A trajectory optimization problem

Question

The following problem comes from information theory. Specifically, for an impossibility bound I need to supply the decoder with side information and the answer will say what policy gives the tightest bound. But the optimization problem is self-contained.

We are given two points on the plane: an origin $(x_0,y_0)$ and destination $(x_1,y_1)$. It is known that $|y_0,y_1| \leq 1$ and that $0 \leq x_0 \leq x_1 \leq 1$. I need to minimize the integral of a cost function over the trajectories. The trajectory is constrained to be increasing in x, thus I can write the problem as an integral over x: minimize $$ \int_{x_0}^{x_1} \frac{[1-y'(x)]^2}{1-y^2(x)} x^4 dx $$ where $y'(x) = dy(x) / dx$, $|y(x)| \leq 1$ always, and of course $y(x_0)=y_0,y(x_1)=y_1$. In case it helps, the interesting case seems to be $x_1=y_1=1$.

Answer 1

As discussed in the comments, this sort of problem can be addressed using variational calculus, and the Euler–Lagrange equation in this case is

$$ \frac{y''}{1-y'}=\frac4x+\frac{(1+y')y}{1-y^2}\;. $$

I don’t see how to solve this differential equation analytically. If it weren’t for the term $\frac4x$, we could multiply by $\frac{y'}{1+y'}$ and then integrate both sides with respect to $x$, but that doesn’t work with the additional term.

However, we can analyze the behaviour near the point of interest, $(x,y)=(1,1)$. The second term on the right diverges at this point, and this divergence must somehow be compensated to satisfy the equation. Unless $y'$ or $y''$ diverges, the numerator on the left must also go to $0$, that is, $y'(1)=1$.

We could now try to use the known values of $y$ and $y'$ at $x=1$ to determine $y''$ from the differential equation. However, using L’Hôpital’s rule merely yields

\begin{eqnarray*} y''(1) &=& \lim_{x\to1}\left(\left(1-y'\right)\left(\frac4x+\frac{(1+y')y}{1-y^2}\right)\right) \\ &=& \lim_{x\to1}\frac{1-y'^2}{1-y^2} \\ &=& \lim_{x\to1}\frac{\left(1-y'^2\right)'}{\left(1-y^2\right)'} \\ &=& \lim_{x\to1}\frac{-2y'y''}{-2yy'} \\[5pt] &=& y''(1)\;. \end{eqnarray*}

Thus, the differential equation doesn’t constrain $y''(1)$, and we can freely chose it as an initial value instead of the missing initial value $y'(1)$ that we’re not free to choose. (For another instance of this phenomenon, see this answer.)

Here are some numerical results obtained using Runge-Kutta integration from $x=1$ to $x=0$ for various values of $y''(1)$ equally spaced between $-2$ and $2$:

Here’s the Java code I used.

Note that $y=x$ is a solution (which is clear when you multiply the differential equation by $1-y'$).

If you want the curve that starts at a given point $(x_0,y_0)$, you can integrate from $(1,1)$ with some value of $y''(1)$ and then adjust this value until you hit the desired origin.