#!/usr/bin/env python
# coding: utf-8

# # Tutorial: One dimensional Helmholtz equation using Periodic Boundary Conditions
# 
# [![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/mathLab/PINA/blob/master/tutorials/tutorial9/tutorial.ipynb)
# 
# This tutorial presents how to solve with Physics-Informed Neural Networks (PINNs)
# a one dimensional Helmholtz equation with periodic boundary conditions (PBC).
# We will train with standard PINN's training by augmenting the input with
# periodic expansion as presented in [*An expert’s guide to training
# physics-informed neural networks*](
# https://arxiv.org/abs/2308.08468).
# 
# First of all, some useful imports.

# In[1]:


## routine needed to run the notebook on Google Colab
try:
  import google.colab
  IN_COLAB = True
except:
  IN_COLAB = False
if IN_COLAB:
  get_ipython().system('pip install "pina-mathlab"')

import torch
import matplotlib.pyplot as plt
plt.style.use('tableau-colorblind10')
from pina import Condition, Plotter
from pina.problem import SpatialProblem
from pina.operators import laplacian
from pina.model import FeedForward
from pina.model.layers import PeriodicBoundaryEmbedding  # The PBC module
from pina.solvers import PINN
from pina.trainer import Trainer
from pina.domain import CartesianDomain
from pina.equation import Equation


# ## The problem definition
# 
# The one-dimensional Helmholtz problem is mathematically written as:
# $$
# \begin{cases}
# \frac{d^2}{dx^2}u(x) - \lambda u(x) -f(x) &=  0 \quad x\in(0,2)\\
# u^{(m)}(x=0) - u^{(m)}(x=2) &= 0 \quad m\in[0, 1, \cdots]\\
# \end{cases}
# $$
# In this case we are asking the solution to be $C^{\infty}$ periodic with
# period $2$, on the infinite domain $x\in(-\infty, \infty)$. Notice that the
# classical PINN would need infinite conditions to evaluate the PBC loss function,
# one for each derivative, which is of course infeasible... 
# A possible solution, diverging from the original PINN formulation,
# is to use *coordinates augmentation*. In coordinates augmentation you seek for
# a coordinates transformation $v$ such that $x\rightarrow v(x)$ such that
# the periodicity condition $ u^{(m)}(x=0) - u^{(m)}(x=2) = 0 \quad m\in[0, 1, \cdots] $ is
# satisfied.
# 
# For demonstration purposes, the problem specifics are $\lambda=-10\pi^2$,
# and $f(x)=-6\pi^2\sin(3\pi x)\cos(\pi x)$ which give a solution that can be
# computed analytically $u(x) = \sin(\pi x)\cos(3\pi x)$.

# In[ ]:


class Helmholtz(SpatialProblem):
    output_variables = ['u']
    spatial_domain = CartesianDomain({'x': [0, 2]})

    def Helmholtz_equation(input_, output_):
        x = input_.extract('x')
        u_xx = laplacian(output_, input_, components=['u'], d=['x'])
        f = - 6.*torch.pi**2 * torch.sin(3*torch.pi*x)*torch.cos(torch.pi*x)
        lambda_ = - 10. * torch.pi ** 2
        return u_xx - lambda_ * output_ - f

    # here we write the problem conditions
    conditions = {
        'phys_cond': Condition(domain=spatial_domain,
                       equation=Equation(Helmholtz_equation)),
    }

    def Helmholtz_sol(self, pts):
        return torch.sin(torch.pi * pts) * torch.cos(3. * torch.pi * pts)
    
    truth_solution = Helmholtz_sol

problem = Helmholtz()

# let's discretise the domain
problem.discretise_domain(200, 'grid', domains=['phys_cond'])


# As usual, the Helmholtz problem is written in **PINA** code as a class. 
# The equations are written as `conditions` that should be satisfied in the
# corresponding domains. The `truth_solution`
# is the exact solution which will be compared with the predicted one. We used
# Latin Hypercube Sampling for choosing the collocation points.

