PINN as a PDE Solver
Table of Contents
$${\partial^4 y \over \partial x^4} + 1 = 0, \qquad \text{where} \quad x \in [0,1]$$
$$y(0) = 0$$
$$y'(0) = 0$$
$$y''(1) = 0, \quad y'''(1) = 0$$
$$y(x) = -{1 \over 24}x^4 + {1 \over 6}x^3 - {1 \over 4}x^2$$
!pip install deepxde
# from deepxde.backend.set_default_backend import set_default_backend
# set_default_backend("tensorflow")
import deepxde as dde
import numpy as np
def dy(x, y):
return dde.grad.jacobian(y, x)
def ddy(x, y):
return dde.grad.hessian(y, x)
def dddy(x, y):
return dde.grad.jacobian(ddy(x, y), x)
def pde(x, y):
dy_xx = ddy(x, y)
dy_xxxx = dde.grad.hessian(dy_xx, x)
return dy_xxxx + 1
def boundary_left(x, on_boundary):
return on_boundary and np.isclose(x[0], 0)
def boundary_right(x, on_boundary):
return on_boundary and np.isclose(x[0], 1)
geom = dde.geometry.Interval(0, 1)
bc1 = dde.DirichletBC(geom, lambda x: 0, boundary_left) # u(0) = 0
bc2 = dde.OperatorBC(geom, lambda x, y, _: dy(x, y), boundary_left) # u'(0) = 0
bc3 = dde.OperatorBC(geom, lambda x, y, _: ddy(x, y), boundary_right) # u''(1) = 0
bc4 = dde.OperatorBC(geom, lambda x, y, _: dddy(x, y), boundary_right) # u'''(1) = 0
# Reference solution to compute the error
def true_solution(x):
return -(x ** 4) / 24 + x ** 3 / 6 - x ** 2 / 4
data = dde.data.PDE(geom,
pde,
[bc1, bc2, bc3, bc4],
num_domain = 10,
num_boundary = 2,
solution = true_solution,
num_test = 100)
layer_size = [1] + [20] * 3 + [1]
activation = "tanh"
initializer = "Glorot uniform"
net = dde.maps.FNN(layer_size, activation, initializer)
model = dde.Model(data, net)
model.compile("adam", lr = 0.001)
losshistory, train_state = model.train(epochs = 5000)
dde.saveplot(losshistory, train_state, issave = False, isplot = True)
# !pip install deepxde
import deepxde as dde
import numpy as np
import matplotlib.pyplot as plt
# from deepxde.backend.set_default_backend import set_default_backend
# set_default_backend("tensorflow")
$$\rho = 1\operatorname{kg/m^3}, \quad \mu = 1\operatorname{N\cdot s/m^2}, \quad D = 2h = 1\operatorname{m}, \quad
L = 2\operatorname{m}, \quad u_{in} = 1\operatorname{m/s}, \quad \nu = \frac{\mu}{\rho}$$
$$\quad D_h = \lim\limits_{b\to\infty} {4(2bh) \over {2b+4h}} = 4h = 2\operatorname{m}$$
$$Re = \frac{\rho u_{in} D_h}{\mu} = 2 $$
$$
\begin{align*}
u{\partial u \over \partial x} + v{\partial u \over \partial y} + {1 \over \rho}{\partial p \over \partial x} - \nu \ \left({\partial^2 u \over {\partial x^2}} + {\partial^2 u \over {\partial y^2}}\right) &= 0\\\\
u{\partial v \over \partial x} + v{\partial v \over \partial y} + {1 \over \rho}{\partial p \over \partial y} - \nu \ \left({\partial^2 v \over {\partial x^2}} + {\partial^2 v \over {\partial y^2}}\right) &= 0\\\\
{\partial u \over \partial x} + {\partial v \over \partial y} &= 0
\end{align*}
$$
$$u(x,y) = 0, \quad v(x,y) = 0 \qquad \text{at} \quad y = \frac{D}{2} \ \; \text{or} \; -\frac{D}{2}$$
$$u(-1,y) = u_{\text{in}}, \quad v(-1,y) = 0$$
$$p(1,y) = 0, \quad v(1,y) = 0$$