Implementing piece-wise PDEs

[andyxzhu]

Hi @lululxvi, thanks for the excellent library.

I am interested in creating a PDE to be learned by the neural network that depends on an input condition.

For instance, from example ode_system.py:

def ode_system(x, y):
    """ODE system.
    dy1/dx = y2
    dy2/dx = -y1
    """
    y1, y2 = y[:, 0:1], y[:, 1:]
    dy1_x = dde.grad.jacobian(y, x, i=0)
    dy2_x = dde.grad.jacobian(y, x, i=1)
    return [dy1_x - y2, dy2_x + y1]

This solves dy1/dx = y2 and dy2/dx = -y1.

What if I wanted it to solve those equations if x in [0,10], then solve for a different condition like dy1/dx = 0 and dy2/dx = 1 if x in (10, 100] using y1(10) and y2(10) as the new "initial" conditions?

Is there a way to get just one network to learn this instead of training multiple networks for each piece? I tried implementing this by changing my PDE function, but eager execution is disabled even though I'm running TF 2.

Many thanks.

[lululxvi]

You need to combine these two cases in one ODE as dy1/dt = f1(t, y1, y2) and dy2/dt = f2(t, y1, y2), and then everything is the same. There might be more than one solutions to define f1 and f2, such that f1 and f2 depend on t. One way is that you can use tf.math.sign (https://www.tensorflow.org/api_docs/python/tf/math/sign) to define f1 and f2.

[andyxzhu]

Could you clarify how to use tf.sign() to define f1 and f2 on arbitrary connected intervals, such as [0,10] and (10, 20]? Right now I have this:

def ode_system(x, y):
    """ODE system.
    dy1/dx = y2
    dy2/dx = -y1
    """
    y1, y2 = y[:, 0:1], y[:, 1:]
    dy1_x = dde.grad.jacobian(y, x, i=0)
    dy2_x = dde.grad.jacobian(y, x, i=1)
    signs = tf.sign(x)
    return [dy1_x - signs*y2, dy2_x + signs*y1]

The only way I can see this being useful is if I wanted to define dy1/dx = y2 and dy2/dx = -y1 on a positive interval like (0, 10), then the negative version dy1/dx = -y2 and dy2/dx = y1 on a negative interval like (-10, 0) (this is what I got when I solved it from (-10, 10) with IC f1(-10) = sin(-10) and f2(-10) = -cos(-10) ).

Can I instead use tf.sign() to make it solve any function on any interval, though? Thanks!

[lululxvi]

You can use tf.sign(x-a) if the break point is a.

[andyxzhu]

Thank you for the advice, but I've found an alternate approach with tf.where. Cheers!

[engsbk]

@azhu529 Did you have to change the last layer of your network to be [2] instead [1] for two different outputs since you have a piecewise function in the output?

as in:
net = dde.maps.FNN([2] + [35] * 4 + [2], "tanh", "Glorot normal")
instead of:
net = dde.maps.FNN([2] + [35] * 4 + [1], "tanh", "Glorot normal")

[andyxzhu]

Yes, the output layer was [2].