[FIXED] How to Replace the Variable from Sympy Computation so it can be Plotted with Matplotlib and Numpy?

Issue

I have this computation (integral and then summation)

The result is s it is in term of independent variables x and t.

But, when I try to define it as u(x,t) with return s, it cannot replace the x variable with the one I define as x = np.linspace(0, 20, 50) and the t variable as t = np.linspace(0, 10, 50). I think it is still following x that is defined by sympy at the beginning, the same happens with t.

I want to draw a 3D plot / the wireframe of this. the Z axis is the u(x,t). Hopefully anyone can help me on this.

My MWE:

import numpy as np
import sympy as sm
from sympy import *
#from spb import *
#from mpmath import nsum, inf

x = sm.symbols("x")
t = sm.symbols("t")
n = sm.symbols("n", integer=True)

L = 20
f1 = (2/L)*x*sin(n*np.pi*x/20)
f2 = (2/L)*(20-x)*sin(n*np.pi*x/20)
fint1 = sm.integrate(f1,(x,0,10))
fint2 = sm.integrate(f2,(x,10,20))

D = 0.475
g = (fint1+fint2)*sin(n*np.pi*x/20)*exp(-(n**2)*(np.pi**2)*D*t/400).nsimplify()
print('')

sm.pretty_print(g)

#gn = g.subs({n:1})

s = 0
for c in range(10):
    s += g.subs({n:c})
    print(s)

print('')
print('The function u(x,t) : ')
print('')
sm.pretty_print(s)

from mpl_toolkits import mplot3d
import numpy as np
import matplotlib.pyplot as plt

# defining surface and axes
x = np.linspace(0, 20, 50)
t = np.linspace(0, 10, 50)

def u(x, t):
   return s

X, T = np.meshgrid(x, t)
Z = u(X, T)

print('')
print('u(x,t)')

print(Z)

#fig = plt.figure()
# syntax for 3-D plotting
#ax = plt.axes(projection='3d')
# syntax for plotting
#ax.plot_wireframe(X,T,Z, cmap='viridis',edgecolor='green')
#ax.set_title('Wireframe')
#plt.show()

Solution

First option: use sm.lambdify to convert s to a numerical function which can later be evaluated.

xx, tt = np.mgrid[0:20:n, 0:10:n]
f = sm.lambdify((x, t), s)

fig = plt.figure()
n=30j
ax = fig.add_subplot(projection="3d")
ax.plot_wireframe(xx, tt, f(xx, tt))
ax.set_xlabel("x")
ax.set_ylabel("t")

Second option: use SymPy Plotting Backend. With it, you can avoid using sm.lambdify and Numpy. Everything is dealt by the module.

plot3d(
    s, (x, 0, 20), (t, 0, 10), {"alpha": 0}, # hide the surface
    wireframe=True, wf_n1=20, wf_n2=10,
    wf_rendering_kw={"color": "tab:blue"} # optional step to customize the wireframe lines
)

The advantage of this module is that you can use different plotting libraries. For example, Plotly is better at 3D visualizations:

plot3d(
    s, (x, 0, 20), (t, 0, 10),
    wireframe=True, wf_n1=20, wf_n2=10,
    backend=PB
)

Answered By - Davide_sd