The flexibility of neural networks is also one of their main drawbacks: there are many hyperparameters to tweak. Not only can you use any imaginable network topology (how neurons/nodes are interconnected), but even in a simple FFNN you can change the number of layers, the number of neurons per layer, the type of activation function to use in each layer, the weight initialization logic, the stochastic gradient optmized and much more. How do you know what combination of hyperparameters is the best for your task?
However,since there are many hyperparameters to tune, and since training a neural network on a large dataset takes a lot of time, you will only be able to explore a tiny part of the hyperparameter space.
For many problems you can start with just one or two hidden layers and it will work just fine. For the MNIST data set you ca easily get a high accuracy using just one hidden layer with a few hundred neurons. You can reach for this data set above 98% accuracy using two hidden layers with the same total amount of neurons, in roughly the same amount of training time.
For more complex problems, you can gradually ramp up the number of hidden layers, until you start overfitting the training set. Very complex tasks, such as large image classification or speech recognition, typically require networks with dozens of layers and they need a huge amount of training data. However, you will rarely have to train such networks from scratch: it is much more common to reuse parts of a pretrained state-of-the-art network that performs a similar task.
The Back propagation algorithm we derived above works by going from the output layer to the input layer, propagating the error gradient on the way. Once the algorithm has computed the gradient of the cost function with regards to each parameter in the network, it uses these gradients to update each parameter with a Gradient Descent (GD) step.
Unfortunately for us, the gradients often get smaller and smaller as the algorithm progresses down to the first hidden layers. As a result, the GD update leaves the lower layer connection weights virtually unchanged, and training never converges to a good solution. This is known in the literature as the vanishing gradients problem.
In other cases, the opposite can happen, namely the the gradients can grow bigger and bigger. The result is that many of the layers get large updates of the weights the algorithm diverges. This is the exploding gradients problem, which is mostly encountered in recurrent neural networks. More generally, deep neural networks suffer from unstable gradients, different layers may learn at widely different speeds
Although this unfortunate behavior has been empirically observed for quite a while (it was one of the reasons why deep neural networks were mostly abandoned for a long time), it is only around 2010 that significant progress was made in understanding it.
A paper titled Understanding the Difficulty of Training Deep Feedforward Neural Networks by Xavier Glorot and Yoshua Bengio found that the problems with the popular logistic sigmoid activation function and the weight initialization technique that was most popular at the time, namely random initialization using a normal distribution with a mean of 0 and a standard deviation of 1.
They showed that with this activation function and this initialization scheme, the variance of the outputs of each layer is much greater than the variance of its inputs. Going forward in the network, the variance keeps increasing after each layer until the activation function saturates at the top layers. This is actually made worse by the fact that the logistic function has a mean of 0.5, not 0 (the hyperbolic tangent function has a mean of 0 and behaves slightly better than the logistic function in deep networks).
Looking at the logistic activation function, when inputs become large (negative or positive), the function saturates at 0 or 1, with a derivative extremely close to 0. Thus when backpropagation kicks in, it has virtually no gradient to propagate back through the network, and what little gradient exists keeps getting diluted as backpropagation progresses down through the top layers, so there is really nothing left for the lower layers.
In their paper, Glorot and Bengio propose a way to significantly alleviate this problem. We need the signal to flow properly in both directions: in the forward direction when making predictions, and in the reverse direction when backpropagating gradients. We don’t want the signal to die out, nor do we want it to explode and saturate. For the signal to flow properly, the authors argue that we need the variance of the outputs of each layer to be equal to the variance of its inputs, and we also need the gradients to have equal variance before and after flowing through a layer in the reverse direction.
One of the insights in the 2010 paper by Glorot and Bengio was that the vanishing/exploding gradients problems were in part due to a poor choice of activation function. Until then most people had assumed that if Nature had chosen to use roughly sigmoid activation functions in biological neurons, they must be an excellent choice. But it turns out that other activation functions behave much better in deep neural networks, in particular the ReLU activation function, mostly because it does not saturate for positive values (and also because it is quite fast to compute).
The ReLU activation function suffers from a problem known as the dying ReLUs: during training, some neurons effectively die, meaning they stop outputting anything other than 0.
In some cases, you may find that half of your network’s neurons are dead, especially if you used a large learning rate. During training, if a neuron’s weights get updated such that the weighted sum of the neuron’s inputs is negative, it will start outputting 0. When this happen, the neuron is unlikely to come back to life since the gradient of the ReLU function is 0 when its input is negative.
To solve this problem, nowadays practitioners use a variant of the ReLU function, such as the leaky ReLU discussed above or the so-called exponential linear unit (ELU) function
$$
ELU(z) = \left\{\begin{array}{cc} \alpha\left( \exp{(z)}-1\right) & z < 0,\\ z & z \ge 0.\end{array}\right.
$$
In general it seems that the ELU activation function is better than the leaky ReLU function (and its variants), which is better than ReLU. ReLU performs better than \( \tanh \) which in turn performs better than the logistic function.
If runtime performance is an issue, then you may opt for the leaky ReLU function over the ELU function If you don’t want to tweak yet another hyperparameter, you may just use the default \( \alpha \) of \( 0.01 \) for the leaky ReLU, and \( 1 \) for ELU. If you have spare time and computing power, you can use cross-validation or bootstrap to evaluate other activation functions.
In most cases you can use the ReLU activation function in the hidden layers (or one of its variants).
It is a bit faster to compute than other activation functions, and the gradient descent optimization does in general not get stuck.
For the output layer:Batch Normalization aims to address the vanishing/exploding gradients problems, and more generally the problem that the distribution of each layer’s inputs changes during training, as the parameters of the previous layers change.
The technique consists of adding an operation in the model just before the activation function of each layer, simply zero-centering and normalizing the inputs, then scaling and shifting the result using two new parameters per layer (one for scaling, the other for shifting). In other words, this operation lets the model learn the optimal scale and mean of the inputs for each layer. In order to zero-center and normalize the inputs, the algorithm needs to estimate the inputs’ mean and standard deviation. It does so by evaluating the mean and standard deviation of the inputs over the current mini-batch, from this the name batch normalization.
It is a fairly simple algorithm: at every training step, every neuron (including the input neurons but excluding the output neurons) has a probability \( p \) of being temporarily dropped out, meaning it will be entirely ignored during this training step, but it may be active during the next step.
The hyperparameter \( p \) is called the dropout rate, and it is typically set to 50%. After training, the neurons are not dropped anymore. It is viewed as one of the most popular regularization techniques.
A popular technique to lessen the exploding gradients problem is to simply clip the gradients during backpropagation so that they never exceed some threshold (this is mostly useful for recurrent neural networks).
This technique is called Gradient Clipping.
In general however, Batch Normalization is preferred.
You may find this website very useful. Thx a million to Ghadi for sharing.
The first thing we would like to do is divide the data into two or three parts. A training set, a validation or dev (development) set, and a test set. The test set is the data on which we want to make predictions. The dev set is a subset of the training data we use to check how well we are doing out-of-sample, after training the model on the training dataset. We use the validation error as a proxy for the test error in order to make tweaks to our model. It is crucial that we do not use any of the test data to train the algorithm. This is a cardinal sin in ML. Then:
If the validation and test sets are drawn from the same distributions, then a good performance on the validation set should lead to similarly good performance on the test set.
