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import matplotlib.pyplot as plt
%matplotlib inline
import matplotlib_inline
import seaborn as sns

!pip show equinox || echo equinox not found. Installing... && pip install equinox 2> /dev/null
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Name: equinox
Version: 0.11.2
Summary: Elegant easy-to-use neural networks in JAX.
Author-email: Patrick Kidger <contact@kidger.site>
License: Apache License
                           Version 2.0, January 2004


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Location: /Users/ibilion/.pyenv/versions/3.11.6/lib/python3.11/site-packages
Requires: jax, jaxtyping, typing-extensions
Required-by: diffrax, lineax, optimistix, orthojax
Requirement already satisfied: equinox in /Users/ibilion/.pyenv/versions/3.11.6/lib/python3.11/site-packages (0.11.2)
Requirement already satisfied: jax>=0.4.13 in /Users/ibilion/.pyenv/versions/3.11.6/lib/python3.11/site-packages (from equinox) (0.4.19)
Requirement already satisfied: jaxtyping>=0.2.20 in /Users/ibilion/.pyenv/versions/3.11.6/lib/python3.11/site-packages (from equinox) (0.2.25)
Requirement already satisfied: typing-extensions>=4.5.0 in /Users/ibilion/.pyenv/versions/3.11.6/lib/python3.11/site-packages (from equinox) (4.8.0)
Requirement already satisfied: ml-dtypes>=0.2.0 in /Users/ibilion/.pyenv/versions/3.11.6/lib/python3.11/site-packages (from jax>=0.4.13->equinox) (0.3.1)
Requirement already satisfied: numpy>=1.22 in /Users/ibilion/.pyenv/versions/3.11.6/lib/python3.11/site-packages (from jax>=0.4.13->equinox) (1.25.2)
Requirement already satisfied: opt-einsum in /Users/ibilion/.pyenv/versions/3.11.6/lib/python3.11/site-packages (from jax>=0.4.13->equinox) (3.3.0)
Requirement already satisfied: scipy>=1.9 in /Users/ibilion/.pyenv/versions/3.11.6/lib/python3.11/site-packages (from jax>=0.4.13->equinox) (1.11.3)
Requirement already satisfied: typeguard<3,>=2.13.3 in /Users/ibilion/.pyenv/versions/3.11.6/lib/python3.11/site-packages (from jaxtyping>=0.2.20->equinox) (2.13.3)
Hide code cell source
import jax
import equinox as eqx

class FourierEncoding(eqx.Module):
    B: jax.Array

    def num_fourier_features(self) -> int:
        return self.B.shape[0]

    def in_size(self) -> int:
        return self.B.shape[1]
    def out_size(self) -> int:
        return self.B.shape[0] * 2

    def __init__(self, 
                 in_size: int, 
                 num_fourier_features: int, 
                 key: jax.random.PRNGKey, 
                 sigma: float = 1.0):
        self.B = jax.random.normal(
            key, shape=(num_fourier_features, in_size),
            dtype=jax.numpy.float32) * sigma
    def __call__(self, x: jax.Array, **kwargs) -> jax.Array:
        return jax.numpy.concatenate(
            [jax.numpy.cos(jax.numpy.dot(self.B, x)),
             jax.numpy.sin(jax.numpy.dot(self.B, x))],

def train_pinn(

    # this is new
    def new_loss(diff_model, static_model, x, y):
        comb_model = eqx.combine(diff_model, static_model)
        return loss(comb_model, x, y)

    def step(opt_state, model, xs, ys):
        # added this line
        diff_model, static_model = eqx.partition(model, filter_spec)
        # changed the loss to the new loss
        value, grads = eqx.filter_value_and_grad(new_loss)(diff_model, static_model, xs, ys)
        updates, opt_state = optimizer.update(grads, opt_state)
        model = eqx.apply_updates(model, updates)
        return model, opt_state, value
    opt_state = optimizer.init(eqx.filter(fourier_mlp, eqx.is_inexact_array))
    losses = []
    for i in range(num_iter):
        key1, key2, key = jrandom.split(key, 3)
        xb = jrandom.uniform(key1, (num_collocation_residual,), maxval=Lx)
        yb = jrandom.uniform(key2, (num_collocation_residual,), maxval=Ly)
        fourier_mlp, opt_state, value = step(opt_state, fourier_mlp, xb, yb)
        if i % freq == 0:
            print(f"Step {i}, residual loss {value:.3e}")
    return fourier_mlp, losses

Energy Functionals#

In this notebook, we are still working with the steady-state heat equation:

\begin{aligned} -k\Delta u(x, y) &= f(x, y) \quad \text{in} \quad \Omega = [0, L_x]\times [0, L_y], \ u(x, y) &= 0 \quad \text{on} \quad \partial \Omega. \end{aligned}

Before, we parameterized the solution so that it satisfied the boundary conditions. Then, we constructed a loss function using the residual of the PDE.

