Solving Differential Equations with Deep Learning
Contents
15. Solving Differential Equations with Deep Learning¶
The Universal Approximation Theorem states that a neural network can approximate any function at a single hidden layer along with one input and output layer to any given precision.
An ordinary differential equation (ODE) is an equation involving functions having one variable.
In general, an ordinary differential equation looks like
where \(g(x)\) is the function to find, and \(g^{(n)}(x)\) is the \(n\)-th derivative of \(g(x)\).
The \(f\left(x, g(x), g'(x), g''(x), \, \dots \, , g^{(n)}(x)\right)\) is just a way to write that there is an expression involving \(x\) and \(g(x), \ g'(x), \ g''(x), \, \dots \, , \text{ and } g^{(n)}(x)\) on the left side of the equality sign in (1). The highest order of derivative, that is the value of \(n\), determines to the order of the equation. The equation is referred to as a \(n\)-th order ODE. Along with (1), some additional conditions of the function \(g(x)\) are typically given for the solution to be unique.
Let the trial solution \(g_t(x)\) be
where \(h_1(x)\) is a function that makes \(g_t(x)\) satisfy a given set of conditions, \(N(x,P)\) a neural network with weights and biases described by \(P\) and \(h_2(x, N(x,P))\) some expression involving the neural network. The role of the function \(h_2(x, N(x,P))\), is to ensure that the output from \(N(x,P)\) is zero when \(g_t(x)\) is evaluated at the values of \(x\) where the given conditions must be satisfied. The function \(h_1(x)\) should alone make \(g_t(x)\) satisfy the conditions.
But what about the network \(N(x,P)\)?
As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation.
For the minimization to be defined, we need to have a cost function at hand to minimize.
It is given that \(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\) should be equal to zero in (1). We can choose to consider the mean squared error as the cost function for an input \(x\). Since we are looking at one input, the cost function is just \(f\) squared. The cost function \(c\left(x, P \right)\) can therefore be expressed as
If \(N\) inputs are given as a vector \(\boldsymbol{x}\) with elements \(x_i\) for \(i = 1,\dots,N\), the cost function becomes
The neural net should then find the parameters \(P\) that minimizes the cost function in (3) for a set of \(N\) training samples \(x_i\).
To perform the minimization using gradient descent, the gradient of \(C\left(\boldsymbol{x}, P\right)\) is needed. It might happen so that finding an analytical expression of the gradient of \(C(\boldsymbol{x}, P)\) from (3) gets too messy, depending on which cost function one desires to use.
Luckily, there exists libraries that makes the job for us through automatic differentiation. Automatic differentiation is a method of finding the derivatives numerically with very high precision.
15.1. Example: Exponential decay¶
An exponential decay of a quantity \(g(x)\) is described by the equation
with \(g(0) = g_0\) for some chosen initial value \(g_0\).
The analytical solution of (4) is
Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of (4).
The program will use a neural network to solve
where \(g(0) = g_0\) with \(\gamma\) and \(g_0\) being some chosen values.
In this example, \(\gamma = 2\) and \(g_0 = 10\).
To begin with, a trial solution \(g_t(t)\) must be chosen. A general trial solution for ordinary differential equations could be
with \(h_1(x)\) ensuring that \(g_t(x)\) satisfies some conditions and \(h_2(x,N(x, P))\) an expression involving \(x\) and the output from the neural network \(N(x,P)\) with \(P \) being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer.
In this network, there are no weights and bias at the input layer, so \(P = \{ P_{\text{hidden}}, P_{\text{output}} \}\). If there are \(N_{\text{hidden} }\) neurons in the hidden layer, then \(P_{\text{hidden}}\) is a \(N_{\text{hidden} } \times (1 + N_{\text{input}})\) matrix, given that there are \(N_{\text{input}}\) neurons in the input layer.
The first column in \(P_{\text{hidden} }\) represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer. If there are \(N_{\text{output} }\) neurons in the output layer, then \(P_{\text{output}} \) is a \(N_{\text{output} } \times (1 + N_{\text{hidden} })\) matrix.
Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron.
It is given that \(g(0) = g_0\). The trial solution must fulfill this condition to be a proper solution of (6). A possible way to ensure that \(g_t(0, P) = g_0\), is to let \(F(N(x,P)) = x \cdot N(x,P)\) and \(A(x) = g_0\). This gives the following trial solution:
15.2. Reformulating the problem¶
We wish that our neural network manages to minimize a given cost function.
A reformulation of out equation, (6), must therefore be done, such that it describes the problem a neural network can solve for.
The neural network must find the set of weights and biases \(P\) such that the trial solution in (7) satisfies (6).
The trial solution
has been chosen such that it already solves the condition \(g(0) = g_0\). What remains, is to find \(P\) such that
is fulfilled as best as possible.
The left hand side and right hand side of (8) must be computed separately, and then the neural network must choose weights and biases, contained in \(P\), such that the sides are equal as best as possible. This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero. In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to \(P\) of the neural network.
This gives the following cost function our neural network must solve for:
(the notation \(\min_{P}\{ f(x, P) \}\) means that we desire to find \(P\) that yields the minimum of \(f(x, P)\))
or, in terms of weights and biases for the hidden and output layer in our network:
for an input value \(x\).
If the neural network evaluates \(g_t(x, P)\) at more values for \(x\), say \(N\) values \(x_i\) for \(i = 1, \dots, N\), then the total error to minimize becomes
Letting \(\boldsymbol{x}\) be a vector with elements \(x_i\) and \(C(\boldsymbol{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2\) denote the cost function, the minimization problem that our network must solve, becomes
In terms of \(P_{\text{hidden} }\) and \(P_{\text{output} }\), this could also be expressed as
For simplicity, it is assumed that the input is an array \(\boldsymbol{x} = (x_1, \dots, x_N)\) with \(N\) elements. It is at these points the neural network should find \(P\) such that it fulfills (9).
