Csdrhrt

Suppose I have a parameterized path $gamma:[0,infty)rightarrowmathbb C$ with initial condition $gamma(0)=1$ and which satisfies the following:
$$gamma'(t)=uBig(igamma(t)-gamma(t)Big)=expBig(frac{3pi}{4}iBig)frac{gamma(t)}{|gamma(t)|}$$
where $u:mathbb C^*rightarrowmathbb D$ denotes the unit vector function. In other words, $gamma(t)$ is always moving at constant unit speed towards its rotation in the plane by $pi/2$. I found computationally that
$$lim_{trightarrowinfty}gamma(t)=0$$

Questions:

1. Is there a formula for $gamma(t)$?




2. Does $gamma$ reach the origin in finite time? In other words, does there exist $T<infty$ such that $lim_{trightarrow T}gamma(t)=0$?

Edit: Originally we had $gamma'(t)=s(i-1)frac{gamma(y)}{|gamma(t)|}$ but without loss of generality set $s=1/sqrt{2}$ so that $gamma'(t)$ has unit magnitude. Also I realize that the path length $int_0^infty|gamma'(t)|dt$ must be infinite since $|gamma'(t)|$ is constant.

It appears that this equation can be well-addressed via a transformation to polar coordinates; in polars, we write

$gamma(t) = r(t)e^{itheta(t)}; tag 1$

then

$dot gamma = dot r e^{itheta} + r i dot theta e^{itheta}; tag 2$

also,

$vert gamma vert = r, tag 3$

whence

$dfrac{gamma}{vert gamma vert} = e^{itheta}; tag 4$

assembling all this into the given equation

$dot gamma = s(i - 1) dfrac{gamma}{vert gamma vert} tag 5$

yields

$dot r e^{itheta} + r i dot theta e^{itheta} = s(i - 1) e^{itheta}; tag 6$

we divide out $e^{itheta}$:

$dot r + ridot theta = s(i - 1) = -s + is; tag 7$

equating real and imaginary parts,

$dot r = -s, tag 8$

$ridot theta = is; tag 9$

$rdot theta = s; tag{10}$

we may easily integrate (8) 'twixt $t_0$ and $t$, assuming $r(t_0) = r_0$:

$r - r_0 = -s(t - t_0), tag{11}$

$r = r_0 - s(t - t_0); tag{12}$

we observe that the coordinate constraint $r > 0$, along with the condition $s > 0$, implies that $t$ must remain less than $r_0/s + t_0$; we shall soon find that the polar solution circumvents this limit on $t$; inserting (12) into (10):

$(r_0 - s(t - t_0))dot theta = s; tag{13}$

$dot theta = dfrac{s}{r_0 - s(t - t_0)}, tag{14}$

which is also easily integrated 'twixt $t_0$ and $t$:

$theta - theta_0 = -ln(r_0 - s(t - t_0)) + ln r_0; tag{15}$

$theta = theta_0 - ln(r_0 - s(t - t_0)) + ln r_0; tag{16}$

using (12), we may express $r$ in term of $theta$:

$theta = theta_0 - ln r + ln r_0; tag{17}$

$e^theta = e^{theta_0} dfrac{r_0}{r}; tag{18}$

$r = r_0 e^{theta_0 - theta}; tag{19}$

the above presents the general solution for $t = t_0$, $r(t_0) = r_0$, $theta(t_0) = theta_0$; we specialize to the case at hand:

$t_0 = 0, ; gamma(0) = 1 + i; tag{20}$

$r_0 = vert gamma(0) vert = sqrt 2; tag{21}$

$theta_0 = dfrac{pi}{4}; tag{22}$

from (19):

$r = sqrt 2 e^{pi/4 - theta}; tag{23}$

from (12) and (16):

$r(t) = sqrt 2 - st; tag{24}$

$theta(t) = dfrac{pi}{4} - ln(sqrt 2 - s(t - t_0)) + ln sqrt 2. tag{25}$

The path length is perhaps easiest found by returning to the expressions for $r(t)$ and $theta(t)$, in terms of which

$vert dot gamma vert = sqrt{ (dot r)^2 + r^2 (dot theta)^2} = sqrt{s^2 + s^2} = sqrt 2 s; tag{26}$

since $t$ is restricted to $[0, r_0/s]$, the length of the curve is

$displaystyle int_0^{r_0/s} vert dot gamma vert ; dt = displaystyle int_0^{r_0/s} sqrt 2 s ; dt = sqrt 2 r_0. tag{27}$

As concerns our OP M. Nestor's final remarks:

The curve $gamma(t)$ (1), expressed as a function of the independent variable $theta$ is, in polar coordiates, given by (23); it is easy to see this curve is a spiral which approaches the origin as $theta to infty$.

Now, as for item (1), there is indeed a formula for $gamma(t)$; in polar coordinates the equations are given in (12) and (16); from these we may derive
cartesian equations for $x(t)$, $y(t)$ by setting

$x(t) = r(t) cos theta (t), ; y(t) = r(t) sin theta(t); tag{28}$

and as for item (2), we have shown above ca. (12) that

$displaystyle lim_{t to t_0 + r_0/s} r(t) = 0, tag{29}$

where $gamma(t_0) = (r_0, theta_0)$; but it should be noted that $gamma(t)$ never "reaches the origin", though it does become arbitrarily close as $t to (t_0 + r_0 / s)^-$.

Finally, in response to M. Nestor's final edit, we remark that the integral

$displaystyle int_{t_0}^infty vert gamma'(t) vert ; dt tag{30}$

is not defined, since $gamma(t)$ can't be extended past $t = t_0 + r_0/s$; the length may instead be found via (27).