*As a scientist I have the privilege to always learn and discover new things. I have recently started a new blog called Microbe Food, about single cell food production, you might be interested in. This leaves my programming blog mostly dormant for the moment but I hope the content is still useful.*

SciPy features two different interfaces to solve differential equations: `odeint`

and `solve_ivp`

. The newer one is `solve_ivp`

and it is recommended but `odeint`

is still widespread, probably because of its simplicity. Here I will go through the difference between both with a focus on moving to the more modern `solve_ivp`

interface. The primary advantage is that `solve_ivp`

offers several methods for solving differential equations whereas `odeint`

is restricted to one. We get started by setting up our system of differential equations and some parameters of the simulation.

UPDATE 07.02.2021: Note that I am focusing here on usage of the different interfaces and not on benchmarking accuracy or performance. Faruk Krecinic has contacted me and noted that assessing accuracy would require a system with a known solution as a benchmark. This is beyond the scope of this blog post. He also pointed out to me that the hmax parameter of odeint is important for the extent to which both interfaces give similar results. You can learn more about the parameters of odeint in the docs.

```
import numpy as np
from scipy.integrate import odeint, solve_ivp
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
def lorenz(t, state, sigma, beta, rho):
x, y, z = state
dx = sigma * (y - x)
dy = x * (rho - z) - y
dz = x * y - beta * z
return [dx, dy, dz]
sigma = 10.0
beta = 8.0 / 3.0
rho = 28.0
p = (sigma, beta, rho) # Parameters of the system
y0 = [1.0, 1.0, 1.0] # Initial state of the system
```

We will be using the Lorenz system. We can directly move on the solving the system with both `odeint`

and `solve_ivp`

.

```
t_span = (0.0, 40.0)
t = np.arange(0.0, 40.0, 0.01)
result_odeint = odeint(lorenz, y0, t, p, tfirst=True)
result_solve_ivp = solve_ivp(lorenz, t_span, y0, args=p)
fig = plt.figure()
ax = fig.add_subplot(1, 2, 1, projection='3d')
ax.plot(result_odeint[:, 0],
result_odeint[:, 1],
result_odeint[:, 2])
ax.set_title("odeint")
ax = fig.add_subplot(1, 2, 2, projection='3d')
ax.plot(result_solve_ivp.y[0, :],
result_solve_ivp.y[1, :],
result_solve_ivp.y[2, :])
ax.set_title("solve_ivp")
```

The first thing that sticks out is that the `solve_ivp`

solution is less smooth. That is because it is calculated at fewer time points, which in turn has to do with the difference between `t_span`

and `t`

. The `odeint`

interface expects `t`

, an array of time points for which we want to calculate the solution. The temporal resolution of the system is given by the interval between time points. The `solve_ivp`

interface on the other hand expects `t_span`

, a tuple that gives the start and end of the simulation interval. `solve_ivp`

determines the temporal resolution by itself, depending on the integration method and the desired accuracy of the solution. We can confirm that the temporal resolution of `solve_ivp`

is lower in this example by inspecting the output of both functions.

```
t.shape
# (4000,)
result_odeint.shape
# (4000, 3)
result_solve_ivp.t.shape
# (467,)
```

The t array has 4000 elements and therefore the result of odeint has 4000 rows, each row being a time point defined by t. The result of `solve_ivp`

is different. It has its own time array as an attribute and it has 1989 elements. This tells us that `solve_ivp `

indeed calculated fewer time points affection temporal resolution. So how can we increase the the number of time points in `solve_ivp`

? There are three ways: 1. We can manually define the time points to integrate, similar to `odeint`