However, sometimes the training data and test data differ in subtle ways because, for example, they are collected using slightly different methods, or because it is cheaper to collect data in one way versus another. In this case, there can be a mismatch between the training and test data. This can lead to the neural network overfitting these small differences between the test and training sets, and a poor performance on the test set despite having a good performance on the validation set. To rectify this, Andrew Ng suggests making two validation or dev sets, one constructed from the training data and one constructed from the test data. The difference between the performance of the algorithm on these two validation sets quantifies the train-test mismatch. This can serve as another important diagnostic when using DNNs for supervised learning.
Like all statistical methods, supervised learning using neural networks has important limitations. This is especially important when one seeks to apply these methods, especially to physics problems. Like all tools, DNNs are not a universal solution. Often, the same or better performance on a task can be achieved by using a few hand-engineered features (or even a collection of random features).
Here we list some of the important limitations of supervised neural network based models.
Some of these remarks are particular to DNNs, others are shared by all supervised learning methods. This motivates the use of unsupervised methods which in part circumvent these problems.
For ordinary differential equations, see for example the lectures at http://compphysics.github.io/ComputationalPhysics/doc/pub/ode/html/ode-reveal.html.
After the discussion of ordinary differential equations, we will give a brief review of partial differential equations (see below).
In our discussions of ordinary differential equations we will also study the usage of Autograd in computing gradients for 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 Introduction to Neural Network Methods for Differential Equations, by Yadav and Kumar.
The lectures on differential equations were developed by Kristine Baluka Hein, now PhD student at IFI. A great thank you to Kristine.
An ordinary differential equation (ODE) is an equation involving functions having one variable.
In general, an ordinary differential equation looks like
$$
\begin{equation} \tag{1}
f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) = 0
\end{equation}
$$
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
$$
\begin{equation}
g_t(x) = h_1(x) + h_2(x,N(x,P))
\tag{2}
\end{equation}
$$
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
$$
C\left(x, P\right) = \big(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\big)^2
$$
If \( N \) inputs are given as a vector \( \boldsymbol{x} \) with elements \( x_i \) for \( i = 1,\dots,N \), the cost function becomes
$$
\begin{equation} \tag{3}
C\left(\boldsymbol{x}, P\right) = \frac{1}{N} \sum_{i=1}^N \big(f\left(x_i, \, g(x_i), \, g'(x_i), \, g''(x_i), \, \dots \, , \, g^{(n)}(x_i)\right)\big)^2
\end{equation}
$$
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.
An exponential decay of a quantity \( g(x) \) is described by the equation
$$
\begin{equation} \tag{4}
g'(x) = -\gamma g(x)
\end{equation}
$$
with \( g(0) = g_0 \) for some chosen initial value \( g_0 \).
The analytical solution of (4) is
$$
\begin{equation}
g(x) = g_0 \exp\left(-\gamma x\right)
\tag{5}
\end{equation}
$$
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
$$
\begin{equation} \tag{6}
g'(x) = -\gamma g(x)
\end{equation}
$$
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
$$
g_t(x, P) = h_1(x) + h_2(x, N(x, P))
$$
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:
$$
\begin{equation} \tag{7}
g_t(x, P) = g_0 + x \cdot N(x, P)
\end{equation}
$$
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
$$
g_t(x, P) = g_0 + x \cdot N(x, P)
$$
has been chosen such that it already solves the condition \( g(0) = g_0 \). What remains, is to find \( P \) such that
$$
\begin{equation} \tag{8}
g_t'(x, P) = - \gamma g_t(x, P)
\end{equation}
$$
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:
$$
\min_{P}\Big\{ \big(g_t'(x, P) - ( -\gamma g_t(x, P) \big)^2 \Big\}
$$
(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:
$$
\min_{P_{\text{hidden} }, \ P_{\text{output} }}\Big\{ \big(g_t'(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) - ( -\gamma g_t(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) \big)^2 \Big\}
$$
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
$$
\begin{equation} \tag{9}
\min_{P}\Big\{\frac{1}{N} \sum_{i=1}^N \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \Big\}
\end{equation}
$$
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
$$
\min_{P} C(\boldsymbol{x}, P)
$$
In terms of \( P_{\text{hidden} } \) and \( P_{\text{output} } \), this could also be expressed as
$$
\min_{P_{\text{hidden} }, \ P_{\text{output} }} C(\boldsymbol{x}, \{P_{\text{hidden} }, P_{\text{output} }\})
$$
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:
$$
\begin{aligned}
z_{i,j}^{\text{hidden}} &= b_i^{\text{hidden}} + w_i^{\text{hidden}}x_j \\
&=
\begin{pmatrix}
b_i^{\text{hidden}} & w_i^{\text{hidden}}
\end{pmatrix}
\begin{pmatrix}
1 \\
x_j
\end{pmatrix}
\end{aligned}
$$
The result after weighting the inputs at the \( i \)-th hidden neuron can be written as a vector:
$$
\begin{aligned}
\boldsymbol{z}_{i}^{\text{hidden}} &= \Big( b_i^{\text{hidden}} + w_i^{\text{hidden}}x_1 , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_2, \ \dots \, , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_N\Big) \\
&=
\begin{pmatrix}
b_i^{\text{hidden}} & w_i^{\text{hidden}}
\end{pmatrix}
\begin{pmatrix}
1 & 1 & \dots & 1 \\
x_1 & x_2 & \dots & x_N
\end{pmatrix} \\
&= \boldsymbol{p}_{i, \text{hidden}}^T X
\end{aligned}
$$
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:
$$
f(z) = \frac{1}{1 + \exp{(-z)}}
$$
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:
$$
\boldsymbol{x}_i^{\text{hidden} } = f\big( \boldsymbol{z}_{i}^{\text{hidden}} \big)
$$
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.
$$
\begin{aligned}
z_{1,j}^{\text{output}} & =
\begin{pmatrix}
b_1^{\text{output}} & \boldsymbol{w}_1^{\text{output}}
\end{pmatrix}
\begin{pmatrix}
1 \\
\boldsymbol{x}_j^{\text{hidden}}
\end{pmatrix}
\end{aligned}
$$
Expressing \( z_{1,j}^{\text{output}} \) as a vector gives the following way of weighting the inputs from the hidden layer:
$$
\boldsymbol{z}_{1}^{\text{output}} =
\begin{pmatrix}
b_1^{\text{output}} & \boldsymbol{w}_1^{\text{output}}
\end{pmatrix}
\begin{pmatrix}
1 & 1 & \dots & 1 \\
\boldsymbol{x}_1^{\text{hidden}} & \boldsymbol{x}_2^{\text{hidden}} & \dots & \boldsymbol{x}_N^{\text{hidden}}
\end{pmatrix}
$$
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
$$
C(\boldsymbol{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2
$$
In order to minimize the cost function, an optimization method must be chosen.
Here, gradient descent with a constant step size has been chosen.
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:
$$
\boldsymbol{\omega}_{\text{new} } = \boldsymbol{\omega} - \lambda \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega})
$$
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
$$
\begin{aligned}
P_{\text{hidden},\text{new}} &= P_{\text{hidden}} - \lambda \nabla_{P_{\text{hidden}}} C(\boldsymbol{x}, P) \\
P_{\text{output},\text{new}} &= P_{\text{output}} - \lambda \nabla_{P_{\text{output}}} C(\boldsymbol{x}, P)
\end{aligned}
$$
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()
It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers.
