PINN解偏微分方程实例4
- 一、正问题
- 1. Diffusion equation
- 2. Burgers’ equation
- 3. Allen–Cahn equation
- 4. Wave equation
 
- 二、反问题
- 1. Burgers’ equation
- 3. 部分代码示例
 
本文使用 PINN解偏微分方程实例1中展示的代码求解了以四个具体的偏微分方程,包括Diffusion,Burgers, Allen–Cahn和Wave方程,另外重新写了一个求解反问题的代码,以burger方程为例。
一、正问题
1. Diffusion equation
一维扩散方程:
  
      
       
        
         
          
           
            
             
              
              
                ∂ 
               
              
                u 
               
              
              
              
                ∂ 
               
              
                t 
               
              
             
            
              = 
             
             
              
               
               
                 ∂ 
                
               
                 2 
                
               
              
                u 
               
              
              
              
                ∂ 
               
               
               
                 x 
                
               
                 2 
                
               
              
             
            
              + 
             
             
             
               e 
              
              
              
                − 
               
              
                t 
               
              
             
             
             
               ( 
              
             
               − 
              
             
               sin 
              
             
                
              
             
               ( 
              
             
               π 
              
             
               x 
              
             
               ) 
              
             
               + 
              
              
              
                π 
               
              
                2 
               
              
             
               sin 
              
             
                
              
             
               ( 
              
             
               π 
              
             
               x 
              
             
               ) 
              
             
               ) 
              
             
            
              , 
             
             
            
              x 
             
            
              ∈ 
             
            
              [ 
             
            
              − 
             
            
              1 
             
            
              , 
             
            
              1 
             
            
              ] 
             
            
              , 
             
            
              t 
             
            
              ∈ 
             
            
              [ 
             
            
              0 
             
            
              , 
             
            
              1 
             
            
              ] 
             
            
           
          
         
         
          
           
            
            
              u 
             
            
              ( 
             
            
              x 
             
            
              , 
             
            
              0 
             
            
              ) 
             
            
              = 
             
            
              sin 
             
            
               
             
            
              ( 
             
            
              π 
             
            
              x 
             
            
              ) 
             
            
           
          
         
         
          
           
            
            
              u 
             
            
              ( 
             
            
              − 
             
            
              1 
             
            
              , 
             
            
              t 
             
            
              ) 
             
            
              = 
             
            
              u 
             
            
              ( 
             
            
              1 
             
            
              , 
             
            
              t 
             
            
              ) 
             
            
              = 
             
            
              0 
             
            
           
          
         
        
       
         \begin{array}{l} \frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}}+e^{-t}\left(-\sin (\pi x)+\pi^{2} \sin (\pi x)\right), \quad x \in[-1,1], t \in[0,1] \\ u(x, 0)=\sin (\pi x) \\ u(-1, t)=u(1, t)=0 \end{array} 
        
       
     ∂t∂u=∂x2∂2u+e−t(−sin(πx)+π2sin(πx)),x∈[−1,1],t∈[0,1]u(x,0)=sin(πx)u(−1,t)=u(1,t)=0
 其中  
     
      
       
       
         u 
        
       
      
        u 
       
      
    u 是扩散物质的浓度。精确解是  
     
      
       
       
         u 
        
       
         ( 
        
       
         x 
        
       
         , 
        
       
         t 
        
       
         ) 
        
       
         = 
        
       
         s 
        
       
         i 
        
       
         n 
        
       
         ( 
        
       
         π 
        
       
         x 
        
       
         ) 
        
        
        
          e 
         
         
         
           − 
          
         
           t 
          
         
        
       
      
        u(x,t)=sin(\pi x)e^{-t} 
       
      
    u(x,t)=sin(πx)e−t 表示。

2. Burgers’ equation
Burgers方程的定义为:
  
      
       
        
         
          
           
            
             
              
              
                ∂ 
               
              
                u 
               
              
              
              
                ∂ 
               
              
                t 
               
              
             
            
              + 
             
            
              u 
             
             
              
              
                ∂ 
               
              
                u 
               
              
              
              
                ∂ 
               
              
                x 
               
              
             
            
              = 
             
            
              v 
             
             
              
               
               
                 ∂ 
                
               
                 2 
                
               
              
                u 
               
              
              
              
                ∂ 
               
               
               
                 x 
                
               
                 2 
                
               
              
             
            
              , 
             
             
            
              x 
             
            
              ∈ 
             
            
              [ 
             
            
              − 
             
            
              1 
             
            
              , 
             
            
              1 
             
            
              ] 
             
            
              , 
             
            
              t 
             
            
              ∈ 
             
            
              [ 
             
            
              0 
             
            
              , 
             
            
              1 
             
            
              ] 
             
            
              , 
             
            
           
          
         
         
          
           
            
