<!-- dom:TITLE: Week 43: Solving Differential Equations with Deep Learning -->
# Week 43: Solving Differential Equations with Deep Learning
<!-- dom:AUTHOR: Morten Hjorth-Jensen at Department of Physics, University of Oslo & Department of Physics and Astronomy and Facility for Rare ion Beams, Michigan State University -->
<!-- Author: -->  
**Morten Hjorth-Jensen**, Department of Physics, University of Oslo and Department of Physics and Astronomy and Facility for Rare ion Beams, Michigan State University

Date: **Apr 23, 2021**

Copyright 1999-2021, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license



## Ordinary Differential Equations

An ordinary differential equation (ODE) is an equation involving functions having one variable.

In general, an ordinary differential equation looks like

<!-- Equation labels as ordinary links -->
<div id="ode"></div>

$$
\begin{equation} \label{ode} \tag{1}
f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) = 0
\end{equation}
$$

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](#ode)).
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](#ode)), some additional conditions of the function $g(x)$ are typically given
for the solution to be unique.

## The trial solution

Let the trial solution $g_t(x)$ be

<!-- Equation labels as ordinary links -->
<div id="_auto1"></div>

$$
\begin{equation}
	g_t(x) = h_1(x) + h_2(x,N(x,P))
\label{_auto1} \tag{2}
\end{equation}
$$

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.


## Minimization process

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](#ode)).
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

<!-- Equation labels as ordinary links -->
<div id="cost"></div>

$$
\begin{equation} \label{cost} \tag{3}
	C\left(\boldsymbol{x}, P\right) = \frac{1}{N} \sum_{i=1}^N \big(f\left(x_i, \, g(x_i), \, g'(x_i), \, g''(x_i), \, \dots \, , \, g^{(n)}(x_i)\right)\big)^2
\end{equation}
$$

The neural net should then find the parameters $P$ that minimizes the cost function in
([3](#cost)) for a set of $N$ training samples $x_i$.

## Minimizing the cost function using gradient descent and automatic differentiation

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](#cost)) 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.


## Example: Exponential decay

An exponential decay of a quantity $g(x)$ is described by the equation

<!-- Equation labels as ordinary links -->
<div id="solve_expdec"></div>

$$
\begin{equation} \label{solve_expdec} \tag{4}
  g'(x) = -\gamma g(x)
\end{equation}
$$

with $g(0) = g_0$ for some chosen initial value $g_0$.

The analytical solution of ([4](#solve_expdec)) is

<!-- Equation labels as ordinary links -->
<div id="_auto2"></div>

$$
\begin{equation}
  g(x) = g_0 \exp\left(-\gamma x\right)
\label{_auto2} \tag{5}
\end{equation}
$$

Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of ([4](#solve_expdec)).


## The function to solve for

The program will use a neural network to solve

<!-- Equation labels as ordinary links -->
<div id="solveode"></div>

$$
\begin{equation} \label{solveode} \tag{6}
g'(x) = -\gamma g(x)
\end{equation}
$$

where $g(0) = g_0$ with $\gamma$ and $g_0$ being some chosen values.

In this example, $\gamma = 2$ and $g_0 = 10$.

## The trial solution
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.

## Setup of Network

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](#solveode)). 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:

<!-- Equation labels as ordinary links -->
<div id="trial"></div>

$$
\begin{equation} \label{trial} \tag{7}
g_t(x, P) = g_0 + x \cdot N(x, P)
\end{equation}
$$

## Reformulating the problem

We wish that our neural network manages to minimize a given cost function.

A reformulation of out equation, ([6](#solveode)), 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](#trial)) satisfies ([6](#solveode)).

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

<!-- Equation labels as ordinary links -->
<div id="nnmin"></div>

$$
\begin{equation} \label{nnmin} \tag{8}
g_t'(x, P) = - \gamma g_t(x, P)
\end{equation}
$$

is fulfilled as *best as possible*.

## More technicalities

The left hand side and right hand side of ([8](#nnmin)) 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 $x$.