The number of neurons within each hidden layer are given as a list of integers in the program below.
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))
# The neural network with one input layer and one output layer,
# but with number of hidden layers specified by the user.
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 consists 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
# The trial solution using the deep neural network:
def g_trial_deep(x,params, g0 = 10):
return g0 + x*deep_neural_network(params, x)
# The right side of the ODE:
def g(x, g_trial, gamma = 2):
return -gamma*g_trial
# The same cost function as before, but calls deep_neural_network instead.
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 neural network
d_net_out = elementwise_grad(deep_neural_network,1)(P,x)
# Find the derivative w.r.t x of the trial function
d_g_t = elementwise_grad(g_trial_deep,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 and one output layer,
# but with specified number of hidden layers from the user.
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
# The number of elements in the list num_hidden_neurons thus represents
# the number of hidden layers.
# Find the number of hidden layers:
N_hidden = np.size(num_neurons)
## Set up initial weights 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 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_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
def g_analytic(x, gamma = 2, g0 = 10):
return g0*np.exp(-gamma*x)
# Solve the given problem
if __name__ == '__main__':
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 = np.array([10,10])
num_iter = 10000
lmb = 0.001
P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)
res = g_trial_deep(x,P)
res_analytical = g_analytic(x)
plt.figure(figsize=(10,10))
plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution')
plt.plot(x, res_analytical)
plt.plot(x, res[0,:])
plt.legend(['analytical','dnn'])
plt.ylabel('g(x)')
plt.show()
A logistic model of population growth assumes that a population converges toward an equilibrium. The population growth can be modeled by
$$
\begin{equation} \tag{10}
g'(t) = \alpha g(t)(A - g(t))
\end{equation}
$$
where \( g(t) \) is the population density at time \( t \), \( \alpha > 0 \) the growth rate and \( A > 0 \) is the maximum population number in the environment. Also, at \( t = 0 \) the population has the size \( g(0) = g_0 \), where \( g_0 \) is some chosen constant.
In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability and high execution time (this might be more apparent in the examples solving PDEs), using a library like TensorFlow is recommended. Here, we stay with a more simple approach and implement for comparison, the simple forward Euler method.
Here, we will model a population \( g(t) \) in an environment having carrying capacity \( A \). The population follows the model
$$
\begin{equation} \tag{11}
g'(t) = \alpha g(t)(A - g(t))
\end{equation}
$$
where \( g(0) = g_0 \).
In this example, we let \( \alpha = 2 \), \( A = 1 \), and \( g_0 = 1.2 \).
We will get a slightly different trial solution, as the boundary conditions are different compared to the case for exponential decay.
A possible trial solution satisfying the condition \( g(0) = g_0 \) could be
$$
h_1(t) = g_0 + t \cdot N(t,P)
$$
with \( N(t,P) \) being the output from the neural network with weights and biases for each layer collected in the set \( P \).
The analytical solution is
$$
g(t) = \frac{Ag_0}{g_0 + (A - g_0)\exp(-\alpha A t)}
$$
The network will be the similar as for the exponential decay example, but with some small modifications for our problem.
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))
# Function to get the parameters.
# Done such that one can easily change the paramaters after one's liking.
def get_parameters():
alpha = 2
A = 1
g0 = 1.2
return alpha, A, g0
def deep_neural_network(P, x):
# N_hidden is the number of hidden layers
N_hidden = np.size(P) - 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 = P[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 = P[-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 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
d_g_t = elementwise_grad(g_trial_deep,0)(x,P)
# The right side of the ODE
func = f(x, g_t)
err_sqr = (d_g_t - func)**2
cost_sum = np.sum(err_sqr)
return cost_sum / np.size(err_sqr)
# The right side of the ODE:
def f(x, g_trial):
alpha,A, g0 = get_parameters()
return alpha*g_trial*(A - g_trial)
# The trial solution using the deep neural network:
def g_trial_deep(x, params):
alpha,A, g0 = get_parameters()
return g0 + x*deep_neural_network(params,x)
# The analytical solution:
def g_analytic(t):
alpha,A, g0 = get_parameters()
return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t))
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
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)')
plt.show()
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 \( x \):
$$
f(x + \Delta x) \approx f(x) + \Delta x f'(x)
$$
In our case, using Euler's method to approximate the value of \( g \) at a step \( \Delta t \) from \( t \) yields
$$
\begin{aligned}
g(t + \Delta t) &\approx g(t) + \Delta t g'(t) \\
&= g(t) + \Delta t \big(\alpha g(t)(A - g(t))\big)
\end{aligned}
$$
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
$$
\begin{aligned}
t_i &= i\Delta t \\
&= (i - 1)\Delta t + \Delta t \\
&= t_{i-1} + \Delta t
\end{aligned}
$$
Now, if \( g_i = g(t_i) \) then
$$
\begin{equation}
\begin{aligned}
g_i &= g(t_i) \\
&= g(t_{i-1} + \Delta t) \\
&\approx g(t_{i-1}) + \Delta t \big(\alpha g(t_{i-1})(A - g(t_{i-1}))\big) \\
&= g_{i-1} + \Delta t \big(\alpha g_{i-1}(A - g_{i-1})\big)
\end{aligned}
\end{equation} \tag{12}
$$
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()
The Poisson equation for \( g(x) \) in one dimension is
$$
\begin{equation} \tag{13}
-g''(x) = f(x)
\end{equation}
$$
where \( f(x) \) is a given function for \( x \in (0,1) \).
The conditions that \( g(x) \) is chosen to fulfill, are
$$
\begin{align*}
g(0) &= 0 \\
g(1) &= 0
\end{align*}
$$
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
$$
-g''(x) = f(x),\qquad x \in (0,1)
$$
where \( f(x) \) is a given function, along with the chosen conditions
$$
\begin{aligned}
g(0) = g(1) = 0
\end{aligned}\tag{14}
$$
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
$$
g_t(x) = x \cdot (1-x) \cdot N(P,x)
$$
The analytical solution for this problem is
$$
g(x) = x(1 - x)\exp(x)
$$
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()
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
$$
g''(x) = \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} + E_{\Delta x}(x)
$$
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:
$$
\begin{equation} \tag{15}
g''(x) \approx \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2}
\end{equation}
$$
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
$$
\begin{aligned}
g''(x_i) &\approx \frac{g(x_i + \Delta x) - 2g(x_i) + g(x_i -\Delta x)}{\Delta x^2} \\
&= \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2}
\end{aligned}
$$
Since we know from our problem that
$$
\begin{aligned}
-g''(x) &= f(x) \\
&= (3x + x^2)\exp(x)
\end{aligned}
$$
along with the conditions \( g(0) = g(1) = 0 \), the following scheme can be used to find an approximate solution for \( g(x) \) numerically:
$$
\begin{equation}
\begin{aligned}
-\Big( \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} \Big) &= f(x_i) \\
-g_{i+1} + 2g_i - g_{i-1} &= \Delta x^2 f(x_i)
\end{aligned}
\end{equation} \tag{16}
$$
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:
$$
\begin{aligned}
\begin{pmatrix}
2 & -1 & 0 & \dots & 0 \\
-1 & 2 & -1 & \dots & 0 \\
\vdots & & \ddots & & \vdots \\
0 & \dots & -1 & 2 & -1 \\
0 & \dots & 0 & -1 & 2\\
\end{pmatrix}
\begin{pmatrix}
g_1 \\
g_2 \\
\vdots \\
g_{N_x - 3} \\
g_{N_x - 2}
\end{pmatrix}
&=
\Delta x^2
\begin{pmatrix}
f(x_1) \\
f(x_2) \\
\vdots \\
f(x_{N_x - 3}) \\
f(x_{N_x - 2})
\end{pmatrix} \\
\boldsymbol{A}\boldsymbol{g} &= \boldsymbol{f},
\end{aligned}
$$
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()
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
$$
\begin{equation} \tag{17}
f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) = 0
\end{equation}
$$
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.
