2D Poisson equation with variable coefficient

[tangqi]

Hi, I am interested in solving some practical problems with your solver. As a first test, I would like to solve -div(coeff*grad y) = 1, where coeff=1+x1+x2. (a similar 1D case works fine.) The computational domain is [0,1]^2.

Here is the simple code:

def main():
    def pde(x, y):
        x1=x[:,0]
        x2=x[:,1]
        coeff=1+x1+x2
        dy_x = tf.gradients(y, x)[0]
        dy_x, dy_y = dy_x[:, 0:1], dy_x[:, 1:]
        dy_xx = tf.gradients(coeff*dy_x, x)[0][:, 0:1]
        dy_yy = tf.gradients(coeff*dy_y, x)[0][:, 1:]
        return -dy_xx-dy_yy-1

    def boundary(x, on_boundary):
        return on_boundary

    def func(x):
        return np.zeros([len(x), 1])

    geom = dde.geometry.Rectangle([0,0],[1,1])
    bc = dde.DirichletBC(geom, func, boundary)

    data = dde.data.PDE(
        geom, 1, pde, bc, num_domain=1200, num_boundary=120, num_test=1500
    )
    net = dde.maps.FNN([2] + [50] * 4 + [1], "tanh", "Glorot uniform")
    model = dde.Model(data, net)

    model.compile("adam", lr=0.001)
    model.train(epochs=50000)
    model.compile("L-BFGS-B")
    losshistory, train_state = model.train()
    dde.saveplot(losshistory, train_state, issave=True, isplot=True)

But the solver converges to the wrong solution. One obvious mistake is the boundary condition is not 0 and also the solution is quite off. I also tried the non conserved version of the problem, it give me an even worse solution.

The solver has no issue to solve a problem of -dy_xx-dy_yy-dy_x-dy_y=1, which is very close to the problem I failed. Therefore, I do not quite understand why the solver fails so badly for the problem I set up. I am new to tensorflow, so it is likely I may miss something obvious. Any suggestions?

Thanks!
Qi

[lululxvi]

Could you change x1=x[:,0] and x2=x[:,1] to x1=x[:, 0:1] and x2=x[:, 1:], so that x1 and x2 are column vectors?

[tangqi]

Thanks a lot! That appears to be the bug.

Do you happen to have a 3D case or a time-dependent 2D example by any chance, or have you ever tried it?

-Qi

[lululxvi]

Yes, we have time-dependent 2D example, but the code is not online.

[tangqi]

Great! Looking forward to it. Thanks again for the help.

[tangqi]

Hi Lu, is there any chance to share a simple 2D time dependent case, like the one in the arxiv paper for us to try? Thanks, -Qi

[lululxvi]

I will upload these days. 2D time case should be very similar to 1D case.

[tangqi]

Thank you! That would be great!

[Fayaud]

Hello Lulu!

Please where can I find an example of a time dependent 2D domain.

best regards,

[Fayaud]

Hello Lulu!

Please is this the right syntax to fix the time-periodicity condition for a PDE problem in 2D in space?

Spatial_domain = dde.geometry.Rectangle(xmin=[-1, -1], xmax=[1, 1])
Temporal_domain = dde.geometry.TimeDomain(0, 2*np.pi)
Spatio_temporal_domain = dde.geometry.GeometryXTime(Spatial_domain, Temporal_domain)
boundary_condition_u = dde.DirichletBC(Spatio_temporal_domain, u_func, lambda _, on_boundary: on_boundary, component=0)
boundary_condition_v = dde.DirichletBC(Spatio_temporal_domain, v_func, lambda _, on_boundary: on_boundary, component=1)

periodic_condition_u = dde.PeriodicBC(Spatio_temporal_domain, 1, lambda _, on_boundary: on_boundary, component=0)???
periodic_condition_v = dde.PeriodicBC(Spatio_temporal_domain, 1, lambda _, on_boundary: on_boundary, component=1)???

I only want to check the two last line for: u(x, y, 0) = u(x, y, 2pi); v(x, y, 0) = v(x, y, 2pi)

thank you very much!

best,

[Fayaud]

Or should I instead use this syntax for the time-periodicity for a problem in 2D in space ?

periodic_condition_u = dde.PeriodicBC(Spatio_temporal_domain, 1, lambda _, on_periodic: on_periodic, component=0)???
periodic_condition_v = dde.PeriodicBC(Spatio_temporal_domain, 1, lambda _, on_periodic: on_periodic, component=1)???

Thank you very much!

best,

[lululxvi]

For the periodic in x and y, why "u(x, y, 0) = u(x, y, 2pi); v(x, y, 0) = v(x, y, 2pi)"? Should it be

• u(-1, y, t) = u(1, y, t), v(-1, y, t) = v(1, y, t)
• u(x, -1, 0) = u(x, 1, t), v(x, -1, 0) = v(x, 1, t)
  ?

[Fayaud]

Hello Professor!

Thank you for replying. I want to enforce the periodic in time. So, my problem is in 2 dimension in space and the third dimension is the time as: (x, y, t). I actually want to write:

u(x, y, 0) = u(x, y, 2pi); v(x, y, 0) = v(x, y, 2pi) for all (x, y).

best regards!

Fayaud!

[Fayaud]

For more precision, I am solving the following time-dependent Navier-Stokes equation:

File.pdf

best,