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[wpimath] Add time-varying RKDP (#7362)

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This makes the ground truth for the Taylor series AQ discretization more
accurate.
This commit is contained in:
Tyler Veness
2024-11-07 23:46:52 -08:00
committed by GitHub
parent 01f85abcfe
commit 661bae568f
10 changed files with 369 additions and 174 deletions
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154
wpimath/src/main/java/edu/wpi/first/math/system/NumericalIntegration.java
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@@ -107,6 +107,32 @@ public final class NumericalIntegration {
return x.plus((k1.plus(k2.times(2.0)).plus(k3.times(2.0)).plus(k4)).times(h / 6.0));
}
/**
* Performs 4th order Runge-Kutta integration of dx/dt = f(t, y) for dt.
*
* @param <Rows> Rows in y.
* @param <Cols> Columns in y.
* @param f The function to integrate. It must take two arguments t and y.
* @param t The initial value of t.
* @param y The initial value of y.
* @param dtSeconds The time over which to integrate.
* @return the integration of dx/dt = f(x) for dt.
*/
public static <Rows extends Num, Cols extends Num> Matrix<Rows, Cols> rk4(
BiFunction<Double, Matrix<Rows, Cols>, Matrix<Rows, Cols>> f,
double t,
Matrix<Rows, Cols> y,
double dtSeconds) {
final var h = dtSeconds;
Matrix<Rows, Cols> k1 = f.apply(t, y);
Matrix<Rows, Cols> k2 = f.apply(t + dtSeconds * 0.5, y.plus(k1.times(h * 0.5)));
Matrix<Rows, Cols> k3 = f.apply(t + dtSeconds * 0.5, y.plus(k2.times(h * 0.5)));
Matrix<Rows, Cols> k4 = f.apply(t + dtSeconds, y.plus(k3.times(h)));
return y.plus((k1.plus(k2.times(2.0)).plus(k3.times(2.0)).plus(k4)).times(h / 6.0));
}
/**
* Performs adaptive Dormand-Prince integration of dx/dt = f(x, u) for dt. By default, the max
* error is 1e-6.
@@ -252,4 +278,132 @@ public final class NumericalIntegration {
return x;
}
/**
* Performs adaptive Dormand-Prince integration of dx/dt = f(t, y) for dt.
*
* @param <Rows> Rows in y.
* @param <Cols> Columns in y.
* @param f The function to integrate. It must take two arguments t and y.
* @param t The initial value of t.
* @param y The initial value of y.
* @param dtSeconds The time over which to integrate.
* @param maxError The maximum acceptable truncation error. Usually a small number like 1e-6.
* @return the integration of dx/dt = f(x, u) for dt.
*/
@SuppressWarnings("overloads")
public static <Rows extends Num, Cols extends Num> Matrix<Rows, Cols> rkdp(
BiFunction<Double, Matrix<Rows, Cols>, Matrix<Rows, Cols>> f,
double t,
Matrix<Rows, Cols> y,
double dtSeconds,
double maxError) {
// See https://en.wikipedia.org/wiki/Dormand%E2%80%93Prince_method for the
// Butcher tableau the following arrays came from.
// final double[6][6]
final double[][] A = {
{1.0 / 5.0},
{3.0 / 40.0, 9.0 / 40.0},
{44.0 / 45.0, -56.0 / 15.0, 32.0 / 9.0},
{19372.0 / 6561.0, -25360.0 / 2187.0, 64448.0 / 6561.0, -212.0 / 729.0},
{9017.0 / 3168.0, -355.0 / 33.0, 46732.0 / 5247.0, 49.0 / 176.0, -5103.0 / 18656.0},
{35.0 / 384.0, 0.0, 500.0 / 1113.0, 125.0 / 192.0, -2187.0 / 6784.0, 11.0 / 84.0}
};
// final double[7]
final double[] b1 = {
35.0 / 384.0, 0.0, 500.0 / 1113.0, 125.0 / 192.0, -2187.0 / 6784.0, 11.0 / 84.0, 0.0
};
// final double[7]
final double[] b2 = {
5179.0 / 57600.0,
0.0,
7571.0 / 16695.0,
393.0 / 640.0,
-92097.0 / 339200.0,
187.0 / 2100.0,
1.0 / 40.0
};
// final double[6]
final double[] c = {1.0 / 5.0, 3.0 / 10.0, 4.0 / 5.0, 8.0 / 9.0, 1.0, 1.0};
Matrix<Rows, Cols> newY;
double truncationError;
double dtElapsed = 0.0;
double h = dtSeconds;
// Loop until we've gotten to our desired dt
while (dtElapsed < dtSeconds) {
do {
// Only allow us to advance up to the dt remaining
h = Math.min(h, dtSeconds - dtElapsed);
var k1 = f.apply(t, y);
var k2 = f.apply(t + h * c[0], y.plus(k1.times(A[0][0]).times(h)));
var k3 = f.apply(t + h * c[1], y.plus(k1.times(A[1][0]).plus(k2.times(A[1][1])).times(h)));
var k4 =
f.apply(
t + h * c[2],
y.plus(k1.times(A[2][0]).plus(k2.times(A[2][1])).plus(k3.times(A[2][2])).times(h)));
var k5 =
f.apply(
t + h * c[3],
y.plus(
k1.times(A[3][0])
.plus(k2.times(A[3][1]))
.plus(k3.times(A[3][2]))
.plus(k4.times(A[3][3]))
.times(h)));
var k6 =
f.apply(
t + h * c[4],
y.plus(
k1.times(A[4][0])
.plus(k2.times(A[4][1]))
.plus(k3.times(A[4][2]))
.plus(k4.times(A[4][3]))
.plus(k5.times(A[4][4]))
.times(h)));
// Since the final row of A and the array b1 have the same coefficients
// and k7 has no effect on newY, we can reuse the calculation.
newY =
y.plus(
k1.times(A[5][0])
.plus(k2.times(A[5][1]))
.plus(k3.times(A[5][2]))
.plus(k4.times(A[5][3]))
.plus(k5.times(A[5][4]))
.plus(k6.times(A[5][5]))
.times(h));
var k7 = f.apply(t + h * c[5], newY);
truncationError =
(k1.times(b1[0] - b2[0])
.plus(k2.times(b1[1] - b2[1]))
.plus(k3.times(b1[2] - b2[2]))
.plus(k4.times(b1[3] - b2[3]))
.plus(k5.times(b1[4] - b2[4]))
.plus(k6.times(b1[5] - b2[5]))
.plus(k7.times(b1[6] - b2[6]))
.times(h))
.normF();
if (truncationError == 0.0) {
h = dtSeconds - dtElapsed;
} else {
h *= 0.9 * Math.pow(maxError / truncationError, 1.0 / 5.0);
}
} while (truncationError > maxError);
dtElapsed += h;
y = newY;
}
return y;
}
}