In the Natural Sciences we often encounter problems with many variables constrained by boundary conditions and initial values. Many of these problems can be modelled as partial differential equations. One case which arises in many situations is the so-called wave equation whose one-dimensional form reads
$$
\begin{equation}
\tag{18}
\frac{\partial^2 u}{\partial x^2}=A\frac{\partial^2 u}{\partial t^2},
\end{equation}
$$
where \( A \) is a constant. The solution \( u \) depends on both spatial and temporal variables, viz. \( u=u(x,t) \).
In two dimension we have \( u=u(x,y,t) \). We will, unless otherwise stated, simply use \( u \) in our discussion below. Familiar situations which this equation can model are waves on a string, pressure waves, waves on the surface of a fjord or a lake, electromagnetic waves and sound waves to mention a few. For e.g., electromagnetic waves we have the constant \( A=c^2 \), with \( c \) the speed of light. It is rather straightforward to extend this equation to two or three dimension. In two dimensions we have
$$
\begin{equation*}
\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=A\frac{\partial^2 u}{\partial t^2},
\end{equation*}
$$
The diffusion equation whose one-dimensional version reads
$$
\begin{equation}
\tag{19}
\frac{\partial^2 u}{\partial x^2}=A\frac{\partial u}{\partial t},
\end{equation}
$$
and \( A \) is in this case called the diffusion constant. It can be used to model a wide selection of diffusion processes, from molecules to the diffusion of heat in a given material.
Another familiar equation from electrostatics is Laplace's equation, which looks similar to the wave equation in Eq. (18) except that we have set \( A=0 \)
$$
\begin{equation}
\tag{20}
\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0,
\end{equation}
$$
or if we have a finite electric charge represented by a charge density \( \rho(\mathbf{x}) \) we have the familiar Poisson equation
$$
\begin{equation}
\tag{21}
\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=-4\pi \rho(\mathbf{x}).
\end{equation}
$$
Other famous partial differential equations are the Helmholtz (or eigenvalue) equation, here specialized to two dimensions only
$$
\begin{equation}
\tag{22}
-\frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial y^2}=\lambda u,
\end{equation}
$$
the linear transport equation (in \( 2+1 \) dimensions) familiar from Brownian motion as well
$$
\begin{equation}
\tag{23}
\frac{\partial u}{\partial t} +\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y }=0,
\end{equation}
$$
Schroedinger's equation
$$
\begin{equation*}
-\frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial y^2}+f(x,y)u = \imath\frac{\partial u}{\partial t}.
\end{equation*}
$$
Important systems of linear partial differential equations are the famous Maxwell equations
$$
\frac{\partial \mathbf{E}}{\partial t} = \mathrm{curl}\mathbf{B},
$$
and
$$
-\mathrm{curl} \mathbf{E} = \mathbf{B}
$$
and
$$
\mathrm{div} \mathbf{E} = \mathrm{div}\mathbf{B} = 0.
$$
Similarly, famous systems of non-linear partial differential equations are for example Euler's equations for incompressible, inviscid flow
$$
\begin{equation*}
\frac{\partial \mathbf{u}}{\partial t} +\mathbf{u}\nabla\mathbf{u}= -Dp; \hspace{1cm} \mathrm{div} \mathbf{u} = 0,
\end{equation*}
$$
with \( p \) being the pressure and
$$
\begin{equation*}
\nabla = \frac{\partial}{\partial x}e_x+\frac{\partial}{\partial y}e_y,
\end{equation*}
$$
in the two dimensions. The unit vectors are \( e_x \) and \( e_y \).
Another example is the set of Navier-Stokes equations for incompressible, viscous flow
$$
\begin{equation*}
\frac{\partial \mathbf{u}}{\partial t} +\mathbf{u}\nabla\mathbf{u}-\Delta \mathbf{u}= -Dp; \hspace{1cm} \mathrm{div} \mathbf{u} = 0.
\end{equation*}
$$
A general partial differential equation with two given dimensions reads
$$
\begin{equation*}
A(x,y)\frac{\partial^2 u}{\partial x^2}+B(x,y)\frac{\partial^2 u}{\partial x\partial y}
+C(x,y)\frac{\partial^2 u}{\partial y^2}=F(x,y,u,\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}),
\end{equation*}
$$
and if we set
$$
\begin{equation*}
B=C=0,
\end{equation*}
$$
we recover the \( 1+1 \)-dimensional diffusion equation which is an example of a so-called parabolic partial differential equation. With
$$
\begin{equation*}
B=0, \hspace{1cm} AC < 0
\end{equation*}
$$
we get the \( 2+1 \)-dim wave equation which is an example of a so-called elliptic PDE, where more generally we have \( B^2 > AC \). For \( B^2 < AC \) we obtain a so-called hyperbolic PDE, with the Laplace equation in Eq. (20) as one of the classical examples. These equations can all be easily extended to non-linear partial differential equations and \( 3+1 \) dimensional cases.
The diffusion equation describes in typical applications the evolution in time of the density \( u \) of a quantity like the particle density, energy density, temperature gradient, chemical concentrations etc.
The basis is the assumption that the flux density \( \mathbf{\rho} \) obeys the Gauss-Green theorem
$$
\begin{equation*}
\int_V \mathrm{div}\mathbf{\rho} dx = \int_{\partial V} \mathbf{\rho}\mathbf{n}dS,
\end{equation*}
$$
where \( n \) is the unit outer normal field and \( V \) is a smooth region with the space where we seek a solution. The Gauss-Green theorem leads to
$$
\begin{equation*}
\mathrm{div} \mathbf{\rho} = 0.
\end{equation*}
$$
Assuming that the flux is proportional to the gradient \( \mathbf{\nabla} u \) but pointing in the opposite direction since the flow is from regions of high concetration to lower concentrations, we obtain
$$
\begin{equation*}
\mathbf{\rho} = -D\mathbf{\nabla} u,
\end{equation*}
$$
resulting in
$$
\begin{equation*}
\mathrm{div} \mathbf{\nabla} u = D\Delta u = 0,
\end{equation*}
$$
which is Laplace's equation. The constant \( D \) can be coupled with various physical constants, such as the diffusion constant or the specific heat and thermal conductivity discussed below.
If we let \( u \) denote the concetration of a particle species, this results in Fick's law of diffusion. If it denotes the temperature gradient, we have Fourier'slaw of heat conduction and if it refers to the electrostatic potential we have Ohm's law of electrical conduction.
