My research has led me to a nonlinear system of differential equations which should yield a teardrop shape in the $x-y$ plane. The equations, parameterized by $t$ are
$$\frac{x'''}{x'}=\frac{y'''}{y'}$$ $$x'^2+y'^2=1$$
Some obvious solutions to these equations are lines and circles, but I am looking for a teardrop like shape satisfying $x(0)=x(1)$, $x'(0)=1$, $x'(1)=-1$, $y(0)=y(1)$, $y'(0)=0$, $y'(1)=0$.
I would love analytic solutions, but I would even appreciate numerical methods for solving this strange boundary-value problem.
Sorry, lines and circles is all you get.
If $x' = u$ and $y' = v$, you have $u^2 + v^2 = 1$ so let $u = \cos(\theta)$ and $v = \sin(\theta)$. Then the differential equation $x'''/x' = y'''/y'$ simplifies to $\theta''= 0$. Thus the general solution has $\theta = A t + B$, $u = \cos(At+B)$, $v = \sin(At+B)$, where $A$ and $B$ are constants. If $A \ne 0$ you get $x = x_0 + \sin(At+B)/A$, $y = y_0 - \cos(At+B)/A$, thus a circle. If $A = 0$, you get a straight line.