This is a strange spiral, which is from the differential equations as follows $$ \left\{\begin{aligned} \dot x-\dot y=\ddot x\sqrt{\dot x^2+\dot y^2} \\ \dot x+\dot y=\ddot y\sqrt{\dot x^2+\dot y^2} \end{aligned}\right. $$ I don't know how to solve the equations. However, the strange spiral has an interesting property, which is the difference in its radius of curvature between two points on it is proportional to its arc-length between these two points.
This property is from a problem of physics. We consider a particle with mass $m$, and the velocity of it is $1$ along with the x-axis. There are two forces whose magnitudes are constant on the particle. One is always along with the velocity of the particle, and the other one is always perpendicular to the velocity of the particle. And we assume that the cross product of these two forces is constant.
If we set $m=1$ and $F_1=F_2=1$, we can easily get the differential equations above. I have calculated it by numerical integration as the picture shows.
Actually, I think it is like a spiral, but I don't know what kind of spiral it is. And we can calculate its arc between two points $$ s(t)-s(t_0)=\int^{t}_{t_0}v(t)\text dt=v(t_0)(t-t_0)+\frac{(t-t_0)^2}2 $$ If we set $t_0=0$ and $v(t_0)=0$, we will get $s(t)\propto t^2$. Actually we cannot do this because $v(0)=0$ isn't what we want, but we can make an approximation that $t\gg t_0$ and $t=F_1t/m\gg v(t_0)$ which can make $s(t)\propto t^2$ approximately correct. Notice that the magnitude of $F_2$ is constant and it is always perpendicular to the velocity, therefore $$ \frac{F_2}m=a_\perp=\frac{v^2}r=\text{const} $$ where $r$ is the radius of curvature. Also, we can get that $v^2\propto t^2$, which means $$ \frac{s(t)}{r(t)}=\text{const} $$ Is it really a spiral? And what more properties does it have? Or are these differential equations solvable?
Answer：The spiral is a logarithmic spiral (or a part of a logarithmic spiral) whose equation is $$ \rho = Ce^{\pm a\theta } $$ in polar coordinates where $$ a=2\frac{a_{//} }{a_{\perp } } $$
The particle has an acceleration a∥ which is always along with the velocity of the particle, and an acceleration a⊥ which is always perpendicular to the velocity of the particle. Let α be the angle between the direction of the velocity and a certain axis (You may assume the axis is x axis or others). Then ω is the angular velocity of the direction of the particle’s velocity i.e. ω=dα/dt. So $$ \omega =\frac{v}{r} =\frac{v^2}{r} \frac{1}{v} =\frac{a_{\perp }}{a_{//}t} \propto \frac{1}{t} $$ For simplicity, we assume that the initial velocity is zero i.e. $$ v\mid _{t=0} \, =0 $$ Use integration we can easily find that for any finite nonzero t0： $$\begin{align} \int_{0}^{t_{0} } \omega \mathrm{d}t = 2\pi \alpha \mid _{0}^{t_{0} } = + \infty && \int_{t_{0} }^{+ \infty} \omega \mathrm{d}t = 2\pi \alpha \mid _{t_{0} }^{+ \infty } = + \infty \end{align}$$ which means the particle will spiral for infinite cycles both when the particle get close to the origin point and when the particle goes to infinite far away. Thus, we can understand that the spiral has no asymptote when the curve extends to infinite far away, and has infinite curvature in the origin point. Look at the appendix of a textbook of calculus, you can find some common curves, and you may find that the logarithmic spiral is likely to be the answer. You can try to test and verify it.
