Physics-Informed Neural Networks (PINNs) - Forward Problems

Name: equinox
Version: 0.11.2
Summary: Elegant easy-to-use neural networks in JAX.
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Author-email: Patrick Kidger <contact@kidger.site>
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Physics-Informed Neural Networks (PINNs) - Forward Problems#

We can use PINNs to solve forward problems in place of traditional numerical methods. This is not recommended, as PINNs are not yet as efficient as traditional numerical methods. But, it is a good way to learn how to use PINNs and to understand their limitations. Throughout this section we will be following closely the methodology of Wang et al. (2023).

The toy problem - Steady-state heat equation#

We are going to solve a Poisson’s equation with a source term:

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

Making an exact solution#

We will use a common trick to construct an exact solution. We will use the following function:

\[ u(x, y) = u_0 \sin^2(\pi x/L_x) \sin^2(10\pi y/L_y). \]

The boundary conditions are satisfied by construction. The source term is:

\[\displaystyle \frac{2 \pi^{2} k u_{0} \left(- L_{x}^{2} \sin^{2}{\left(\frac{\pi x}{L_{x}} \right)} \cos{\left(\frac{2 \pi y}{L_{y}} \right)} - L_{y}^{2} \sin^{2}{\left(\frac{\pi y}{L_{y}} \right)} \cos{\left(\frac{2 \pi x}{L_{x}} \right)}\right)}{L_{x}^{2} L_{y}^{2}}\]

This is a common trick for constructing exact solutions. Remember it!

Let’s use the following parameters:

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

Enforcing the boundary conditions#

We are going to solve this boundary value problem using PINNS. Our model will be:

\[ u(x,y) = x(1-x)y(1-y)\text{MLP}(x, y), \]

where \(\text{MLP}(x, y)\) is a multi-layer perceptron, and \(x(1-x)y(1-y)\) is a polynomial that satisfies the boundary conditions. This is also a common trick for enforcing boundary conditions in PINNS.

Multi-layer perceptron#

We will be using a simple multi-layer perceptron to represent the solution of a PDE. The architecture of the network is given by:

\[ \text{MLP}(\mathbf{x}) = z^{(L)}, \]

where

\[ z^{(0)} = \mathbf{x} = (x,y), \quad z^{(l)} = g^{(l)}(W^{(l)}z^{(l-1)} + b^{(l)}), \quad l=1,\ldots,L-1, \]

The terms \(W^{(l)}\) and \(b^{(l)}\) are the weights and biases of the \(l\)-th layer, and \(g^{(l)}\) is the activation function of the \(l\)-th layer. The parameters \(\theta\) are the weights and biases of the network:

\[ \theta = \left\{\left(W^{(l)}, b^{(l)}\right)\right\}_{l=1}^L. \]

For PINNS applications, it is recommended that:

We use the tanh activation function all layers.
We use 128 to 512 neurons per layer.
We use 3 to 5 layers.
Use the Glorot initialization (this is the default in equinox).

Let’s make the model:

import equinox as eqx
import jax.numpy as jnp
import jax.random as jrandom

key = jrandom.PRNGKey(0)
key, subkey = jrandom.split(key)
# MLP parameters
width_size = 128
depth = 4
mlp = eqx.nn.MLP(2, 1, width_size, depth, jnp.tanh, key=subkey)

# This is the parameterization of the solution that satisfies the boundary conditions
u_hat = lambda x, y, mlp: x * (Lx - x) * y * (Ly - y) * mlp(jnp.array([x, y]))[0]

Let’s see how it looks like before we train it:

../_images/5ca77cae60422370412875c4f75537468590b542d32570ac33471754f3b02508.svg

Setting up the loss function#

We are going to train this by minimizing the following loss function:

\[ \mathcal{L}(\theta) = \int_{\Omega} \left[ \Delta u(x, y) + f(x, y) \right]^2 \, dx \, dy. \]

