问如何用solve_ivp解谱方法求解偏微分方程？

EN

import numpy as np
from scipy.integrate import solve_ivp
from scipy.fftpack import diff as psdiff

#RHS of equations
def f(t,u):
    uxx= psdiff(u[N:],period=L,order=2)
    du1dt=u[:N]
    du2dt =a**2*uxx
    dudt=np.append(du1dt,du2dt)
    return dudt

a=1
amin=-40;bmax=40
L=bmax-amin;N=256
deltax=L/N
x=np.arange(amin,bmax,deltax)

u01 = 2*np.cosh(x)**(-1)
u02=np.zeros(N)
# y0
inital=np.append(u01,u02)

sola1 = solve_ivp(f, t_span=[0,40],y0=inital,args=(a,))

import matplotlib.pyplot as plt
fig, ax = plt.subplots()
ax.plot(x,sola1.y[:N,5])
plt.show()

import matplotlib.pyplot as plt
import numpy as np
from scipy.integrate import odeint
from scipy.fftpack import diff as psdiff
from itertools import chain 
def lambdifide_odes(y,t,a):
    # uxx =- (1j)**2*k**2*u[:N]
    u1=y[::2]
    u2=y[1::2]
    dudt=np.empty_like(y)
    du1dt=dudt[::2]
    du2dt=dudt[1::2]

    du1dt=u2
    uxx=psdiff(u1,order=2,period=L)
    du2dt=a**2*uxx
    return dudt
a=1
amin=-40;bmax=40
L=bmax-amin;N=256
deltax=L/N
x=np.arange(amin,bmax,deltax)
u01 = 2*np.cosh(x)**(-1)
u02=np.zeros(N)
initial=np.array(list(chain.from_iterable(zip(u01,u02))))
t0=np.linspace(0,40,100)
sola1 = odeint(lambdifide_odes,y0=initial,t=t0,args=(a,))
fig, ax = plt.subplots()
ax.plot(x,sola1[20,::2])
plt.show()