D_pred = pinn.D.detach().cpu().numpy() K_pred = pinn.K.detach().cpu().numpy() 什么意思
时间: 2024-06-06 18:08:03 浏览: 21
这个问题属于技术问题,D_pred和K_pred分别表示pinn模型中的某个参数D和K的预测结果,detach()是一个pytorch中的函数,用于将张量从当前计算图中分离出来,detach()之后的张量不再与计算图有任何关系,cpu()用于将计算结果放到CPU上。numpy()用于将张量转换为numpy数组,以便后续的处理。
相关问题
pytorch用PINN方法求解初值条件为x=[-1,-0.5]时,u=2;x=[-0.5,0.5]时,u=1;x=[0.51.5]时=0的burgers的代码的代码
下面是使用 PyTorch 实现的 PINN 的代码,用于求解 Burgers 方程的初值问题。
```python
import torch
import torch.nn as nn
import torch.nn.functional as F
import numpy as np
import matplotlib.pyplot as plt
# 定义 Burgers 方程
def burgers(u, x, nu):
# u: 初始解
# x: 空间坐标
# nu: 粘性系数
# 计算 u 的梯度
du_dx = torch.autograd.grad(u, x, grad_outputs=torch.ones_like(u), create_graph=True)[0]
# 计算 u 的二阶导数
d2u_dx2 = torch.autograd.grad(du_dx, x, grad_outputs=torch.ones_like(u), create_graph=True)[0]
# 计算 Burgers 方程右侧的函数 f
f = -u * du_dx + nu * d2u_dx2
return f
class PINN(nn.Module):
def __init__(self):
super(PINN, self).__init__()
self.fc1 = nn.Linear(1, 50)
self.fc2 = nn.Linear(50, 50)
self.fc3 = nn.Linear(50, 50)
self.fc4 = nn.Linear(50, 1)
def forward(self, x):
x = torch.tanh(self.fc1(x))
x = torch.tanh(self.fc2(x))
x = torch.tanh(self.fc3(x))
x = self.fc4(x)
return x
# 定义初始条件
x_1 = torch.tensor([-1.0]).requires_grad_()
u_1 = torch.tensor([2.0]).requires_grad_()
x_2 = torch.tensor([-0.5]).requires_grad_()
u_2 = torch.tensor([1.0]).requires_grad_()
x_3 = torch.tensor([0.5]).requires_grad_()
u_3 = torch.tensor([1.0]).requires_grad_()
x_4 = torch.tensor([1.0]).requires_grad_()
u_4 = torch.tensor([0.0]).requires_grad_()
# 定义模型
model = PINN()
optimizer = torch.optim.Adam(model.parameters(), lr=0.001)
# 定义损失函数
loss_fn = nn.MSELoss()
# 训练模型
for epoch in range(10000):
optimizer.zero_grad()
# 预测解
u_1_pred = model(x_1)
u_2_pred = model(x_2)
u_3_pred = model(x_3)
u_4_pred = model(x_4)
# 计算损失函数
loss = loss_fn(u_1_pred, u_1) + loss_fn(u_2_pred, u_2) + loss_fn(u_3_pred, u_3) + loss_fn(u_4_pred, u_4)
# 计算 Burgers 方程的残差
f_1 = burgers(u_1_pred, x_1, nu=0.01)
f_2 = burgers(u_2_pred, x_2, nu=0.01)
f_3 = burgers(u_3_pred, x_3, nu=0.01)
f_4 = burgers(u_4_pred, x_4, nu=0.01)
# 计算残差的损失函数
loss += loss_fn(f_1, torch.tensor([0.0])) + loss_fn(f_2, torch.tensor([0.0])) + loss_fn(f_3, torch.tensor([0.0])) + loss_fn(f_4, torch.tensor([0.0]))
# 反向传播求解梯度
loss.backward()
# 更新模型参数
optimizer.step()
if epoch % 1000 == 0:
print(f"Epoch {epoch}, Loss {loss.item():.6f}")
# 绘制预测解
x = np.linspace(-2, 2, 1000)
u_pred = model(torch.tensor(x).float().unsqueeze(1)).detach().numpy()
fig, ax = plt.subplots()
ax.plot(x, u_pred, label="PINN")
ax.set_xlabel("x")
ax.set_ylabel("u")
ax.legend()
plt.show()
```
在这个代码中,首先定义了 Burgers 方程的函数 `burgers`,用于计算方程右侧的函数 $f$。然后定义了一个包含四个全连接层的神经网络 `PINN`,用于预测解。在训练过程中,首先预测初始条件的解,然后计算 Burgers 方程的残差并使用残差的损失函数进行优化。最后,绘制了预测解的图像。请注意,这个代码只是一个简单的实现,可能需要针对具体问题进行修改。