## More details

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

<!-- Equation labels as ordinary links -->
<div id="min"></div>

$$
\begin{equation} \label{min} \tag{9}
\min_{P}\Big\{\frac{1}{N} \sum_{i=1}^N  \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \Big\}
\end{equation}
$$

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

$$
\min_{P_{\text{hidden} }, \ P_{\text{output} }} C(\boldsymbol{x}, \{P_{\text{hidden} }, P_{\text{output} }\})
$$

## A possible implementation of a neural network

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](#min)).

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 $N_{\text{hidden} }$.

## Technicalities

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:

$$
\begin{aligned}
z_{i,j}^{\text{hidden}} &= b_i^{\text{hidden}} + w_i^{\text{hidden}}x_j \\
&=
\begin{pmatrix}
b_i^{\text{hidden}} & w_i^{\text{hidden}}
\end{pmatrix}
\begin{pmatrix}
1 \\
x_j
\end{pmatrix}
\end{aligned}
$$

## Final technicalities I

The result after weighting the inputs at the $i$-th hidden neuron can be written as a vector:

$$
\begin{aligned}
\boldsymbol{z}_{i}^{\text{hidden}} &= \Big( b_i^{\text{hidden}} + w_i^{\text{hidden}}x_1 , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_2, \ \dots \, , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_N\Big)  \\
&=
\begin{pmatrix}
 b_i^{\text{hidden}}  & w_i^{\text{hidden}}
\end{pmatrix}
\begin{pmatrix}
1  & 1 & \dots & 1 \\
x_1 & x_2 & \dots & x_N
\end{pmatrix} \\
&= \boldsymbol{p}_{i, \text{hidden}}^T X
\end{aligned}
$$

## Final technicalities II

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](#min)).

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.

## Final technicalities III


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.

$$
\begin{aligned}
z_{1,j}^{\text{output}} & =
\begin{pmatrix}
b_1^{\text{output}} & \boldsymbol{w}_1^{\text{output}}
\end{pmatrix}
\begin{pmatrix}
1 \\
\boldsymbol{x}_j^{\text{hidden}}
\end{pmatrix}
\end{aligned}
$$

## Final technicalities IV

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.

## Back propagation

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.

## Gradient descent

The idea of the gradient descent algorithm is to update parameters in
a direction where the cost function decreases goes to a minimum.

In general, the update of some parameters $\boldsymbol{\omega}$ given a cost
function defined by some weights $\boldsymbol{\omega}$, $C(\boldsymbol{x},
\boldsymbol{\omega})$, goes as follows:

$$
\boldsymbol{\omega}_{\text{new} } = \boldsymbol{\omega} - \lambda \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega})
$$

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

$$
\begin{aligned}
P_{\text{hidden},\text{new}} &= P_{\text{hidden}} - \lambda \nabla_{P_{\text{hidden}}} C(\boldsymbol{x}, P)  \\
P_{\text{output},\text{new}} &= P_{\text{output}} - \lambda \nabla_{P_{\text{output}}} C(\boldsymbol{x}, P)
\end{aligned}
$$

