CODEWITHSUNDEEP
pythonscipyodeorbital-mechanicsintegrator
Vishu Singh
Vishu Singh

Reputation: 21

How to input time steps in scipy.integrate.RK45

The implementation of scipy.integrate.RK45 does not specify where to mention the times at which the integration should be performed.

The input option "t_bound" seems to be the final epoch.

To integrate, one must use the "step" option, which has no input, so one can't specify the time stamp/s there.

Compared to the other integrator: scipy.integrate.solve_ivp, which has the option "t_eval" where one can mention all the epochs (as an array) where the solution is needed.

The following code works:

check=solve_ivp(fun_integrator, t_span, y0, \
             method='RK45', t_eval=t, max_step=1800, rtol=10**(-11), atol=10**(-12))

with "fun_integrator" as my function with input (t,y) t_span = [0 max(t)] and

t = np.linspace(0,0+0.3*(pts-1)*86400,20)

^= epochs at which the answer is needed

However, Matlab ode113 gives better results than scipy solve_ivp, so I wanted to try scipy.integrate.RK45 to check if I get better results with it.

check45=RK45(fun_integrator, 0, y0, max(t), max_step=1800, \
             rtol=1e-11, atol=1e-12)
for i in range(len(t)):
    dt=0+0.3*i*86400
    check45.step() #step has no input
    x_E45=check45.y
    print(x_E45[0:3])

BUT how to input step size in the above code?

For more info: Function and y0:

def fun_integrator(t,y):
    G=6.67428*10**(-11)/(10**3)**3 #Km**3/(Kg s**2) 
    m_E=5.97218639014246e+24 #Kg
    m_M=7.34581141668673e+22 #Kg
    m_S=1.98841586057223e+30 #Kg
#    c=2.997924580000000e+05 # [Km/s], speed of light, IERS TN36 p. 18
    x_E=y[0:3]
    x_M=y[6:9]
    x_S=y[12:15]
    ##
#    v_E=y[0,3:6]
#    v_M=y[0,9:12]
#    v_S=y[0,15:18]
    ##
    diff_EM=x_E-x_M
    diff_SM=x_S-x_M
    diff_ES=x_E-x_S
    d_EM=math.sqrt(diff_EM[0]**2 + diff_EM[1]**2 + diff_EM[2]**2)
    d_SM=math.sqrt(diff_SM[0]**2 + diff_SM[1]**2 + diff_SM[2]**2)
    d_ES=math.sqrt(diff_ES[0]**2 + diff_ES[1]**2 + diff_ES[2]**2)
    ##
    dydt= y[3:6] #E vel
    two=G*((m_M*(-diff_EM)/d_EM**3) + (m_S*(-diff_ES)/d_ES**3)) # E acc
    three=y[9:12] #M vel
    four=G*((m_S*(diff_SM)/d_SM**3) + (m_E*(diff_EM)/d_EM**3)) # M acc
    five=y[15:18] #S vel
    six=G*((m_E*(diff_ES)/d_ES**3) + (m_M*(-diff_SM)/d_SM**3)) # S acc
    dydt=np.append(dydt,two)
    dydt=np.append(dydt,three)
    dydt=np.append(dydt,four)
    dydt=np.append(dydt,five)
    dydt=np.append(dydt,six)
    return dydt

y0=np.array([0.18030617557041088626562258966E+08, #Epos
-0.13849983879561028969707483658E+09,
-0.60067586558743159549398380427E+08,
0.29095339700433727181771510027E+02, #Evel
0.30306447236947439878791802685E+01,
0.13145956648460951506322034682E+01,
0.17909715946785370737211415301E+08, #Mpos
-0.13879823119464902933064808964E+09,
-0.60230238740754862546525906740E+08,
0.30136092114725188798503502634E+02, #Mvel
0.27407201232607066962011074680E+01,
0.11664485102480685879418310910E+01,
0.67356572699027185845872823900E+06, #Spos
0.11475300015697844979868843500E+06,
0.39801697980815553718511148000E+05,
-0.60903913908114150920444000000E-03, #Svel
0.89648366457310610847232000000E-02,
0.38595942076153950302606500000E-02])

Upvotes: 2

Views: 3761

Answers (1)

Matthew Flamm
Matthew Flamm

Reputation: 620

To answer the actual question and not the underlying question:

You can use the dense_output method to calculate the value at a particular desired time point. Otherwise you will lose a lot of accuracy in your solution if you, for example, try to use your own interpolation method.

But, I would note that solve_ivp (using RK45 method) is essentially an easy to use wrapper for this code, so I wouldn't expect anything significantly different this way.

https://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.RK45.dense_output.html#scipy.integrate.RK45.dense_output

Upvotes: 2

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