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tutorials/tutorial9/tutorial.ipynb
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tutorials/tutorial9/tutorial.py
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tutorials/tutorial9/tutorial.py
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#!/usr/bin/env python
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# coding: utf-8
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# # Tutorial: One dimensional Helmholtz equation using Periodic Boundary Conditions
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#
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# [](https://colab.research.google.com/github/mathLab/PINA/blob/master/tutorials/tutorial9/tutorial.ipynb)
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#
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# This tutorial presents how to solve with Physics-Informed Neural Networks (PINNs)
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# a one dimensional Helmholtz equation with periodic boundary conditions (PBC).
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# We will train with standard PINN's training by augmenting the input with
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# periodic expansion as presented in [*An expert’s guide to training
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# physics-informed neural networks*](
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# https://arxiv.org/abs/2308.08468).
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#
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# First of all, some useful imports.
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# In[ ]:
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## routine needed to run the notebook on Google Colab
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try:
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import google.colab
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IN_COLAB = True
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except:
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IN_COLAB = False
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if IN_COLAB:
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get_ipython().system('pip install "pina-mathlab"')
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import torch
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import matplotlib.pyplot as plt
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import warnings
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from pina import Condition, Trainer
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from pina.problem import SpatialProblem
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from pina.operator import laplacian
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from pina.model import FeedForward
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from pina.model.block import PeriodicBoundaryEmbedding # The PBC module
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from pina.solver import PINN
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from pina.domain import CartesianDomain
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from pina.equation import Equation
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from pina.callback import MetricTracker
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warnings.filterwarnings("ignore")
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# ## The problem definition
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#
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# The one-dimensional Helmholtz problem is mathematically written as:
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# $$
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# \begin{cases}
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# \frac{d^2}{dx^2}u(x) - \lambda u(x) -f(x) &= 0 \quad x\in(0,2)\\
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# u^{(m)}(x=0) - u^{(m)}(x=2) &= 0 \quad m\in[0, 1, \cdots]\\
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# \end{cases}
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# $$
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# In this case we are asking the solution to be $C^{\infty}$ periodic with
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# period $2$, on the infinite domain $x\in(-\infty, \infty)$. Notice that the
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# classical PINN would need infinite conditions to evaluate the PBC loss function,
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# one for each derivative, which is of course infeasible...
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# A possible solution, diverging from the original PINN formulation,
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# is to use *coordinates augmentation*. In coordinates augmentation you seek for
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# a coordinates transformation $v$ such that $x\rightarrow v(x)$ such that
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# the periodicity condition $ u^{(m)}(x=0) - u^{(m)}(x=2) = 0 \quad m\in[0, 1, \cdots] $ is
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# satisfied.
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#
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# For demonstration purposes, the problem specifics are $\lambda=-10\pi^2$,
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# and $f(x)=-6\pi^2\sin(3\pi x)\cos(\pi x)$ which give a solution that can be
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# computed analytically $u(x) = \sin(\pi x)\cos(3\pi x)$.
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# In[15]:
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def helmholtz_equation(input_, output_):
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x = input_.extract("x")
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u_xx = laplacian(output_, input_, components=["u"], d=["x"])
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f = (
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-6.0
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* torch.pi**2
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* torch.sin(3 * torch.pi * x)
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* torch.cos(torch.pi * x)
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)
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lambda_ = -10.0 * torch.pi**2
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return u_xx - lambda_ * output_ - f
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class Helmholtz(SpatialProblem):
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output_variables = ["u"]
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spatial_domain = CartesianDomain({"x": [0, 2]})
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# here we write the problem conditions
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conditions = {
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"phys_cond": Condition(
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domain=spatial_domain, equation=Equation(helmholtz_equation)
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),
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}
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def solution(self, pts):
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return torch.sin(torch.pi * pts) * torch.cos(3.0 * torch.pi * pts)
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problem = Helmholtz()
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# let's discretise the domain
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problem.discretise_domain(200, "grid", domains=["phys_cond"])
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# As usual, the Helmholtz problem is written in **PINA** code as a class.
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# The equations are written as `conditions` that should be satisfied in the
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# corresponding domains. The `solution`
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# is the exact solution which will be compared with the predicted one. We used
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# Latin Hypercube Sampling for choosing the collocation points.
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# ## Solving the problem with a Periodic Network
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# Any $\mathcal{C}^{\infty}$ periodic function
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# $u : \mathbb{R} \rightarrow \mathbb{R}$ with period
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# $L\in\mathbb{N}$ can be constructed by composition of an
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# arbitrary smooth function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ and a
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# given smooth periodic function $v : \mathbb{R} \rightarrow \mathbb{R}^n$ with
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# period $L$, that is $u(x) = f(v(x))$. The formulation is generalizable for
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# arbitrary dimension, see [*A method for representing periodic functions and
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# enforcing exactly periodic boundary conditions with
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# deep neural networks*](https://arxiv.org/pdf/2007.07442).