Coupling the rate of change (temporal dependence) of \( u \) with the flux density we have
$$
\begin{equation*}
\frac{\partial u}{\partial t} = -\mathrm{div}\mathbf{\rho},
\end{equation*}
$$
which results in
$$
\begin{equation*}
\frac{\partial u}{\partial t}= D \mathrm{div} \mathbf{\nabla} u = D \Delta u,
\end{equation*}
$$
the diffusion equation, or heat equation.
If we specialize to the heat equation, we assume that the diffusion of heat through some material is proportional with the temperature gradient \( T(\mathbf{x},t) \) and using conservation of energy we arrive at the diffusion equation
$$
\begin{equation*}
\frac{\kappa}{C\rho}\nabla^2 T(\mathbf{x},t) =\frac{\partial T(\mathbf{x},t)}{\partial t}
\end{equation*}
$$
where \( C \) is the specific heat and \( \rho \) the density of the material. Here we let the density be represented by a constant, but there is no problem introducing an explicit spatial dependence, viz.,
$$
\begin{equation*}
\frac{\kappa}{C\rho(\mathbf{x},t)}\nabla^2 T(\mathbf{x},t) =
\frac{\partial T(\mathbf{x},t)}{\partial t}.
\end{equation*}
$$
Setting all constants equal to the diffusion constant \( D \), i.e.,
$$
\begin{equation*}
D=\frac{C\rho}{\kappa},
\end{equation*}
$$
we arrive at
$$
\begin{equation*}
\nabla^2 T(\mathbf{x},t) =
D\frac{\partial T(\mathbf{x},t)}{\partial t}.
\end{equation*}
$$
Specializing to the \( 1+1 \)-dimensional case we have
$$
\begin{equation*}
\frac{\partial^2 T(x,t)}{\partial x^2}=D\frac{\partial T(x,t)}{\partial t}.
\end{equation*}
$$
We note that the dimension of \( D \) is time/length$^2$. Introducing the dimensionless variables \( \alpha\hat{x}=x \) we get
$$
\begin{equation*}
\frac{\partial^2 T(x,t)}{\alpha^2\partial \hat{x}^2}=
D\frac{\partial T(x,t)}{\partial t},
\end{equation*}
$$
and since \( \alpha \) is just a constant we could define \( \alpha^2D= 1 \) or use the last expression to define a dimensionless time-variable \( \hat{t} \). This yields a simplified diffusion equation
$$
\begin{equation*}
\frac{\partial^2 T(\hat{x},\hat{t})}{\partial \hat{x}^2}=
\frac{\partial T(\hat{x},\hat{t})}{\partial \hat{t}}.
\end{equation*}
$$
It is now a partial differential equation in terms of dimensionless variables. In the discussion below, we will however, for the sake of notational simplicity replace \( \hat{x}\rightarrow x \) and \( \hat{t}\rightarrow t \). Moreover, the solution to the \( 1+1 \)-dimensional partial differential equation is replaced by \( T(\hat{x},\hat{t})\rightarrow u(x,t) \).
In one dimension we have the following equation
$$
\begin{equation*}
\nabla^2 u(x,t) =\frac{\partial u(x,t)}{\partial t},
\end{equation*}
$$
or
$$
\begin{equation*}
u_{xx} = u_t,
\end{equation*}
$$
with initial conditions, i.e., the conditions at \( t=0 \),
$$
\begin{equation*}
u(x,0)= g(x) \hspace{0.5cm} 0 < x < L
\end{equation*}
$$
with \( L=1 \) the length of the \( x \)-region of interest.
The boundary conditions are
$$
\begin{equation*}
u(0,t)= a(t) \hspace{0.5cm} t \ge 0,
\end{equation*}
$$
and
$$
\begin{equation*}
u(L,t)= b(t) \hspace{0.5cm} t \ge 0,
\end{equation*}
$$
where \( a(t) \) and \( b(t) \) are two functions which depend on time only, while \( g(x) \) depends only on the position \( x \). Our next step is to find a numerical algorithm for solving this equation. Here we recur to our familiar equal-step methods and introduce different step lengths for the space-variable \( x \) and time \( t \) through the step length for \( x \)
$$
\begin{equation*}
\Delta x=\frac{1}{n+1}
\end{equation*}
$$
and the time step length \( \Delta t \). The position after \( i \) steps and time at time-step \( j \) are now given by
$$
\begin{equation*}
\begin{array}{cc} t_j=j\Delta t & j \ge 0 \\
x_i=i\Delta x & 0 \le i \le n+1\end{array}
\end{equation*}
$$
If we use standard approximations for the derivatives we obtain
$$
\begin{equation*}
u_t\approx \frac{u(x,t+\Delta t)-u(x,t)}{\Delta t}=\frac{u(x_i,t_j+\Delta t)-u(x_i,t_j)}{\Delta t}
\end{equation*}
$$
with a local approximation error \( O(\Delta t) \) and
$$
\begin{equation*}
u_{xx}\approx \frac{u(x+\Delta x,t)-2u(x,t)+u(x-\Delta x,t)}{\Delta x^2},
\end{equation*}
$$
or
$$
\begin{equation*}
u_{xx}\approx \frac{u(x_i+\Delta x,t_j)-2u(x_i,t_j)+u(x_i-\Delta x,t_j)}{\Delta x^2},
\end{equation*}
$$
with a local approximation error \( O(\Delta x^2) \). Our approximation is to higher order in coordinate space. This can be justified since in most cases it is the spatial dependence which causes numerical problems.
These equations can be further simplified as
$$
\begin{equation*}
u_t\approx \frac{u_{i,j+1}-u_{i,j}}{\Delta t},
\end{equation*}
$$
and
$$
\begin{equation*}
u_{xx}\approx \frac{u_{i+1,j}-2u_{i,j}+u_{i-1,j}}{\Delta x^2}.
\end{equation*}
$$
The one-dimensional diffusion equation can then be rewritten in its discretized version as
$$
\begin{equation*}
\frac{u_{i,j+1}-u_{i,j}}{\Delta t}=\frac{u_{i+1,j}-2u_{i,j}+u_{i-1,j}}{\Delta x^2}.
\end{equation*}
$$
Defining \( \alpha = \Delta t/\Delta x^2 \) results in the explicit scheme
$$
\begin{equation}
\tag{24}
u_{i,j+1}= \alpha u_{i-1,j}+(1-2\alpha)u_{i,j}+\alpha u_{i+1,j}.
\end{equation}
$$
Since all the discretized initial values
$$
\begin{equation*}
u_{i,0} = g(x_i),
\end{equation*}
$$
are known, then after one time-step the only unknown quantity is \( u_{i,1} \) which is given by
$$
\begin{equation*}
u_{i,1}= \alpha u_{i-1,0}+(1-2\alpha)u_{i,0}+\alpha u_{i+1,0}=
\alpha g(x_{i-1})+(1-2\alpha)g(x_{i})+\alpha g(x_{i+1}).
\end{equation*}
$$
We can then obtain \( u_{i,2} \) using the previously calculated values \( u_{i,1} \) and the boundary conditions \( a(t) \) and \( b(t) \). This algorithm results in a so-called explicit scheme, since the next functions \( u_{i,j+1} \) are explicitely given by Eq. (24).