Because it’s a spiral, it’s much more convenient to use use polar coordinates instead of cartesian coordinate system to study it. We will use polar coordinates to find the solution. The differential equation of this curve comes from the formula of the curvature radius:$$ \frac{\left ( \rho ^2+\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta } \right ) ^2\right ) ^{\frac{3}{2} } }{ \rho ^2+2\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta } \right ) ^2-\rho \frac{\mathrm{d}^2\rho }{\mathrm{d}\theta ^2} }=r $$ where ρ and θ are radial coordinate and angular coordinate, respectively. The expression of r is: $$ r=\frac{v^2}{a_{\perp } } =\frac{\left ( a_{//} t \right ) ^2}{a_{\perp } }=\frac{a_{//}}{a_{\perp }} a_{//}t^2=2\frac{a_{//}}{a_{\perp }}s $$ where s is the distance traveled from t=0 and v=0. The expression of s in polar coordinates is: $$ s=\int\limits_{0}^{t} \mathrm{d}s=\int\limits_{0}^{t} \sqrt{\rho ^2+\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta } \right)^2} \mathrm{d}\theta $$ From the three equations above, we have: $$ \frac{\left ( \rho ^2+\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta } \right ) ^2\right ) ^{\frac{3}{2} } }{ \rho ^2+2\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta } \right ) ^2-\rho \frac{ \mathrm{d}^2\rho }{ \mathrm{d}\theta ^2} }=r=2\frac{a_{//}}{a_{\perp }}s=2\frac{a_{//}}{a_{\perp }}\int\limits_{0}^{t} \sqrt{\rho ^2+\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta } \right)^2} \mathrm{d}\theta $$ You can test the solution by simply substitute ρ=Ce^(±aθ) into the integral equation above, hence find the dimensionless parameter a is equal to the ratio between the accelerations i.e. a=2a∥/a⊥. $$\because \rho = Ce^{\pm a\theta }\therefore \left ( \frac{\mathrm{d} \rho }{\mathrm{d} \theta } \right ) ^2=a^2\rho ^2;\frac{\mathrm{d} \rho }{\mathrm{d} \theta ^2} =a^2\rho$$ Substitute it into the equation, we have: $$ \rho =2\frac{a_{//} }{a_{\perp } } \int\limits_{0}^{t} \rho \mathrm{d}\theta \Leftrightarrow Ce^{\pm a\theta } =\frac{2}{\pm a} \frac{a_{//} }{a_{\perp } }\int_{-\infty }^{\pm a\theta } Ce^{\pm a\theta }\mathrm{d}\left ( \pm a\theta \right ) \Leftrightarrow \pm a=2\frac{a_{//} }{a_{\perp } }$$ So the solution fit this equation very well.
Or you can solve the equation directly, it looks difficult, but it’s not such difficult. $$\frac{\mathrm{d} } {\mathrm{d} \theta } \left [ \frac{\left ( \rho ^2+\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta } \right ) ^2\right ) ^{\frac{3}{2} } }{ \rho ^2+2\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta } \right ) ^2-\rho \frac{ \mathrm{d}^2\rho }{ \mathrm{d}\theta ^2} } \right ] =2\frac{a_{//}}{a_{\perp }} \sqrt{\rho ^2+\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta } \right)^2} $$ By differentiation，we have： $$\frac{3\left ( \rho {\rho }'+ {\rho }'{\rho }'' \right )\left ( \rho ^2+2 {\rho}' ^2-\rho{ \rho }'' \right ) -\left ( \rho ^2+{\rho}' ^2 \right ) \left ( 2\rho {\rho }'+4{\rho }'{\rho }''-{\rho }'{\rho }''-\rho {\rho }'''\right ) }{\left ( \rho ^2+2 {\rho}' ^2-\rho{ \rho }'' \right ) ^2} \sqrt{\rho ^2+{\rho}' ^2}=2\frac{a_{//}}{a_{\perp }} \sqrt{\rho ^2+{\rho}' ^2}$$ Where ρ’ denotes dρ/dθ; ρ”=dρ’/dθ; ρ”’=dρ”/dθ
Rearranging: $$(\rho ^3{\rho }'+4\rho {\rho }'^3 +3{\rho }'^3{\rho }''-3\rho^2{\rho }'{\rho }''-3\rho{\rho }'{\rho }''^2+\rho ^3{\rho }'''+\rho {\rho }'^2{\rho }''') \sqrt{\rho ^2+{\rho}' ^2}=2\frac{a_{//}}{a_{\perp }} \sqrt{\rho ^2+{\rho}' ^2}\left ( \rho ^2+2 {\rho}' ^2-\rho{ \rho }'' \right ) ^2$$ $$ (\rho ^3{\rho }'+4\rho {\rho }'^3 +3{\rho }'^3{\rho }''-3\rho^2{\rho }'{\rho }''-3\rho{\rho }'{\rho }''^2+\rho ^3{\rho }'''+\rho {\rho }'^2{\rho }''')^2 \left ( \rho ^2+{\rho}' ^2 \right ) =4\left ( \frac{a_{//}}{a_{\perp }} \right )^2 \left ( \rho ^2+{\rho}' ^2 \right )\left ( \rho ^2+2 {\rho}' ^2-\rho{ \rho }'' \right ) ^4$$ This is a high order nonlinear homogeneous differential equation, so its solutions have a form of $$ \rho =Ce^{\lambda \theta } $$ where C can be any constant. Substitute it into the equation, we got its Characteristic equation： $$ \left ( \lambda +4\lambda ^3 +3\lambda ^5-3\lambda ^3-3\lambda ^5+\lambda ^3+\lambda ^5 \right )^2\left ( 1+\lambda ^2 \right ) =4\left ( \frac{a_{//}}{a_{\perp }} \right )^2 \left ( 1+\lambda ^2 \right )\left ( 