The first step, is to turn the loss function into an expectation. Construct the random variable \(\mathbf{X} = (X,Y)\) uniformly distributed in \(\Omega\). Then, the loss function can be written as:

\[ \mathcal{L}(\theta) = |\Omega|\mathbb{E}\left[ \left[ \Delta u(\mathbf{X}) + f(\mathbf{X}) \right]^2 \right]. \]

Notice that the volume of \(\Omega\), \(\Omega|\), appears as a constant factor. Where did it come from? Recall the probability density function of \(\mathbf{X}\) is:

\[ p(\mathbf{x}) = \frac{1}{|\Omega|} \quad \text{if} \quad \mathbf{x} \in \Omega, \quad \text{and} \quad 0 \quad \text{otherwise}. \]

So:

\[ \int_{\Omega} \left[ \Delta u(\mathbf{x}) + f(\mathbf{x}) \right]^2 \, d\mathbf{x} = \int_{\Omega} \left[ \Delta u(\mathbf{x}) + f(\mathbf{x}) \right]^2 |\Omega|p(\mathbf{x}) \, d\mathbf{x} = |\Omega|\mathbb{E}\left[ \left[ \Delta u(\mathbf{X}) + f(\mathbf{X}) \right]^2 \right]. \]

We can approximate this expectation by sampling a finite number of points from \(\Omega\) and averaging the loss function over these points. This is what we will be doing in each iteration of the training process.

from jax import grad

# First order derivatives
u_x = grad(u_hat, argnums=0)
u_y = grad(u_hat, argnums=1)

# Second order deerivatives
u_xx = grad(u_x, argnums=0)
u_yy = grad(u_y, argnums=1)

# The Laplacian
D2_u = lambda x, y, mlp: u_xx(x, y, mlp) + u_yy(x, y, mlp)

# The source term
source_term = 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)

# The PDE residual (vectorize)
pde_residual = vmap(
    lambda x, y, mlp: D2_u(x, y, mlp) + source_term(x, y),
    in_axes=(0, 0, None))

# Finally the loss function:
loss = lambda mlp, x, y: Lx * Ly * jnp.mean(jnp.square(pde_residual(x, y, mlp)))

Training the model#

We will use the ADAM optimizer with standard parameters. This is the training algorithm:

def train(
        loss,
        mlp,
        key,
        optimizer,
        Lx=1.0,
        Ly=1.0,
        num_collocation_residual=512,
        num_iter=10_000,
        freq=1,
    ):

    @eqx.filter_jit
    def step(opt_state, mlp, xs, ys):
        value, grads = eqx.filter_value_and_grad(loss)(mlp, xs, ys)
        updates, opt_state = optimizer.update(grads, opt_state)
        mlp = eqx.apply_updates(mlp, updates)
        return mlp, opt_state, value
    
    opt_state = optimizer.init(eqx.filter(mlp, eqx.is_inexact_array))
    
    losses = []
    for i in range(num_iter):
        key, subkey = jrandom.split(key)
        xb = jrandom.uniform(subkey, (num_collocation_residual,), maxval=Lx)
        key, subkey = jrandom.split(key)
        yb = jrandom.uniform(subkey, (num_collocation_residual,), maxval=Ly)
        mlp, opt_state, value = step(opt_state, mlp, xb, yb)
        if i % freq == 0:
            losses.append(value)
            print(f"Step {i}, residual loss {value:.3e}")
    return mlp, losses

And this is the actual trainning:

import optax

optimizer = optax.adam(1e-3)
trained_mlp, losses = train(loss, mlp, key, optimizer, num_collocation_residual=256, num_iter=1_000, freq=100, Lx=Lx, Ly=Ly)

Step 0, residual loss 1.694e+12
Step 100, residual loss 1.799e+12
Step 200, residual loss 1.611e+12
Step 300, residual loss 1.676e+12
Step 400, residual loss 1.715e+12
Step 500, residual loss 1.520e+12
Step 600, residual loss 1.874e+12
Step 700, residual loss 2.111e+12
Step 800, residual loss 1.660e+12
Step 900, residual loss 1.762e+12