We specialize to the case \( a(t)=b(t)=0 \) which results in \( u_{0,j}=u_{n+1,j}=0 \). We can then reformulate our partial differential equation through the vector \( V_j \) at the time \( t_j=j\Delta t \)
$$
\begin{equation*}
V_j=\begin{bmatrix}u_{1,j}\\ u_{2,j} \\ \dots \\ u_{n,j}\end{bmatrix}.
\end{equation*}
$$
This results in a matrix-vector multiplication
$$
\begin{equation*}
V_{j+1} = \mathbf{A}V_{j}
\end{equation*}
$$
with the matrix \( \mathbf{A} \) given by
$$
\begin{equation*}
\mathbf{A}=\begin{bmatrix}1-2\alpha&\alpha&0& 0\dots\\
\alpha&1-2\alpha&\alpha & 0\dots \\
\dots & \dots & \dots & \dots \\
0\dots & 0\dots &\alpha& 1-2\alpha\end{bmatrix}
\end{equation*}
$$
which means we can rewrite the original partial differential equation as a set of matrix-vector multiplications
$$
\begin{equation*}
V_{j+1} = \mathbf{A}V_{j}=\dots = \mathbf{A}^{j+1}V_0,
\end{equation*}
$$
where \( V_0 \) is the initial vector at time \( t=0 \) defined by the initial value \( g(x) \). In the numerical implementation one should avoid to treat this problem as a matrix vector multiplication since the matrix is triangular and at most three elements in each row are different from zero.
It is rather easy to implement this matrix-vector multiplication as seen in the following piece of code
// First we set initialise the new and old vectors
// Here we have chosen the boundary conditions to be zero.
// n+1 is the number of mesh points in x
// Armadillo notation for vectors
u(0) = unew(0) = u(n) = unew(n) = 0.0;
for (int i = 1; i < n; i++) {
x = i*step;
// initial condition
u(i) = func(x);
// intitialise the new vector
unew(i) = 0;
}
// Time integration
for (int t = 1; t <= tsteps; t++) {
for (int i = 1; i < n; i++) {
// Discretized diff eq
unew(i) = alpha * u(i-1) + (1 - 2*alpha) * u(i) + alpha * u(i+1);
}
// note that the boundaries are not changed.
However, although the explicit scheme is easy to implement, it has a very weak stability condition, given by
$$
\begin{equation*}
\Delta t/\Delta x^2 \le 1/2.
\end{equation*}
$$
This means that if \( \Delta x = 0.01 \) (a rather frequent choice), then \( \Delta t= 5\times 10^{-5} \). This has obviously bad consequences if our time interval is large. In order to derive this relation we need some results from studies of iterative schemes. If we require that our solution approaches a definite value after a certain amount of time steps we need to require that the so-called spectral radius \( \rho(\mathbf{A}) \) of our matrix \( \mathbf{A} \) satisfies the condition
$$
\begin{equation}
\tag{25}
\rho(\mathbf{A}) < 1.
\end{equation}
$$
The spectral radius is defined as
$$
\begin{equation*}
\rho(\mathbf{A}) = \hspace{0.1cm}\mathrm{max}\left\{|\lambda|:\mathrm{det}(\mathbf{A}-\lambda\hat{I})=0\right\},
\end{equation*}
$$
which is interpreted as the smallest number such that a circle with radius centered at zero in the complex plane contains all eigenvalues of \( \mathbf{A} \). If the matrix is positive definite, the condition in Eq. (25) is always satisfied.
We can obtain closed-form expressions for the eigenvalues of \( \mathbf{A} \). To achieve this it is convenient to rewrite the matrix as
$$
\begin{equation*}
\mathbf{A}=\hat{I}-\alpha\hat{B},
\end{equation*}
$$
with
$$
\begin{equation*}
\hat{B} =\begin{bmatrix}2&-1&0& 0 &\dots\\
-1&2&-1& 0&\dots \\
\dots & \dots & \dots & \dots & -1 \\
0 & 0 &\dots &-1&2\end{bmatrix}.
\end{equation*}
$$
The eigenvalues of \( \mathbf{A} \) are \( \lambda_i=1-\alpha\mu_i \), with \( \mu_i \) being the eigenvalues of \( \hat{B} \). To find \( \mu_i \) we note that the matrix elements of \( \hat{B} \) are
$$
\begin{equation*}
b_{ij} = 2\delta_{ij}-\delta_{i+1j}-\delta_{i-1j},
\end{equation*}
$$
meaning that we have the following set of eigenequations for component \( i \)
$$
\begin{equation*}
(\hat{B}\hat{x})_i = \mu_ix_i,
\end{equation*}
$$
resulting in
$$
\begin{equation*}
(\hat{B}\hat{x})_i=\sum_{j=1}^n\left(2\delta_{ij}-\delta_{i+1j}-\delta_{i-1j}\right)x_j =
2x_i-x_{i+1}-x_{i-1}=\mu_ix_i.
\end{equation*}
$$
If we assume that \( x \) can be expanded in a basis of \( x=(\sin{(\theta)}, \sin{(2\theta)},\dots, \sin{(n\theta)}) \) with \( \theta = l\pi/n+1 \), where we have the endpoints given by \( x_0 = 0 \) and \( x_{n+1}=0 \), we can rewrite the last equation as
$$
\begin{equation*}
2\sin{(i\theta)}-\sin{((i+1)\theta)}-\sin{((i-1)\theta)}=\mu_i\sin{(i\theta)},
\end{equation*}
$$
or
$$
\begin{equation*}
2\left(1-\cos{(\theta)}\right)\sin{(i\theta)}=\mu_i\sin{(i\theta)},
\end{equation*}
$$
which is nothing but
$$
\begin{equation*}
2\left(1-\cos{(\theta)}\right)x_i=\mu_ix_i,
\end{equation*}
$$
with eigenvalues \( \mu_i = 2-2\cos{(\theta)} \).
Our requirement in Eq. (25) results in
$$
\begin{equation*}
-1 < 1-\alpha2\left(1-\cos{(\theta)}\right) < 1,
\end{equation*}
$$
which is satisfied only if \( \alpha < \left(1-\cos{(\theta)}\right)^{-1} \) resulting in \( \alpha \le 1/2 \) or \( \Delta t/\Delta x^2 \le 1/2 \).
A more general tridiagonal matrix
$$
\begin{equation*}
\mathbf{A} =\begin{bmatrix}a&b&0& 0 &\dots\\
c&a&b& 0&\dots \\
\dots & \dots & \dots & \dots & b \\
0 & 0 &\dots &c&a\end{bmatrix},
\end{equation*}
$$
has eigenvalues \( \mu_i=a+s\sqrt{bc}\cos{(i\pi/n+1)} \) with \( i=1:n \).
In deriving the equations for the explicit scheme we started with the so-called forward formula for the first derivative, i.e., we used the discrete approximation
$$
\begin{equation*}
u_t\approx \frac{u(x_i,t_j+\Delta t)-u(x_i,t_j)}{\Delta t}.
\end{equation*}
$$
However, there is nothing which hinders us from using the backward formula
$$
\begin{equation*}
u_t\approx \frac{u(x_i,t_j)-u(x_i,t_j-\Delta t)}{\Delta t},
\end{equation*}
$$
still with a truncation error which goes like \( O(\Delta t) \).
