture of the vector field. Identify the matrix as a complex number. Deduce that $e^{it}=\cos t+i\sin t$ is little more than the observation that exponentiation is about solving ODE's by Euler's method taken to the limit.

I would like very much to understand this very well. I've actually done most of it and perhaps the problem is that I haven't actually taken the ODE's class yet, but I've read ahead enough to understand most of what's being said.

I drew the vector field (by hand) and got some lovely circley looking things. When he says, "identify the matrix as a complex number," I understand that he is referring to the fact that in the book he defines a complex number $a+bi$ to be the matrix $$\begin{bmatrix} a & -b\\ b & a \end{bmatrix}$$and so $A$ is $i$. I also managed to do the exponentiation $e^{tA}$ and got

$$\begin{bmatrix} \cos t & -\sin t\\ \sin t & \cos t \end{bmatrix}$$

Which is, of course, the complex number $\cos t + i\sin t$.

So so far so good, I've shown that $e^{it}=\cos t + i\sin t$.

I'm just having problems understanding the last little bit, and maybe that's cause I haven't taken the ODE's class yet, but I see that there are tangent lines to radiuses of circles somewhere in there since the vector field makes tangent lines to circles around the origin and $e^{At}$ ends up being the rotation matrix with angle $t$, so we have some notion of a radius rotating around the origin somewhere? How does this relate to Euler's method for solving ODE's? Is the idea that there is an ODE which produces that vector field as a direction field, and $e^{it}$ gives solutions? I get a little bit lost at this point, maybe someone can help me finish putting the pieces together.

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Let me quote a few results and see if we can find a good interpretation of your post. Let $\vec{x} = [x_1,x_2,\dots , x_n ]^T$ and suppose $A$ is a constant $n \times n$ matrix. A system of linear ODEs in standard form is simply $\frac{d\vec{x}}{dt} = A\vec{x}$.

The complete solution of this problem involves a good amount of eigenvector theory and to do it right we need to know at least the basics of the Jordan form. Hopefully you can read on those things in your course... for now, just observe that the derivative of $e^{tA}$ is just $Ae^{tA}$. Why? Just differentiate term-by-term, tilt head and squint:

$$ \begin{align}\frac{d}{dt} e^{tA} &= \frac{d}{dt}(I+tA+\frac{1}{2}t^2A^2+ \cdots \frac{1}{k!}t^kA^k+ \cdots ) \\ &= A+tA^2+ \cdots + \frac{k}{k!}t^{k-1}A^k + \cdots \\ &=A(I+tA+\frac{1}{2}t^2A^2+ \cdots + \frac{1}{(k-1)!}t^{k-1}A^{k-1}+ \cdots ) \\ &= Ae^{tA} \end{align}$$

Each column of the matrix $e^{tA}$ is a solution to $\frac{d\vec{x}}{dt} = A\vec{x}$ because $\frac{d}{dt}e^{tA}=Ae^{tA}$ (just rip-apart this equation one column at a time).

The column vectors of the matrix exponential provide solutions $\frac{d\vec{x}}{dt} = A\vec{x}$. In the case you examine, these happen to be circles. But, they could be many other things.

For example, if $A=I$ then $e^{tI} = \left[ \begin{array}{cc} e^t & 0 \\ 0 & e^t \end{array} \right]$. Column one is $x=e^t$ and $y=0$ and column two is $x=0$ and $y=e^t$. The solutions are horizontal and vertical lines. But, the theory allows a linear combination hence the general solution is a line. My point is just that your example is special. Circles are connected to the imaginary exponential and those $2 \times 2$ matrices and that differential equation. This is more about the algebra of complex numbers than the general structure of DEqns.

Let me return to the geometric question at the heart of your inquiry: given $$ \frac{dx}{dt} = -y \qquad \frac{dy}{dt} = x$$ The solution is a parametric curve $\vec{r}(t) = (x(t),y(t))$ which has tangent vector field $\frac{d\vec{r}}{dt}=\langle -y,x \rangle$. You rightly understand this suggests circular trajectories. Moreover, because you explicitly calculated the matrix exponential you possess the solutions which make explicit the circular solutions of the given ODE.

In my example,with $A=I$, $$ \frac{dx}{dt} = x \qquad \frac{dy}{dt} = y$$ The solution is a parametric curve $\vec{r}(t) = (x(t),y(t))$ which has tangent vector field $\frac{d\vec{r}}{dt}=\langle x,y \rangle$. So, if we have initial point $(x_o,y_o)$ on the solution then the direction in which the curve continues is $\langle x_o,y_o \rangle$

You mention direction fields, perhaps you saw these in connection with exact equations in your calculus course; $Mdx+Ndy=0$. How are $M,N$ related to our systems? One formal method is to solve for $dt$. For your example, $$ dt = -dx/y = dy/x $$ Hence $xdx+ydy=0$ which integrates to give circles! For my silly example, $$ dt = dx/x = dy/y $$ Hence $\ln|x|=\ln|y|+C$ which yields $y = mx$. It turns out $\langle M,N \rangle$ is normal to the solutions of $Mdx+Ndy=0$. Or, we could solve for $\frac{dy}{dx} = \frac{-M}{N}$ to read off the tangent vector field $\langle -M,N \rangle$

Perhaps I have answered your question.

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Thank you very much for your detailed answer! – crf Sep 6 '12 at 23:48

Euler's method is about making successive linear approximations in short steps. I believe the comment (which isn't especially clear to me, either) is saying that "exponentiation as rotation in the complex plane" can be seen in the same way. It may be more clear if you formulate it as

$$e^{tA}=\lim_{n\to\infty} \left(I+\frac{tA}{n}\right)^n,$$

i.e. apply a tiny fraction of the $tA$ transformation many, many times. This is related to Lie algebras; see and specifically for your case.

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To be explicit, we have the ODE $$\frac{d}{d\tau}z(\tau) = iz(\tau)$$ which we want to integrate over $\tau \in [0,t]$. Divide the interval into $n$ steps of length $t/n$ each. Euler's method gives $$z_{k+1} = z_k + \frac tn\cdot iz_k = \left(1 + \frac{it}n\right) z_k$$ and so $$\quad z_n = \left(1+\frac{it}n\right)z_{n-1} = \left(1+\frac{it}n\right)^2z_{n-2} = \cdots = \left(1+\frac{it}n\right)^nz_0.$$ Does that look familiar?

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