            
              u 
             
            
              ( 
             
            
              x 
             
            
              , 
             
            
              0 
             
            
              ) 
             
            
              = 
             
            
              − 
             
            
              sin 
             
            
               
             
            
              ( 
             
            
              π 
             
            
              x 
             
            
              ) 
             
            
              , 
             
            
           
          
         
         
          
           
            
            
              u 
             
            
              ( 
             
            
              − 
             
            
              1 
             
            
              , 
             
            
              t 
             
            
              ) 
             
            
              = 
             
            
              u 
             
            
              ( 
             
            
              1 
             
            
              , 
             
            
              t 
             
            
              ) 
             
            
              = 
             
            
              0 
             
            
              , 
             
            
           
          
         
        
       
         \begin{array}{l} \frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}=v \frac{\partial^{2} u}{\partial x^{2}}, \quad x \in[-1,1], t \in[0,1], \\ u(x, 0)=-\sin (\pi x), \\ u(-1, t)=u(1, t)=0, \end{array} 
        
       
     ∂t∂u+u∂x∂u=v∂x2∂2u,x∈[−1,1],t∈[0,1],u(x,0)=−sin(πx),u(−1,t)=u(1,t)=0,
 其中, 
     
      
       
       
         u 
        
       
      
        u 
       
      
    u 为流速, 
     
      
       
       
         ν 
        
       
      
        ν 
       
      
    ν 为流体的粘度。在本文中, 
     
      
       
       
         ν 
        
       
      
        ν 
       
      
    ν 设为  
     
      
       
       
         0.01 
        
       
         / 
        
       
         π 
        
       
      
        0.01/\pi 
       
      
    0.01/π。
 
3. Allen–Cahn equation
Allen–Cahn方程的形式如下:
  
      
       
        
         
          
           
            
             
              
              
                ∂ 
               
              
                u 
               
              
              
              
                ∂ 
               
              
                t 
               
              
             
            
              = 
             
            
              D 
             
             
              
               
               
                 ∂ 
                
               
                 2 
                
               
              
                u 
               
              
              
              
                ∂ 
               
               
               
                 x 
                
               
                 2 
                
               
              
             
            
              + 
             
            
              5 
             
             
             
               ( 
              
             
               u 
              
             
               − 
              
              
              
                u 
               
              
                3 
               
              
             
               ) 
              
             
            
              , 
             
             
            
              x 
             
            
              ∈ 
             
            
              [ 
             
            
              − 
             
            
              1 
             
            
              , 
             
            
              1 
             
            
              ] 
             
            
              , 
             
            
              t 
             
            
              ∈ 
             
            
              [ 
             
            
              0 
             
            
              , 
             
            
              1 
             
            
              ] 
             
            
              , 
             
            
           
          
         
         
          
           
            
            
              u 
             
            
              ( 
             
            
              x 
             
            
              , 
             
            
              0 
             
            
              ) 
             
            
              = 
             
             
             
               x 
              
             
               2 
              
             
            
              cos 
             
            
               
             
            
              ( 
             
            
              π 
             
            
              x 
             
            
              ) 
             
            
              , 
             
            
           
          
         
         
          
           
            
            
              u 
             
            
              ( 
             
            
              − 
             
            
              1 
             
            
              , 
             
            
              t 
             
            
              ) 
             
            
              = 
             
            
              u 
             
            
              ( 
             
            
              1 
             
            
              , 
             
            
              t 
             
            
              ) 
             
            
              = 
             
            
              − 
             
            
              1 
             
            
              , 
             
            
           
          
         
        
       
         \begin{array}{l} \frac{\partial u}{\partial t}=D \frac{\partial^{2} u}{\partial x^{2}}+5\left(u-u^{3}\right), \quad x \in[-1,1], t \in[0,1], \\ u(x, 0)=x^{2} \cos (\pi x), \\ u(-1, t)=u(1, t)=-1, \end{array} 
        
       
     ∂t∂u=D∂x2∂2u+5(u−u3),x∈[−1,1],t∈[0,1],u(x,0)=x2cos(πx),u(−1,t)=u(1,t)=−1,
 其中,扩散系数  
     
      
       
       
         D 
        
       
         = 
        
       
         0.001 
        
       
      
        D=0.001 
       
      
    D=0.001 .