We could also have used a midpoint approximation for the first derivative, resulting in
$$
\begin{equation*}
u_t\approx \frac{u(x_i,t_j+\Delta t)-u(x_i,t_j-\Delta t)}{2\Delta t},
\end{equation*}
$$
with a truncation error \( O(\Delta t^2) \). Here we will stick to the backward formula and come back to the latter below. For the second derivative we use however
$$
\begin{equation*}
u_{xx}\approx \frac{u(x_i+\Delta x,t_j)-2u(x_i,t_j)+u(x_i-\Delta x,t_j)}{\Delta x^2},
\end{equation*}
$$
and define again \( \alpha = \Delta t/\Delta x^2 \).
We obtain now
$$
\begin{equation*}
u_{i,j-1}= -\alpha u_{i-1,j}+(1-2\alpha)u_{i,j}-\alpha u_{i+1,j}.
\end{equation*}
$$
Here \( u_{i,j-1} \) is the only unknown quantity. Defining the matrix \( \mathbf{A} \)
$$
\begin{equation*}
\mathbf{A}=\begin{bmatrix}1+2\alpha&-\alpha&0& 0 &\dots\\
-\alpha&1+2\alpha&-\alpha & 0 & \dots \\
\dots & \dots & \dots & \dots &\dots \\
\dots & \dots & \dots & \dots & -\alpha \\
0 & 0 &\dots &-\alpha& 1+2\alpha\end{bmatrix},
\end{equation*}
$$
we can reformulate again the problem as a matrix-vector multiplication
$$
\begin{equation*}
\mathbf{A}V_{j} = V_{j-1}
\end{equation*}
$$
It means that we can rewrite the problem as
$$
\begin{equation*}
V_{j} = \mathbf{A}^{-1}V_{j-1}=\mathbf{A}^{-1}\left(\mathbf{A}^{-1}V_{j-2}\right)=\dots = \mathbf{A}^{-j}V_0.
\end{equation*}
$$
This is an implicit scheme since it relies on determining the vector \( u_{i,j-1} \) instead of \( u_{i,j+1} \). If \( \alpha \) does not depend on time \( t \), we need to invert a matrix only once. Alternatively we can solve this system of equations using our methods from linear algebra. These are however very cumbersome ways of solving since they involve \( \sim O(N^3) \) operations for a \( N\times N \) matrix. It is much faster to solve these linear equations using methods for tridiagonal matrices, since these involve only \( \sim O(N) \) operations.
The implicit scheme is always stable since the spectral radius satisfies \( \rho(\mathbf{A}) < 1 \). We could have inferred this by noting that the matrix is positive definite, viz. all eigenvalues are larger than zero. We see this from the fact that \( \mathbf{A}=\hat{I}+\alpha\hat{B} \) has eigenvalues \( \lambda_i = 1+\alpha(2-2cos(\theta)) \) which satisfy \( \lambda_i > 1 \). Since it is the inverse which stands to the right of our iterative equation, we have \( \rho(\mathbf{A}^{-1}) < 1 \) and the method is stable for all combinations of \( \Delta t \) and \( \Delta x \).
We show here parts of a simple example of how to solve the one-dimensional diffusion equation using the implicit scheme discussed above. The program uses the function to solve linear equations with a tridiagonal matrix.
// parts of the function for backward Euler
void backward_euler(int n, int tsteps, double delta_x, double alpha)
{
double a, b, c;
vec u(n+1); // This is u of Au = y
vec y(n+1); // Right side of matrix equation Au=y, the solution at a previous step
// Initial conditions
for (int i = 1; i < n; i++) {
y(i) = u(i) = func(delta_x*i);
}
// Boundary conditions (zero here)
y(n) = u(n) = u(0) = y(0);
// Matrix A, only constants
a = c = - alpha;
b = 1 + 2*alpha;
// Time iteration
for (int t = 1; t <= tsteps; t++) {
// here we solve the tridiagonal linear set of equations,
tridag(a, b, c, y, u, n+1);
// boundary conditions
u(0) = 0;
u(n) = 0;
// replace previous time solution with new
for (int i = 0; i <= n; i++) {
y(i) = u(i);
}
// You may consider printing the solution at regular time intervals
.... // print statements
} // end time iteration
...
}
It is possible to combine the implicit and explicit methods in a slightly more general approach. Introducing a parameter \( \theta \) (the so-called \( \theta \)-rule) we can set up an equation
$$
\begin{equation}
\tag{26}
\frac{\theta}{\Delta x^2}\left(u_{i-1,j}-2u_{i,j}+u_{i+1,j}\right)+
\frac{1-\theta}{\Delta x^2}\left(u_{i+1,j-1}-2u_{i,j-1}+u_{i-1,j-1}\right)=
\frac{1}{\Delta t}\left(u_{i,j}-u_{i,j-1}\right),
\end{equation}
$$
which for \( \theta=0 \) yields the forward formula for the first derivative and the explicit scheme, while \( \theta=1 \) yields the backward formula and the implicit scheme. These two schemes are called the backward and forward Euler schemes, respectively. For \( \theta = 1/2 \) we obtain a new scheme after its inventors, Crank and Nicolson. This scheme yields a truncation in time which goes like \( O(\Delta t^2) \) and it is stable for all possible combinations of \( \Delta t \) and \( \Delta x \).
To derive the Crank-Nicolson equation, we start with the forward Euler scheme and Taylor expand \( u(x,t+\Delta t) \), \( u(x+\Delta x, t) \) and \( u(x-\Delta x,t) \)
$$
\begin{align}
u(x+\Delta x,t)&=u(x,t)+\frac{\partial u(x,t)}{\partial x} \Delta x+\frac{\partial^2 u(x,t)}{2\partial x^2}\Delta x^2+\mathcal{O}(\Delta x^3),
\tag{27}\\ \nonumber
u(x-\Delta x,t)&=u(x,t)-\frac{\partial u(x,t)}{\partial x}\Delta x+\frac{\partial^2 u(x,t)}{2\partial x^2} \Delta x^2+\mathcal{O}(\Delta x^3),\\ \nonumber
u(x,t+\Delta t)&=u(x,t)+\frac{\partial u(x,t)}{\partial t}\Delta t+ \mathcal{O}(\Delta t^2).
\tag{28}
\end{align}
$$
With these Taylor expansions the approximations for the derivatives takes the form
$$
\begin{align}
&\left[\frac{\partial u(x,t)}{\partial t}\right]_{\text{approx}} =\frac{\partial u(x,t)}{\partial t}+\mathcal{O}(\Delta t) ,
\tag{29}\\ \nonumber
&\left[\frac{\partial^2 u(x,t)}{\partial x^2}\right]_{\text{approx}}=\frac{\partial^2 u(x,t)}{\partial x^2}+\mathcal{O}(\Delta x^2).
\end{align}
$$
It is easy to convince oneself that the backward Euler method must have the same truncation errors as the forward Euler scheme.
