4cfd90b904
* update docs/tests * tutorial and device fix --------- Co-authored-by: Dario Coscia <dariocoscia@Dario-Coscia.local> Co-authored-by: Dario Coscia <dariocoscia@dhcp-015.eduroam.sissa.it> Co-authored-by: Dario Coscia <dariocoscia@Dario-Coscia.lan> Co-authored-by: Dario Coscia <dariocoscia@Dario-Coscia.Home>
192 lines
7.2 KiB
Python
Vendored
192 lines
7.2 KiB
Python
Vendored
#!/usr/bin/env python
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# coding: utf-8
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# # Tutorial: One dimensional Helmotz equation using Periodic Boundary Conditions
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# This tutorial presents how to solve with Physics-Informed Neural Networks (PINNs)
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# a one dimensional Helmotz 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 expasion 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[1]:
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import torch
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import matplotlib.pyplot as plt
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from pina import Condition, Plotter
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from pina.problem import SpatialProblem
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from pina.operators import laplacian
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from pina.model import FeedForward
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from pina.model.layers import PeriodicBoundaryEmbedding # The PBC module
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from pina.solvers import PINN
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from pina.trainer import Trainer
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from pina.geometry import CartesianDomain
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from pina.equation import Equation
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# ## The problem definition
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#
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# The one-dimensional Helmotz 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 inifite domain $x\in(-\infty, \infty)$. Notice that the
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# classical PINN would need inifinite conditions to evaluate the PBC loss function,
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# one for each derivative, which is of course infeasable...
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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 porpuses 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 gives 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[2]:
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class Helmotz(SpatialProblem):
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output_variables = ['u']
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spatial_domain = CartesianDomain({'x': [0, 2]})
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def helmotz_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 = - 6.*torch.pi**2 * torch.sin(3*torch.pi*x)*torch.cos(torch.pi*x)
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lambda_ = - 10. * torch.pi ** 2
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return u_xx - lambda_ * output_ - f
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# here we write the problem conditions
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conditions = {
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'D': Condition(location=spatial_domain,
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equation=Equation(helmotz_equation)),
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}
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def helmotz_sol(self, pts):
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return torch.sin(torch.pi * pts) * torch.cos(3. * torch.pi * pts)
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truth_solution = helmotz_sol
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problem = Helmotz()
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# let's discretise the domain
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problem.discretise_domain(200, 'grid', locations=['D'])
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# As usual the Helmotz 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 `truth_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[3]:
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# we encapsulate all modules in a torch.nn.Sequential container
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model = torch.nn.Sequential(PeriodicBoundaryEmbedding(input_dimension=1,
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periods=2),
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FeedForward(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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# As simple as that! Notice 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 sole the problem as usually with the `PINN` and `Trainer` class.
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# In[5]:
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pinn = PINN(problem=problem, model=model)
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trainer = Trainer(pinn, max_epochs=5000, accelerator='cpu', enable_model_summary=False) # we train on CPU and avoid model summary at beginning of training (optional)
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trainer.train()
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# We are going to plot the solution now!
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# In[6]:
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pl = Plotter()
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pl.plot(pinn)
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# Great, they overlap perfectly! This seeams 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[7]:
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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.truth_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.truth_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. Obviusly 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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# Nice you have completed the one dimensional Helmotz 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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