4. Wave equation
一维波动方程如下:
  
      
       
        
         
          
           
            
             
              
               
               
                 ∂ 
                
               
                 2 
                
               
              
                u 
               
              
              
              
                ∂ 
               
               
               
                 t 
                
               
                 2 
                
               
              
             
            
              − 
             
            
              4 
             
             
              
               
               
                 ∂ 
                
               
                 2 
                
               
              
                u 
               
              
              
              
                ∂ 
               
               
               
                 x 
                
               
                 2 
                
               
              
             
            
              = 
             
            
              0 
             
            
              , 
             
             
            
              x 
             
            
              ∈ 
             
            
              [ 
             
            
              0 
             
            
              , 
             
            
              1 
             
            
              ] 
             
            
              , 
             
            
              t 
             
            
              ∈ 
             
            
              [ 
             
            
              0 
             
            
              , 
             
            
              1 
             
            
              ] 
             
            
              , 
             
            
           
          
         
         
          
           
            
            
              u 
             
            
              ( 
             
            
              0 
             
            
              , 
             
            
              t 
             
            
              ) 
             
            
              = 
             
            
              u 
             
            
              ( 
             
            
              1 
             
            
              , 
             
            
              t 
             
            
              ) 
             
            
              = 
             
            
              0 
             
            
              , 
             
             
            
              t 
             
            
              ∈ 
             
            
              [ 
             
            
              0 
             
            
              , 
             
            
              1 
             
            
              ] 
             
            
              , 
             
            
           
          
         
         
          
           
            
            
              u 
             
            
              ( 
             
            
              x 
             
            
              , 
             
            
              0 
             
            
              ) 
             
            
              = 
             
            
              sin 
             
            
               
             
            
              ( 
             
            
              π 
             
            
              x 
             
            
              ) 
             
            
              + 
             
             
             
               1 
              
             
               2 
              
             
            
              sin 
             
            
               
             
            
              ( 
             
            
              4 
             
            
              π 
             
            
              x 
             
            
              ) 
             
            
              , 
             
             
            
              x 
             
            
              ∈ 
             
            
              [ 
             
            
              0 
             
            
              , 
             
            
              1 
             
            
              ] 
             
            
              , 
             
            
           
          
         
         
          
           
            
             
              
              
                ∂ 
               
              
                u 
               
              
              
              
                ∂ 
               
              
                t 
               
              
             
            
              ( 
             
            
              x 
             
            
              , 
             
            
              0 
             
            
              ) 
             
            
              = 
             
            
              0 
             
            
              , 
             
             
            
              x 
             
            
              ∈ 
             
            
              [ 
             
            
              0 
             
            
              , 
             
            
              1 
             
            
              ] 
             
            
              , 
             
            
           
          
         
        
       
         \begin{array}{l} \frac{\partial^{2} u}{\partial t^{2}}-4 \frac{\partial^{2} u}{\partial x^{2}}=0, \quad x \in[0,1], t \in[0,1], \\ u(0, t)=u(1, t)=0, \quad t \in[0,1], \\ u(x, 0)=\sin (\pi x)+\frac{1}{2} \sin (4 \pi x), \quad x \in[0,1], \\ \frac{\partial u}{\partial t}(x, 0)=0, \quad x \in[0,1], \end{array} 
        
       
     ∂t2∂2u−4∂x2∂2u=0,x∈[0,1],t∈[0,1],u(0,t)=u(1,t)=0,t∈[0,1],u(x,0)=sin(πx)+21sin(4πx),x∈[0,1],∂t∂u(x,0)=0,x∈[0,1],
 精确解为:
  
      
       
        
        
          u 
         
        
          ( 
         
        
          x 
         
        
          , 
         
        
          t 
         
        
          ) 
         
        
          = 
         
        
          sin 
         
        
           
         
        
          ( 
         
        
          π 
         
        
          x 
         
        
          ) 
         
        
          cos 
         
        
           
         
        
          ( 
         
        
          2 
         
        
          π 
         
        
          t 
         
        
          ) 
         
        
          + 
         
         
         
           1 
          
         
           2 
          
         
        
          sin 
         
        
           
         
        
          ( 
         
        
          4 
         
        
          π 
         
        
          x 
         
        
          ) 
         
        
          cos 
         
        
           
         
        
          ( 
         
        
          8 
         
        
          π 
         
        
          t 
         
        
          ) 
         
        
          . 
         
        
       
         u(x, t)=\sin (\pi x) \cos (2 \pi t)+\frac{1}{2} \sin (4 \pi x) \cos (8 \pi t) . 
        
       
     u(x,t)=sin(πx)cos(2πt)+21sin(4πx)cos(8πt).
二、反问题
1. Burgers’ equation
Burgers方程的定义为:
  
      
       
        
         
          
           
            
             
              
              
                ∂ 
               
              
                u 
               
              
              
              
                ∂ 
               
              
                t 
               
              
             
            
              + 
             
            
              u 
             
             
              
              
                ∂ 
               
              
                u 
               
              
              
              
                ∂ 
               
              
                x 
               
              
             
            
              = 
             
            
              v 
             
             
              
               
               
                 ∂ 
                
               
                 2 
                
               
              
                u 
               
              
              
              
                ∂ 
               
               
               
                 x 
                
               
                 2 
                
               
              
             
            