First, the neural network must feed forward the inputs. This means that \(\boldsymbol{x}s\) must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further. The input layer will consist of \(N_{\text{input} }\) neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be \(N_{\text{hidden} }\).
For the \(i\)-th in the hidden layer with weight \(w_i^{\text{hidden} }\) and bias \(b_i^{\text{hidden} }\), the weighting from the \(j\)-th neuron at the input layer is:
The result after weighting the inputs at the \(i\)-th hidden neuron can be written as a vector:
The vector \(\boldsymbol{p}_{i, \text{hidden}}^T\) constitutes each row in \(P_{\text{hidden} }\), which contains the weights for the neural network to minimize according to (9).
After having found \(\boldsymbol{z}_{i}^{\text{hidden}} \) for every \(i\)-th neuron within the hidden layer, the vector will be sent to an activation function \(a_i(\boldsymbol{z})\).
In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron:
It is possible to use other activations functions for the hidden layer also.
The output \(\boldsymbol{x}_i^{\text{hidden}}\) from each \(i\)-th hidden neuron is:
The outputs \(\boldsymbol{x}_i^{\text{hidden} } \) are then sent to the output layer.
The output layer consists of one neuron in this case, and combines the output from each of the neurons in the hidden layers. The output layer combines the results from the hidden layer using some weights \(w_i^{\text{output}}\) and biases \(b_i^{\text{output}}\). In this case, it is assumes that the number of neurons in the output layer is one.
The procedure of weighting the output neuron \(j\) in the hidden layer to the \(i\)-th neuron in the output layer is similar as for the hidden layer described previously.
Expressing \(z_{1,j}^{\text{output}}\) as a vector gives the following way of weighting the inputs from the hidden layer:
In this case we seek a continuous range of values since we are approximating a function. This means that after computing \(\boldsymbol{z}_{1}^{\text{output}}\) the neural network has finished its feed forward step, and \(\boldsymbol{z}_{1}^{\text{output}}\) is the final output of the network.
The next step is to decide how the parameters should be changed such that they minimize the cost function.
The chosen cost function for this problem is
In order to minimize the cost function, an optimization method must be chosen.
Here, gradient descent with a constant step size has been chosen.
15.3. Gradient descent¶
The idea of the gradient descent algorithm is to update parameters in a direction where the cost function decreases goes to a minimum.
In general, the update of some parameters \(\boldsymbol{\omega}\) given a cost function defined by some weights \(\boldsymbol{\omega}\), \(C(\boldsymbol{x}, \boldsymbol{\omega})\), goes as follows:
for a number of iterations or until \( \big|\big| \boldsymbol{\omega}_{\text{new} } - \boldsymbol{\omega} \big|\big|\) becomes smaller than some given tolerance.
The value of \(\lambda\) decides how large steps the algorithm must take in the direction of \( \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega})\). The notation \(\nabla_{\boldsymbol{\omega}}\) express the gradient with respect to the elements in \(\boldsymbol{\omega}\).
In our case, we have to minimize the cost function \(C(\boldsymbol{x}, P)\) with respect to the two sets of weights and biases, that is for the hidden layer \(P_{\text{hidden} }\) and for the output layer \(P_{\text{output} }\) .
This means that \(P_{\text{hidden} }\) and \(P_{\text{output} }\) is updated by
15.4. The code for solving the ODE¶
%matplotlib inline
import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt
def sigmoid(z):
return 1/(1 + np.exp(-z))
# Assuming one input, hidden, and output layer
def neural_network(params, x):
# Find the weights (including and biases) for the hidden and output layer.
# Assume that params is a list of parameters for each layer.
# The biases are the first element for each array in params,
# and the weights are the remaning elements in each array in params.
w_hidden = params[0]
w_output = params[1]
# Assumes input x being an one-dimensional array
num_values = np.size(x)
x = x.reshape(-1, num_values)
# Assume that the input layer does nothing to the input x
x_input = x
## Hidden layer:
# Add a row of ones to include bias
x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0)
z_hidden = np.matmul(w_hidden, x_input)
x_hidden = sigmoid(z_hidden)
## Output layer:
# Include bias:
x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0)
z_output = np.matmul(w_output, x_hidden)
x_output = z_output
return x_output
# The trial solution using the deep neural network:
def g_trial(x,params, g0 = 10):
return g0 + x*neural_network(params,x)
# The right side of the ODE:
def g(x, g_trial, gamma = 2):
return -gamma*g_trial
# The cost function:
def cost_function(P, x):
# Evaluate the trial function with the current parameters P
g_t = g_trial(x,P)
# Find the derivative w.r.t x of the neural network
d_net_out = elementwise_grad(neural_network,1)(P,x)
# Find the derivative w.r.t x of the trial function
d_g_t = elementwise_grad(g_trial,0)(x,P)
# The right side of the ODE
func = g(x, g_t)
err_sqr = (d_g_t - func)**2
cost_sum = np.sum(err_sqr)
return cost_sum / np.size(err_sqr)
# Solve the exponential decay ODE using neural network with one input, hidden, and output layer
def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb):
## Set up initial weights and biases
# For the hidden layer
p0 = npr.randn(num_neurons_hidden, 2 )
# For the output layer
p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included
P = [p0, p1]
print('Initial cost: %g'%cost_function(P, x))
## Start finding the optimal weights using gradient descent
# Find the Python function that represents the gradient of the cost function
# w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
cost_function_grad = grad(cost_function,0)
# Let the update be done num_iter times
for i in range(num_iter):
