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