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  Nonlinear Systems-Newtons Method

Nonlinear Systems - Newton’s Method

An Example:

The LORAN (Long Range Navigation) system computes the position of a boat at sea using signals from fixed transmitters. From the time divergences of the incoming signals the boat obtains differences of distances to the transmitters. This leads to two equations every representing hyperbolas defined by the differences of distance of two points (foci). An illustration of such equations is:

(x2/1862)− {y2/(3002− 18620} = 1 and
{(y − 500)2/2792} – {(x − 300)2/(5002− 2792) }= 1 .

Resolving two quadratic equations with two unknowns would require solving a 4 degree polynomial equation. We could do this through hand but for a navigational system to work well it should do the calculations automatically and numerically. We note down that the Global Positioning System (GPS) works on similar principles and should do similar computations.

Vector Notation:

Generally we are able to usually find solutions to a system of equations when the number of unknowns matches the number of equations. Therefore we wish to find solutions to systems that have the form:

f1(x1, x2, x3, . . . , xn) = 0
f2(x1, x2, x3, . . . , xn) = 0
f3(x1, x2, x3, . . . , xn) = 0
.......
fn(x1, x2, x3, . . . , xn) = 0.

For ease we can think of (x1, x2, x3, . . . , xn) as a vector x and (f1, f2, . . . , fn) as a vector-valued function f . With this notation we are able to write the system of equations simply as:

f(x) = 0,

That is we wish to find a vector that makes the vector function equal to the zero vector.

While in Newton’s method for one variable we need to start with an initial guess x0. Theoretically the more variables one has the harder it is to find a good initial guess. In fact this should be overcome by using physically reasonable assumptions about the possible values of a solution that is take advantage of engineering knowledge of the problem. Formerly x0 is chosen let

Δx = x1− x0.

Linear Approximation for Vector Functions:

In the single changeable case Newton’s method was derived by considering the linear approximation of the function f at the initial guess x0. From Calculus the subsequent is the linear approximation of f at x0, for vectors and vector-valued functions:

f (x) ≈ f (x0) + Df (x0)(x − x0).

Here Df (x0) is an n × n matrix whose access are the various partial derivative of the components of f. Purposely:

549_partial derrivative of components.jpg

Newton’s Method:

We wish to find x that makes f equivalent to the zero vectors so let’s choose x1 so that:

f (x0) + Df (x0)(x1− x0) = 0.

Since Df (x0) is a square matrix we can solve this equation by:

x1 = x0− (Df (x0))−1f (x0),

Offered that the inverse exists. The formula is the vector equal of the Newton’s method formula we learned before. Nevertheless in practice we never use the inverse of a matrix for computations so we cannot use this formula directly. Rather we are able to do the following First solve the equation:

Df (x0) Δx = −f (x0).

Since Df (x0) is a known matrix and −f (x0) is a known vector, this equation is just a system of linear equations which can be solved efficiently and accurately. Once we have the solution vector ?x we can acquire our improved estimate x1 by

x1 = x0 + Δx.

For following steps we have the following process:

• Solve Df (xi)Δx = −f (xi) for Δx.
• Let xi+1 = xi + Δx

2389_graphs of equation.jpg

Graphs of the equations x3 + y = 1 also y3− x = −1. There is one and only one intersection at (x, y) = (1, 0).

An Experiment:

We will solve the subsequent set of equations:

x3 + y = 1
y3− x = −1.

You can simply check that (x, y) = (1, 0) is a solution of this system. Through graphing both of the equations you can as well see that (1, 0) is the only solution.

We are able to put these equations into vector form by letting x1 = x, x2 = y and

f1(x1, x2) = x31+ x2− 1
f2(x1, x2) = x32− x1 + 1.

or

563_equations.jpg

Now that we comprise the equation in the vector form write the following script program format long

f = inline(’[ x(1)^3+x(2)-1 ; x(2)^3-x(1)+1 ]’);
x = [.5;.5]
x = fsolve(f,x)

Save this program as myfsolve.m as well as run it. You will observe that the internal Mat lab solving command fsolve approximates the solution however only to about 7 decimal places. While that would be close sufficient for most applications one would expect that we could do better on such a easy problem.

Next we will execute Newton’s method for this problem. Change your myfsolve program to

% mymultnewton

format long

n=8 % set some number of iterations may need adjusting

f = inline(’[x(1)^3+x(2)-1 ; x(2)^3-x(1)+1]’); % the vector function

Df = inline(’[3*x(1)^2, 1 ; -1, 3*x(2)^2]’); % the matrix of partial derivatives

x = [.5;.5] % starting guess

for i = 1:n

Dx = -Df(x)\f(x); % solve for increment

x = x + Dx % add on to get new guess

f(x) % see if f(x) is really zero

end

Save as well as run this program (as mymultnewton) and you will see that it finds the root exactly (to machine precision) in only 6 iterations. Why is this easy program able to do better than Mat lab’s built-in program?

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