深度学习求解微分方程系列一:PINN求解框架

下面我将介绍内嵌物理知识神经网络(PINN)求解微分方程。首先介绍PINN基本方法,并基于Pytorch框架实现求解一维Poisson方程。

1.PINN简介

神经网络作为一种强大的信息处理工具在计算机视觉、生物医学、 油气工程领域得到广泛应用, 引发多领域技术变革.。深度学习网络具有非常强的学习能力, 不仅能发现物理规律, 还能求解偏微分方程.。近年来,基于深度学习的偏微分方程求解已是研究新热点。内嵌物理知识神经网络(PINN)是一种科学机器在传统数值领域的应用方法,能够用于解决与偏微分方程 (PDE) 相关的各种问题,包括方程求解、参数反演、模型发现、控制与优化等。

2.PINN方法

PINN的主要思想如图1,先构建一个输出结果为

u

^

\hat{u}

u^的神经网络,将其作为PDE解的代理模型,将PDE信息作为约束,编码到神经网络损失函数中进行训练。
在这里插入图片描述

损失函数主要包括4部分:偏微分结构损失(PDE loss),边值条件损失(BC loss)、初值条件损失(IC loss)以及真实数据条件损失(Data loss)。特别的,考虑下面这个的PDE问题,其中PDE的解

u

(

x

)

u(x)

u(x)

Ω

R

d

\Omega \subset \mathbb{R}^{d}

ΩRd定义,其中

x

=

(

x

1

,

,

x

d

)

\mathbf{x}=\left(x_{1}, \ldots, x_{d}\right)

x=(x1,,xd)

f

(

x

;

u

x

1

,

,

u

x

d

;

2

u

x

1

x

1

,

,

2

u

x

1

x

d

)

=

0

,

x

Ω

f\left(\mathbf{x} ; \frac{\partial u}{\partial x_{1}}, \ldots, \frac{\partial u}{\partial x_{d}} ; \frac{\partial^{2} u}{\partial x_{1} \partial x_{1}}, \ldots, \frac{\partial^{2} u}{\partial x_{1} \partial x_{d}} \right)=0, \quad \mathbf{x} \in \Omega

f(x;x1u,,xdu;x1x12u,,x1xd2u)=0,xΩ
同时,满足下面的边界

B

(

u

,

x

)

=

0

 on 

Ω

\mathcal{B}(u, \mathbf{x})=0 \quad \text { on } \quad \partial \Omega

B(u,x)=0 on Ω
为了衡量神经网络

u

^

\hat{u}

u^和约束之间的差异,考虑损失函数定义:

L

(

θ

)

=

w

f

L

P

D

E

(

θ

;

T

f

)

+

w

i

L

I

C

(

θ

;

T

i

)

+

w

b

L

B

C

(

θ

,

;

T

b

)

+

w

d

L

D

a

t

a

(

θ

,

;

T

d

a

t

a

)

\mathcal{L}\left(\boldsymbol{\theta}\right)=w_{f} \mathcal{L}_{PDE}\left(\boldsymbol{\theta}; \mathcal{T}_{f}\right)+w_{i} \mathcal{L}_{IC}\left(\boldsymbol{\theta} ; \mathcal{T}_{i}\right)+w_{b} \mathcal{L}_{BC}\left(\boldsymbol{\theta},; \mathcal{T}_{b}\right)+w_{d} \mathcal{L}_{Data}\left(\boldsymbol{\theta},; \mathcal{T}_{data}\right)

L(θ)=wfLPDE(θ;Tf)+wiLIC(θ;Ti)+wbLBC(θ,;Tb)+wdLData(θ,;Tdata)
式中:

L

P

D

E

(

θ

;

T

f

)

=

1

T

f

x

T

f

f

(

x

;

u

^

x

1

,

,

u

^

x

d

;

2

u

^

x

1

x

1

,

,

2

u

^

x

1

x

d

)

2

2

L

I

C

(

θ

;

T

i

)

=

1

T

i

x

T

i

u

^

(

x

)

u

(

x

)

2

2

L

B

C

(

θ

;

T

b

)

=

1

T

b

x

T

b

B

(

u

^

,

x

)

2

2

L

D

a

t

a

(

θ

;

