我有这样的计算（积分，然后求和）
结果是s，它是关于自变量x和t的。
但是，当我尝试用return s将其定义为u(x,t)时，它无法将x变量替换为我定义为x = np.linspace(0, 20, 50)的变量，并将t变量替换为t = np.linspace(0, 10, 50)。我认为它仍然遵循sympy在开始时定义的x，t也是如此。
我想画一个3D图/线框图。Z轴是u(x,t)。希望任何人都能在这方面帮助我。
我的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()