Let’s visualize first the evolution of the loss function:

../_images/1ce1e39b17d4d7970c48b924e717fc9d21044495950c9a95f80ce11540d946c0.svg

It’s not working! The loss function is not decreasing. Here is also the trained “solution:”

../_images/7501020c1de36e2ca47cecbc25e00d452a3ddcdd779772865edc540c6d8ed16a.svg

Non-dimensionalization of Partial Differential Equations#

The problem is that the initialization of the network weights is completely off. Glorot initialization is assuming that the input data and the output data are scaled to be around zero (scale of about 1). To get our problem to that scale, we need to non-dimensionalize it.

We start by picking characteristic lengthscales. Here, it makes sense to pick \(L_x\) and \(L_y\) as characteristic lengthscales. We also need to pick a characteristic value for the temperature, say \(u_s\).

\[ \begin{aligned} \tilde{x} = \frac{x}{L_x}, \quad \tilde{y} = \frac{y}{L_y}, \quad \tilde{u} = \frac{u}{u_s}. \end{aligned} \]

The next step is to find the partial derivatives of \(u\) with respect to \(x\) and \(y\) in terms of the non-dimensional variables:

\[ \frac{\partial u}{\partial x} = \frac{u_s}{L_x}\frac{\partial \tilde{u}}{\partial \tilde{x}}, \quad \frac{\partial u}{\partial y} = \frac{u_s}{L_y}\frac{\partial \tilde{u}}{\partial \tilde{y}}. \]

Here, we just used the chain rule.

The second derivatives are:

\[ \frac{\partial^2 u}{\partial x^2} = \frac{u_s}{L_x^2}\frac{\partial^2 \tilde{u}}{\partial \tilde{x}^2}, \quad \frac{\partial^2 u}{\partial y^2} = \frac{u_s}{L_y^2}\frac{\partial^2 \tilde{u}}{\partial \tilde{y}^2}. \]

Plugging these into the original PDE, we get:

\[ -k\left(\frac{u_s}{L_x^2}\frac{\partial^2 \tilde{u}}{\partial \tilde{x}^2} + \frac{u_s}{L_y^2}\frac{\partial^2 \tilde{u}}{\partial \tilde{y}^2}\right) = f. \]

Define now the non-dimensional source term:

\[ \tilde{f} = \frac{f}{f_s}. \]

And the non-dimensional thermal conductivity:

\[ \tilde{k}_x = \frac{ku_s}{L_x^2 f_s}, \quad \tilde{k}_y = \frac{ku_s}{L_y^2 f_s}. \]

The equation becomes:

\[ -\left(\tilde{k_x}\frac{\partial^2 \tilde{u}}{\partial \tilde{x}^2} + \tilde{k_y}\frac{\partial^2 \tilde{u}}{\partial \tilde{y}^2}\right) = \tilde{f}, \]

defined on the non-dimensional domain \(\tilde{\Omega} = [0, 1]\times [0, 1]\).

Now we need to pick \(u_s\) and \(f_s\) so that both sides of the equation are of order one. We can easily make the lef-hand side to be of order one by picking:

\[ f_s = \|f\|_{\infty} = \max_{(x, y) \in \Omega} |f(x, y)|. \]

For the left-hand side, we can pick \(u_s\) so that:

\[ \left\|-\left(\tilde{k_x}\frac{\partial^2 \tilde{u}}{\partial \tilde{x}^2} + \tilde{k_y}\frac{\partial^2 \tilde{u}}{\partial \tilde{y}^2}\right)\right\| = 1. \]

Solving for \(u_s\), we get that \(u_s\) should be of the order:

\[ u_s = \frac{f_s}{k\max\{1/L_x^2, 1/L_y^2\}\max\left\{\left\|\frac{\partial^2\tilde{u}}{\partial \tilde{x}^2}\right\|, \left\|\frac{\partial^2\tilde{u}}{\partial \tilde{y}^2}\right\|\right\}}. \]

Do not take the equality sign too seriously. All we need is that \(u_s\) is of the order of the right-hand side of the equation.