1+2\lambda ^2-\lambda ^2 \right ) ^4 $$ $$ \left ( \lambda +2\lambda ^3+\lambda ^5 \right )^2\left ( 1+\lambda ^2 \right ) =4\left ( \frac{a_{//}}{a_{\perp }} \right )^2 \left ( 1+\lambda ^2 \right )\left ( 1+\lambda ^2 \right ) ^4 $$ $$\therefore \lambda ^2\left ( 1+\lambda ^2 \right )^5 =4\left ( \frac{a_{//}}{a_{\perp }} \right )^2 \left ( 1+\lambda ^2 \right ) ^5$$ Then we got the solution: $$ \therefore \left | \lambda \right | =2\frac{a_{//}}{a_{\perp }}\quad or \quad \lambda =\pm i $$ $$\therefore \rho =Ce^{\pm a\theta } ,a=2\frac{a_{//} }{a_{\perp } } \quad or \quad \rho =C_{1} \cos\theta +C_{2} \sin\theta $$ The latter solution are not suitable for the integral equation above, one of the reasons is that it doesn’t contain the parameter 2a∥/a⊥.
So we got the solution ρ=Ce^(±aθ). Remember that we got this solution under the condition that the initial velocity is equal to zero, so the geometric center of the spiral is the origin point of the coordinate. If the initial velocity isn’t zero, the trajectory of the particle will still be a part of a logarithmic spiral, you can simply get the trajectory by translation and rotation of the trajectory of the particle which has the same a∥/a⊥ and the same direction . The shape of a logarithmic spiral is only related to the magnitude of a∥/a⊥. The direction of a logarithmic spiral is only related to the sign of $$\frac{\overrightarrow{a_{//}}\times \overrightarrow{a_{\perp}}}{\left | \overrightarrow{a_{//}} \right |\left | \overrightarrow{a_{\perp}} \right | } $$ The geometric center of the spiral can be obtained from the momentary position of the particle, the direction of the momentary velocity, the angle between the line connecting the momentary position and the direction of the velocity (which can be easily obtained from the magnitude of a∥/a⊥) and the linear distance between the momentary position and the center (which can be obtained from the magnitude of v²/a∥ where v represents the momentary velocity) . Then we get all of the information about the spiral.
The logarithmic spiral has a lot of interesting properties： It’s also called equiangular spiral，because the line connecting any of a point on the spiral and the origin point have the same included angle of arccot(a) with the spiral. You can easily prove it by calculating dρ/ρdθ . That’s why we can know the direction of the geometric center of the spiral. If two logarithmic spiral are similar to each other, the two spiral are actually congruent to each other.
Also, the particle has a ‘momentary rotation center’ when it spirals. The center is located in the perpendicular line of the tangent line of the spiral, and the distance between them is equal to the square of the velocity divided by the perpendicular acceleration i.e. r=v²/a⊥. Hence find the position of the ‘momentary rotation center’. You can easily prove that r is proportional to ρ for a certain spiral hence find the ratio between them. Then you can find the trajectory of the ‘momentary rotation center’. It is very interesting that the trajectory is also a logarithmic spiral which has the same shape and same direction of the trajectory of the particle. (You can prove it by simply using geometric methods. Then you can find the relation between the two spirals by using The Law of Cosines.)You can call this trajectory ‘adjoint spiral’ or ‘companion spiral’ of the logarithmic spiral. Moreover, another interesting property is that the triangle which the vertexes are the geometric center, the particle and the ‘momentary rotation center’ (i.e. the center of curvature) is always a Rectangular triangle of which the geometric center is the Rectangular vertex. (You can easily prove it by using The Law of Cosines or other methods.) You can calculate the angle difference △θ between the two spirals. Specially, when $$\frac{\ln{a} }{a} =\frac{\pi }{2} +2k\pi ,k\in Z,k< 0$$ (where k can be any negative integer i.e. k∈Z,k<0) i.e. △θ=2kπ, the ‘adjoint spiral’ and the original spiral coincide.
You can learn more and detailed information about logarithmic spiral from Wikipedia. Further, you can search 'list of spirals' in Wikipedia for other interesting spirals. (If you are in Mainland China and cannot open the pages of Wikipedia, you can Search ‘wanweibaike’ i.e. https://en.wanweibaike.com/ )