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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 pltsin = torch.sin
cos = torch.cos
exp = torch.exp
pi = torch.piepochs = 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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