For the Crank-Nicolson scheme we also need to Taylor expand \( u(x+\Delta x, t+\Delta t) \) and \( u(x-\Delta x, t+\Delta t) \) around \( t'=t+\Delta t/2 \).
$$
\begin{align}
u(x+\Delta x, t+\Delta t)&=u(x,t')+\frac{\partial u(x,t')}{\partial x}\Delta x+\frac{\partial u(x,t')}{\partial t} \frac{\Delta t}{2} +\frac{\partial^2 u(x,t')}{2\partial x^2}\Delta x^2+\frac{\partial^2 u(x,t')}{2\partial t^2}\frac{\Delta t^2}{4} +\notag
\tag{30}\\ \nonumber
&\frac{\partial^2 u(x,t')}{\partial x\partial t}\frac{\Delta t}{2} \Delta x+ \mathcal{O}(\Delta t^3)\\ \nonumber
u(x-\Delta x, t+\Delta t)&=u(x,t')-\frac{\partial u(x,t')}{\partial x}\Delta x+\frac{\partial u(x,t')}{\partial t} \frac{\Delta t}{2} +\frac{\partial^2 u(x,t')}{2\partial x^2}\Delta x^2+\frac{\partial^2 u(x,t')}{2\partial t^2}\frac{\Delta t^2}{4} - \notag\\ \nonumber
&\frac{\partial^2 u(x,t')}{\partial x\partial t}\frac{\Delta t}{2} \Delta x+ \mathcal{O}(\Delta t^3)\\
u(x+\Delta x,t)&=u(x,t')+\frac{\partial u(x,t')}{\partial x}\Delta x-\frac{\partial u(x,t')}{\partial t} \frac{\Delta t}{2} +\frac{\partial^2 u(x,t')}{2\partial x^2}\Delta x^2+\frac{\partial^2 u(x,t')}{2\partial t^2}\frac{\Delta t^2}{4} -\notag
\tag{31}\\ \nonumber
&\frac{\partial^2 u(x,t')}{\partial x\partial t}\frac{\Delta t}{2} \Delta x+ \mathcal{O}(\Delta t^3)\\ \nonumber
u(x-\Delta x,t)&=u(x,t')-\frac{\partial u(x,t')}{\partial x}\Delta x-\frac{\partial u(x,t')}{\partial t} \frac{\Delta t}{2} +\frac{\partial^2 u(x,t')}{2\partial x^2}\Delta x^2+\frac{\partial^2 u(x,t')}{2\partial t^2}\frac{\Delta t^2}{4} +\notag \\ \nonumber
&\frac{\partial^2 u(x,t')}{\partial x\partial t}\frac{\Delta t}{2} \Delta x+ \mathcal{O}(\Delta t^3)\\ \nonumber
u(x,t+\Delta t)&=u(x,t')+\frac{\partial u(x,t')}{\partial t}\frac{\Delta_t}{2} +\frac{\partial ^2 u(x,t')}{2\partial t^2}\Delta t^2 + \mathcal{O}(\Delta t^3)\\ \nonumber
u(x,t)&=u(x,t')-\frac{\partial u(x,t')}{\partial t}\frac{\Delta t}{2}+\frac{\partial ^2 u(x,t')}{2\partial t^2}\Delta t^2 + \mathcal{O}(\Delta t^3)
\tag{32}
\end{align}
$$
We now insert these expansions in the approximations for the derivatives to find
$$
\begin{align}
&\left[\frac{\partial u(x,t')}{\partial t}\right]_{\text{approx}} =\frac{\partial u(x,t')}{\partial t}+\mathcal{O}(\Delta t^2) ,
\tag{33}\\ \nonumber
&\left[\frac{\partial^2 u(x,t')}{\partial x^2}\right]_{\text{approx}}=\frac{\partial^2 u(x,t')}{\partial x^2}+\mathcal{O}(\Delta x^2).
\end{align}
$$
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
$$
\begin{align*}
g_t(x_1,\dots,x_N) = h_1(x_1,\dots,x_N) + h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))
\end{align*}
$$
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.
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
$$
\begin{equation*}
C\left(x_1, \dots, x_N, P\right) = \left( f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) \right)^2
\end{equation*}
$$
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:
$$
C\left(\boldsymbol{x}, P\right) = f\left( \left( \boldsymbol{x}, \frac{\partial g(\boldsymbol{x}) }{\partial x_1}, \dots , \frac{\partial g(\boldsymbol{x}) }{\partial x_N}, \frac{\partial g(\boldsymbol{x}) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\boldsymbol{x}) }{\partial x_N^n} \right) \right)^2
$$
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
$$
\begin{equation*}
C\left(X, P \right) = \sum_{i=1}^M f\left( \left( \boldsymbol{x}_i, \frac{\partial g(\boldsymbol{x}_i) }{\partial x_1}, \dots , \frac{\partial g(\boldsymbol{x}_i) }{\partial x_N}, \frac{\partial g(\boldsymbol{x}_i) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\boldsymbol{x}_i) }{\partial x_N^n} \right) \right)^2.
\end{equation*}
$$
In one spatial dimension, the equation reads
$$
\begin{equation*}
\frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2}
\end{equation*}
$$
where a possible choice of conditions are
$$
\begin{align*}
g(0,t) &= 0 ,\qquad t \geq 0 \\
g(1,t) &= 0, \qquad t \geq 0 \\
g(x,0) &= u(x),\qquad x\in [0,1]
\end{align*}
$$
with \( u(x) \) being some given function.
For this case, we want to find \( g(x,t) \) such that
$$
\begin{equation}
\frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2}
\end{equation} \tag{34}
$$
and
$$
\begin{align*}
g(0,t) &= 0 ,\qquad t \geq 0 \\
g(1,t) &= 0, \qquad t \geq 0 \\
g(x,0) &= u(x),\qquad x\in [0,1]
\end{align*}
$$
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
$$
g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P)
$$
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:
$$
h_1(x,t) = (1-t)\Big(u(x) - \big((1-x)u(0) + x u(1)\big)\Big) = (1-t)u(x) = (1-t)\sin(\pi x)
$$
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
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
$$
g(x,t) = \exp(-\pi^2 t)\sin(\pi x)
$$
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()
The wave equation is
$$
\begin{equation*}
\frac{\partial^2 g(x,t)}{\partial t^2} = c^2\frac{\partial^2 g(x,t)}{\partial x^2}
\end{equation*}
$$
with \( c \) being the specified wave speed.
Here, the chosen conditions are
$$
\begin{align*}
g(0,t) &= 0 \\
g(1,t) &= 0 \\
g(x,0) &= u(x) \\
\frac{\partial g(x,t)}{\partial t} \Big |_{t = 0} &= v(x)
\end{align*}
$$
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
$$
\begin{equation} \tag{35}
\frac{\partial^2 g(x,t)}{\partial t^2} = c^2 \frac{\partial^2 g(x,t)}{\partial x^2}
\end{equation}
$$
where \( c \) is the given wave speed. The chosen conditions for this equation are
$$
\begin{aligned}
g(0,t) &= 0, &t \geq 0 \\
g(1,t) &= 0, &t \geq 0 \\
g(x,0) &= u(x), &x\in[0,1] \\
\frac{\partial g(x,t)}{\partial t}\Big |_{t = 0} &= v(x), &x \in [0,1]
\end{aligned} \tag{36}
$$
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 (36) 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
$$
g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P)
$$
where
$$
h_1(x,t) = (1-t^2)u(x) + tv(x)
$$
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
$$
g(x,t) = \sin(\pi x)\cos(\pi t) - \sin(\pi x)\sin(\pi t)
$$
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()