# Evaluate the gradient at the current weights and biases in P.
# The cost_grad consist now of two arrays;
# one for the gradient w.r.t P_hidden and
# one for the gradient w.r.t P_output
cost_grad = cost_function_grad(P, x)
P[0] = P[0] - lmb * cost_grad[0]
P[1] = P[1] - lmb * cost_grad[1]
print('Final cost: %g'%cost_function(P, x))
return P
def g_analytic(x, gamma = 2, g0 = 10):
return g0*np.exp(-gamma*x)
# Solve the given problem
if __name__ == '__main__':
# Set seed such that the weight are initialized
# with same weights and biases for every run.
npr.seed(15)
## Decide the vales of arguments to the function to solve
N = 10
x = np.linspace(0, 1, N)
## Set up the initial parameters
num_hidden_neurons = 10
num_iter = 10000
lmb = 0.001
# Use the network
P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb)
# Print the deviation from the trial solution and true solution
res = g_trial(x,P)
res_analytical = g_analytic(x)
print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical)))
# Plot the results
plt.figure(figsize=(10,10))
plt.title('Performance of neural network solving an ODE compared to the analytical solution')
plt.plot(x, res_analytical)
plt.plot(x, res[0,:])
plt.legend(['analytical','nn'])
plt.xlabel('x')
plt.ylabel('g(x)')
plt.show()
Initial cost: 367.01
Final cost: 0.0666807
Max absolute difference: 0.0437499
15.6. Using forward Euler to solve the ODE¶
A straightforward way of solving an ODE numerically, is to use Euler’s method.
Euler’s method uses Taylor series to approximate the value at a function \(f\) at a step \(\Delta x\) from \(x\):
In our case, using Euler’s method to approximate the value of \(g\) at a step \(\Delta t\) from \(t\) yields
along with the condition that \(g(0) = g_0\).
Let \(t_i = i \cdot \Delta t\) where \(\Delta t = \frac{T}{N_t-1}\) where \(T\) is the final time our solver must solve for and \(N_t\) the number of values for \(t \in [0, T]\) for \(i = 0, \dots, N_t-1\).
For \(i \geq 1\), we have that
Now, if \(g_i = g(t_i)\) then
for \(i \geq 1\) and \(g_0 = g(t_0) = g(0) = g_0\).
Equation (12) could be implemented in the following way, extending the program that uses the network using Autograd:
# Assume that all function definitions from the example program using Autograd
# are located here.
if __name__ == '__main__':
npr.seed(4155)
## Decide the vales of arguments to the function to solve
Nt = 10
T = 1
t = np.linspace(0,T, Nt)
## Set up the initial parameters
num_hidden_neurons = [100,50,25]
num_iter = 1000
lmb = 1e-3
P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)
g_dnn_ag = g_trial_deep(t,P)
g_analytical = g_analytic(t)
# Find the maximum absolute difference between the solutons:
diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))
print("The max absolute difference between the solutions is: %g"%diff_ag)
plt.figure(figsize=(10,10))
plt.title('Performance of neural network solving an ODE compared to the analytical solution')
plt.plot(t, g_analytical)
plt.plot(t, g_dnn_ag[0,:])
plt.legend(['analytical','nn'])
plt.xlabel('t')
plt.ylabel('g(t)')
## Find an approximation to the funtion using forward Euler
alpha, A, g0 = get_parameters()
dt = T/(Nt - 1)
# Perform forward Euler to solve the ODE
g_euler = np.zeros(Nt)
g_euler[0] = g0
for i in range(1,Nt):
g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1]))
# Print the errors done by each method
diff1 = np.max(np.abs(g_euler - g_analytical))
diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical))
print('Max absolute difference between Euler method and analytical: %g'%diff1)
print('Max absolute difference between deep neural network and analytical: %g'%diff2)
# Plot results
plt.figure(figsize=(10,10))
plt.plot(t,g_euler)
plt.plot(t,g_analytical)
plt.plot(t,g_dnn_ag[0,:])
plt.legend(['euler','analytical','dnn'])
plt.xlabel('Time t')
plt.ylabel('g(t)')
plt.show()
Initial cost: 0.221805
/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/numpy/core/fromnumeric.py:3245: VisibleDeprecationWarning: Creating an ndarray from ragged nested sequences (which is a list-or-tuple of lists-or-tuples-or ndarrays with different lengths or shapes) is deprecated. If you meant to do this, you must specify 'dtype=object' when creating the ndarray.
return asarray(a).size
Final cost: 0.000417932
The max absolute difference between the solutions is: 0.00424909
Max absolute difference between Euler method and analytical: 0.011225
Max absolute difference between deep neural network and analytical: 0.00424909
15.7. Solving the one dimensional Poisson equation¶
The Poisson equation for \(g(x)\) in one dimension is
where \(f(x)\) is a given function for \(x \in (0,1)\).
The conditions that \(g(x)\) is chosen to fulfill, are
This equation can be solved numerically using programs where e.g Autograd and TensorFlow are used. The results from the networks can then be compared to the analytical solution. In addition, it could be interesting to see how a typical method for numerically solving second order ODEs compares to the neural networks.
Here, the function \(g(x)\) to solve for follows the equation
where \(f(x)\) is a given function, along with the chosen conditions
In this example, we consider the case when \(f(x) = (3x + x^2)\exp(x)\).
For this case, a possible trial solution satisfying the conditions could be
The analytical solution for this problem is
import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt
def sigmoid(z):
return 1/(1 + np.exp(-z))
def deep_neural_network(deep_params, x):
# N_hidden is the number of hidden layers
N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
# Assumes input x being an one-dimensional array
num_values = np.size(x)
x = x.reshape(-1, num_values)
# Assume that the input layer does nothing to the input x
x_input = x
# Due to multiple hidden layers, define a variable referencing to the
# output of the previous layer:
x_prev = x_input
## Hidden layers:
for l in range(N_hidden):
# From the list of parameters P; find the correct weigths and bias for this layer
w_hidden = deep_params[l]
# Add a row of ones to include bias
x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)
z_hidden = np.matmul(w_hidden, x_prev)
x_hidden = sigmoid(z_hidden)
# Update x_prev such that next layer can use the output from this layer
x_prev = x_hidden
## Output layer:
# Get the weights and bias for this layer
w_output = deep_params[-1]
# Include bias:
x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)
z_output = np.matmul(w_output, x_prev)
x_output = z_output
return x_output
def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
# num_hidden_neurons is now a list of number of neurons within each hidden layer
# Find the number of hidden layers:
N_hidden = np.size(num_neurons)
## Set up initial weigths and biases
# Initialize the list of parameters:
P = [None]*(N_hidden + 1) # + 1 to include the output layer
P[0] = npr.randn(num_neurons[0], 2 )
for l in range(1,N_hidden):
P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
# For the output layer
P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
print('Initial cost: %g'%cost_function_deep(P, x))
## Start finding the optimal weigths using gradient descent
# Find the Python function that represents the gradient of the cost function
# w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
cost_function_deep_grad = grad(cost_function_deep,0)
# Let the update be done num_iter times
for i in range(num_iter):