If we forget about the parameterization for a moment, we can think of the integrated squared residual of the PDE as a function that takes a candidate solution to the PDE and returns a scalar:

\[ R[u] = \int_{\Omega} \left[ -k\Delta u(x, y) - f(x, y) \right]^2 \, dx \, dy. \]

Such a function is called a \emph{functional}.

It turns out that for the steady-state heat equation, there is a much much better functional. It is called the \emph{energy functional} and is defined as:

\[ E[u] = \int_{\Omega} \left[ \frac{k}{2}|\nabla u(x, y)|^2 - f(x, y)u(x, y) \right] \, dx \, dy. \]

Here are some good things about this functional:

  • A function that minimizes \(E[u]\) is a solution to the PDE. We will show this.

  • Only a first derivative of \(u\) appears in the integrand. This means that our function lives in the Sobolev space \(W^{1, 2}(\Omega)\) instead of \(W^{2, 2}(\Omega)\). This means that we could even use a ReLU network to represent the solution to the PDE.

  • The functional \(E[u]\) has a unique minimizer in \(W^{1, 2}(\Omega)\), among the functions that satisfy the boundary conditions.


The Sobolev space \(W^{1, 2}(\Omega)\) is the space of functions whose first derivatives are square-integrable. The space \(W^{2, 2}(\Omega)\) is the space of functions whose second derivatives are square-integrable. Learn more about Sobolev spaces here.

To show that a function that minimizes \(E[u]\) is a solution to the PDE, we will use the concept of the functional derivative. A functional derivative is a generalization of the derivative of a function to functionals. It is defined as:

\[ \frac{\delta E[u]}{\delta v} = \lim_{\epsilon \to 0} \frac{E[u + \epsilon v] - E[u]}{\epsilon} = \frac{d}{d\epsilon}E[u + \epsilon v] \bigg|_{\epsilon = 0} \]

where \(v\) is a function in \(W_0^{1, 2}(\Omega)\). The subscript \(0\) in \(W_0^{1, 2}(\Omega)\) means that the functions in the space are zero on the boundary of \(\Omega\). We also call \(\frac{\delta E[u]}{\delta v}\) the \emph{functional derivative} of \(E[u]\) with respect to \(u\) in the direction of \(v\). Another name for the functional derivative is the \emph{first variation} of \(E[u]\) at \(u\) in the direction of \(v\).

There is a theorem that says that if a function \(u\) minimizes \(E[u]\), then the functional derivative of \(E[u]\) at \(u\) in the direction of \(v\) is zero for all \(v\) in \(W_0^{1, 2}(\Omega)\). We prove this in the lecture. Here, let’s find the first variation of \(E[u]\) at \(u\) in the direction of \(v\). We have:

\begin{align*} \frac{\delta E[u]}{\delta v} &= \frac{d}{d\epsilon}E[u + \epsilon v] \bigg|{\epsilon = 0} \ &= \frac{d}{d\epsilon}\int{\Omega} \left[ \frac{k}{2}|\nabla (u + \epsilon v)|^2 - f(u + \epsilon v) \right] , dx , dy \bigg|{\epsilon = 0} \ &= \int{\Omega} \frac{d}{d\epsilon}\left[ \frac{k}{2}|\nabla (u + \epsilon v)|^2 - f(u + \epsilon v) \right]\bigg|{\epsilon = 0} , dx , dy \ &= \int{\Omega} \frac{d}{d\epsilon}\left[ \frac{k}{2}(\nabla u + \epsilon \nabla v)\cdot (\nabla u + \epsilon \nabla v) - f(u + \epsilon v) \right]\bigg|{\epsilon = 0} , dx , dy \ &= \int{\Omega} \frac{d}{d\epsilon}\left[ \frac{k}{2}(\nabla u\cdot \nabla u + 2\epsilon \nabla u\cdot \nabla v + \epsilon^2 \nabla v\cdot \nabla v) - f(u + \epsilon v) \right]\bigg|{\epsilon = 0} , dx , dy \ &= \int{\Omega} \left[ k\nabla u\cdot \nabla v - fv \right] , dx , dy. \ \end{align*}

