def test_ab2():
def dydt(t, y):
return y
tmax = 1.0
# Oscillate dt to check variable timestep maths is ok.
dt_base = 1e-4
input_dts = it.cycle([dt_base, 15 * dt_base])
# Starting values
ts = [0.0, 1e-6]
ys = map(exp, ts)
dts = [1e-6]
while ts[-1] < tmax:
dtprev = dts[-1]
dt = input_dts.next()
y_np1 = ab2_step(dt, ys[-1], dydt(ts[-1], ys[-1]), dtprev,
dydt(ts[-2], ys[-2]))
ys.append(y_np1)
ts.append(ts[-1] + dt)
dts.append(dt)
# plt.plot(ts, ys, 'x', ts, map(exp, ts))
# plt.show()
utils.assert_almost_equal(ys[-1], exp(ts[-1]), 1e-5)
def test_ab2():
def dydt(t, y):
return y
tmax = 1.0
# Oscillate dt to check variable timestep maths is ok.
dt_base = 1e-4
input_dts = it.cycle([dt_base, 15*dt_base])
# Starting values
ts = [0.0, 1e-6]
ys = map(exp, ts)
dts = [1e-6]
while ts[-1] < tmax:
dtprev = dts[-1]
dt = input_dts.next()
y_np1 = ab2_step(dt, ys[-1], dydt(ts[-1], ys[-1]),
dtprev, dydt(ts[-2], ys[-2]))
ys.append(y_np1)
ts.append(ts[-1] + dt)
dts.append(dt)
# plt.plot(ts, ys, 'x', ts, map(exp, ts))
# plt.show()
utils.assert_almost_equal(ys[-1], exp(ts[-1]), 1e-5)
def _get_initial_dy(self, base_residual, ts, ys):
"""Calculate a step with midpoint to get dydt at y_n.
"""
# We want to ignore the most recent two steps (the one being solved
# for "now" outside this function and the one we are computing the
# derivative at). We also want to ensure nothing is modified in the
# solutions list:
temp_ts = copy.deepcopy(ts[:-2])
temp_ys = copy.deepcopy(ys[:-2])
# Timestep should be double the timestep used for the previous
# step, so that the midpoint is at y_n.
dt_n = 2 * (ts[-2] - ts[-3])
# Calculate time step
temp_ys, temp_ts = _odeint(base_residual, temp_ys, temp_ts, dt_n,
temp_ts[-1] + dt_n, midpoint_residual)
# Check that we got the right times: the midpoint should be at
# the step before the most recent time.
utils.assert_almost_equal((temp_ts[-1] + temp_ts[-2]) / 2, ts[-2])
# Now invert midpoint to get the derivative
dy_nph = (temp_ys[-1] - temp_ys[-2]) / dt_n
# Fill in dummys (as many as we have y values) followed by the
# derivative we just calculated.
self.dys = [float('nan')] * (len(ys) - 1)
self.dys[-1] = dy_nph
def _get_initial_dy(self, base_residual, ts, ys):
"""Calculate a step with midpoint to get dydt at y_n.
"""
# We want to ignore the most recent two steps (the one being solved
# for "now" outside this function and the one we are computing the
# derivative at). We also want to ensure nothing is modified in the
# solutions list:
temp_ts = copy.deepcopy(ts[:-2])
temp_ys = copy.deepcopy(ys[:-2])
# Timestep should be double the timestep used for the previous
# step, so that the midpoint is at y_n.
dt_n = 2*(ts[-2] - ts[-3])
# Calculate time step
temp_ys, temp_ts = _odeint(base_residual, temp_ys, temp_ts, dt_n,
temp_ts[-1] + dt_n, midpoint_residual)
# Check that we got the right times: the midpoint should be at
# the step before the most recent time.
utils.assert_almost_equal((temp_ts[-1] + temp_ts[-2])/2, ts[-2])
# Now invert midpoint to get the derivative
dy_nph = (temp_ys[-1] - temp_ys[-2])/dt_n
# Fill in dummys (as many as we have y values) followed by the
# derivative we just calculated.
self.dys = [float('nan')] * (len(ys)-1)
self.dys[-1] = dy_nph
def check_vector_timestepper(method, tol):
def residual(t, y, dydt):
return sp.array([-1.0 * sin(t), y[1]]) - dydt
tmax = 1.0
ys, ts = odeint(residual, [cos(0.0), exp(0.0)], tmax, dt=0.001,
method=method)
utils.assert_almost_equal(ys[-1][0], cos(tmax), tol[0])
utils.assert_almost_equal(ys[-1][1], exp(tmax), tol[1])
def check_vector_timestepper(method, tol):
def residual(t, y, dydt):
return sp.array([-1.0 * sin(t), y[1]]) - dydt
tmax = 1.0
ys, ts = odeint(residual, [cos(0.0), exp(0.0)],
tmax,
dt=0.001,
method=method)
utils.assert_almost_equal(ys[-1][0], cos(tmax), tol[0])
utils.assert_almost_equal(ys[-1][1], exp(tmax), tol[1])
def check_exp_timestepper(method, tol):
def residual(t, y, dydt):
return y - dydt
tmax = 1.0
dt = 0.001
ys, ts = odeint(exp_residual, [exp(0.0)], tmax, dt=dt, method=method)
# plt.plot(ts,ys)
# plt.plot(ts, map(exp,ts), '--r')
# plt.show()
utils.assert_almost_equal(ys[-1], exp(tmax), tol)
def check_exp_timestepper(method, tol):
def residual(t, y, dydt):
return y - dydt
tmax = 1.0
dt = 0.001
ys, ts = odeint(exp_residual, [exp(0.0)], tmax, dt=dt,
method=method)
# plt.plot(ts,ys)
# plt.plot(ts, map(exp,ts), '--r')
# plt.show()
utils.assert_almost_equal(ys[-1], exp(tmax), tol)
def check_zeeman(m, H, ans):
"""Helper function for test_zeeman."""
mag_params = utils.MagParameters()
mag_params.Hvec = H
utils.assert_almost_equal(zeeman_energy(m, mag_params), ans(mag_params))
def check_zeeman(m, H, ans):
"""Helper function for test_zeeman."""
mag_params = utils.MagParameters()
mag_params.Hvec = H
utils.assert_almost_equal(zeeman_energy(m, mag_params),
ans(mag_params))