public void Solve()
{
// Example taken from https://tutorial.math.lamar.edu/classes/de/BoundaryValueProblem.aspx
// y'' + 4y = 0, 0 <= t <= pi/4, y(0) = -2, y(pi/4) = 10
DoubleRange t = new DoubleRange(0, Math.PI / 4, increment: 0.1);
double y0 = -2;
double ypi_4 = 10;
//DoubleFunction3 yPP = new DoubleFunction3((t,y,yP) => (-4 * y));
// y`` = 0*y` -4*y + 0
var de = new SecondOrderDE <double>(new DoubleFunction((x) => 0), new DoubleFunction((x) => - 4), new DoubleFunction((x) => 0));
var solver = new LinearShootingBvpSolver();
var(approxy, _) = solver.SolveBoundary(de, t, y0, ypi_4); //discard approximation to y`
var exact = new DoubleFunction((t) => - 2 * Math.Cos(2 * t) + 10 * Math.Sin(2 * t));
SaveCsv("LinearShootingBVP", approxy, exact, t);
}
/// <summary>
/// Solve the boundary value problem where y`` = f(t, y, y`), a <= t <= b, y(a) = y0, y(b) = y1
/// </summary>
/// <param name="de">function of three arguments (t,y,y`)</param>
/// <param name="tRange">range (a,b) of values for t</param>
/// <param name="y0">value of y at a</param>
/// <param name="y1">value of y at b</param>
/// <returns>approximation for the function y and y`</returns>
public (IFunction <double> y, IFunction <double> dy) SolveBoundary(SecondOrderDE <double> de, Range <double> tRange, double y0, double y1)
{
// INPUTS: endpoints a and b for t, boundary conditions y0 and y1, number of subintervals N
var a = tRange.Start; var b = tRange.End;
var p = de.Px;
var q = de.Qx;
var r = de.Rx;
// Step 1
var h = (b - a) / N;
var u1 = new double[N + 1];
var u2 = new double[N + 1];
var v1 = new double[N + 1];
var v2 = new double[N + 1];
u1[0] = y0;
u2[0] = 0d;
v1[0] = 0d;
v2[0] = 1d;
var xs = new double[N + 1];
var w1 = new double[N + 1];
var w2 = new double[N + 1];
// Step 2
for (var i = 0; i <= N - 1; i++)
{
// Step 3
var x = a + i * h;
// Step 4
// Solve for approximation to 'y'
var k11 = h * u2[i];
var k12 = h * (p[x] * u2[i] + q[x] * u1[i] + r[x]);
var k21 = h * (u2[i] + 0.5 * k12);
var k22 = h * (p[x + h / 2] * (u2[i] + 0.5 * k12) + q[x + h / 2] * (u1[i] + 0.5 * k11) + r[x + h / 2]);
var k31 = h * (u2[i] + 0.5 * k22);
var k32 = h * (p[x + h / 2] * (u2[i] + 0.5 * k22) + q[x + h / 2] * (u1[i] + 0.5 * k21) + r[x + h / 2]);
var k41 = h * (u2[i] + k32);
var k42 = h * (p[x + h] * (u2[i] + k32) + q[x + h] * (u1[i] + k31) + r[x + h]);
u1[i + 1] = u1[i] + (1f / 6f) * (k11 + 2 * k21 + 2 * k31 + k41);
u2[i + 1] = u2[i] + (1f / 6f) * (k12 + 2 * k22 + 2 * k32 + k42);
// Solve for approximation to 'y`'
var kp11 = h * v2[i];
var kp12 = h * (p[x] * v2[i] + q[x] * v1[i] + r[x]);
var kp21 = h * (v2[i] + 0.5 * kp12);
var kp22 = h * (p[x + h / 2] * (v2[i] + 0.5 * kp12) + q[x + h / 2] * (v1[i] + 0.5 * kp11) + r[x + h / 2]);
var kp31 = h * (v2[i] + 0.5 * kp22);
var kp32 = h * (p[x + h / 2] * (v2[i] + 0.5 * kp22) + q[x + h / 2] * (v1[i] + 0.5 * kp21) + r[x + h / 2]);
var kp41 = h * (v2[i] + kp32);
var kp42 = h * (p[x + h] * (v2[i] + kp32) + q[x + h] * (v1[i] + kp31) + r[x + h]);
v1[i + 1] = v1[i] + (1f / 6f) * (kp11 + 2 * kp21 + 2 * kp31 + kp41);
v2[i + 1] = v2[i] + (1f / 6f) * (kp12 + 2 * kp22 + 2 * kp32 + kp42);
}
// Step 5
w1[0] = y0;
w2[0] = (y1 - u1[N]) / v1[N];
xs[0] = a;
// Step 6
for (var i = 1; i <= N; i++)
{
w1[i] = u1[i] + w2[0] * v1[i];
w2[i] = u2[i] + w2[0] * v2[i];
xs[i] = a + i * h;
}
// Step 7
return(new DoubleInterpolatedFunction(xs, w1), new DoubleInterpolatedFunction(xs, w2));
}