 # Solving 1D Boundary Value Problem BVP Using Shooting Method

by admin in on April 4, 2019

This code implements the shooting method for solving 1D boundary value problem (abbrev. BVP ODE). It uses the Runge-Kutta method of 4th order for solving ODE and the interval bisection method for finding the alpha parameter.

Shooting method function form:

[x,y] = shooting_method(fun,h,zero,a,b,con,type,init)

The meaning of input and output parameters:

• x – grid of points where the solution have been found
• y – 2D array, values of function (solution) are in first row, values of 1st derivative are in second row
• fun – the function handle, fun contains system of solved differential equations, e.g. let y” + y = sqrt(x+1) be the solved differential equation, then fun has the shape:
function out = fce(x,y)
out(1) = y(2);
out(2) = sqrt(x+1)-y(1);
• h – the step of the Runge-Kutta method (the step of the grid).
• zero – the interval bisection method accuracy, it is recommended to set 1e-6.
• a,b – outer points of the interval.
• BC – values of boundary conditions (2D vector).
• type – to specify type of the boundary condition in the particular point,  it is the string consists of 2 char (‘f’ for function, ‘d’ for derivative), e.g ‘fd’ is meant that in a point the condition is related to function and in the b point to derivative.
• init – 2D vector of initial alpha parameters, it is not required, the implicit value is [-10 10]

The ploting of the solution: The solution (both function and 1st derivative) of BVP ODE is shown graphicaly after enumeration.

Example on using the function:

[x y] = shooting_method(fun,0.001,1e-6,0,1,[1 3],’fd’) – It is meant that will be solved the BVP ODE described in the function fun, on the interval (0,1) with boundary conditions y(0) = 1 and y'(1) = 3. With 0.001 Runge-Kutta step size, and 1e-6 of the interval bisection method accuracy. Share Now!

#### Release Information

• Price
:

\$4.99

• Released
:

April 4, 2019

• Last Updated
:

May 29, 2019