bvp: Boundary Value Problems

In pracma: Practical Numerical Math Functions

Boundary Value Problems

Description

Solves boundary value problems of linear second order differential equations.

Usage

bvp(f, g, h, x, y, n = 50)

Arguments

f, g, h	functions on the right side of the differential equation. If f, g or h is a scalar instead of a function, it is assumed to be a constant coefficient in the differential equation.
x	x[1], x[2] are the interval borders where the solution shall be computed.
y	boundary conditions such that y(x[1]) = y[1], y(x[2]) = y[2].
n	number of intermediate grid points; default 50.

Details

Solves the two-point boundary value problem given as a linear differential equation of second order in the form:

y'' = f(x) y' + g(x) y + h(x)

with the finite element method. The solution y(x) shall exist on the interval [a, b] with boundary conditions y(a) = y_a and y(b) = y_b.

Value

Returns a list list(xs, ys) with the grid points xs and the values ys of the solution at these points, including the boundary points.

Note

Uses a tridiagonal equation solver that may be faster then qr.solve for large values of n.

References

Kutz, J. N. (2005). Practical Scientific Computing. Lecture Notes 98195-2420, University of Washington, Seattle.

See Also

shooting

Examples

##  Solve y'' = 2*x/(1+x^2)*y' - 2/(1+x^2) * y + 1
##  with y(0) = 1.25 and y(4) = -0.95 on the interval [0, 4]:
f1 <- function(x) 2*x / (1 + x^2)
f2 <- function(x)  -2 / (1 + x^2)
f3 <- function(x) rep(1, length(x))     # vectorized constant function 1
x <- c(0.0,   4.0)
y <- c(1.25, -0.95)
sol <- bvp(f1, f2, f3, x, y)
## Not run: 
plot(sol$xs, sol$ys, ylim = c(-2, 2),
     xlab = "", ylab = "", main = "Boundary Value Problem")
# The analytic solution is
sfun <- function(x) 1.25 + 0.4860896526*x - 2.25*x^2 + 
                    2*x*atan(x) - 1/2 * log(1+x^2) + 1/2 * x^2 * log(1+x^2)
xx <- linspace(0, 4)
yy <- sfun(xx)
lines(xx, yy, col="red")
grid()
## End(Not run)

pracma documentation built on Nov. 10, 2023, 1:14 a.m.