Let’s find these numbers:

# These functions convert from x to xt and y to yt and vice versa
to_x = lambda xt: xt * Lx
to_y = lambda yt: yt * Ly
to_xt = lambda x: x / Lx
to_yt = lambda y: y / Ly
Xt = to_xt(X)
Yt = to_yt(Y)

fs = jnp.abs(source_term(X.flatten(), Y.flatten())).max()

# Start a random MLP
tmlp = eqx.nn.MLP(2, 1, width_size, depth, jnp.tanh, key=subkey)
tu_hat = lambda tx, ty, tmlp: tx * (1.0 - tx) * ty * (1.0 - ty) * tmlp(jnp.array([tx, ty]))[0]
# Find the required gradients
tu_x = grad(tu_hat, argnums=0)
tu_y = grad(tu_hat, argnums=1)
tu_xx = grad(tu_x, argnums=0)
tu_yy = grad(tu_y, argnums=1)
v_tu_xx = eqx.filter_jit(vmap(u_xx, in_axes=(0, 0, None)))
v_tu_yy = eqx.filter_jit(vmap(u_yy, in_axes=(0, 0, None)))
max_u_xx = jnp.abs(v_tu_xx(Xt.flatten(), Yt.flatten(), tmlp).reshape(X.shape)).max()
max_u_yy = jnp.abs(v_tu_yy(Xt.flatten(), Yt.flatten(), tmlp).reshape(X.shape)).max()

# Calculate the scale:
us = fs / k / max(max_u_xx, max_u_yy) / max(1/Lx**2, 1/Ly**2)

print(f"Scale factor fs: {fs:.2e}")
print(f"Scale factor us: {us:.2e}")

Scale factor fs: 9.96e+06
Scale factor us: 4.40e+04

Let’s check if the left hand side is comparable to the right hand side:

lhs = lambda tx, ty, tmlp: -k * us / fs * (1.0 / Lx ** 2 * tu_xx(tx, ty, tmlp) + 1.0 / Ly ** 2 * tu_yy(tx, ty, tmlp))
v_lhs = eqx.filter_jit(vmap(lhs, in_axes=(0, 0, None)))

fig, ax = plt.subplots(1,2)
tl = v_lhs(Xt.flatten(), Yt.flatten(), tmlp).reshape(X.shape)
c = ax[0].contourf(Xt, Yt, tl, cmap='viridis')
fig.colorbar(c, ax=ax[0])
ax[0].set_xlabel(r'$\tilde{x}$')
ax[0].set_ylabel(r'$\tilde{y}$')
ax[0].set_title('Left-hand side')

tf = source_term(X.flatten(), Y.flatten()).reshape(X.shape) / fs
c = ax[1].contourf(Xt, Yt, tf, cmap='viridis')
fig.colorbar(c, ax=ax[1])
ax[1].set_xlabel(r'$\tilde{x}$')
ax[1].set_ylabel(r'$\tilde{y}$')
ax[1].set_title('Scaled source term')

sns.despine(trim=True)
plt.tight_layout();

../_images/0593d442c98155d917189fa65538fb023dcff9f3aa873ffebe7f35745d2fe297.svg

This looks pretty good as a starting point!

