因此，正如@jodag在他的评论中所述， ReLU 为空或线性，其梯度是恒定的( 0除外)，这是一种罕见的事件)，因此其二阶导数为零.我将激活功能更改为 Tanh ，最终使我可以计算两次jacobian.

So as @jodag mentioned in his comment, ReLU being null or linear, its gradient is constant (except on 0, which is a rare event), so its second-order derivative is zero. I changed the activation function to Tanh, which finally allows me to compute the jacobian twice.

最终代码是

import torch
import torch.nn as nn

class PINN(torch.nn.Module):
    
    def __init__(self, layers:list):
        super(PINN, self).__init__()
        self.linears = nn.ModuleList([])
        for i, dim in enumerate(layers[:-2]):
            self.linears.append(nn.Linear(dim, layers[i+1]))
            self.linears.append(nn.Tanh())
        self.linears.append(nn.Linear(layers[-2], layers[-1]))
        
    def forward(self, x):
        for layer in self.linears:
            x = layer(x)
        return x
        
    def compute_u_x(self, x):
        self.u_x = torch.autograd.functional.jacobian(self, x, create_graph=True)
        self.u_x = torch.squeeze(self.u_x)
        return self.u_x
    
    def compute_u_xx(self, x):
        self.u_xx = torch.autograd.functional.jacobian(self.compute_u_x, x)
        self.u_xx = torch.squeeze(self.u_xx)
        return self.u_xx

然后在 x.require_grad 设置为 True 的 PINN 实例上调用 compute_u_xx(x)获得我在那里.尽管如何理解如何摆脱 torch.autograd.functional.jacobian 引入的无用尺寸，仍然需要理解...

Then calling compute_u_xx(x) on an instance of PINN with x.require_grad set to True gets me there. How to get rid of useless dimensions introduced by torch.autograd.functional.jacobian remains to be understood though...