PINN求解一维Sod代码
时间: 2023-07-09 09:45:28 浏览: 147
基于PINN物理信息网络求解圆柱坐标中的热方程(python源码数据)
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好的,以下是使用基于物理神经网络(PINN)方法求解一维Sod问题的代码示例。在这个示例中,我们使用Python编写代码,使用TensorFlow库来实现PINN方法:
首先,导入必要的库:
```python
import tensorflow as tf
import numpy as np
import matplotlib.pyplot as plt
```
接下来,我们定义Sod问题的初始条件和边界条件:
```python
# 定义Sod问题的初始条件和边界条件
# 初始条件
rho_left = 1.0 # 左侧密度
u_left = 0.0 # 左侧速度
p_left = 1.0 # 左侧压力
rho_right = 0.125 # 右侧密度
u_right = 0.0 # 右侧速度
p_right = 0.1 # 右侧压力
# 定义计算区域
x_start = 0.0 # 区域起点
x_end = 1.0 # 区域终点
# 边界条件
def bc(x, t, input_x, u_model, rho_model, p_model):
# 左边界条件
if x[0] == x_start:
return np.array([rho_left, u_left, p_left])
# 右边界条件
elif x[0] == x_end:
return np.array([rho_right, u_right, p_right])
# 内部点
else:
x_tf = tf.convert_to_tensor(x)
u = u_model(x_tf, t)
rho = rho_model(x_tf, t)
p = p_model(x_tf, t)
return np.array([rho, u, p])
```
然后,我们定义神经网络模型和PINN方法:
```python
# 定义神经网络模型
class SodNet(tf.keras.Model):
def __init__(self):
super(SodNet, self).__init__()
self.dense1 = tf.keras.layers.Dense(64, activation='tanh')
self.dense2 = tf.keras.layers.Dense(64, activation='tanh')
self.dense_rho = tf.keras.layers.Dense(1)
self.dense_u = tf.keras.layers.Dense(1)
self.dense_p = tf.keras.layers.Dense(1)
def call(self, inputs):
x = self.dense1(inputs)
x = self.dense2(x)
rho = self.dense_rho(x)
u = self.dense_u(x)
p = self.dense_p(x)
return rho, u, p
# 定义PINN方法
class SodPINN:
def __init__(self, net):
self.net = net
# 定义损失函数
def loss_fn(self, x, t):
x_tf = tf.convert_to_tensor(x)
with tf.GradientTape(persistent=True) as tape:
tape.watch(x_tf)
rho, u, p = self.net(x_tf)
u_t = tape.gradient(u, t)
rho_t = tape.gradient(rho, t)
p_t = tape.gradient(p, t)
u_x = tape.gradient(u, x_tf)
rho_x = tape.gradient(rho, x_tf)
p_x = tape.gradient(p, x_tf)
p_x2 = tape.gradient(p_x, x_tf)
a = tf.sqrt(p_x2 / rho + 1e-6)
f_u = rho * a
f_rho = u * rho_x + rho * u_x
f_p = u * p_x + gamma * p * u_x
loss = tf.reduce_mean(tf.square(rho_t + f_rho) + tf.square(u_t + f_u) + tf.square(p_t + f_p))
return loss
# 定义训练函数
def train(self, x_data, t_data, epochs=1000, lr=0.001):
optimizer = tf.optimizers.Adam(lr)
for epoch in range(epochs):
with tf.GradientTape() as tape:
loss = self.loss_fn(x_data, t_data)
grads = tape.gradient(loss, self.net.trainable_variables)
optimizer.apply_gradients(zip(grads, self.net.trainable_variables))
if epoch % 100 == 0:
print('Epoch {}/{}: Loss = {}'.format(epoch, epochs, loss.numpy()))
```
最后,我们使用上述定义的模型和方法来求解Sod问题:
```python
# 定义计算区域
x = np.linspace(x_start, x_end, 1000)
t = np.linspace(0.0, 0.2, 100)
# 创建神经网络模型和PINN方法
net = SodNet()
pinn = SodPINN(net)
# 训练模型
pinn.train(x, t, epochs=5000)
# 使用模型求解Sod问题
x_test = np.linspace(x_start, x_end, 100)
t_test = np.linspace(0.0, 0.2, 100)
x_test, t_test = np.meshgrid(x_test, t_test)
u_pred = net(tf.convert_to_tensor(np.hstack((x_test.flatten()[:, None], t_test.flatten()[:, None])), dtype=tf.float32))[1].numpy().reshape(*x_test.shape)
# 绘制结果
plt.figure(figsize=(10, 6))
plt.pcolor(x_test, t_test, u_pred, cmap='jet')
plt.colorbar()
plt.xlabel('x')
plt.ylabel('t')
plt.title('Sod Problem')
plt.show()
```
这样,就可以使用基于物理神经网络(PINN)方法求解一维Sod问题了。
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