## The code for solving the ODE


```python
%matplotlib inline

import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt

def sigmoid(z):
    return 1/(1 + np.exp(-z))

# Assuming one input, hidden, and output layer
def neural_network(params, x):

    # Find the weights (including and biases) for the hidden and output layer.
    # Assume that params is a list of parameters for each layer.
    # The biases are the first element for each array in params,
    # and the weights are the remaning elements in each array in params.

    w_hidden = params[0]
    w_output = params[1]

    # Assumes input x being an one-dimensional array
    num_values = np.size(x)
    x = x.reshape(-1, num_values)

    # Assume that the input layer does nothing to the input x
    x_input = x

    ## Hidden layer:

    # Add a row of ones to include bias
    x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0)

    z_hidden = np.matmul(w_hidden, x_input)
    x_hidden = sigmoid(z_hidden)

    ## Output layer:

    # Include bias:
    x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0)

    z_output = np.matmul(w_output, x_hidden)
    x_output = z_output

    return x_output

# The trial solution using the deep neural network:
def g_trial(x,params, g0 = 10):
    return g0 + x*neural_network(params,x)

# The right side of the ODE:
def g(x, g_trial, gamma = 2):
    return -gamma*g_trial

# The cost function:
def cost_function(P, x):

    # Evaluate the trial function with the current parameters P
    g_t = g_trial(x,P)

    # Find the derivative w.r.t x of the neural network
    d_net_out = elementwise_grad(neural_network,1)(P,x)

    # Find the derivative w.r.t x of the trial function
    d_g_t = elementwise_grad(g_trial,0)(x,P)

    # The right side of the ODE
    func = g(x, g_t)

    err_sqr = (d_g_t - func)**2
    cost_sum = np.sum(err_sqr)

    return cost_sum / np.size(err_sqr)

# Solve the exponential decay ODE using neural network with one input, hidden, and output layer
def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb):
    ## Set up initial weights and biases

    # For the hidden layer
    p0 = npr.randn(num_neurons_hidden, 2 )

    # For the output layer
    p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included

    P = [p0, p1]

    print('Initial cost: %g'%cost_function(P, x))

    ## Start finding the optimal weights using gradient descent

    # Find the Python function that represents the gradient of the cost function
    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
    cost_function_grad = grad(cost_function,0)

    # Let the update be done num_iter times
    for i in range(num_iter):
        # Evaluate the gradient at the current weights and biases in P.
        # The cost_grad consist now of two arrays;
        # one for the gradient w.r.t P_hidden and
        # one for the gradient w.r.t P_output
        cost_grad =  cost_function_grad(P, x)

        P[0] = P[0] - lmb * cost_grad[0]
        P[1] = P[1] - lmb * cost_grad[1]

    print('Final cost: %g'%cost_function(P, x))

    return P

def g_analytic(x, gamma = 2, g0 = 10):
    return g0*np.exp(-gamma*x)

# Solve the given problem
if __name__ == '__main__':
    # Set seed such that the weight are initialized
    # with same weights and biases for every run.
    npr.seed(15)

    ## Decide the vales of arguments to the function to solve
    N = 10
    x = np.linspace(0, 1, N)

    ## Set up the initial parameters
    num_hidden_neurons = 10
    num_iter = 10000
    lmb = 0.001

    # Use the network
    P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb)

    # Print the deviation from the trial solution and true solution
    res = g_trial(x,P)
    res_analytical = g_analytic(x)

    print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical)))

    # Plot the results
    plt.figure(figsize=(10,10))

    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
    plt.plot(x, res_analytical)
    plt.plot(x, res[0,:])
    plt.legend(['analytical','nn'])
    plt.xlabel('x')
    plt.ylabel('g(x)')
    plt.show()
```

## The network with one input layer, specified number of hidden layers, and one output layer

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.


```python
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()
```

## Example: Population growth, comparing Autograd, TensorFlow, and Euler's scheme

A logistic model of population growth assumes that a population converges toward an equilibrium.
The population growth can be modeled by

<!-- Equation labels as ordinary links -->
<div id="log"></div>

$$
\begin{equation} \label{log} \tag{10}
	g'(t) = \alpha g(t)(A - g(t))
\end{equation}
$$

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),
a network has been constructed using TensorFlow also.
For comparison, the forward Euler method has been implemented in order to see how the networks performs compared to a numerical scheme.

## Setting up the problem

Here, we will model a population $g(t)$ in an environment having carrying capacity $A$.
The population follows the model

<!-- Equation labels as ordinary links -->
<div id="solveode_population"></div>

$$
\begin{equation} \label{solveode_population} \tag{11}
g'(t) = \alpha g(t)(A - g(t))
\end{equation}
$$

where $g(0) = g_0$.

In this example, we let $\alpha = 2$, $A = 1$, and $g_0 = 1.2$.

