The loss value is very low, but the predicted value is incorrect

[Tomandjob]

Dear lulu, I solved a simple partial differential equation using deepxde, but the predicted result is several orders of magnitude different from the true result. I modified the output of the network, but it did not work.

def pde(x, y): dy_xx = dde.grad.hessian(y, x) A=xy dA_x=dde.grad.jacobian(A, x) return 0.5dy_xx + dA_x

def func(x): return np.exp(-x*x)/(math.sqrt(np.pi))

def boundary_l(x, on_boundary): return on_boundary and np.isclose(x[0], -2.2)

def boundary_r(x, on_boundary): return on_boundary and np.isclose(x[0], 2.2)

geom = dde.geometry.Interval(-2.2, 2.2) bc1 = dde.icbc.DirichletBC(geom, lambda x: 0, boundary_l) bc2 = dde.icbc.DirichletBC(geom, lambda x: 0, boundary_r)

data = dde.data.PDE(geom, pde, [bc1,bc2], 8000, 2, solution=func, num_test=1000)

True value:

Predictive value:

[lululxvi]

See FAQ "Q: I failed to train the network or get the right solution, e.g., large training loss, unbalanced losses." and "Q: Implement new losses or constraints."

[Tomandjob]

See FAQ "Q: I failed to train the network or get the right solution, e.g., large training loss, unbalanced losses." and "Q: Implement new losses or constraints."

I reviewed the code. I didn't put the normalization conditions (∫(u)dx=1) in the code. I put “0.05*u(x)-1” into the pde. The predicted shape is the same as the actual shape, but the value is several times different. I changed the output, but it didn't work. I referred to Volterra_IDE, but still didn't know how to write the integral into pde.

[praksharma]

@Tomandjob First of all, please do not paste codes with errors. A=xy is not valid statement in Python. return 0.5dy_xx + dA_x is also not valid.

Also, when you paste a code use this tool . It helps is preserving the indentation.

I looked into it. This is what I get when I run the same stuff. The predictions are zero. Which is correct. It satisfies the BC as well as the PDE at each point.

Here, is the loss stat:

Training model...

Step      Train loss                        Test loss                         Test metric
0         [1.58e-02, 2.15e-02, 2.15e-02]    [1.58e-02, 2.15e-02, 2.15e-02]    []  
1000      [9.75e-08, 2.80e-09, 1.14e-10]    [9.75e-08, 2.80e-09, 1.14e-10]    []  
2000      [6.61e-08, 3.77e-09, 1.71e-08]    [6.61e-08, 3.77e-09, 1.71e-08]    []  
3000      [1.98e-07, 4.25e-07, 1.19e-08]    [1.98e-07, 4.25e-07, 1.19e-08]    []  
4000      [5.41e-06, 6.38e-06, 7.66e-06]    [5.41e-06, 6.38e-06, 7.66e-06]    []  
5000      [3.61e-07, 3.65e-07, 3.83e-07]    [3.61e-07, 3.65e-07, 3.83e-07]    []  

Best model at step 2000:
  train loss: 8.69e-08
  test loss: 8.69e-08
  test metric: []

'train' took 83.282071 s

DeepXDE always works on well-posed 1D problems. Are you sure the problem is well-posed i.e. the solution is unique?

[Tomandjob]

@praksharma Thank you for your reply. The right codes should be A=x*y and 0.5*dy_xx + dA_x.

The predicted solution of the neural network is very small and fluctuates around 0. The code is missing a normalization conditions (∫ydx=1). I put “0.05*y-1” into the pde. The shape of the predicted value curve is the same as the true value curve，but the predicted value is many times larger than the real value.

[praksharma]

@Tomandjob I don’t think the implementation of the normalisation condition is correct.

[Tomandjob]

@praksharma Yes, this is an approximate method. I don't know how to define this condition precisely yet.

[lululxvi]

@Tomandjob For the normalization, see https://github.com/lululxvi/deepxde/issues/174