Now, we integrate by parts to get:

\begin{align*} \frac{\delta E[u]}{\delta v} &= \int_{\Omega} \left[ k\nabla u\cdot \nabla v - fv \right] , dx , dy \ &= \int_{\Omega} \left[ -\nabla \cdot (k\nabla u) - f \right]v , dx , dy \ &= \int_{\Omega} \left[ -k\Delta u - f \right]v , dx , dy. \end{align*}

If this is zero for all \(v\) in \(W_0^{1, 2}(\Omega)\), the part of the integrand that multiplies \(v\) must be zero. So, \(u\) must satisfy the PDE.

To show that the functional \(E[u]\) has a unique minimizer in \(W^{1, 2}(\Omega)\), we would have to find the second variation of \(E[u]\) at \(u\) in the direction of \(v\) and show that it is strongly positive. We will do this in the lecture.

Non-dimensionalizing the Energy Functional#

We scale everything like we did before:

\begin{align*} x &\mapsto \tilde{x} = \frac{x}{L_x}, \ y &\mapsto \tilde{y} = \frac{y}{L_y}, \ u &\mapsto \tilde{u} = \frac{u}{u_s}, \ f &\mapsto \tilde{f} = \frac{f}{f_s}. \ \end{align*}

We can now directly make the change on the energy functional:

\begin{align*} E[u] &= L_xL_y\int_{\tilde{\Omega}} \left{ \frac{1}{2}\left[\frac{ku_s^2}{L_x^2}\left(\frac{\partial \tilde{u}}{\partial \tilde{x}}\right)^2 + \frac{ku_s^2}{L_y^2}\left(\frac{\partial \tilde{u}}{\partial \tilde{y}}\right)^2\right] - f_su_s\tilde{f}\tilde{u} \right} , d\tilde{x} , d\tilde{y} \ &= L_xL_yf_su_s \int_{\tilde{\Omega}} \left{ \frac{1}{2}\left[\frac{ku_s}{L_x^2f_s}\left(\frac{\partial \tilde{u}}{\partial \tilde{x}}\right)^2 + \frac{ku_s}{L_y^2f_s}\left(\frac{\partial \tilde{u}}{\partial \tilde{y}}\right)^2\right] - \tilde{f}\tilde{u} \right} , d\tilde{x} , d\tilde{y} \ &= L_xL_yf_su_s \int_{\tilde{\Omega}} \left{ \frac{1}{2}\left[k_x\left(\frac{\partial \tilde{u}}{\partial \tilde{x}}\right)^2 + k_y\left(\frac{\partial \tilde{u}}{\partial \tilde{y}}\right)^2\right] - \tilde{f}\tilde{u} \right} , d\tilde{x} , d\tilde{y}, \ &= L_xL_yf_su_s \tilde{E}[\tilde{u}], \end{align*}

where we have defined \(k_x = \frac{ku_s}{L_x^2f_s}\) and \(k_y = \frac{ku_s}{L_y^2f_s}\).

We pick

Now, we will pick \(u_s\) to make the two terms inside \(\tilde{E}[\tilde{u}]\) have the same order of magnitude. So, we pick:

\begin{align*} &\left|k_x\left(\frac{\partial \tilde{u}}{\partial \tilde{x}}\right)^2+ k_y\left(\frac{\partial \tilde{u}}{\partial \tilde{y}}\right)^2\right|{\infty} = |\tilde{f}\tilde{u}|{\infty}, \ &\implies \frac{ku_s}{f_s} \left|\left(L_x^{-2}\frac{\partial \tilde{u}}{\partial \tilde{x}}\right)^2+ L_y^{-2}\left(\frac{\partial \tilde{u}}{\partial \tilde{y}}\right)^2\right|{\infty} = |\tilde{f}\tilde{u}|{\infty}, \ &\implies u_s = \frac{f_s}{k}|\tilde{f}\tilde{u}|{\infty}\left|L_x^{-2}\left(\frac{\partial \tilde{u}}{\partial \tilde{x}}\right)^2+ L_y^{-2}\left(\frac{\partial \tilde{u}}{\partial \tilde{y}}\right)^2\right|{\infty}^{-1}. \end{align*}

At the same time, we want to make both terms to be of order one. This can be achieved by picking:

\[ \|\tilde{f}\tilde{u}\|_{\infty} = 1. \]


\[ f_s = \|f(\tilde{x}L_x, \tilde{y}L_y)\tilde{u}\|_{\infty}. \]

Okay, let’s do it.