## The trial solution

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 program using Autograd

The network will be the similar as for the exponential decay example, but with some small modifications for our problem.


```python
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()
```

## Using forward Euler to solve the ODE

A straight-forward 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

<!-- Equation labels as ordinary links -->
<div id="odenum"></div>

$$
\begin{equation}
  \begin{aligned}
  g_i &= g(t_i) \\
  &= g(t_{i-1} + \Delta t) \\
  &\approx g(t_{i-1}) + \Delta t \big(\alpha g(t_{i-1})(A - g(t_{i-1}))\big) \\
  &= g_{i-1} + \Delta t \big(\alpha g_{i-1}(A - g_{i-1})\big)
  \end{aligned}
\end{equation} \label{odenum} \tag{12}
$$

for $i \geq 1$ and $g_0 = g(t_0) = g(0) = g_0$.

Equation ([12](#odenum)) could be implemented in the following way,
extending the program that uses the network using Autograd:


```python
# 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()
```

## Using TensorFlow to model logistic population growth

TensorFlow is a library widely used in the machine learning community.

Tensorflow is an open source library machine learning library
developed by the Google Brain team for internal use. It was released
under the Apache 2.0 open source license in November 9, 2015.

Tensorflow is a computational framework that allows you to construct
machine learning models at different levels of abstraction, from
high-level, object-oriented APIs like Keras, down to the C++ kernels
that Tensorflow is built upon. The higher levels of abstraction are
simpler to use, but less flexible, and our choice of implementation
should reflect the problems we are trying to solve.

[Tensorflow uses](https://www.tensorflow.org/guide/graphs) so-called graphs to represent your computation
in terms of the dependencies between individual operations, such that you first build a Tensorflow *graph*
to represent your model, and then create a Tensorflow *session* to run the graph.

In this guide we will analyze the same data as we did in our NumPy and
scikit-learn tutorial, gathered from the MNIST database of images. We
will give an introduction to the lower level Python Application
Program Interfaces (APIs), and see how we use them to build our graph.
Then we will build (effectively) the same graph in Keras, to see just
how simple solving a machine learning problem can be.

To install tensorflow on Unix/Linux systems, use pip as


```python
pip3 install tensorflow
```

and/or if you use **anaconda**, just write (or install from the graphical user interface)
(current release of CPU-only TensorFlow)


```python
conda create -n tf tensorflow
conda activate tf
```

To install the current release of GPU TensorFlow


```python
conda create -n tf-gpu tensorflow-gpu
conda activate tf-gpu
```

## Using Keras

Keras is a high level [neural network](https://en.wikipedia.org/wiki/Application_programming_interface)
that supports Tensorflow, CTNK and Theano as backends.  
If you have Anaconda installed you may run the following command


```python
conda install keras
```

You can look up the [instructions here](https://keras.io/) for more information.



## The general program flow in TensorFlow

Usually, a program in TensorFlow is divided into two parts; the
*construction phase* and the *execution phase*.  In the construction
phase, the computational graph that TensorFlow uses to perform its
calculations are set up.  In the execution phase, TensorFlow evaluates
any procedure that was defined in the construction phase.

## Program flow in TensorFlow - Construction phase

Here, the architecture for the neural network will be set up, along
with the cost function and an optimizer class used during training of
the network.  Note that TensorFlow uses a different convention for the
weighting done in each neuron in each layer within the network than in
the implementation using Autograd.  The matrix-vector multiplication
between the input from the previous layer and the weighting at the
neuron at current layer in the program using Autograd, is the
transpose of the convention used in TensorFlow. But it will not affect
that much our construction, as TensorFlow takes care of most of the
computations. The only thing we have to be aware of, is how the
dimensions are for our inputs.

## Program flow in TensorFlow - Execution phase

The computation graph has been defined, and is ready to be evaluated.
In order to get access to the graph, it has to be initialized and be runned within a Session.