# ## Solving the problem with a Periodic Network

# Any $\mathcal{C}^{\infty}$ periodic function
# $u : \mathbb{R} \rightarrow \mathbb{R}$ with period
# $L\in\mathbb{N}$ can be constructed by composition of an
# arbitrary smooth function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ and a
# given smooth periodic function $v : \mathbb{R} \rightarrow \mathbb{R}^n$ with
# period $L$, that is $u(x) = f(v(x))$. The formulation is generalizable for
# arbitrary dimension, see [*A method for representing periodic functions and
# enforcing exactly periodic boundary conditions with
# deep neural networks*](https://arxiv.org/pdf/2007.07442).
# 
# In our case, we rewrite
# $v(x) = \left[1, \cos\left(\frac{2\pi}{L} x\right),
# \sin\left(\frac{2\pi}{L} x\right)\right]$, i.e
# the coordinates augmentation, and $f(\cdot) = NN_{\theta}(\cdot)$ i.e. a neural
# network. The resulting neural network obtained by composing $f$ with $v$ gives
# the PINN approximate solution, that is
# $u(x) \approx u_{\theta}(x)=NN_{\theta}(v(x))$.
# 
# In **PINA** this translates in using the `PeriodicBoundaryEmbedding` layer for $v$, and any
# `pina.model` for $NN_{\theta}$. Let's see it in action! 
# 

# In[3]:


# we encapsulate all modules in a torch.nn.Sequential container
model = torch.nn.Sequential(PeriodicBoundaryEmbedding(input_dimension=1,
                                         periods=2),
                            FeedForward(input_dimensions=3, # output of PeriodicBoundaryEmbedding = 3 * input_dimension
                                        output_dimensions=1,
                                        layers=[10, 10]))


# As simple as that! Notice that in higher dimension you can specify different periods
# for all dimensions using a dictionary, e.g. `periods={'x':2, 'y':3, ...}`
# would indicate a periodicity of $2$ in $x$, $3$ in $y$, and so on...
# 
# We will now solve the problem as usually with the `PINN` and `Trainer` class.

# In[ ]:


pinn = PINN(problem=problem, model=model)
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)
trainer.train()


# We are going to plot the solution now!

# In[6]:


pl = Plotter()
pl.plot(pinn)


# 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 $(-\infty, \infty)$.

# In[7]:


# plotting solution
with torch.no_grad():
    # Notice here we put [-4, 4]!!!
    new_domain = CartesianDomain({'x' : [0, 4]})
    x = new_domain.sample(1000, mode='grid')
    fig, axes = plt.subplots(1, 3, figsize=(15, 5))
    # Plot 1
    axes[0].plot(x, problem.truth_solution(x), label=r'$u(x)$', color='blue')
    axes[0].set_title(r'True solution $u(x)$')
    axes[0].legend(loc="upper right")
    # Plot 2
    axes[1].plot(x, pinn(x), label=r'$u_{\theta}(x)$', color='green')
    axes[1].set_title(r'PINN solution $u_{\theta}(x)$')
    axes[1].legend(loc="upper right")
    # Plot 3
    diff = torch.abs(problem.truth_solution(x) - pinn(x))
    axes[2].plot(x, diff, label=r'$|u(x) - u_{\theta}(x)|$', color='red')
    axes[2].set_title(r'Absolute difference $|u(x) - u_{\theta}(x)|$')
    axes[2].legend(loc="upper right")
    # Adjust layout
    plt.tight_layout()
    # Show the plots
    plt.show()


# 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!
# 
# ## What's next?
# 
# Congratulations on completing the one dimensional Helmholtz tutorial of **PINA**! There are multiple directions you can go now:
# 
# 1. Train the network for longer or with different layer sizes and assert the finaly accuracy
# 
# 2. Apply the `PeriodicBoundaryEmbedding` layer for a time-dependent problem (see reference in the documentation)
# 
# 3. Exploit extrafeature training ?
# 
# 4. Many more...

#