Is it possible to solve for $y(x)$ from $\min \int _a^bf(x^2+y^2)\sqrt{1+y'^2}\;dx$

Question

I have the following problem:

Show that if in

$$ \min \int_a^b f(x^2+y(x)^2)\sqrt{1+y'(x)^2}\;dx$$

polar coordinates are used, then the problem will be converted into one that contains no independent variable. Solve it to optimality.

and I have converted the above into polar coordinates: $$\min \int_a^b f(x^2+y^2)\sqrt{1+y'^2}\;dx=\min \int_{\alpha}^{\beta}f(r^2)\sqrt{r(\theta)^2+r'(\theta)^2}\;d\theta.$$

Now my question is about the "Solve it to optimality" in the problem description. Am I now supposed to solve for $y(x)$? Can I solve for $y(x)$ without knowing $f$? What do you think?

In the above, $y=y(x),\;y'=y'(x)=\frac{dy}{dx},\;r^2=x^2+y^2$.

Thank you for any help!

Answer 1

I have no other idea than applying the Euler-Lagrange formalism. With $$ L(r, r', \theta) = f(r^2) \sqrt{r^2 + (r')^2} $$ The problem becomes $$ \min \int\limits_\alpha^\beta L(r, r', \theta) \, d\theta $$ where for optimality the Euler-Lagrange equation $$ 0 = \frac{\partial L}{\partial r} - \frac{d}{d\theta} \frac{\partial L}{\partial r'} $$ has to be solved. This gives $$ \frac{\partial L}{\partial r} = 2 f'(r^2) r \sqrt{r^2 + (r')^2} + f(r^2) \frac{r}{\sqrt{r^2 + (r')^2}} $$ and $$ \frac{d}{d\theta} \frac{\partial L}{\partial r'} = \frac{d}{d\theta} \left( f(r^2) \frac{r'}{\sqrt{r^2 + (r')^2}} \right) = 0 $$

and leads to a differential equation $$ 0 = 2 f'(r^2) \left(r^2 + (r')^2\right) + f(r^2) $$

Answer 2

Need not solve y = f(x) at all. This is because we are taking taking partial derivatives anyway. Euler-Lagrange equation is independent of the choice of coordinate system. If you begin in polar coordinates the result will be in polar coordinates. If begun in Cartesian coordinates the result will be in Cartesian coordinates. The result and its graphs will be identically same, whatever choice.

You get answers to your question yourself once it is applied and seen while working through.

What he means is, for the polar case $ r, r^{'} $ are involved, $ \theta $ is implicitly involved, and for Cartesian case in $ x + y\, y^{'}, x $ is there explicitly.Or,

$$ f(r^2)\sqrt{r(\theta)^2+r'(\theta)^2}\;.$$

does not contain $\theta$ explicitly.

A nice simplification of E-L (due to Beltrami) results if independent variable is not explicitly involved.

Consider the simplest case of minimum distance between two points ( BTW, it is to find maximum / minimum, not optimum as there is no second constraint function).

In polar and Cartesian systems respectively ,

$$ L(r, r', \theta) = \sqrt{r^2 + (r')^2} d\theta ; L(x,y, y') = \sqrt{1 + y'^2} dx $$

one obtains after applying E-L equations

$$ 1/\sqrt{(1+y^{'2})} = const. ; \sqrt{r^2 + (r')^2} - r^{'2}/\sqrt{r^2 + (r')^2} = const. $$

which integrate to $$ y= m x +c ; r sec( \theta + \alpha) = p $$

where $ m,c; \alpha, p $ are arbitrary constants. The solution is the same, that of a straight line.

Due to appearance of $ x^2 + y^2 $ it is simpler calculation using polar coordinates.