## The full program modeling logistic population growth using TensorFlow


```python
# import necessary packages
import numpy as np
import matplotlib.pyplot as plt
import tensorflow as tf


# Just to reset the graph such that it is possible to rerun this in a
# Jupyter cell without resetting the whole kernel.
tf.reset_default_graph()

# Set a seed to ensure getting the same results from every run
tf.set_random_seed(4155)

Nt = 10
T = 1
t = np.linspace(0,T, Nt)

## The construction phase

# Convert the values the trial solution is evaluated at to a tensor.
t_tf = tf.convert_to_tensor(t.reshape(-1,1),dtype=tf.float64)
zeros = tf.reshape(tf.convert_to_tensor(np.zeros(t.shape)),shape=(-1,1))

# Define the parameters of the equation
alpha = tf.constant(2.,dtype=tf.float64)
A = tf.constant(1.,dtype=tf.float64)
g0 = tf.constant(1.2,dtype=tf.float64)

num_iter = 100000

# Define the number of neurons at each hidden layer
num_hidden_neurons = [100,50,25]
num_hidden_layers = np.size(num_hidden_neurons)

# Construct the network.
# tf.name_scope is used to group each step in the construction,
# just for a more organized visualization in TensorBoard
with tf.name_scope('dnn'):

    # Input layer
    previous_layer = t_tf

    # Hidden layers
    for l in range(num_hidden_layers):
        current_layer = tf.layers.dense(previous_layer, num_hidden_neurons[l], name='hidden%d'%(l+1), activation=tf.nn.sigmoid)
        previous_layer = current_layer

    # Output layer
    dnn_output = tf.layers.dense(previous_layer, 1, name='output')

# Define the cost function
with tf.name_scope('cost'):
    g_trial = g0 + t_tf*dnn_output
    d_g_trial = tf.gradients(g_trial,t_tf)

    func = alpha*g_trial*(A - g_trial)
    cost = tf.losses.mean_squared_error(zeros, d_g_trial[0] - func)


# Choose the method to minimize the cost function, along with a learning rate
learning_rate = 1e-2
with tf.name_scope('train'):
    optimizer = tf.train.GradientDescentOptimizer(learning_rate)
    traning_op = optimizer.minimize(cost)

# Set up a referance to the result from the neural network:
g_dnn_tf = None

# Define a node that initializes all of the other nodes in the computational graph
# used by TensorFlow:
init = tf.global_variables_initializer()

## Execution phase

# Start a session where the graph defined from the construction phase can be evaluated at:
with tf.Session() as sess:
    # Initialize the whole graph
    init.run()

    # Evaluate the initial cost:
    print('Initial cost: %g'%cost.eval())

    # The training of the network:
    for i in range(num_iter):
        sess.run(traning_op)

        # If one desires to see how the cost function behaves for each iteration:
        #if i % 1000 == 0:
        #    print(cost.eval())

    # Training is done, and we have an approximate solution to the ODE
    print('Final cost: %g'%cost.eval())

    # Store the result
    g_dnn_tf = g_trial.eval()

# Compare with analytical solution
def get_parameters():
    alpha = 2
    A = 1
    g0 = 1.2
    return alpha, A, g0

def g_analytic(t):
    alpha,A, g0 = get_parameters()
    return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t))

g_analytical = g_analytic(t)
diff_tf = g_dnn_tf - g_analytical.reshape(-1,1)

print('\nMax absolute difference between the analytical solution and solution from TensorFlow DNN: %g'%np.max(np.abs(diff_tf)))

# Plot the result
plt.figure(figsize=(10,10))

plt.title('Numerical solutions of the ODE')

plt.plot(t, g_dnn_tf)
plt.plot(t, g_analytical)

plt.legend(['dnn, tensorflow', 'exact'])
plt.xlabel('Time t')
plt.ylabel('g(t)')

plt.show()
```


```python


# ensure the same random numbers appear every time
np.random.seed(0)

# display images in notebook
%matplotlib inline
plt.rcParams['figure.figsize'] = (12,12)


# download MNIST dataset
digits = datasets.load_digits()

# define inputs and labels
inputs = digits.images
labels = digits.target

print("inputs = (n_inputs, pixel_width, pixel_height) = " + str(inputs.shape))