              , 
             
             
            
              x 
             
            
              ∈ 
             
            
              [ 
             
            
              − 
             
            
              1 
             
            
              , 
             
            
              1 
             
            
              ] 
             
            
              , 
             
            
              t 
             
            
              ∈ 
             
            
              [ 
             
            
              0 
             
            
              , 
             
            
              1 
             
            
              ] 
             
            
              , 
             
            
           
          
         
         
          
           
            
            
              u 
             
            
              ( 
             
            
              x 
             
            
              , 
             
            
              0 
             
            
              ) 
             
            
              = 
             
            
              − 
             
            
              sin 
             
            
               
             
            
              ( 
             
            
              π 
             
            
              x 
             
            
              ) 
             
            
              , 
             
            
           
          
         
         
          
           
            
            
              u 
             
            
              ( 
             
            
              − 
             
            
              1 
             
            
              , 
             
            
              t 
             
            
              ) 
             
            
              = 
             
            
              u 
             
            
              ( 
             
            
              1 
             
            
              , 
             
            
              t 
             
            
              ) 
             
            
              = 
             
            
              0 
             
            
              , 
             
            
           
          
         
        
       
         \begin{array}{l} \frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}=v \frac{\partial^{2} u}{\partial x^{2}}, \quad x \in[-1,1], t \in[0,1], \\ u(x, 0)=-\sin (\pi x), \\ u(-1, t)=u(1, t)=0, \end{array} 
        
       
     ∂t∂u+u∂x∂u=v∂x2∂2u,x∈[−1,1],t∈[0,1],u(x,0)=−sin(πx),u(−1,t)=u(1,t)=0,
 其中, 
     
      
       
       
         u 
        
       
      
        u 
       
      
    u 为流速, 
     
      
       
       
         ν 
        
       
      
        ν 
       
      
    ν 为流体的粘度。
   这里假设  
     
      
       
       
         v 
        
       
      
        v 
       
      
    v 未知,我们同时求解方程的解和v的值。

3. 部分代码示例
import torch
import numpy as np
import matplotlib.pyplot as plt
sin = torch.sin
cos = torch.cos
exp = torch.exp
pi = torch.pi
epochs = 50000    # 训练代数,要为1000的整数倍
h = 100    # 画图网格密度
N = 30    # 内点配置点数
N1 = 10    # 边界点配置点数
N2 = 5000    # 数据点
# error
L2_error = []
L2_error_data = []
L2_error_eq = []
# Training
u = MLP()
opt = torch.optim.Adam(params=u.parameters())
xt, u_real = test_data(x_inf=-1, x_sup=1, t_inf=0, t_sup=1, h=h)
print("**************** equation+data ********************")
for i in range(epochs):
    opt.zero_grad()
    l = l_interior(u) \
        + l_down(u) \
        + l_left(u) \
        + l_right(u) \
        + l_data(u)
    l.backward()
    opt.step()
    if (i+1) % 1000 == 0 or i == 0:
        u_pred = u(xt)
        error = l2_relative_error(u_real, u_pred.detach().numpy())
        L2_error.append(error)
        print("L2相对误差: ", error)
u1 = MLP()
opt = torch.optim.Adam(params=u1.parameters())
print("**************** data ********************")
for i in range(epochs):
    opt.zero_grad()
    l = l_data(u1)
    l.backward()
    opt.step()
    if (i+1) % 1000 == 0 or i == 0:
        u_pred = u1(xt)
        error = l2_relative_error(u_real, u_pred.detach().numpy())
        L2_error_data.append(error)
        print("L2相对误差: ", error)
u2 = MLP()
opt = torch.optim.Adam(params=u2.parameters())
print("**************** equation ********************")
for i in range(epochs):
    opt.zero_grad()
    l = l_interior(u2) \
        + l_down(u2) \
        + l_left(u2) \
        + l_right(u2)
    l.backward()
    opt.step()
    if (i+1) % 1000 == 0 or i == 0:
        u_pred = u2(xt)
        error = l2_relative_error(u_real, u_pred.detach().numpy())
        L2_error_eq.append(error)
        print("L2相对误差: ", error)
print("********************************")
print("PINN相对误差为: ", L2_error[-1])
print("equation相对误差为: ", L2_error_eq[-1])
print("data相对误差为: ", L2_error_data[-1])
print("********************************")
x = range(int(epochs / 1000 + 1))
plt.plot(x, L2_error, c='red', label='pinn')
plt.plot(x, L2_error_data, c='blue', label='only data')
plt.plot(x, L2_error_eq, c='yellow', label='only equation')
plt.scatter(x, L2_error, c='red')
plt.scatter(x, L2_error_data, c='blue')
plt.scatter(x, L2_error_eq, c='yellow')
plt.yscale('log')
plt.legend()
plt.show()
完整代码目录如下:
 















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