# Evaluate the gradient at the current weights and biases in P.
# The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
# in the hidden layers and output layers evaluated at x.
cost_deep_grad = cost_function_deep_grad(P, x)
for l in range(N_hidden+1):
P[l] = P[l] - lmb * cost_deep_grad[l]
print('Final cost: %g'%cost_function_deep(P, x))
return P
## Set up the cost function specified for this Poisson equation:
# The right side of the ODE
def f(x):
return (3*x + x**2)*np.exp(x)
def cost_function_deep(P, x):
# Evaluate the trial function with the current parameters P
g_t = g_trial_deep(x,P)
# Find the derivative w.r.t x of the trial function
d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)
right_side = f(x)
err_sqr = (-d2_g_t - right_side)**2
cost_sum = np.sum(err_sqr)
return cost_sum/np.size(err_sqr)
# The trial solution:
def g_trial_deep(x,P):
return x*(1-x)*deep_neural_network(P,x)
# The analytic solution;
def g_analytic(x):
return x*(1-x)*np.exp(x)
if __name__ == '__main__':
npr.seed(4155)
## Decide the vales of arguments to the function to solve
Nx = 10
x = np.linspace(0,1, Nx)
## Set up the initial parameters
num_hidden_neurons = [200,100]
num_iter = 1000
lmb = 1e-3
P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)
g_dnn_ag = g_trial_deep(x,P)
g_analytical = g_analytic(x)
# Find the maximum absolute difference between the solutons:
max_diff = np.max(np.abs(g_dnn_ag - g_analytical))
print("The max absolute difference between the solutions is: %g"%max_diff)
plt.figure(figsize=(10,10))
plt.title('Performance of neural network solving an ODE compared to the analytical solution')
plt.plot(x, g_analytical)
plt.plot(x, g_dnn_ag[0,:])
plt.legend(['analytical','nn'])
plt.xlabel('x')
plt.ylabel('g(x)')
plt.show()
Initial cost: 457.256
/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/numpy/core/fromnumeric.py:3245: VisibleDeprecationWarning: Creating an ndarray from ragged nested sequences (which is a list-or-tuple of lists-or-tuples-or ndarrays with different lengths or shapes) is deprecated. If you meant to do this, you must specify 'dtype=object' when creating the ndarray.
return asarray(a).size
Final cost: 0.00310113
The max absolute difference between the solutions is: 0.000464088
15.7.1. Comparing with a numerical scheme¶
The Poisson equation is possible to solve using Taylor series to approximate the second derivative.
Using Taylor series, the second derivative can be expressed as
where \(\Delta x\) is a small step size and \(E_{\Delta x}(x)\) being the error term.
Looking away from the error terms gives an approximation to the second derivative:
If \(x_i = i \Delta x = x_{i-1} + \Delta x\) and \(g_i = g(x_i)\) for \(i = 1,\dots N_x - 2\) with \(N_x\) being the number of values for \(x\), (15) becomes
Since we know from our problem that
along with the conditions \(g(0) = g(1) = 0\), the following scheme can be used to find an approximate solution for \(g(x)\) numerically:
for \(i = 1, \dots, N_x - 2\) where \(g_0 = g_{N_x - 1} = 0\) and \(f(x_i) = (3x_i + x_i^2)\exp(x_i)\), which is given for our specific problem.
The equation can be rewritten into a matrix equation:
which makes it possible to solve for the vector \(\boldsymbol{g}\).
We can then compare the result from this numerical scheme with the output from our network using Autograd:
import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt
def sigmoid(z):
return 1/(1 + np.exp(-z))
def deep_neural_network(deep_params, x):
# N_hidden is the number of hidden layers
N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
# Assumes input x being an one-dimensional array
num_values = np.size(x)
x = x.reshape(-1, num_values)
# Assume that the input layer does nothing to the input x
x_input = x
# Due to multiple hidden layers, define a variable referencing to the
# output of the previous layer:
x_prev = x_input
## Hidden layers:
for l in range(N_hidden):
# From the list of parameters P; find the correct weigths and bias for this layer
w_hidden = deep_params[l]
# Add a row of ones to include bias
x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)
z_hidden = np.matmul(w_hidden, x_prev)
x_hidden = sigmoid(z_hidden)
# Update x_prev such that next layer can use the output from this layer
x_prev = x_hidden
## Output layer:
# Get the weights and bias for this layer
w_output = deep_params[-1]
# Include bias:
x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)
z_output = np.matmul(w_output, x_prev)
x_output = z_output
return x_output
def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
# num_hidden_neurons is now a list of number of neurons within each hidden layer
# Find the number of hidden layers:
N_hidden = np.size(num_neurons)
## Set up initial weigths and biases
# Initialize the list of parameters:
P = [None]*(N_hidden + 1) # + 1 to include the output layer
P[0] = npr.randn(num_neurons[0], 2 )
for l in range(1,N_hidden):
P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
# For the output layer
P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
print('Initial cost: %g'%cost_function_deep(P, x))
## Start finding the optimal weigths using gradient descent
# Find the Python function that represents the gradient of the cost function
# w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
cost_function_deep_grad = grad(cost_function_deep,0)
# Let the update be done num_iter times
for i in range(num_iter):
