error value at boundary

[sumantkrsoni]

while solving simple ODE example [ dy/dx = 0, y(0) =0, domain =[0,1] ]

the solution has a nice agreement with the exact solution but the error value for boundary condition is not equal to zero (approx error at that point is 0.00001).

I think, boundary value are well known for Diff eq, and the error should be equal to zero at that boundary point.

[smao-astro]

Hi @sanjusoni , maybe you are interested in this paper https://arxiv.org/abs/1904.06619.

[sumantkrsoni]

Thanks, I'll ask further in case of not getting my answer from this research paper.

[lululxvi]

@smao-astro Thanks for providing the paper.

The idea is actually simple, see the second paragraph on Page 6 in the DeepXDE paper (https://arxiv.org/abs/1907.04502).

@sanjusoni For your problem, by default DeepXDE uses a soft constraint for the IC. If you want a hard constraint, i.e., y(0) = 0, you can multiply the network output N(x) by x, i.e., the final network N'(x) is N'(x) = x N(x). Obviously, N'(0) = 0. In DeepXDE, use the following code:

net.outputs_modify(lambda x, y: x * y)

Here is an another example for the diffusion equation to satisfy both IC and BC exactly:
https://github.com/lululxvi/deepxde/blob/master/examples/diffusion_1d_exactBC.py

[sumantkrsoni]

Now, I am getting the exact boundary condition satisfying at boundaries (for dy/dx = 0 , y(0) = 0 )

But, for second-order linear differential equation, I have got exact boundary values at one end only.

I have used differential equation: d2y/dx2 -1 = 0 , y(0) = 0, y(1) = 1,

for that I have written: net.outputs_modify(lambda x, y: x y)*

because it would satisfy for both condition at 0 and 1.

but result That I got is as:

I think at x = 1, the output value should have error 0.000.

[lululxvi]

Could you try

net.outputs_modify(lambda x, y: x + x * (1 - x) * y)

[sumantkrsoni]

Thannks, It give me an idea to implement.

[ryabs07]

Could you try

net.outputs_modify(lambda x, y: x + x * (1 - x) * y)

I am a little bit confused by this application of hard BC. Although the output transform function satisfies the required conditions when x = 0 or x = 1, wouldnt it also change the outcome for rest of the values of x?

[lululxvi]

See https://doi.org/10.1137/21M1397908