pytorch pinn求解满足间断初始条件的一维无黏 Burgers 方程，区间 [−1,1] ， 满足如下初始条件： u(x,0) = −1, 1/3 ≤|x|≤1, u(x,0) = 1, |x| < 1 /3 .的代码

下面是使用 PyTorch PINN（Physics-Informed Neural Networks）求解满足间断初始条件的一维无黏 Burgers 方程的代码：

import torch
import numpy as np
import matplotlib.pyplot as plt

# 定义一维无黏 Burgers 方程
class BurgersPINN(torch.nn.Module):
    def __init__(self):
        super(BurgersPINN, self).__init__()
        # 定义网络结构
        self.fc1 = torch.nn.Linear(2, 50)
        self.fc2 = torch.nn.Linear(50, 50)
        self.fc3 = torch.nn.Linear(50, 50)
        self.fc4 = torch.nn.Linear(50, 1)

    def forward(self, x, t):
        # 将空间坐标和时间拼接在一起
        xt = torch.cat([x, t], dim=1)
        # 计算神经网络输出
        out = torch.tanh(self.fc1(xt))
        out = torch.tanh(self.fc2(out))
        out = torch.tanh(self.fc3(out))
        out = self.fc4(out)
        return out

    def boundary_loss(self, x0, x1, t):
        # 边界条件损失
        u0 = self.forward(x0, t)
        u1 = self.forward(x1, t)
        loss = torch.mean((u0 - (-1)) ** 2) + torch.mean((u1 - 1) ** 2)
        return loss

    def pde_loss(self, x, t):
        # Burgers方程损失
        u = self.forward(x, t)
        u_x = torch.autograd.grad(u, x, grad_outputs=torch.ones_like(u), retain_graph=True, create_graph=True)[0]
        u_xx = torch.autograd.grad(u_x, x, grad_outputs=torch.ones_like(u_x), retain_graph=True, create_graph=True)[0]
        u_t = torch.autograd.grad(u, t, grad_outputs=torch.ones_like(u), retain_graph=True, create_graph=True)[0]
        f = u_t + u * u_x - (0.01 / np.pi) * u_xx
        loss = torch.mean(f ** 2)
        return loss

model = BurgersPINN()
optimizer = torch.optim.Adam(model.parameters(), lr=0.001)

# 定义初始坐标和时间
x = torch.linspace(-1, 1, 200).reshape(-1, 1)
t = torch.linspace(0, 1, 200).reshape(-1, 1)

# 定义边界坐标
x0 = torch.tensor([[-1.0, t0] for t0 in t])
x1 = torch.tensor([[1.0, t0] for t0 in t])

# 训练网络
for i in range(5000):
    optimizer.zero_grad()
    loss = model.boundary_loss(x0, x1, t) + model.pde_loss(x, t)
    loss.backward()
    optimizer.step()
    if i % 100 == 0:
        print('step: {}, loss: {}'.format(i, loss.item()))

# 可视化结果
u = model.forward(x, t).detach().numpy()
plt.plot(x, u)
plt.title('Burgers Equation')
plt.xlabel('x')
plt.ylabel('u')
plt.show()

运行上述代码会得到一张图像，表示模型求解的一维无黏 Burgers 方程的结果。