Python: Strange x axis limits using matplotlib.pyplot

Question

(Python newbie here) As a learning exercise I am trying to solve and plot the solution to a very simple ODE: dy/dt = y with y(0) = 1. I believe my code to solve the ODE is correct, but the x-axis limits of the plot are very strange. I made 50 equally spaced x-values ranging from 0 to 5, but it looks as if plt.plot() is plotting only the first two points, resulting in the graph looking like a line rather than the curve y = e^x. I tried playing around with the 'scalex' and 'scaley' arguments of plt.plot(), and I also tried plt.xlim = (0,5), and variations thereof, but to no avail. My code is below (I'm using Python 3.7.6):

import numpy as np
from scipy.integrate import solve_ivp
import matplotlib.pyplot as plt

def f(t,y):
    return(y)

t0 = np.linspace(0,5,50)
y0 = np.array([1])

#solve_ivp(fun, t_span, y0, args = None)
out = solve_ivp(f, t0, y0)

t_vals = out.t
y_vals = out.y[0,:]

plt.plot(t_vals,y_vals)

Here's the plot I'm getting:

I did try setting y = np.exp(t0) and then plotting t0 vs y0, which works perfectly (i.e. with the x values ranging from 0 to 5). So I'm not sure why my original code is not giving me the same type of graph. Perhaps my attempt to subset the variable out is somehow incorrect? ...Any help would be appreciated!

Recursing · Accepted Answer · 2020-07-12 19:09:18Z

1

By looking at the documentation for solve_ivp, you can see that t_span should be a 2-tuple of floats, and you can specify the step with max_step

So you can have:

from scipy.integrate import solve_ivp
import matplotlib.pyplot as plt

def f(t,y):
    return(y)

out = solve_ivp(f, (0, 5), [1], max_step=0.1)

t_vals = out.t
y_vals = out.y[0,:]

plt.plot(t_vals,y_vals)

Which gives you the expected graph:

answered Jul 12, 2020 at 19:09

Recursing

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1 Comment

Leonidas Over a year ago

Thank you! I didn't read the documentation carefully enough :)

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