Newton's method with inequality constraint

AI Thread Summary
The discussion centers on using Newton's method to solve for variables s and gamma while imposing an inequality constraint on s, specifically norm(s) <= 1. Participants suggest that modifying the initial system or reformulating the problem with Lagrange multipliers may be necessary. The problem involves projecting a point onto an ellipsoid defined by the equation f(x,y,z) = y^2 + z^2 - (1 - x^2)^2, while ensuring the components remain within the range of -1 to 1. Suggestions include improving the starting guess, changing variable representations, and limiting step sizes to avoid convergence to incorrect roots. The complexity arises from the need to satisfy equilibrium conditions alongside the projection onto the surface, indicating that a simple projection approach may not suffice.
Denise00
Messages
3
Reaction score
0
Dear all,

Consider the system given by : http://www.freeimagehosting.net/image.php?53f7eed9ce.jpg

where we are trying to solve for s and gamma using Newton's method. It turns out to be a simple implementation. Now, what if we need to impose an inequality constraint on the solution s : one of the form norm(s) <= 1 .

Is there a simple way to proceed by modifying a little bit the initial system? Or do we have to reformulate the problem using Lagrange multipliers or other similar methods? Many thanks for your input

Sincerely
 
Mathematics news on Phys.org
It would help if you told us what equations you want to solve. The link seems to be showing how you tried to solve them with Newton's method.

It's often easier to answer a question if you know what the question is, not just an attempt at solving it.
 
HI there,

Here are more information : I am trying to project a point onto an ellipsoid type of shape. The equation is given by f(x,y,z) = y^2 + z^2 - (1- x^2)^2 .

I would like to project any random point (x,y,z) onto this surface, but my only constraint is that I must have the absolute value of each component to be smaller than one : abs(x) <=1 , abs(y) <= 1 and abs(z) <= 1

Here is an illustration of my problem : http://www.freeimagehosting.net/image.php?632fa92cea.png

I do find a point satisfying f(x,y,z) = 0 (on the surface), but not in the appropriate range. How could I modify my Newton's algorithm to converge to the appropriate solution?

Thanks
 
I guess the problem is that in some sense there are multiple roots of your "projection" equations, and Newton-Raphson is converging to the wrong one, outside the range you want.

Some things you could try:

1 Find a better starting guess. For example draw a line from your point to the origin and find the intersection of that line with the surface.

2 Find a different way to represent just the part of the surface that you want. For eaxmple change variables so x = sin t or something similar.

3 Limit the maximum step that you can take in one iteration (say to 0.1, looking at your picture) so you can't "jump" onto the wrong part of the surface.

When it fails, print out all the iterations and see if you can figure out geometrically why it failed (analogous to showing how Newton's method works for a single variable by drawing tangents to the curve representing the function). That might suggest a way to fix it.
 
None of these "fixes" will do the trick. As I mentioned, it's a plasticity problem, the initial step is an "elastic trial" that can be (often is) outside of the ellipsoid right off the bat.
The complexity of the problem is associated both with the fact that we are trying to satisfy equilibrium and land on the surface. They have to be solved simultaneously ... and it can't be simplified to a basic projection.

Anyways, thanks for taking the time to answer
 
Denise00 said:
The complexity of the problem is associated both with the fact that we are trying to satisfy equilibrium and land on the surface. They have to be solved simultaneously ... and it can't be simplified to a basic projection.

You can get a starting point for your iterative solution in any way you like. You don't have to be constrained by the physics of the situation. If starting from a point way off the ellipsoid doesn't work, try starting from close to it - for example pick any point on it, in the correct octant. The only thing that matters in nonlinear equation solution is the final converged solution. It doesn't matter what path you take to get there, so long as you DO get there.

Your equation f(x,y,z) = y^2 + z^2 - (1- x^2)^2 has cylindrical symmetry. Does it help to transform into cyliindrical polar coordiates x, r, theta? If you do that, can you eliminate theta by projection and just iterate to solve for x and r?
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top