[英]Stable solution of a 4th order non linear differential equations
I have solved the following bvp problem using bvp solver in python.我已经在 python 中使用 bvp 求解器解决了以下 bvp 问题。
d4y/dx4= 0.00033*V/(0.000001-y)^(2), y(0)=y'(0)=y(1)=y'(1)=0
In the above eqn 'V' is a parameter which has been varied using the for loop. d4y/dx4= 0.00033*V/(0.000001-y)^(2), y(0)=y'(0)=y(1)=y'(1)=0
在上面的 eqn 'V' 是使用 for 循环改变的参数。 The interesting part is that the solution to the above differential equation should be unstable for V>Vo.有趣的是,对于 V>Vo,上述微分方程的解应该是不稳定的。 The bvp solver still yields some wrong values for V>Vo.
对于 V>Vo,bvp 求解器仍然会产生一些错误的值。 How do I make the solver stop computing as soon as this instability arises?
一旦出现这种不稳定性,如何让求解器停止计算?
For the normalized equation (changed scale of y
and V
)对于归一化方程(改变
y
和V
的比例)
y''''*(1e-6-y)**2 = 3.3e-4*V
(1e6*y)''''*(1-1e6*y)**2 = 3.3e14*V
u = 1e6*y, c = 3.3e14*V
u'''' = c/(1-u)**2
I get a critical value for c=70.099305
, that is, V0=0.2124e-12
.我得到
c=70.099305
的临界值,即V0=0.2124e-12
。 For very small c
the solution is likewise small and close to y(t)=c/24*(t*(1-t))**2
.对于非常小的
c
解决方案同样小并且接近y(t)=c/24*(t*(1-t))**2
。 For c
close to the critical value the grid refinement concentrates at the maximum close to y=0.4
.对于
c
接近临界值,网格细化集中在接近y=0.4
的最大值处。
c=70.099305
def f(t,u): return [u[1],u[2],u[3],c/(1-u[0])**2]
def bc(u0,u1): return [u0[0], u0[1], u1[0], u1[1]]
t = np.linspace(0,1,5);
u = np.zeros([4,len(t)])
res = solve_bvp(f,bc,t,u, tol=1e-4, max_nodes=5000)
print(res.message)
%matplotlib inline
if res.success:
plt.figure(figsize=(10,5))
t = np.linspace(0,1,502)
plt.plot(t, c/24*(t*(1-t))**2,c='y', lw=3)
plt.plot(t,res.sol(t)[0],'b')
plt.plot(res.x,res.y[0],'xr')
plt.grid(); plt.show()
blue - numerical solution, yellow - approximation for small c
蓝色 - 数值解,黄色 - 小
c
的近似值
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