Start with our problem definition:

import jax.numpy as jnp

u0 = 500 # degrees Kelvin
k = 10.0 # thermal conductivity in W/mK
Lx = 0.1 # meters
Ly = 1.0 # meters

f = lambda x, y: 2.0 * jnp.pi ** 2 * k * u0 * (
    -Lx ** 2 * jnp.sin(jnp.pi * x / Lx) ** 2 * jnp.cos(2.0 * jnp.pi * y / Ly)
    -Ly ** 2 * jnp.sin(jnp.pi * y / Ly) ** 2 * jnp.cos(2.0 * jnp.pi * x / Lx)
) / (Lx ** 2 * Ly ** 2)

Some of the scaling:

to_x = lambda xt: xt * Lx
to_y = lambda yt: yt * Ly
to_xt = lambda x: x / Lx
to_yt = lambda y: y / Ly

For \(u_s\), we need to parameterize the field:

# Make our model
import jax
import jax.random as jrandom
key = jrandom.PRNGKey(0)

key1, key2, key = jrandom.split(key, 3)
num_fourier_features = 100
width = 256
depth = 4
model = eqx.nn.Sequential([
    FourierEncoding(2, num_fourier_features, key1),
    eqx.nn.MLP(2 * num_fourier_features, 'scalar', width, depth, jax.nn.tanh, key=key2),
ut = lambda xt, yt, model: xt * (1 - xt) * yt * (1 - yt) * model(jnp.array([xt, yt]))

Some code to calculate infinity norms of functions using points:

jit_vmap = lambda g: jit(vmap(g, in_axes=(0, 0)))
infinity_norm = lambda g, xs, ys: jnp.max(jnp.abs(jit_vmap(g)(xs, ys)))

Now, we can find \(f_s\):

from jax import vmap, jit

xs = jnp.linspace(0, Lx, 100)
ys = jnp.linspace(0, Ly, 100)
X, Y = jnp.meshgrid(xs, ys)
Xt = to_xt(X)
Yt = to_yt(Y)
Xt_flat = Xt.flatten()
Yt_flat = Yt.flatten()

fs = infinity_norm(lambda xt, yt: f(to_x(xt), to_y(yt)) * ut(xt, yt, model), Xt_flat, Yt_flat)

# The scaled version of the source
ft = lambda xt, yt: f(to_x(xt), to_y(yt)) / fs

print(f"fs = {fs:.3e}")
fs = 1.364e+04

And we can find \(u_s\):

term1 = jit(vmap(lambda xt, yt: jnp.abs(ft(xt, yt) * ut(xt, yt, model)), in_axes=(0, 0)))(Xt_flat, Yt_flat).max()

from jax import grad
ut_x = grad(ut, 0)
ut_y = grad(ut, 1)
term2 = infinity_norm(lambda xt, yt: ut_x(xt, yt, model) ** 2 / Lx ** 2 + ut_y(xt, yt, model) **2 / Ly ** 2, Xt_flat, Yt_flat)

us = (fs * term1) / (k * term2)

print(f"us = {us:.3e}")
us = 6.414e+04

Now, we can define the scaled energy thermal conductivity:

kx = (k * us) / (Lx ** 2  * fs)
ky = (k * us) / (Ly ** 2  * fs)

print(f"kx = {kx:.3e}, ky = {ky:.3e}")
kx = 4.702e+03, ky = 4.702e+01

Let’s verify that we have the proper scaling:

tE_term1 = jit(vmap(lambda xt, yt: 0.5 * (kx * ut_x(xt, yt, model) ** 2 + kx * ut_y(xt, yt, model) ** 2), in_axes=(0, 0)))
tE_term2 = jit(vmap(lambda xt, yt: ft(xt, yt) * ut(xt, yt, model), in_axes=(0, 0)))

fig, ax = plt.subplots(1,2)
trm1 = tE_term1(Xt_flat, Yt_flat).reshape(X.shape)
c = ax[0].contourf(Xt, Yt, trm1, cmap='viridis')
fig.colorbar(c, ax=ax[0])
ax[0].set_title('Term 1')

trm2 = tE_term2(Xt_flat, Yt_flat).reshape(X.shape)
c = ax[1].contourf(Xt, Yt, trm2, cmap='viridis')
fig.colorbar(c, ax=ax[1])
ax[1].set_title('Term 2')


We will take this. It looks good. Notice that the scaling is completely different from before.