# Evaluate the gradient at the current weights and biases in P.
# The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
# in the hidden layers and output layers evaluated at x.
cost_deep_grad = cost_function_deep_grad(P, x)
for l in range(N_hidden+1):
P[l] = P[l] - lmb * cost_deep_grad[l]
print('Final cost: %g'%cost_function_deep(P, x))
return P
## Set up the cost function specified for this Poisson equation:
# The right side of the ODE
def f(x):
return (3*x + x**2)*np.exp(x)
def cost_function_deep(P, x):
# Evaluate the trial function with the current parameters P
g_t = g_trial_deep(x,P)
# Find the derivative w.r.t x of the trial function
d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)
right_side = f(x)
err_sqr = (-d2_g_t - right_side)**2
cost_sum = np.sum(err_sqr)
return cost_sum/np.size(err_sqr)
# The trial solution:
def g_trial_deep(x,P):
return x*(1-x)*deep_neural_network(P,x)
# The analytic solution;
def g_analytic(x):
return x*(1-x)*np.exp(x)
if __name__ == '__main__':
npr.seed(4155)
## Decide the vales of arguments to the function to solve
Nx = 10
x = np.linspace(0,1, Nx)
## Set up the initial parameters
num_hidden_neurons = [200,100]
num_iter = 1000
lmb = 1e-3
P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)
g_dnn_ag = g_trial_deep(x,P)
g_analytical = g_analytic(x)
# Find the maximum absolute difference between the solutons:
plt.figure(figsize=(10,10))
plt.title('Performance of neural network solving an ODE compared to the analytical solution')
plt.plot(x, g_analytical)
plt.plot(x, g_dnn_ag[0,:])
plt.legend(['analytical','nn'])
plt.xlabel('x')
plt.ylabel('g(x)')
## Perform the computation using the numerical scheme
dx = 1/(Nx - 1)
# Set up the matrix A
A = np.zeros((Nx-2,Nx-2))
A[0,0] = 2
A[0,1] = -1
for i in range(1,Nx-3):
A[i,i-1] = -1
A[i,i] = 2
A[i,i+1] = -1
A[Nx - 3, Nx - 4] = -1
A[Nx - 3, Nx - 3] = 2
# Set up the vector f
f_vec = dx**2 * f(x[1:-1])
# Solve the equation
g_res = np.linalg.solve(A,f_vec)
g_vec = np.zeros(Nx)
g_vec[1:-1] = g_res
# Print the differences between each method
max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical))
max_diff2 = np.max(np.abs(g_vec - g_analytical))
print("The max absolute difference between the analytical solution and DNN Autograd: %g"%max_diff1)
print("The max absolute difference between the analytical solution and numerical scheme: %g"%max_diff2)
# Plot the results
plt.figure(figsize=(10,10))
plt.plot(x,g_vec)
plt.plot(x,g_analytical)
plt.plot(x,g_dnn_ag[0,:])
plt.legend(['numerical scheme','analytical','dnn'])
plt.show()
Initial cost: 457.256
/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/numpy/core/fromnumeric.py:3245: VisibleDeprecationWarning: Creating an ndarray from ragged nested sequences (which is a list-or-tuple of lists-or-tuples-or ndarrays with different lengths or shapes) is deprecated. If you meant to do this, you must specify 'dtype=object' when creating the ndarray.
return asarray(a).size
Final cost: 0.00310113
The max absolute difference between the analytical solution and DNN Autograd: 0.000464088
The max absolute difference between the analytical solution and numerical scheme: 0.00266858
15.8. Partial Differential Equations¶
A partial differential equation (PDE) has a solution here the function is defined by multiple variables. The equation may involve all kinds of combinations of which variables the function is differentiated with respect to.
In general, a partial differential equation for a function \(g(x_1,\dots,x_N)\) with \(N\) variables may be expressed as
where \(f\) is an expression involving all kinds of possible mixed derivatives of \(g(x_1,\dots,x_N)\) up to an order \(n\). In order for the solution to be unique, some additional conditions must also be given.
15.8.1. Type of problem¶
The problem our network must solve for, is similar to the ODE case. We must have a trial solution \(g_t\) at hand.
For instance, the trial solution could be expressed as
where \(h_1(x_1,\dots,x_N)\) is a function that ensures \(g_t(x_1,\dots,x_N)\) satisfies some given conditions. The neural network \(N(x_1,\dots,x_N,P)\) has weights and biases described by \(P\) and \(h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))\) is an expression using the output from the neural network in some way.
The role of the function \(h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))\), is to ensure that the output of \(N(x_1,\dots,x_N,P)\) is zero when \(g_t(x_1,\dots,x_N)\) is evaluated at the values of \(x_1,\dots,x_N\) where the given conditions must be satisfied. The function \(h_1(x_1,\dots,x_N)\) should alone make \(g_t(x_1,\dots,x_N)\) satisfy the conditions.
15.8.2. Network requirements¶
The network tries then the minimize the cost function following the same ideas as described for the ODE case, but now with more than one variables to consider. The concept still remains the same; find a set of parameters \(P\) such that the expression \(f\) in (17) is as close to zero as possible.
As for the ODE case, the cost function is the mean squared error that the network must try to minimize. The cost function for the network to minimize is
If we let \(\boldsymbol{x} = \big( x_1, \dots, x_N \big)\) be an array containing the values for \(x_1, \dots, x_N\) respectively, the cost function can be reformulated into the following:
If we also have \(M\) different sets of values for \(x_1, \dots, x_N\), that is \(\boldsymbol{x}_i = \big(x_1^{(i)}, \dots, x_N^{(i)}\big)\) for \(i = 1,\dots,M\) being the rows in matrix \(X\), the cost function can be generalized into
15.9. Example: The diffusion equation¶
In one spatial dimension, the equation reads
where a possible choice of conditions are
with \(u(x)\) being some given function.
For this case, we want to find \(g(x,t)\) such that
and
with \(u(x) = \sin(\pi x)\).
First, let us set up the deep neural network. The deep neural network will follow the same structure as discussed in the examples solving the ODEs. First, we will look into how Autograd could be used in a network tailored to solve for bivariate functions.
