Exemplo n.º 1
0
def test_Shooting_2():
    # Full 2PBVP test problem
    # This is calculating the 4th eigenvalue of Mathieu's Equation
    # This problem contains an adjustable parameter.

    def odefun(t, X, p, const):
        return (X[1], -(p[0] - 2 * 5 * np.cos(2 * t)) * X[0])

    def bcfun(t0, X0, q0, tf, Xf, qf, p, ndp, aux):
        return (X0[1], Xf[1], X0[0] - 1)

    algo = Shooting(odefun, None, bcfun)
    solinit = Solution()
    solinit.t = np.linspace(0, np.pi, 30)
    solinit.y = np.vstack(
        (np.cos(4 * solinit.t), -4 * np.sin(4 * solinit.t))).T
    solinit.dynamical_parameters = np.array([15])

    out = algo.solve(solinit)
    assert abs(out.t[-1] - np.pi) < tol
    assert abs(out.y[0][0] - 1) < tol
    assert abs(out.y[0][1]) < tol
    assert abs(out.y[-1][0] - 1) < tol
    assert abs(out.y[-1][1]) < tol
    assert abs(out.dynamical_parameters[0] - 17.09646175) < tol
Exemplo n.º 2
0
def test_Shooting_3():
    # This problem contains a parameter, but it is not explicit in the BCs.
    # Since time is buried in the ODEs, this tests if the BVP solver calculates
    # sensitivities with respect to parameters.
    def odefun(t, X, p, const):
        return 1 * p[0]

    def bcfun(t0, X0, q0, tf, Xf, qf, p, ndp, aux):
        return (X0[0] - 0, Xf[0] - 2)

    algo = Shooting(odefun, None, bcfun)
    solinit = Solution()
    solinit.t = np.linspace(0, 1, 2)
    solinit.y = np.array([[0], [0]])
    solinit.dynamical_parameters = np.array([1])
    out = algo.solve(solinit)
    assert abs(out.dynamical_parameters - 2) < tol
Exemplo n.º 3
0
def test_Shooting_1():
    # Full 2PBVP test problem
    # This is the simplest BVP

    def odefun(t, X, p, const):
        return (X[1], -abs(X[0]))

    def bcfun(t0, X0, q0, tf, Xf, qf, p, ndp, aux):
        return (X0[0], Xf[0]+2)

    algo = Shooting(odefun, None, bcfun)
    solinit = Solution()
    solinit.t = np.linspace(0,4,2)
    solinit.y = np.array([[0,1],[0,1]])
    out = algo.solve(solinit)
    assert out.y[0][0] < tol
    assert out.y[0][1] - 2.06641646 < tol
    assert out.y[-1][0] + 2 < tol
    assert out.y[-1][1] + 2.87588998 < tol
    assert out.t[-1] - 4 < tol
    assert abs(out.y[0][1] - solinit.y[0][1]) > tol
    assert abs(out.y[-1][0] - solinit.y[-1][0]) - 2 < tol
Exemplo n.º 4
0
def test_Shooting_4():
    # This problem contains a quad and tests if the bvp solver correctly
    # integrates the quadfun.

    def odefun(t, x, p, const):
        return -x[1], x[0]

    def quadfun(t, x, p, const):
        return x[0]

    def bcfun(t0, X0, q0, tf, Xf, qf, params, ndp, aux):
        return X0[0], X0[1] - 1, qf[0] - 1.0

    algo = Shooting(odefun, quadfun, bcfun)
    solinit = Solution()
    solinit.t = np.linspace(0, np.pi / 2, 2)
    solinit.y = np.array([[1, 0], [1, 0]])
    solinit.q = np.array([[0], [0]])
    out = algo.solve(solinit)
    assert (out.y[0,0] - 0) < tol
    assert (out.y[0,1] - 1) < tol
    assert (out.q[0,0] - 2) < tol
    assert (out.q[-1,0] - 1) < tol