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#
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# In our case, we rewrite
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# $v(x) = \left[1, \cos\left(\frac{2\pi}{L} x\right),
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# \sin\left(\frac{2\pi}{L} x\right)\right]$, i.e
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# the coordinates augmentation, and $f(\cdot) = NN_{\theta}(\cdot)$ i.e. a neural
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# network. The resulting neural network obtained by composing $f$ with $v$ gives
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# the PINN approximate solution, that is
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# $u(x) \approx u_{\theta}(x)=NN_{\theta}(v(x))$.
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#
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# In **PINA** this translates in using the `PeriodicBoundaryEmbedding` layer for $v$, and any
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# `pina.model` for $NN_{\theta}$. Let's see it in action!
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#
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# In[16]:
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# we encapsulate all modules in a torch.nn.Sequential container
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model = torch.nn.Sequential(
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PeriodicBoundaryEmbedding(input_dimension=1, periods=2),
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FeedForward(
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input_dimensions=3, # output of PeriodicBoundaryEmbedding = 3 * input_dimension
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output_dimensions=1,
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layers=[10, 10],
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),
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)
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# As simple as that! Notice that in higher dimension you can specify different periods
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# for all dimensions using a dictionary, e.g. `periods={'x':2, 'y':3, ...}`
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# would indicate a periodicity of $2$ in $x$, $3$ in $y$, and so on...
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#
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# We will now solve the problem as usually with the `PINN` and `Trainer` class, then we will look at the losses using the `MetricTracker` callback from `pina.callback`.
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# In[17]:
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pinn = PINN(
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problem=problem,
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model=model,
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)
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trainer = Trainer(
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pinn,
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max_epochs=5000,
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accelerator="cpu",
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enable_model_summary=False,
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callbacks=[MetricTracker()],
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train_size=1.0,
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val_size=0.0,
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test_size=0.0,
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)
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trainer.train()
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# In[18]:
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# plot loss
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trainer_metrics = trainer.callbacks[0].metrics
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plt.plot(
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range(len(trainer_metrics["train_loss"])), trainer_metrics["train_loss"]
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)
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# plotting
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plt.xlabel("epoch")
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plt.ylabel("loss")
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plt.yscale("log")
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# We are going to plot the solution now!
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# In[19]:
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pts = pinn.problem.spatial_domain.sample(256, "grid", variables="x")
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predicted_output = pinn.forward(pts).extract("u").tensor.detach()
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true_output = pinn.problem.solution(pts)
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plt.plot(pts.extract(["x"]), predicted_output, label="Neural Network solution")
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plt.plot(pts.extract(["x"]), true_output, label="True solution")
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plt.legend()
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# Great, they overlap perfectly! This seems a good result, considering the simple neural network used to some this (complex) problem. We will now test the neural network on the domain $[-4, 4]$ without retraining. In principle the periodicity should be present since the $v$ function ensures the periodicity in $(-\infty, \infty)$.
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# In[20]:
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# plotting solution
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with torch.no_grad():
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# Notice here we put [-4, 4]!!!
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new_domain = CartesianDomain({"x": [0, 4]})
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x = new_domain.sample(1000, mode="grid")
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fig, axes = plt.subplots(1, 3, figsize=(15, 5))
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# Plot 1
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axes[0].plot(x, problem.solution(x), label=r"$u(x)$", color="blue")
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axes[0].set_title(r"True solution $u(x)$")
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axes[0].legend(loc="upper right")
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# Plot 2
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axes[1].plot(x, pinn(x), label=r"$u_{\theta}(x)$", color="green")
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axes[1].set_title(r"PINN solution $u_{\theta}(x)$")
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axes[1].legend(loc="upper right")
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# Plot 3
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diff = torch.abs(problem.solution(x) - pinn(x))
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axes[2].plot(x, diff, label=r"$|u(x) - u_{\theta}(x)|$", color="red")
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axes[2].set_title(r"Absolute difference $|u(x) - u_{\theta}(x)|$")
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axes[2].legend(loc="upper right")
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# Adjust layout
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plt.tight_layout()
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# Show the plots
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plt.show()
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# It is pretty clear that the network is periodic, with also the error following a periodic pattern. Obviously a longer training and a more expressive neural network could improve the results!
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#
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# ## What's next?
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#
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# Congratulations on completing the one dimensional Helmholtz tutorial of **PINA**! There are multiple directions you can go now:
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#
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# 1. Train the network for longer or with different layer sizes and assert the finaly accuracy
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#
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# 2. Apply the `PeriodicBoundaryEmbedding` layer for a time-dependent problem (see reference in the documentation)
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#
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# 3. Exploit extrafeature training ?
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#
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# 4. Many more...
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Reference in New Issue
Block a user