pytorch 用PINN求解具有分段初值的PDE方程的代码
求解具有分段初值的PDE方程,可以使用PINN(Physics-Informed Neural Networks,物理约束神经网络)方法。下面是使用PyTorch实现的PINN求解具有分段初值的PDE方程的代码示例。
首先,我们需要导入相关的包:
```python
import torch
import torch.nn as nn
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
```
然后,定义PDE方程和初始条件:
```python
def pde(x, t, u):
return torch.sin(np.pi * x) * torch.exp(-t)
def init_cond(x):
return torch.sin(np.pi * x)
```
接着,定义神经网络模型:
```python
class Net(nn.Module):
def __init__(self):
super(Net, self).__init__()
self.fc_layer = nn.Sequential(
nn.Linear(2, 50),
nn.Tanh(),
nn.Linear(50, 50),
nn.Tanh(),
nn.Linear(50, 50),
nn.Tanh(),
nn.Linear(50, 1)
)
def forward(self, x):
u = self.fc_layer(x)
return u
```
然后,定义PINN求解器:
```python
class PINNSolver:
def __init__(self, net):
self.net = net
self.optimizer = torch.optim.Adam(self.net.parameters(), lr=0.001)
self.loss_func = nn.MSELoss()
def train(self, x, t, x_bc, t_bc, u_bc):
x = torch.tensor(x, dtype=torch.float32, requires_grad=True)
t = torch.tensor(t, dtype=torch.float32, requires_grad=True)
x_bc = torch.tensor(x_bc, dtype=torch.float32)
t_bc = torch.tensor(t_bc, dtype=torch.float32)
u_bc = torch.tensor(u_bc, dtype=torch.float32)
for i in range(10000):
self.optimizer.zero_grad()
u = self.net(torch.stack([x, t], dim=1)).squeeze()
u_x = torch.autograd.grad(u, x, grad_outputs=torch.ones_like(u), create_graph=True)[0]
u_t = torch.autograd.grad(u, t, grad_outputs=torch.ones_like(u), create_graph=True)[0]
f = u_t + u * u_x - 0.01 * u_xx
u_bc_pred = self.net(torch.stack([x_bc, t_bc], dim=1)).squeeze()
loss = self.loss_func(u_bc_pred, u_bc) + self.loss_func(f, torch.zeros_like(f))
loss.backward()
self.optimizer.step()
if i % 1000 == 0:
print('step: {}, loss: {}'.format(i, loss.item()))
def predict(self, x, t):
x = torch.tensor(x, dtype=torch.float32)
t = torch.tensor(t, dtype=torch.float32)
u = self.net(torch.stack([x, t], dim=1)).detach().numpy().squeeze()
return u
```
最后,我们可以使用PINN求解器求解具有分段初值的PDE方程:
```python
net = Net()
solver = PINNSolver(net)
x = np.linspace(0, 1, 100)
t = np.linspace(0, 1, 100)
x_bc = np.array([0, 1]).repeat(50)
t_bc = np.linspace(0, 1, 50).repeat(2)
u_bc = init_cond(x_bc)
solver.train(x, t, x_bc, t_bc, u_bc)
x_grid, t_grid = np.meshgrid(x, t)
u_pred = solver.predict(x_grid.reshape(-1), t_grid.reshape(-1)).reshape(x_grid.shape)
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(x_grid, t_grid, u_pred)
ax.set_xlabel('x')
ax.set_ylabel('t')
ax.set_zlabel('u')
plt.show()
```
这里我们使用3D图来展示求解结果。如果想使用2D图,可以使用下面的代码:
```python
plt.imshow(u_pred, origin='lower', extent=[0, 1, 0, 1], cmap='jet')
plt.xlabel('x')
plt.ylabel('t')
plt.colorbar()
plt.show()
```
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