The only change to do here, is to extend our network such that functions of multiple parameters are correctly handled. In this case we have two variables in our function to solve for, that is time \(t\) and position \(x\). The variables will be represented by a one-dimensional array in the program. The program will evaluate the network at each possible pair \((x,t)\), given an array for the desired \(x\)-values and \(t\)-values to approximate the solution at.
def sigmoid(z):
return 1/(1 + np.exp(-z))
def deep_neural_network(deep_params, x):
# x is now a point and a 1D numpy array; make it a column vector
num_coordinates = np.size(x,0)
x = x.reshape(num_coordinates,-1)
num_points = np.size(x,1)
# N_hidden is the number of hidden layers
N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
# Assume that the input layer does nothing to the input x
x_input = x
x_prev = x_input
## Hidden layers:
for l in range(N_hidden):
# From the list of parameters P; find the correct weigths and bias for this layer
w_hidden = deep_params[l]
# Add a row of ones to include bias
x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)
z_hidden = np.matmul(w_hidden, x_prev)
x_hidden = sigmoid(z_hidden)
# Update x_prev such that next layer can use the output from this layer
x_prev = x_hidden
## Output layer:
# Get the weights and bias for this layer
w_output = deep_params[-1]
# Include bias:
x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)
z_output = np.matmul(w_output, x_prev)
x_output = z_output
return x_output[0][0]
The cost function must then iterate through the given arrays containing values for \(x\) and \(t\), defines a point \((x,t)\) the deep neural network and the trial solution is evaluated at, and then finds the Jacobian of the trial solution.
A possible trial solution for this PDE is
with \(A(x,t)\) being a function ensuring that \(g_t(x,t)\) satisfies our given conditions, and \(N(x,t,P)\) being the output from the deep neural network using weights and biases for each layer from \(P\).
To fulfill the conditions, \(A(x,t)\) could be:
since \((0) = u(1) = 0\) and \(u(x) = \sin(\pi x)\).
The Jacobian is used because the program must find the derivative of the trial solution with respect to \(x\) and \(t\).
This gives the necessity of computing the Jacobian matrix, as we want to evaluate the gradient with respect to \(x\) and \(t\) (note that the Jacobian of a scalar-valued multivariate function is simply its gradient).
In Autograd, the differentiation is by default done with respect to the first input argument of your Python function. Since the points is an array representing \(x\) and \(t\), the Jacobian is calculated using the values of \(x\) and \(t\).
To find the second derivative with respect to \(x\) and \(t\), the Jacobian can be found for the second time. The result is a Hessian matrix, which is the matrix containing all the possible second order mixed derivatives of \(g(x,t)\).
# Set up the trial function:
def u(x):
return np.sin(np.pi*x)
def g_trial(point,P):
x,t = point
return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)
# The right side of the ODE:
def f(point):
return 0.
# The cost function:
def cost_function(P, x, t):
cost_sum = 0
g_t_jacobian_func = jacobian(g_trial)
g_t_hessian_func = hessian(g_trial)
for x_ in x:
for t_ in t:
point = np.array([x_,t_])
g_t = g_trial(point,P)
g_t_jacobian = g_t_jacobian_func(point,P)
g_t_hessian = g_t_hessian_func(point,P)
g_t_dt = g_t_jacobian[1]
g_t_d2x = g_t_hessian[0][0]
func = f(point)
err_sqr = ( (g_t_dt - g_t_d2x) - func)**2
cost_sum += err_sqr
return cost_sum
15.9.1. Setting up the network using Autograd; The full program¶
Having set up the network, along with the trial solution and cost function, we can now see how the deep neural network performs by comparing the results to the analytical solution.
The analytical solution of our problem is
A possible way to implement a neural network solving the PDE, is given below. Be aware, though, that it is fairly slow for the parameters used. A better result is possible, but requires more iterations, and thus longer time to complete.
Indeed, the program below is not optimal in its implementation, but rather serves as an example on how to implement and use a neural network to solve a PDE. Using TensorFlow results in a much better execution time. Try it!
import autograd.numpy as np
from autograd import jacobian,hessian,grad
import autograd.numpy.random as npr
from matplotlib import cm
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import axes3d
## Set up the network
def sigmoid(z):
return 1/(1 + np.exp(-z))
def deep_neural_network(deep_params, x):
# x is now a point and a 1D numpy array; make it a column vector
num_coordinates = np.size(x,0)
x = x.reshape(num_coordinates,-1)
num_points = np.size(x,1)
# N_hidden is the number of hidden layers
N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
# Assume that the input layer does nothing to the input x
x_input = x
x_prev = x_input
## Hidden layers:
for l in range(N_hidden):
# From the list of parameters P; find the correct weigths and bias for this layer
w_hidden = deep_params[l]
# Add a row of ones to include bias
x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)
z_hidden = np.matmul(w_hidden, x_prev)
x_hidden = sigmoid(z_hidden)
# Update x_prev such that next layer can use the output from this layer
x_prev = x_hidden
## Output layer:
# Get the weights and bias for this layer
w_output = deep_params[-1]
# Include bias:
x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)
z_output = np.matmul(w_output, x_prev)
x_output = z_output
return x_output[0][0]
## Define the trial solution and cost function
def u(x):
return np.sin(np.pi*x)
def g_trial(point,P):
x,t = point
return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)
# The right side of the ODE:
def f(point):
return 0.
