Python數(shù)值求解微分方程方法(歐拉法,隱式歐拉)
更新時間:2022年09月01日 09:51:13 作者:Chandler_river
這篇文章主要介紹了Python數(shù)值求解微分方程方法(歐拉法,隱式歐拉),文章圍繞主題展開詳細的內(nèi)介紹,具有一定的參考價值,需要的小伙伴可以參考一下
不說什么,先上代碼
這里先求解形如
的微分方程
1.歐拉法
def eluer(rangee,h,fun,x0,y0):
step = int(rangee/h)
x = [x0] + [h * i for i in range(step)]
u = [y0] + [0 for i in range(step)]
for i in range(step):
u[i+1] = u[i] + h * fun(x[i],u[i])
plt.plot(x,u,label = "eluer")
return u2.隱式歐拉法
def implicit_euler(rangee,h,fun,x0,y0):
step = int(rangee/h)
x = [x0] + [h * i for i in range(step)]
u = [y0] + [0 for i in range(step)]
v = ["null"] + [0 for i in range(step)]
for i in range(step):
v[i+1] = u[i] + h * fun(x[i],u[i])
u[i+1] = u[i] + h/2 * (fun(x[i],u[i]) + fun(x[i],v[i+1]))
plt.plot(x,u,label = "implicit eluer")
return u3.三階runge-kutta法
def order_3_runge_kutta(rangee,h,fun,x0,y0):
step = int(rangee/h)
k1,k2,k3 = [[0 for i in range(step)] for i in range(3)]
x = [x0] + [h * i for i in range(step)]
y = [y0] + [0 for i in range(step)]
for i in range(step):
k1[i] = fun(x[i],y[i])
k2[i] = fun(x[i]+0.5*h,y[i]+0.5*h*k1[i])
k3[i] = fun(x[i]+0.5*h,y[i]+2*h*k2[i]-h*k1[i])
y[i+1] = y[i] + 1/6 * h * (k1[i]+4*k2[i]+k3[i])
plt.plot(x,y,label = "order_3_runge_kutta")
return y4.四階runge-kutta法
def order_4_runge_kutta(rangee,h,fun,x0,y0):
step = int(rangee/h)
k1,k2,k3,k4 = [[0 for i in range(step)] for i in range(4)]
x = [x0] + [h * i for i in range(step)]
y = [y0] + [0 for i in range(step)]
for i in range(step):
k1[i] = fun(x[i],y[i])
k2[i] = fun(x[i]+0.5*h,y[i]+0.5*h*k1[i])
k3[i] = fun(x[i]+0.5*h,y[i]+0.5*h*k2[i])
k4[i] = fun(x[i]+h,y[i]+h*k3[i])
y[i+1] = y[i] + 1/6 * h * (k1[i]+2*k2[i]+2*k3[i]+k4[i])
plt.plot(x,y,label = "order_4_runge_kutta")
return y5.上圖

當(dāng)然,想要成功操作,得加上這個
rangee = 1 fun = lambda x,y:y-2*x/y implicit_euler(rangee,0.0001,fun,0,1) order_4_runge_kutta(rangee,0.0001,fun,0,1) order_3_runge_kutta(rangee,0.0001,fun,0,1) eluer(rangee,0.0001,fun,0,1) plt.legend() plt.show()
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