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解双曲型非线性方程的TVD格式Harden-Yee算法
算法如图
该算法可以很好地压制震荡,并且耗散很小。具体算法如图所示
import matplotlib import math matplotlib.use('TkAgg') import numpy as np import matplotlib.pyplot as plt def Phiy(yy,epsi):#phi(y) if abs(yy) >= epsi: phiyy = abs(yy) else: phiyy = (yy*yy + epsi*epsi)/2.*epsi return phiyy def Deltau(u2, u1):#\delta u(n,i+1/2),u2表示右边的 return u2 - u1 def Gi_0(u2,u1,u0): result = Minmod((u1-u0),(u2-u1)) return result def Minmod(aa,bb): if aa*bb<=0.: result = 0. else: result = np.sign(aa)*min(abs(aa),abs(bb)) return result def Lax_Wendroff(u0,u1,u2,corrant): #lax——Wendroff格式 dE1 = Efun(u2)-Efun(u0) dE2 = Efun(u2)-Efun(u1) dE3 = Efun(u1)-Efun(u0) result = u1 - corrant/2.*dE1 + corrant*corrant/2.*((u1+u2)/2.*dE2-(u1+u0)/2.*dE3) return result def Efun(e):# E fundtion result = e*e/2 return result def Hartenyee(u,x,t): corrant = (t[2] - t[1])/( x[2]-x[1])#corrant 数 epsi=0.3 for j in range(0, t.size-1):# j 表示t方向,t从零开始 u[j+1,1]=Lax_Wendroff(u[j,0],u[j,1],u[j,2],corrant)#赋值x1 print(j) for i in range(2, x.size-2): # i 表示x 方向 if Deltau(u[j,i+1],u[j,i]) == 0.: alphai_1plus2 = u[j,i] beta_1plus2 = 0. else: alphai_1plus2 = (Efun(u[j,i+1]) - Efun(u[j,i]))/(u[j,i+1] - u[j,i])# .#E2-E1 beta_1plus2 = (Gi_0(u[j,i+2],u[j,i+1],u[j,i]) - Gi_0(u[j,i+1],u[j,i],u[j,i-1]))/(u[j,i+1]-u[j,i]) if Deltau(u[j,i],u[j,i-1]) == 0.: alphai_1minus2 = u[j,i] beta_1minus2 = 0. else: alphai_1minus2 = (Efun(u[j,i]) - Efun(u[j,i-1]))/(u[j,i] - u[j,i-1])#E2-E1 beta_1minus2 = (Gi_0(u[j,i+1],u[j,i],u[j,i-1]) - Gi_0(u[j,i],u[j,i-1],u[j,i-2]))/(u[j,i]-u[j,i-1]) phini_plus12 = Gi_0(u[j,i+2],u[j,i+1],u[j,i])+Gi_0(u[j,i+1],u[j,i],u[j,i-1])-Phiy(alphai_1plus2+beta_1plus2,epsi)*(u[j,i+1]-u[j,i]) phini_minus12 = Gi_0(u[j,i+1],u[j,i],u[j,i-1])+Gi_0(u[j,i],u[j,i-1],u[j,i-2])-Phiy(alphai_1minus2+beta_1minus2,epsi)*(u[j,i]-u[j,i-1]) u[j+1,i] = u[j,i] - corrant/2.*(u[j,i+1]*u[j,i+1]/2.-u[j,i-1]*u[j,i-1]/2.)-corrant/2*(phini_plus12-phini_minus12) return u def Plot(x, t, result,title): plt.figure() plt.plot(x, result[0,:]) y = [t[int(t.size/5)], t[int(t.size/4)], t[int(t.size/3)], t[int(t.size/2)],t[int(t.size/1.1)]] labels = ['t='f'{num:.2f}' for num in y]#将变量转换为字符串 plt.plot(x, result[int(t.size/5),:], label=labels[0]) plt.plot(x, result[int(t.size/4),:],label=labels[1]) plt.plot(x, result[int(t.size/3),:],label=labels[2]) plt.plot(x, result[int(t.size/2),:],label=labels[3]) plt.plot(x, result[int(t.size/1.1),:],label=labels[4]) plt.legend() plt.xlabel('x') plt.ylabel('t') plt.title(title) plt.show() plt.close() plt.figure() plt.contourf(x, t, result, 50, cmap = 'jet') plt.colorbar() plt.savefig('CN.png') plt.xlabel('x') plt.ylabel('t') plt.title(title) plt.show() plt.close() return 0 x = np.linspace(0,4,400) t = np.linspace(0,4.0,400) u = np.zeros((t.size,x.size),dtype=float)#注意,这里u不是必须二维,我用二维主要为了测试和画图方便,当t比较大的时候会占用大量内存 u[0,0:int(x.size/2)] = 1.0 u[:,0] = 1.0 u[:,-1] = 0.0 corrant = (t[2] - t[1])/( x[2]-x[1])#corrant 数 '''#Lax_Wendroff 方法做对比 for j in range (1, t.size-1): print(j) for i in range (1,x.size-1): u[j,i] = Lax_Wendroff(u[j-1,i-1],u[j-1,i],u[j-1,i+1],corrant) ''' u2=Hartenyee(u,x,t) Plot(x,t,u2,'Harten-Yee solver')
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