T

d

a

t

a

)

=

1

T

d

a

t

a

x

T

d

a

t

a

u

^

(

x

)

u

(

x

)

2

2

\begin{aligned} \mathcal{L}_{PDE}\left(\boldsymbol{\theta} ; \mathcal{T}_{f}\right) &=\frac{1}{\left|\mathcal{T}_{f}\right|} \sum_{\mathbf{x} \in \mathcal{T}_{f}}\left\|f\left(\mathbf{x} ; \frac{\partial \hat{u}}{\partial x_{1}}, \ldots, \frac{\partial \hat{u}}{\partial x_{d}} ; \frac{\partial^{2} \hat{u}}{\partial x_{1} \partial x_{1}}, \ldots, \frac{\partial^{2} \hat{u}}{\partial x_{1} \partial x_{d}} \right)\right\|_{2}^{2} \\ \mathcal{L}_{IC}\left(\boldsymbol{\theta}; \mathcal{T}_{i}\right) &=\frac{1}{\left|\mathcal{T}_{i}\right|} \sum_{\mathbf{x} \in \mathcal{T}_{i}}\|\hat{u}(\mathbf{x})-u(\mathbf{x})\|_{2}^{2} \\ \mathcal{L}_{BC}\left(\boldsymbol{\theta}; \mathcal{T}_{b}\right) &=\frac{1}{\left|\mathcal{T}_{b}\right|} \sum_{\mathbf{x} \in \mathcal{T}_{b}}\|\mathcal{B}(\hat{u}, \mathbf{x})\|_{2}^{2}\\ \mathcal{L}_{Data}\left(\boldsymbol{\theta}; \mathcal{T}_{data}\right) &=\frac{1}{\left|\mathcal{T}_{data}\right|} \sum_{\mathbf{x} \in \mathcal{T}_{data}}\|\hat{u}(\mathbf{x})-u(\mathbf{x})\|_{2}^{2} \end{aligned}

LPDE(θ;Tf)LIC(θ;Ti)LBC(θ;Tb)LData(θ;Tdata)=Tf1xTf

f(x;x1u^,,xdu^;x1x12u^,,x1xd2u^)

22=Ti1xTiu^(x)u(x)22=Tb1xTbB(u^,x)22=Tdata1xTdatau^(x)u(x)22

w

f

w_{f}

wf

w

i

w_{i}

wi

w

b

w_{b}

wb

w

d

w_{d}

wd是权重。

T

f

\mathcal{T}_{f}

Tf

T

i

\mathcal{T}_{i}

Ti

T

b

\mathcal{T}_{b}

Tb

T

d

a

t

a

\mathcal{T}_{data}

Tdata表示来自PDE,初值、边值以及真值的residual points。这里的

T

f

Ω

\mathcal{T}_{f} \subset \Omega

TfΩ是一组预定义的点来衡量神经网络输出

u

^

\hat{u}

u^与PDE的匹配程度。

3.求解问题定义

d

2

u

 

d

x

2

=

0.49

sin

(

0.7

x

)

2.25

cos

(

1.5

x

)

u

(

10

)

=

sin

(

7

)

+

cos

(

15

)

+

1

u

(

10

)

=

sin

(

7

)

+

cos

(

15

)

1

\begin{aligned} \frac{\mathrm{d}^2 u}{\mathrm{~d} x^2} &=-0.49 \cdot \sin (0.7 x)-2.25 \cdot \cos (1.5 x) \\ u(-10) &=-\sin (7)+\cos (15)+1 \\ u(10) &=\sin (7)+\cos (15)-1 \end{aligned}

 dx2d2uu(10)u(10)=0.49sin(0.7x)2.25cos(1.5x)=sin(7)+cos(15)+1=sin(7)+cos(15)1
真实解为

u

:

=

sin

(

0.7

x

)

+

cos

(

1.5

x

)

