`bvp4c`

Solves boundary value problems for ordinary differential equations.

Category: Mathematics > Numerical Integration and Differential Equations

Syntax

sol = bvp4c(odefun, bcfun, solinit)

Solves y' = odefun(x, y) subject to the boundary conditions bcfun(y(a), y(b)) = 0, starting from the guess solinit.

sol = bvp4c(odefun, bcfun, solinit, options)

Also applies options created with bvpset.

Description

Write the equation as a first-order system, then give the boundary conditions as a function of the values at both ends. The solver adapts the mesh until the residual is below the tolerance, so sol.x is generally finer than solinit.x.

Use deval(sol, x) to evaluate the solution at any point, since the mesh points are not chosen by you. Boundary value problems can have several solutions or none, and the initial guess decides which one is found.

Input arguments

odefun: Function handle dydx = odefun(x, y) returning a column vector.
bcfun: Function handle res = bcfun(ya, yb) returning the residuals of the boundary conditions.
solinit: Initial mesh and guess, built with bvpinit.
options: Structure from bvpset (tolerances, maximum mesh size, and so on).

Output arguments

sol: Structure with the mesh sol.x, the solution sol.y (one row per component), the derivative sol.yp and sol.stats.

Example

y'' + y = 0 with y(0) = 0, y(1) = 1

odefun = @(x, y) [y(2); -y(1)];
bcfun = @(ya, yb) [ya(1); yb(1) - 1];
solinit = bvpinit(linspace(0, 1, 5), [0; 1]);
sol = bvp4c(odefun, bcfun, solinit);
y = deval(sol, 0.5);
fprintf('%.4f %.4f\n', y(1), sin(0.5)/sin(1))

Output

0.5697 0.5697

bvp4c