print("labels = (n_inputs) = " + str(labels.shape))


# flatten the image
# the value -1 means dimension is inferred from the remaining dimensions: 8x8 = 64
n_inputs = len(inputs)
inputs = inputs.reshape(n_inputs, -1)
print("X = (n_inputs, n_features) = " + str(inputs.shape))


# choose some random images to display
indices = np.arange(n_inputs)
random_indices = np.random.choice(indices, size=5)

for i, image in enumerate(digits.images[random_indices]):
    plt.subplot(1, 5, i+1)
    plt.axis('off')
    plt.imshow(image, cmap=plt.cm.gray_r, interpolation='nearest')
    plt.title("Label: %d" % digits.target[random_indices[i]])
plt.show()
```


```python
from tensorflow.keras.layers import Input
from tensorflow.keras.models import Sequential      #This allows appending layers to existing models
from tensorflow.keras.layers import Dense           #This allows defining the characteristics of a particular layer
from tensorflow.keras import optimizers             #This allows using whichever optimiser we want (sgd,adam,RMSprop)
from tensorflow.keras import regularizers           #This allows using whichever regularizer we want (l1,l2,l1_l2)
from tensorflow.keras.utils import to_categorical   #This allows using categorical cross entropy as the cost function

from sklearn.model_selection import train_test_split

# one-hot representation of labels
labels = to_categorical(labels)

# split into train and test data
train_size = 0.8
test_size = 1 - train_size
X_train, X_test, Y_train, Y_test = train_test_split(inputs, labels, train_size=train_size,
                                                    test_size=test_size)
```


```python

epochs = 100
batch_size = 100
n_neurons_layer1 = 100
n_neurons_layer2 = 50
n_categories = 10
eta_vals = np.logspace(-5, 1, 7)
lmbd_vals = np.logspace(-5, 1, 7)
def create_neural_network_keras(n_neurons_layer1, n_neurons_layer2, n_categories, eta, lmbd):
    model = Sequential()
    model.add(Dense(n_neurons_layer1, activation='sigmoid', kernel_regularizer=regularizers.l2(lmbd)))
    model.add(Dense(n_neurons_layer2, activation='sigmoid', kernel_regularizer=regularizers.l2(lmbd)))
    model.add(Dense(n_categories, activation='softmax'))
    
    sgd = optimizers.SGD(lr=eta)
    model.compile(loss='categorical_crossentropy', optimizer=sgd, metrics=['accuracy'])
    
    return model
```


```python
DNN_keras = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
        
for i, eta in enumerate(eta_vals):
    for j, lmbd in enumerate(lmbd_vals):
        DNN = create_neural_network_keras(n_neurons_layer1, n_neurons_layer2, n_categories,
                                         eta=eta, lmbd=lmbd)
        DNN.fit(X_train, Y_train, epochs=epochs, batch_size=batch_size, verbose=0)
        scores = DNN.evaluate(X_test, Y_test)
        
        DNN_keras[i][j] = DNN
        
        print("Learning rate = ", eta)
        print("Lambda = ", lmbd)
        print("Test accuracy: %.3f" % scores[1])
        print()
```


```python
# optional
# visual representation of grid search
# uses seaborn heatmap, could probably do this in matplotlib
import seaborn as sns

sns.set()

train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))

for i in range(len(eta_vals)):
    for j in range(len(lmbd_vals)):
        DNN = DNN_keras[i][j]

        train_accuracy[i][j] = DNN.evaluate(X_train, Y_train)[1]
        test_accuracy[i][j] = DNN.evaluate(X_test, Y_test)[1]

        
fig, ax = plt.subplots(figsize = (10, 10))
sns.heatmap(train_accuracy, annot=True, ax=ax, cmap="viridis")
ax.set_title("Training Accuracy")
ax.set_ylabel("$\eta$")
ax.set_xlabel("$\lambda$")
plt.show()

fig, ax = plt.subplots(figsize = (10, 10))
sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis")
ax.set_title("Test Accuracy")
ax.set_ylabel("$\eta$")
ax.set_xlabel("$\lambda$")
plt.show()
```

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2021-09-04T16:21:16.000Z
2021-09-04T16:21:16.000Z
Qwen/Qwen-72B
1. YES
2. YES