Here is what the scaled conductivity looks like:

tkx = (k * us) / (Lx ** 2 * fs)
tky = (k * us) / (Ly ** 2 * fs)
print(f"tkx = {tkx:.3e}, tky = {tky:.3e}") 

tkx = 4.415e+00, tky = 4.415e-02

Let’s now try training the scaled version of the problem:

tmlp = eqx.nn.MLP(2, 1, width_size, depth, jnp.tanh, key=subkey)
tu_hat = lambda tx, ty, tmlp: tx * (1.0 - tx) * ty * (1.0 - ty) * tmlp(jnp.array([tx, ty]))[0]
tu_x = grad(tu_hat, argnums=0)
tu_y = grad(tu_hat, argnums=1)
tu_xx = grad(tu_x, argnums=0)
tu_yy = grad(tu_y, argnums=1)
tilde_source_term = lambda tx, ty: source_term(to_x(tx), to_y(ty)) / fs
tpde_residual = vmap(
    lambda tx, ty, tmlp: tkx * tu_xx(tx, ty, tmlp) + tky * tu_yy(tx, ty, tmlp) + tilde_source_term(tx, ty),
    in_axes=(0, 0, None))
tloss = lambda tmlp, tx, ty: jnp.mean(jnp.square(tpde_residual(tx, ty, tmlp)))

optimizer = optax.adam(1e-3)
trained_tmlp, losses = train(tloss, tmlp, key, optimizer, num_collocation_residual=256, num_iter=2_000, freq=100, Lx=1.0, Ly=1.0)

Step 0, residual loss 1.776e-01
Step 100, residual loss 2.152e-02
Step 200, residual loss 1.609e-02
Step 300, residual loss 9.821e-03
Step 400, residual loss 8.162e-03
Step 500, residual loss 1.964e-03
Step 600, residual loss 1.231e-03
Step 700, residual loss 3.521e-04
Step 800, residual loss 2.916e-04
Step 900, residual loss 1.718e-04
Step 1000, residual loss 2.290e-04
Step 1100, residual loss 2.774e-04
Step 1200, residual loss 1.921e-04
Step 1300, residual loss 1.431e-04
Step 1400, residual loss 1.637e-04
Step 1500, residual loss 2.281e-04
Step 1600, residual loss 1.966e-04
Step 1700, residual loss 1.857e-04
Step 1800, residual loss 1.171e-04
Step 1900, residual loss 4.257e-04

Note

If you are running this on Google Colab Pro, try using an A100 GPU. On my M1 Macbook Pro, the training takes about 17.5s. On an A100 GPU, it takes about 8s.

I save this loss for the next notebook.

import numpy as np
np.savez("mlp_losses.npz", losses=losses)

Here is the loss function:

fig, ax = plt.subplots()
ax.plot(losses)
ax.set_xlabel('Iteration x 100')
ax.set_ylabel('Residual loss')
sns.despine(trim=True);

../_images/c8f150f76b8b287d9515a43f24e14ba7b90a0fdccdc20762b3e744c4c6ac89f1.svg

And here is the solution scaled back to the original domain:

v_tu_hat = vmap(tu_hat, in_axes=(0, 0, None))
fig, ax = plt.subplots(1,2)
u_pred = v_tu_hat(to_xt(X.flatten()), to_yt(Y.flatten()), trained_tmlp).reshape(X.shape)
c = ax[0].contourf(X, Y, u_pred * us, cmap='viridis',
    levels=jnp.linspace(-1, 510.0, 10))
fig.colorbar(c, ax=ax[0])
ax[0].set_xlabel('x')
ax[0].set_ylabel('y')
ax[0].set_title('Trained solution')
sns.despine(trim=True);
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_xlabel('x')
ax[1].set_ylabel('y')
ax[1].set_title('Scaled true solution')
sns.despine(trim=True)
plt.tight_layout();

../_images/c3646ee90adeab0b55a2e7807675f46d48ce4e5401b138a32483821a339f6025.svg

The point wise error is here:

fig, ax = plt.subplots()
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_title('Trained solution')
c = ax.contourf(to_xt(X), to_yt(Y), jnp.abs(u_pred * us - u_true), cmap='viridis')
plt.colorbar(c, ax=ax)
sns.despine(trim=True);

../_images/d18a6583b4eea3720d5955f31393c1d4120ddc858ecf2f3e83c28c18c0cd01aa.svg

It is very hard to achieve better error with plain MLPs.