Matrix Exponential

We seek a solution of a homogeneous first order linear system of differential equations. For illustration purposes we consider the

case:

First, write the system in vector and matrix form

.

Then, find the eigenvalues and eigenvectors of the matrix , denote the eigenpairs of A by

and .

Assumption. Assume that there are two linearly independent eigenvectors , which correspond to the eigenvalues , respectively. Then two linearly independent solution to are

, and
.

Definition (Fundamental Matrix Solution) The fundamental matrix solution , is formed by using the two column vectors .

(1) .

The general solution to is the linear combination

(2) [Graphics:Images/MatrixExponentialMod_gr_18.gif] .

It can be written in matrix form using the fundamental matrix solution as follows

.

Notation. When we introduce the notation

,
and

The fundamental matrix solution can be written as

(3) .
or
(4) .

The initial condition

If we desire to have the initial condition , then this produces the equation

.

The vector of constant can be solved as follows

.

The solution with the prescribed initial conditions is

.

Observe that where is the identity matrix. This leads us to make the following important definition

Definition (Matrix Exponential) If is a fundamental matrix solution to , then the matrix exponential is defined to be

.

Notation. This can be written as

(5) ,
or
(6) .

Fact. For a system, the initial condition is

,

and the solution with the initial condition is

,
or
.

Theorem (Matrix Diagonalization) The eigen decomposition of a square matrix A is

,

which exists when A has a full set of eigenpairs for , and d is the diagonal matrix

and

is the augmented matrix whose columns are the eigenvectors of A.

.

Comments

Runge-Kutta-Fehlberg Method

One way to guarantee accuracy in the solution of an I.V.P. is to solve the problem twice using step sizes h and and compare answers at the mesh points corresponding to the larger step size. But this requires a significant amount of computation for the smaller step size and must be repeated if it is determined that the agreement is not good enough. The Runge-Kutta-Fehlberg method (denoted RKF45) is one way to try to resolve this problem. It has a procedure to determine if the proper step size h is being used. At each step, two different approximations for the solution are made and compared. If the two answers are in close agreement, the approximation is accepted. If the two answers do not agree to a specified accuracy, the step size is reduced. If the answers agree to more significant digits than required, the step size is increased. Each Runge-Kutta-Fehlberg step requires the use of the following six values: Then an approximation to the solution of the I.V.P....

Van Der Pol System

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Powell's Method

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Newton-Raphson Method

This method is generally used to improve the result obtained by one of the previous methods. Let x o be an approximate root of f(x) = 0 and let x 1 = x o + h be the correct root so that f(x 1 ) = 0. Expanding f(x o + h) by Taylor ’s series, we obtain f(x o ) + hf’(x o ) + h 2 f’’(x o )/2! +-----= 0 Neglecting the second and higher order derivatives, we have f(x o ) + hf’(x o ) = 0 Which gives h = - f(x o )/f’(x o ) A better approximation than x o is therefore given be x 1 , where x 1 = x o – f(x o )/f’(x o ) Successive approximations are given by x 2 , x 3 ,----,x n+1 , where x n+1 = x n – f(x n )/f’(x n ) Which is the Newton-Raphson formula. Example: - Find a real root of the equation x 3 -5x + 3 = 0. Sol n : - Let, f(x) = x 3 -5x + 3 = 0 f’(x) = 3x 2 - 5 Choosing x o = 1 Step-1: f(x o ) = -1 f(x o ) = -2 So, x 1 =1 – ½ = 0.5 Step-2: f(x 1 ) = 0.625 f’(x 1 ) = -4.25 x 2 = 0.5 + 0.625/4.25 = 0.647 S...

Numerical Analysis

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