PINN Deep Learning to Solve Differential Equations Series 1: Solving Framework

Below I will introduce the embedded physics knowledge neural network (PINN) to solve differential equations. Firstly, the basic method of PINN is introduced, and the one-dimensional Poisson equation is solved based on the Pytorch framework.
Embedded Physical Knowledge Neural Network (PINN) Introduction and Related Papers
Deep Learning to Solve Differential Equations Series 1: PINN Solution Framework (Poisson 1d)
Deep Learning to Solve Differential Equations Series 2: PINN to Solve Burger Equation Forward Problems
Deep Learning to Solve Differential Equations Series 3: PINN solves the inverse problem of the Burger equation
Deep learning to solve the differential equation series four: Based on the adaptive activation function PINN solves the inverse problem of the Burger equation

1. Introduction to PINN

As a powerful information processing tool, neural network has been widely used in the fields of computer vision, biomedicine, and oil and gas engineering, triggering technological changes in many fields. The deep learning network has a very strong learning ability, not only can discover physical laws, but also solve partial differential equations. In recent years, the solution of partial differential equations based on deep learning has become a new research hotspot. Embedded physics-informed neural network (PINN) is an application of scientific machines in the traditional numerical field, which can be used to solve various problems related to partial differential equations (PDE), including equation solving, parameter inversion, model discovery, control and optimization etc.

2. PINN method

The main idea of ​​PINN is shown in Figure 1, first construct an output result as $\hat{u}$ 's neural network, which is used as a proxy model for the PDE solution, and the PDE information is used as a constraint, encoded into the neural network loss function for training.

The loss function mainly includes four parts: partial differential structure loss (PDE loss), boundary value condition loss (BC loss), initial value condition loss (IC loss) and real data condition loss (Data loss). In particular, consider the following PDE problem, where the solution of the PDE $u(x)$在$Oh\subset {\mathbb{R}}^d$ definition, where$\mathbf{x}=\left(x_1,\dots ,x_d\right)$：
$$f\left(\mathbf{x};\frac{\partial u}{\partial x_1},\dots ,\frac{\partial u}{\partial x_d};\frac{{\partial }^2}{\partial x_1\partial x_1},\dots ,\frac{{\partial }^2}{\partial x_1\partial x_d}\right)=0,x\in \Omega At$$
the same time, satisfy the following boundary
$$\mathcal{B}(u,\mathbf{x})=0\text{ on }\partial \Omega$$
in order to measure the neural network$\hat{u}$和约束之间的差异，考虑损失函数定义：
$$\mathcal{L}\left(i\right)=w_f{\mathcal{L}}_{PDE}\left(i;{\mathcal{T}}_f\right)+w_i{L}_{IC}\left(i;T_i\right)+w_b{L}_{BC}\left(i,;T_b\right)+w_d{L}_{Data}\left(i,;T_{data}\right)$$
where:
$$\begin{array}{rl}{\mathcal{L}}_{PDE}\left(i;{\mathcal{T}}_f\right)& =\frac{1}{\left|{\mathcal{T}}_f\right|}\sum _{\mathbf{x}\in {\mathcal{T}}_f}{\Vert f\left(\mathbf{x};\frac{\partial \hat{u}}{\partial x_1},\dots ,\frac{\partial \hat{u}}{\partial x_d};\frac{{\partial }^2\hat{u}}{\partial x_1\partial x_1},\dots ,\frac{{\partial }^2\hat{u}}{\partial x_1\partial x_d}\right)\Vert }_2^2\\ L_{IC}\left(i;T_i\right)& =\frac{1}{|T_i|}\sum _{x\in T_i}\Vert \hat{u}(\mathbf{x})-u(\mathbf{x}){\Vert }_2^2\\ L_{BC}\left(i;T_b\right)& =\frac{1}{|T_b|}\sum _{x\in T_b}\Vert \mathcal{B}(\hat{u},\mathbf{x}){\Vert }_2^2\\ L_{Data}\left(i;T_{data}\right)& =\frac{1}{|T_{data}|}\sum _{x\in T_{data}}\Vert \hat{u}\left(x\right)-u\left(x\right){\Vert }_2^2\end{array}$$
$w_f$，$w_i$、$w_b$and $w_d$is the weight. ${\mathcal{T}}_f$，$T_i$、$T_b$和$T_{data}$Represents residual points from PDE, initial value, boundary value and true value. Here $T_f\subset \Omega$ is a predefined set of points to measure the neural network output$\hat{u}$ Extent of match to PDE.

3. Solving problem definition

$\begin{array}{rl}\frac{{\mathrm{d}}^2}{\mathrm{d}{x}^2}& =-0.49\cdot \sin (0.7x)-2.25\cdot \cos (1.5x)\\ u(-10)& =-\sin (7)+\cos (15)+1\\ u(10)& =\sin \left(7\right)+\cos \left(15\right)-1\end{array}$
The real solution is
$u:=\sin (0.7x)+\cos (1.5x)-0.1x$

4. Results display