# The cost function:
def cost_function(P, x, t):
cost_sum = 0
g_t_jacobian_func = jacobian(g_trial)
g_t_hessian_func = hessian(g_trial)
for x_ in x:
for t_ in t:
point = np.array([x_,t_])
g_t = g_trial(point,P)
g_t_jacobian = g_t_jacobian_func(point,P)
g_t_hessian = g_t_hessian_func(point,P)
g_t_dt = g_t_jacobian[1]
g_t_d2x = g_t_hessian[0][0]
func = f(point)
err_sqr = ( (g_t_dt - g_t_d2x) - func)**2
cost_sum += err_sqr
return cost_sum /( np.size(x)*np.size(t) )
## For comparison, define the analytical solution
def g_analytic(point):
x,t = point
return np.exp(-np.pi**2*t)*np.sin(np.pi*x)
## Set up a function for training the network to solve for the equation
def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):
## Set up initial weigths and biases
N_hidden = np.size(num_neurons)
## Set up initial weigths and biases
# Initialize the list of parameters:
P = [None]*(N_hidden + 1) # + 1 to include the output layer
P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias
for l in range(1,N_hidden):
P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
# For the output layer
P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
print('Initial cost: ',cost_function(P, x, t))
cost_function_grad = grad(cost_function,0)
# Let the update be done num_iter times
for i in range(num_iter):
cost_grad = cost_function_grad(P, x , t)
for l in range(N_hidden+1):
P[l] = P[l] - lmb * cost_grad[l]
print('Final cost: ',cost_function(P, x, t))
return P
if __name__ == '__main__':
### Use the neural network:
npr.seed(15)
## Decide the vales of arguments to the function to solve
Nx = 10; Nt = 10
x = np.linspace(0, 1, Nx)
t = np.linspace(0,1,Nt)
## Set up the parameters for the network
num_hidden_neurons = [100, 25]
num_iter = 250
lmb = 0.01
P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)
## Store the results
g_dnn_ag = np.zeros((Nx, Nt))
G_analytical = np.zeros((Nx, Nt))
for i,x_ in enumerate(x):
for j, t_ in enumerate(t):
point = np.array([x_, t_])
g_dnn_ag[i,j] = g_trial(point,P)
G_analytical[i,j] = g_analytic(point)
# Find the map difference between the analytical and the computed solution
diff_ag = np.abs(g_dnn_ag - G_analytical)
print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag))
## Plot the solutions in two dimensions, that being in position and time
T,X = np.meshgrid(t,x)
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection='3d')
ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))
s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel('Time $t$')
ax.set_ylabel('Position $x$');
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection='3d')
ax.set_title('Analytical solution')
s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel('Time $t$')
ax.set_ylabel('Position $x$');
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection='3d')
ax.set_title('Difference')
s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel('Time $t$')
ax.set_ylabel('Position $x$');
## Take some slices of the 3D plots just to see the solutions at particular times
indx1 = 0
indx2 = int(Nt/2)
indx3 = Nt-1
t1 = t[indx1]
t2 = t[indx2]
t3 = t[indx3]
# Slice the results from the DNN
res1 = g_dnn_ag[:,indx1]
res2 = g_dnn_ag[:,indx2]
res3 = g_dnn_ag[:,indx3]
# Slice the analytical results
res_analytical1 = G_analytical[:,indx1]
res_analytical2 = G_analytical[:,indx2]
res_analytical3 = G_analytical[:,indx3]
# Plot the slices
plt.figure(figsize=(10,10))
plt.title("Computed solutions at time = %g"%t1)
plt.plot(x, res1)
plt.plot(x,res_analytical1)
plt.legend(['dnn','analytical'])
plt.figure(figsize=(10,10))
plt.title("Computed solutions at time = %g"%t2)
plt.plot(x, res2)
plt.plot(x,res_analytical2)
plt.legend(['dnn','analytical'])
plt.figure(figsize=(10,10))
plt.title("Computed solutions at time = %g"%t3)
plt.plot(x, res3)
plt.plot(x,res_analytical3)
plt.legend(['dnn','analytical'])
plt.show()
/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/numpy/core/fromnumeric.py:3245: VisibleDeprecationWarning: Creating an ndarray from ragged nested sequences (which is a list-or-tuple of lists-or-tuples-or ndarrays with different lengths or shapes) is deprecated. If you meant to do this, you must specify 'dtype=object' when creating the ndarray.
return asarray(a).size
Initial cost: 41.05505310046363
---------------------------------------------------------------------------
KeyboardInterrupt Traceback (most recent call last)
Input In [9], in <cell line: 129>()
140 num_iter = 250
141 lmb = 0.01
--> 143 P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)
145 ## Store the results
146 g_dnn_ag = np.zeros((Nx, Nt))
Input In [9], in solve_pde_deep_neural_network(x, t, num_neurons, num_iter, lmb)
118 # Let the update be done num_iter times
119 for i in range(num_iter):
--> 120 cost_grad = cost_function_grad(P, x , t)
122 for l in range(N_hidden+1):
123 P[l] = P[l] - lmb * cost_grad[l]
File ~/miniforge3/envs/myenv/lib/python3.9/site-packages/autograd/wrap_util.py:20, in unary_to_nary.<locals>.nary_operator.<locals>.nary_f(*args, **kwargs)
18 else:
19 x = tuple(args[i] for i in argnum)
---> 20 return unary_operator(unary_f, x, *nary_op_args, **nary_op_kwargs)
File ~/miniforge3/envs/myenv/lib/python3.9/site-packages/autograd/differential_operators.py:29, in grad(fun, x)
26 if not vspace(ans).size == 1:
27 raise TypeError("Grad only applies to real scalar-output functions. "
28 "Try jacobian, elementwise_grad or holomorphic_grad.")
---> 29 return vjp(vspace(ans).ones())
File ~/miniforge3/envs/myenv/lib/python3.9/site-packages/autograd/core.py:14, in make_vjp.<locals>.vjp(g)
---> 14 def vjp(g): return backward_pass(g, end_node)
File ~/miniforge3/envs/myenv/lib/python3.9/site-packages/autograd/core.py:23, in backward_pass(g, end_node)
21 ingrads = node.vjp(outgrad[0])
22 for parent, ingrad in zip(node.parents, ingrads):
---> 23 outgrads[parent] = add_outgrads(outgrads.get(parent), ingrad)
24 return outgrad[0]
KeyboardInterrupt:
15.10. Solving the wave equation with Neural Networks¶
The wave equation is
with \(c\) being the specified wave speed.
