使用scipy.integrate.odeint在调用（可能是错误的Dfun类型）上完成的过多工作

我'm writing a code in Python that predicts the energy levels of Hydrogen which I will use as a template for research into quarkonium energy levels. I' m使用 scipy.integrate.odeint() 函数来解决Shroedinger方程，它可以在低至1818992_的较低能量水平下正常工作 . 我不需要超越它，但odeint会返回 Excess work done on this call (perhaps wrong Dfun type). ，这只会鼓励我扩展我能预测的内容 .

我正在使用的Shroedinger方程替换是：

u'' - (l*(l+1)/r**2 - 2mu_e(E-V_emag(r))) * u = 0
=>
u' = v
v' = ((l*(l+1))/(r**2) - 2.0*mu_e*(E - V_emag(r)))*u

然后我在它上面使用 scipy.integrate.odeint() 并迭代能量并使用我定义的其他函数来评估结果中的转折点和节点 . 我找到能量水平的方法是找到尽可能低的E值，其中转折点和节点的数量与它应该的相匹配;然后将 L 递增1并找到新的地面能量，例如如果 L=0 我会找到 n=1 能量，如果 L=3 ，我会找到 n=2 能量 .

一旦代码增加到 L=7 ，它就不会返回任何有用的内容 . r 的范围已经延长，但我已经阅读了关于雅各比人的内容 . 我已经确定了它们是什么，或者我还需要它们 . 有任何想法吗？

def v_emag(r):
    v = -alpha/r
    return v

def s_e(y,r,l,E): #Shroedinger equation for electromagntism
    x = numpy.zeros_like(y)
    x[0] = y[1]
    x[1] = ((l*(l+1))/(r**2) - 2.0*mu_e*(E - V_emag(r)))*y[0]
    return x

def i_s_e(l,E,start=0.001,stop=None,step=(0.005*bohr)):
    if stop is None:
        stop = ((l+1)*30-10)*bohr
    r = numpy.arange(start,stop,step)
    y = odeint(s_e,y0,r,args=(l,E))
    return y

def inormalise_e(l,E,start=0.001,stop=None,step=(0.005*bohr)):
    if stop is None:
        stop = ((l+1)*30-10)*bohr
    r = numpy.arange(start,stop,step)
    f = i_s_e(l,E,start,stop,step)[:,0]
    f2 = f**2
    area = numpy.trapz(f2,x=r)
    return f/(numpy.sqrt(area))

def inodes_e(l,E,start=0.001,stop=None,step=(0.005*bohr)):
    if stop is None:
        stop = ((l+1)*30-10)*bohr
    x = i_s_e(l,E,start,stop,step)
    r = numpy.arange(start,stop,step)
    k=0
    for i in range(len(r)-1):
        if x[i,0]*x[i+1,0] < 0: #If u value times next u value <0,
            k+=1               #crossing of u=0 has occured, therefore count node
    return k

def iturns_e(l,E,start=0.001,stop=None,step=(0.005*bohr)):
    if stop is None:
        stop = ((l+1)*30-10)*bohr
    x = i_s_e(l,E,start,stop,step)
    r = numpy.arange(start,stop,step)
    k = 0
    for i in range(len(r)-1):
        if x[i,1]*x[i+1,1] < 0: #If du/dr value times next du/dr value <0,
            k=k+1               #crossing of du/dr=0, therefore a maximum/minimum
    return k

l = 0
while l < 10:    #The ground state for a system with a non-zero angular momentum will
    E1 = -1.5e-08    #be the energy level of principle quantum number l-1, therefore
    E3 = 0           #by changing l, we can find n by searching for the ground state
    E2 = 0.5*(E1+E3)
    i = 0
    while i < 40:
        N1 = inodes_e(l,E1)
        N2 = inodes_e(l,E2)
        N3 = inodes_e(l,E3)
        T1 = iturns_e(l,E1)
        T2 = iturns_e(l,E2)
        T3 = iturns_e(l,E3)
        if N1 != N2:# and T1 != T2: #Looks in lower half first, therefore will tend to ground state
            E3 = E2
            E2 = 0.5*(E1+E3)
        elif N2 != N3:# and T2 != T3:
            E1 = E2
            E2 = 0.5*(E1+E3)
        else:
            print "Can't find satisfactory E in range"
            break

        i += 1
    x = inormalise_e(l,E2)
    if x[((l+1)**2)/0.005] > (x[2*((l+1)**2)/0.005]) and iturns_e(l,E2+1e-20)==1:
        print 'Energy of state: n =',(l+1),'is: ',(E2*(10**9)),'eV'
        l += 1
    else:
        E1 = E2+10e-20