Training the Network#

Let’s make our new loss:

energy_loss = vmap(
    lambda xt, yt, model: 0.5 * (kx * ut_x(xt, yt, model) ** 2 + ky * ut_y(xt, yt, model) ** 2) - ft(xt, yt) * ut(xt, yt, model),
    in_axes=(0, 0, None),
pinn_loss = lambda model, xt, yt: jnp.mean(energy_loss(xt, yt, model))
import optax
import jax.tree_util as jtu

# Here, I am actually optimizing the Fourier coefficients
filter_spec = jtu.tree_map(lambda _: True, model)

key, subkey = jax.random.split(key)
optimizer = optax.adam(1e-3)
trained_model, losses = train_pinn(
    pinn_loss, model, subkey, optimizer, filter_spec,
    num_collocation_residual=256, num_iter=5_000, freq=100, Lx=1.0, Ly=1.0)
Step 0, residual loss 1.035e-01
Step 100, residual loss -2.370e-01
Step 200, residual loss -2.250e-01
Step 300, residual loss -1.002e-01
Step 400, residual loss -3.273e-01
Step 500, residual loss -2.182e-01
Step 600, residual loss -1.557e-01
Step 700, residual loss -2.311e-01
Step 800, residual loss -2.145e-01
Step 900, residual loss -2.937e-01
Step 1000, residual loss -3.032e-02
Step 1100, residual loss -3.340e-01
Step 1200, residual loss -2.046e-01
Step 1300, residual loss -1.865e-01
Step 1400, residual loss -1.772e-01
Step 1500, residual loss -3.224e-01
Step 1600, residual loss -3.415e-01
Step 1700, residual loss -1.926e-01
Step 1800, residual loss -2.110e-01
Step 1900, residual loss -3.176e-01
Step 2000, residual loss -4.067e-01
Step 2100, residual loss -2.817e-01
Step 2200, residual loss -2.407e-01
Step 2300, residual loss -2.483e-01
Step 2400, residual loss -3.020e-01
Step 2500, residual loss -2.907e-01
Step 2600, residual loss -2.233e-01
Step 2700, residual loss -1.766e-01
Step 2800, residual loss -2.577e-01
Step 2900, residual loss -5.195e-01
Step 3000, residual loss -2.384e-01
Step 3100, residual loss -1.722e-01
Step 3200, residual loss -2.518e-01
Step 3300, residual loss -2.413e-01
Step 3400, residual loss -4.015e-01
Step 3500, residual loss -3.214e-01
Step 3600, residual loss -4.055e-01
Step 3700, residual loss -2.391e-01
Step 3800, residual loss -1.856e-01
Step 3900, residual loss -1.745e-01
Step 4000, residual loss -3.876e-01
Step 4100, residual loss -2.380e-01
Step 4200, residual loss -1.656e-01
Step 4300, residual loss -2.280e-01
Step 4400, residual loss -2.837e-01
Step 4500, residual loss -2.168e-01
Step 4600, residual loss -1.857e-01
Step 4700, residual loss -3.263e-01
Step 4800, residual loss -2.783e-01
Step 4900, residual loss -3.873e-01

Let’s look at the evolution of the energy functional during training:

fig, ax = plt.subplots()
ax.set_xlabel('Iteration x 100')
ax.set_ylabel('Energy loss')
u_true = u0 * jnp.sin(jnp.pi * X / Lx) ** 2 * jnp.sin(jnp.pi * Y / Ly) ** 2

vtu = jit(vmap(lambda xt, yt: ut(xt, yt, trained_model)))
u_pred = vtu(Xt_flat, Yt_flat).reshape(X.shape) * us

fig, ax = plt.subplots(1,2)
c = ax[0].contourf(X, Y, u_pred, cmap='viridis',
    levels=jnp.linspace(-1, 510.0, 10))
fig.colorbar(c, ax=ax[0])
ax[0].set_title('Trained solution')

c = ax[1].contourf(X, Y, u_true, cmap='viridis',
    levels=jnp.linspace(-1, 510.0, 10))
fig.colorbar(c, ax=ax[1])
ax[1].set_title('Scaled true solution')
fig, ax = plt.subplots()
ax.set_title('Trained solution')
c = ax.contourf(to_xt(X), to_yt(Y), jnp.abs(u_pred - u_true), cmap='viridis')
plt.colorbar(c, ax=ax)

It is actually not as accurate as the residual loss. It probably needs more iterations to converge. The structure and optimization parameters were optimized for the integrated residual loss.