Matlab Code For Keller Box Method TOP

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How to Solve Nonlinear Problems with Matlab Code for Keller Box Method

Nonlinear problems are common in science and engineering, but they are often difficult to solve analytically. One numerical method that can handle nonlinear problems is the Keller-box method, which is based on finite differences and Newton's method. In this article, we will explain the basic idea of the Keller-box method and show how to implement it in Matlab code. We will also give some examples of nonlinear problems that can be solved by this method, such as fluid flow problems governed by nonlinear equations.

What is the Keller-box Method

The Keller-box method is a numerical technique for solving boundary value problems of ordinary differential equations (ODEs) or partial differential equations (PDEs) that are nonlinear. The method was developed by Herbert B. Keller in the 1960s and 1970s[^1^]. The main idea of the method is to discretize the domain of the problem into a finite number of points (called nodes) and approximate the derivatives of the unknown functions at each node by finite differences. Then, the original problem is transformed into a system of nonlinear algebraic equations, which can be solved by Newton's method or other iterative methods.

The advantage of the Keller-box method is that it can handle problems with complex geometries, nonlinearities, singularities, and multiple solutions. The method is also easy to implement and modify for different types of problems. The disadvantage of the method is that it requires a good initial guess for the unknown functions and their derivatives at each node, otherwise it may not converge or converge to a wrong solution. The method also requires a careful choice of the number and location of the nodes, as well as the step size for the finite differences.

How to Implement Matlab Code for Keller Box Method

To illustrate how to implement Matlab code for Keller box method, we will use a simple example of a boundary value problem of an ODE:

$$f''(x)+f(x)f'(x)=0,\\quad f(0)=0,\\quad f(1)=1$$

This problem has an exact solution:

$$f(x)=\\tan(\\frac{\\pi}{4}x)$$

To solve this problem by the Keller-box method, we need to do the following steps:

Discretize the domain [0,1] into N+1 equally spaced nodes: x0=0, x1=h, x2=2h,..., xN=Nh, xN+1=1, where h=1/(N+1) is the step size.

Define two unknown functions u(x) and v(x) such that u(x)=f(x) and v(x)=f'(x). Then, the original ODE can be written as:

$$v'(x)+u(x)v(x)=0,\\quad u(0)=0,\\quad u(1)=1$$

At each node xi, we approximate the derivatives of u(x) and v(x) by central finite differences:

$$u'(x_i)\\approx \\frac{u_{i+1}-u_{i-1}}{2h},\\quad v'(x_i)\\approx \\frac{v_{i+1}-v_{i-1}}{2h}$$

where ui=u(xi) and vi=v(xi). Then, we obtain a system of 2N nonlinear algebraic equations:

$$\\frac{v_{i+1}-v_{i-1}}{2h}+u_iv_i=0,\\quad i=1,...,N$$

$$u_0=0,\\quad u_{N+1}=1$$

We can write this system in matrix form as:

$$F(U,V)=0$$

where U=[u0,u1,...,uN+1], V=[v0,v1,...,vN+1] 061ffe29dd