. 2. We can decrease the error of the solution we are willing to tolerate. 3. We can change to a more accurate integration method. We will first change the integration method. The default integration method of `solve_ivp`

is `RK45`

and we will compare it to the default method of `odeint`

, which is `LSODA`

.

```
solve_ivp_rk45 = solve_ivp(lorenz, t_span, y0, args=p,
method='RK45')
solve_ivp_lsoda = solve_ivp(lorenz, t_span, y0, args=p,
method='LSODA')
fig = plt.figure()
ax = fig.add_subplot(1, 2, 1, projection='3d')
ax.plot(solve_ivp_rk45.y[0, :],
solve_ivp_rk45.y[1, :],
solve_ivp_rk45.y[2, :])
ax.set_title("RK45")
ax = fig.add_subplot(1, 2, 2, projection='3d')
ax.plot(solve_ivp_lsoda.y[0, :],
solve_ivp_lsoda.y[1, :],
solve_ivp_lsoda.y[2, :])
ax.set_title("LSODA")
```

The `LSODA`

method is already more accurate but we can make it even more accurate but it is still not as accurate as the solution we got from `odeint`

. That is because we made odeint solve at even higher temporal resolution when we passed it t. To get the exact same result from `solve_ivp`

we got from `odeint`

, we must pass it the exact time points we want to solve with the `t_eval `

parameter.

```
t = np.arange(0.0, 40.0, 0.01)
result_odeint = odeint(lorenz, y0, t, p, tfirst=True)
result_solve_ivp = solve_ivp(lorenz, t_span, y0, args=p,
method='LSODA', t_eval=t)
fig = plt.figure()
ax = fig.add_subplot(1, 2, 1, projection='3d')
ax.plot(result_odeint[:, 0],
result_odeint[:, 1],
result_odeint[:, 2])
ax.set_title("odeint")
ax = fig.add_subplot(1, 2, 2, projection='3d')
ax.plot(result_solve_ivp.y[0, :],
result_solve_ivp.y[1, :],
result_solve_ivp.y[2, :])
ax.set_title("solve_ivp LSODA")
```

Now both solutions have identical temporal resolution. But their solution is still not identical. I was unable to confirm why that is the case but I suspect very small floating point errors. The Lorenz attractor is a chaotic system and even small errors can make it diverge. The following plot shows the first variable of the system for `odeint`

and `solve_ivp`

from the above simulation. It confirms my suspicion that floating point accuracy is to blame but was unable to confirm the source.

There is one more way to make the system more smooth: decrease the tolerated error. We will not go into that here because it does not relate to the difference between odeint and `solve_ivp`

. It can be used in both interfaces. There are some more subtle differences. For example, by default, `odeint`

expects the first parameter of the problem to be the state and the second parameter to be time `t`

. This is the other way around for `solve_ivp`

. To make `odeint`

accept a problem definition where time is the first parameter we set the parameter `tfirst=True`

.

In summary, `solve_ivp`

offers several integration methods while `odeint`

only uses `LSODA`

.

Excellent way of explanation. It helps me lot to understand the concept. Dr., can you explain, if we take t_span = (-20.0, 40.0) instead of t_span = (0.0, 40.0) with the same initial condition y0 = [1.0, 1.0, 1.0]. Is solver (solve_ivp) will interpret as x[0]= 1.0, y[0]=1.0, z[0]=1.0 or x[-20]= 1.0, y[-20]=1.0, z[-20]=1.0 ? Thanks.

LikeLike

Hey Suresh, glad you like the blog. To answer your question, the indices do not correspond to time. Indices always need to be intergers to using them for time would create a lot of issues. Regardless of t_span, your initial condition will always be at the zeroth index. However, indexing with negative integers works in Python in a way that it wraps around the other end of the sequence such that x[-20] == x[len(x)-20]. This is still completely independent of t_span though.

LikeLike

Thank you very much Daniel. I would like to know if is it possible to change the boundy contions at initial point? I mean, suppose that the boundary conditions is x(0)=1 and y(1)=3, for dx=x+y, dy=x-y. Thank you for tou answer.

LikeLike

Hey Iván,

Both odeint and solve_ivp solve initial value problems so I don’t think will be well suited for your boundary problem.

LikeLike