Here, the chosen conditions are
where \(\frac{\partial g(x,t)}{\partial t} \Big |_{t = 0}\) means the derivative of \(g(x,t)\) with respect to \(t\) is evaluated at \(t = 0\), and \(u(x)\) and \(v(x)\) being given functions.
The wave equation to solve for, is
where \(c\) is the given wave speed. The chosen conditions for this equation are
In this example, let \(c = 1\) and \(u(x) = \sin(\pi x)\) and \(v(x) = -\pi\sin(\pi x)\).
Setting up the network is done in similar matter as for the example of solving the diffusion equation. The only things we have to change, is the trial solution such that it satisfies the conditions from (20) and the cost function.
The trial solution becomes slightly different since we have other conditions than in the example of solving the diffusion equation. Here, a possible trial solution \(g_t(x,t)\) is
where
Note that this trial solution satisfies the conditions only if \(u(0) = v(0) = u(1) = v(1) = 0\), which is the case in this example.
The analytical solution for our specific problem, is
import autograd.numpy as np
from autograd import hessian,grad
import autograd.numpy.random as npr
from matplotlib import cm
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import axes3d
## Set up the trial function:
def u(x):
return np.sin(np.pi*x)
def v(x):
return -np.pi*np.sin(np.pi*x)
def h1(point):
x,t = point
return (1 - t**2)*u(x) + t*v(x)
def g_trial(point,P):
x,t = point
return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point)
## Define the cost function
def cost_function(P, x, t):
cost_sum = 0
g_t_hessian_func = hessian(g_trial)
for x_ in x:
for t_ in t:
point = np.array([x_,t_])
g_t_hessian = g_t_hessian_func(point,P)
g_t_d2x = g_t_hessian[0][0]
g_t_d2t = g_t_hessian[1][1]
err_sqr = ( (g_t_d2t - g_t_d2x) )**2
cost_sum += err_sqr
return cost_sum / (np.size(t) * np.size(x))
## The neural network
def sigmoid(z):
return 1/(1 + np.exp(-z))
def deep_neural_network(deep_params, x):
# x is now a point and a 1D numpy array; make it a column vector
num_coordinates = np.size(x,0)
x = x.reshape(num_coordinates,-1)
num_points = np.size(x,1)
# N_hidden is the number of hidden layers
N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
# Assume that the input layer does nothing to the input x
x_input = x
x_prev = x_input
## Hidden layers:
for l in range(N_hidden):
# From the list of parameters P; find the correct weigths and bias for this layer
w_hidden = deep_params[l]
# Add a row of ones to include bias
x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)
z_hidden = np.matmul(w_hidden, x_prev)
x_hidden = sigmoid(z_hidden)
# Update x_prev such that next layer can use the output from this layer
x_prev = x_hidden
## Output layer:
# Get the weights and bias for this layer
w_output = deep_params[-1]
# Include bias:
x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)
z_output = np.matmul(w_output, x_prev)
x_output = z_output
return x_output[0][0]
## The analytical solution
def g_analytic(point):
x,t = point
return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t)
def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):
## Set up initial weigths and biases
N_hidden = np.size(num_neurons)
## Set up initial weigths and biases
# Initialize the list of parameters:
P = [None]*(N_hidden + 1) # + 1 to include the output layer
P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias
for l in range(1,N_hidden):
P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
# For the output layer
P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
print('Initial cost: ',cost_function(P, x, t))
cost_function_grad = grad(cost_function,0)
# Let the update be done num_iter times
for i in range(num_iter):
cost_grad = cost_function_grad(P, x , t)
for l in range(N_hidden+1):
P[l] = P[l] - lmb * cost_grad[l]
print('Final cost: ',cost_function(P, x, t))
return P
if __name__ == '__main__':
### Use the neural network:
npr.seed(15)
## Decide the vales of arguments to the function to solve
Nx = 10; Nt = 10
x = np.linspace(0, 1, Nx)
t = np.linspace(0,1,Nt)
## Set up the parameters for the network
num_hidden_neurons = [50,20]
num_iter = 1000
lmb = 0.01
P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)
## Store the results
res = np.zeros((Nx, Nt))
res_analytical = np.zeros((Nx, Nt))
for i,x_ in enumerate(x):
for j, t_ in enumerate(t):
point = np.array([x_, t_])
res[i,j] = g_trial(point,P)
res_analytical[i,j] = g_analytic(point)
diff = np.abs(res - res_analytical)
print("Max difference between analytical and solution from nn: %g"%np.max(diff))
## Plot the solutions in two dimensions, that being in position and time
T,X = np.meshgrid(t,x)
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection='3d')
ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))
s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel('Time $t$')
ax.set_ylabel('Position $x$');
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection='3d')
ax.set_title('Analytical solution')
s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel('Time $t$')
ax.set_ylabel('Position $x$');
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection='3d')
ax.set_title('Difference')
s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel('Time $t$')
ax.set_ylabel('Position $x$');
## Take some slices of the 3D plots just to see the solutions at particular times
indx1 = 0
indx2 = int(Nt/2)
indx3 = Nt-1
t1 = t[indx1]
t2 = t[indx2]
t3 = t[indx3]
# Slice the results from the DNN
res1 = res[:,indx1]
res2 = res[:,indx2]
res3 = res[:,indx3]
# Slice the analytical results
res_analytical1 = res_analytical[:,indx1]
res_analytical2 = res_analytical[:,indx2]
res_analytical3 = res_analytical[:,indx3]
# Plot the slices
plt.figure(figsize=(10,10))
plt.title("Computed solutions at time = %g"%t1)
plt.plot(x, res1)
plt.plot(x,res_analytical1)
plt.legend(['dnn','analytical'])
plt.figure(figsize=(10,10))
plt.title("Computed solutions at time = %g"%t2)
plt.plot(x, res2)
plt.plot(x,res_analytical2)
plt.legend(['dnn','analytical'])
plt.figure(figsize=(10,10))
plt.title("Computed solutions at time = %g"%t3)
plt.plot(x, res3)
plt.plot(x,res_analytical3)
plt.legend(['dnn','analytical'])
plt.show()