0.1

x

u:=\sin (0.7 x)+\cos (1.5 x)-0.1 x

u:=sin(0.7x)+cos(1.5x)0.1x

4.Python求解代码

第一步,首先定义神经网络

# torch version 1.4.0    
import torch
import torch.optim
from collections import OrderedDict
import torch.nn as nn

class Net(nn.Module):
    def __init__(self, seq_net, name='MLP', activation=torch.tanh):
        super().__init__()
        self.features = OrderedDict()
        for i in range(len(seq_net) - 1):
            self.features['{}_{}'.format(name, i)] = nn.Linear(seq_net[i], seq_net[i + 1], bias=True)
        self.features = nn.ModuleDict(self.features)
        self.active = activation

        # initial_bias
        for m in self.modules():
            if isinstance(m, nn.Linear):
                nn.init.constant_(m.bias, 0)

    def forward(self, x):
        length = len(self.features)
        i = 0
        for name, layer in self.features.items():
            x = layer(x)
            if i == length - 1: break
            i += 1
            x = self.active(x)
        return x

第二步,PINN求解框架

from net import Net
import os
import matplotlib.pyplot as plt
import torch
from torch.autograd import grad

def d(f, x):
    return grad(f, x, grad_outputs=torch.ones_like(f), create_graph=True, only_inputs=True)[0]

def PDE(u, x):
    return d(d(u, x), x) + 0.49 * torch.sin(0.7 * x) + 2.25 * torch.cos(1.5 * x)

def Ground_true(x):
    return torch.sin(0.7 * x) + torch.cos(1.5 * x) - 0.1 * x

def train():
    lr = 0.001
    n_pred = 100
    n_f = 200
    epochs = 8000

    os.environ['CUDA_VISIBLE_DEVICES'] = '0'
    device = torch.device('cuda:0') if torch.cuda.is_available() else torch.cuda('cpu')

    PINN = Net([1, 50, 50, 50, 1]).to(device)

    x_left, x_right = -10, 10
    optimizer = torch.optim.Adam(PINN.parameters(), lr)
    criterion = torch.nn.MSELoss()

    loss_history = []

    for epoch in range(epochs):
        optimizer.zero_grad()
        # inside
        x_f = ((x_left + x_right) / 2 + (x_right - x_left) *
               (torch.rand(size=(n_f, 1), dtype=torch.float, device=device) - 0.5)
               ).requires_grad_(True)
        u_f = PINN(x_f)
        PDE_ = PDE(u_f, x_f)
        mse_PDE = criterion(PDE_, torch.zeros_like(PDE_))

        # boundary
        x_b = torch.tensor([[-10.], [10.]]).requires_grad_(True).to(device)
        u_b = PINN(x_b)
        true_b = Ground_true(x_b)
        mse_BC = criterion(u_b, true_b)

        # predict
        x_pred = ((x_left + x_right) / 2 + (x_right - x_left) *
                  (torch.rand(size=(n_pred, 1), dtype=torch.float, device=device) - 0.5)
                  ).requires_grad_(True)
        u_f = PINN(x_pred)
        true_f = Ground_true(x_pred)
        mse_pred = criterion(u_f, true_f)

        loss = 1 * mse_PDE + 1 * mse_BC
        loss_history.append([mse_PDE.item(), mse_BC, mse_pred])

        if epoch % 100 == 0:
            print(
                'epoch:{:05d}, EoM: {:.08e}, BC: {:.08e}, loss: {:.08e}'.format(
                    epoch, mse_PDE.item(), mse_BC.item(), loss.item()
                )
            )

        loss.backward()
        optimizer.step()

        xx = torch.linspace(-10, 10, 10000).reshape((-1, 1)).to(device)

        if (epoch + 1) % 300 == 0:
            yy = PINN(xx)
            zz = Ground_true(xx)
            xx = xx.reshape((-1)).data.detach().cpu().numpy()
            yy = yy.reshape((-1)).data.detach().cpu().numpy()
            zz = zz.reshape((-1)).data.detach().cpu().numpy()
            plt.cla()

            plt.plot(xx, yy, label='PINN')
            plt.plot(xx, zz, label='True', color='r')
            plt.legend()

            plt.title('PINN(epoch{}))'.format(epoch + 1))
            plt.show()

    plt.plot(loss_history)
    plt.legend(('PDE loss', 'BC loss', 'Pred loss'), loc='best')
    plt.show()

if name == ‘main’:
train()

5.结果展示

请添加图片描述

请添加图片描述

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