Hi,


I have an elementary question concerning an extremal problem:


Given a manifold $M$:={$(x\_1, x\_2, x\_3) $ | $x\_1^2 + x\_2^2 + x\_3^2 = c$ and $\frac{x\_1^2}{a\_1^2} + \frac{x\_2^2}{a\_2^2} + \frac{x\_3^2}{a\_3^2}= d$ and max($a\_1, a\_2, a\_3$) $> \sqrt{c} >$ min($a\_1, a\_2, a\_3$) } consider an arbitrary point P($\tilde{x}\_1, \tilde{x}\_2, \tilde{x}\_3$) with $M \cap P = \emptyset$.


Now I'm interested in the minimal distance between P and M. My idea was to use Lagrange:


Minimize f:= $||(\tilde{x}\_1, \tilde{x}\_2, \tilde{x}\_3)-(x\_1, x\_2, x\_3)||^2$ with the following boundary conditions: 


1) $\varphi\_1 := x\_1^2 + x\_2^2 + x\_3^2 - c =0$


2) $\varphi\_2 := \frac{x\_1^2}{a\_1^2} + \frac{x\_2^2}{a\_2^2} + \frac{x\_3^2}{a\_3^2} - d = 0$


=> $\Lambda (x\_1, x\_2, x\_3, \lambda\_1, \lambda\_2)$ = f + $\lambda\_1 \varphi\_1 + \lambda\_2 \varphi\_2$.


Differentiating yields:


i) $2x\_1 - 2\tilde{x\_1} + 2\lambda\_1 x\_1 + 2 \lambda\_2 \frac{x\_1^2}{a\_1^2} = 0$


ii) $2x\_2 - 2\tilde{x\_2} + 2\lambda\_1 x\_2 + 2 \lambda\_2 \frac{x\_2^2}{a\_2^2} = 0$


iii) $2x\_3 - 2\tilde{x\_3} +2\lambda\_1 x\_3 + 2 \lambda\_2 \frac{x\_3^2}{a\_3^2} = 0$


iv) $x\_1^2 + x\_2^2 + x\_3^2 = c$ 


v) $\frac{x\_1^2}{a\_1^2} + \frac{x\_2^2}{a\_2^2} + \frac{x\_3^2}{a\_3^2}= d$


vi) max($a\_1, a\_2, a\_3$) $> \sqrt{c} >$ min($a\_1, a\_2, a\_3$)


vii) $P \cap M$ = $\emptyset$


Rearranging i)-iii) yields


$x\_i\cdot (1+\lambda\_1 + \frac{\lambda\_2}{a\_i^2})=\tilde{x\_i}$ , thus
$x\_i = \frac{\tilde{x\_i}}{1+\lambda\_1 + \frac{\lambda\_2}{a\_i^2}}$


Putting this into iv) and v) you get $\sum\limits\_{i=1}^3 (\frac{\tilde{x\_i}}{1+\lambda\_1 + \frac{\lambda\_2}{a\_i^2}})^2 = c$ and
$\sum\limits\_{i=1}^{3} ({\frac{\frac{\tilde{x\_i}}{a\_i^2}}{1+\lambda\_1 + \frac{\lambda\_2}{a\_i^2} } })^2 = d$


Now I wish to solve this for the variables $\lambda\_1, \lambda\_2, x\_1, x\_2, x\_3$, whereat these variables depend on $a\_1, a\_2, a\_3, c, d, \tilde{x}\_1, \tilde{x}\_2$ and $\tilde{x}\_3$.


I had no idea how to calculate an exact solution, so I wanted to use Newton's Method in order to calculate $\lambda\_1$ and $\lambda\_2$, but I unfortunately did not succeed in achieving the minimum of f, because there are many different solutions for the $\lambda\_i$ and I don't know how I can be sure that Newton's Method found the minimal distance between P and M.



Minimal distance between a point and a manifold M (M is the intersection of a ball and an ellipsoid)
Lagrange method is based on a fact that extrema can occur only at points where the gradient of $f$ is a linear combination of gradients of the constraints -- i.e. at points where $\{\nabla f,\nabla g\_1, \nabla g\_2 \}$ is linearly dependent with coefficient in front of $\nabla f$ nontrivial. If the constraints are independent ($\nabla g\_1 \neq c \nabla g\_2$) then you can test this linear depenedence by determinant of the matrix $\begin{pmatrix} \partial\_x f &\partial\_y f &\partial\_z f \\\ \partial\_x g\_1 &\partial\_y g\_1 &\partial\_z g\_1 \\\ \partial\_x g\_2 &\partial\_y g\_2 &\partial\_z g\_2\end{pmatrix}$. This gives you the third equations to your constaints $g\_1=0$, $g\_2=0$. This may be more amenable to numerical solutions. I've tried solving the resulting equations in Maple and it did produce a result, but it is too long to display. For example $z$ variable is a solution of a polynomial of degree 8 with coefficients being polynomial expressions of degree up to 16. One can go further and try to find Groebner basis for this system of polynomial equations, but I don't think that it'll lead to anything more useful than direct numerical solution.


