Calculating particle paths for a two-dimensional flow

Question

I'm having trouble considering the velocity field $\mathbf{u}=(u,v)=(y,-x)$. I have been asked (as part of a homework assignment) to determine the time-dependent position of a particle in this field that is initially at $\mathbf{x}=(x_0,y_0)$.

I know that in a problem that has an $x$-component independent of $y$ and a $y$-component independent of $x$ that I would form the equations:

$$\frac{dx}{dt}=u,\frac{dy}{dt}=v$$

and integrate directly, then substitute in the initial conditions to determine the constant of integration. However, in this case I'm confused as the equations I obtain are:

$$\frac{dx}{dt}=y,\frac{dy}{dt}=-x$$

How does one solve this system? I attempted to formulate an expression for $\dfrac{dy}{dx}=-\dfrac{x}{y}$, but while solving this is simple, the solution is not time-dependent.

Answer 1

Hint: If you consider the variable $z=x+iy$ your equations become $\frac {dz}{dt}=-iz$

Answer 2

Hint: $$\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}\begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix} x' \\ y' \end{pmatrix}.$$

Answer 3

Along similar lines to Ross' answer, let $w=\binom{x}{y}$ and $A=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$, then you have $\dot{w} = A w$, for which the solution is $w(t) = e^{(t-t_0)A}w(t_0)$.

If we let $U = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ i & -i \end{bmatrix}$, then $A = U \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} U^{-1}$, hence $e^{At} = U \begin{bmatrix} e^{it} & 0 \\ 0 & e^{-it} \end{bmatrix} U^{-1}$, and expanding gives $e^{At}=\begin{bmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{bmatrix} $.

Hence the solution is $\binom{x(t)}{y(t)} = \begin{bmatrix} \cos (t-t_0) & \sin (t-t_0) \\ -\sin (t-t_0) & \cos (t-t_0) \end{bmatrix} \binom{x(t_0)}{y(t_0)}$.