# Indirect method for solving non-linear optimal control of a non-rectilinear motion of a rocket with variable mass

Mohamed Aliane, Nacima Moussouni, Kahina Louadj, Nicolas Boizot

► To cite this version:

Mohamed Aliane, Nacima Moussouni, Kahina Louadj, Nicolas Boizot. Indirect method for solving non-linear optimal control of a non-rectilinear motion of a rocket with variable mass. Boletim da Sociedade Paranaense de Matematica, 2024, 42, pp.1-10. 10.5269/bspm.62778. hal-04594709

HAL Id: hal-04594709

https://univ-tln.hal.science/hal-04594709

Submitted on 30 May 2024

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# Indirect Method for Solving Non-Linear Optimal Control of a Non-Rectilinear Motion of a Rocket With Variable Mass

Mohamed Aliane*, Nacima Moussouni, Kahina Louadj, Nicolas Boizot

**ABSTRACT:** In this paper, an optimal trajectory of the rocket angle with a variable mass will be calculated by considering the aerodynamic forces, the acceleration of gravity and moves with a non-rectilinear motion from an initial state to a final state with a known altitude. The aim is to optimize the lateral offset of the rocket. For this, we formulate an optimal control problem where the rocket angle is the control. In order to solve the problem, let applied Shooting method's based on the Pontryagin's maximum principle, and study the precision and a duration time. Finally, we validate the results by using MATLAB software.

**Key Words:** Optimal control, shooting method, aerodynamic forces, rocket.

## Contents

<table><tr><td>1 Introduction</td><td>1</td></tr><tr><td>2 Problem Statement</td><td>1</td></tr><tr><td>3 Solution of Problem by using Indirect Method</td><td>3</td></tr><tr><td>3.1 Necessary conditions of optimality</td><td>3</td></tr><tr><td>3.1.1 The trانversality conditions</td><td>5</td></tr><tr><td>3.2 Numerical solution by the Shooting method</td><td>5</td></tr><tr><td>4 Conclusion</td><td>9</td></tr></table>

## 1. Introduction

In this paper, we have presented a nonlinear optimal control model with free final time to maximize the lateral offset of a rocket moving from an initial position to a final position, where the control represents the flight bath angle of the rocket. In order to solve the considered problem, first the Pontryagin's maximum principle [15] is applied, then a numerical solution is found using the shooting method.

The problem of maximizing the lateral offset of a rocket which a non-rectilinear motion is formulated in Section 2. The resolution is detailed in Section 3. In particular, numerical results obtained with the help of the software MATLAB are given in Section 3.2. Finally, Section 4 concludes the article.

## 2. Problem Statement

Let consider $m(t)$, $t \in [0,T]$ the mass of the Rocket which moves from an initial position $M_0 = (x_{10}, x_{20})$ to a final position $M_f = (x_{1f}, x_{2f})$, where $x_{1f}$ is free.
$x(t) = (x_1(t), x_2(t))$ and $v(t) = (v_1(t), v_2(t))$ are respectively the position and speed of the rocket at the instant $t$. The motion equations are given by:

$$ \begin{cases} \dot{x}_1(t) = v_1(t), \\ \dot{x}_2(t) = v_2(t), \quad t \in [0, T]. \end{cases} \tag{2.1} $$

Let $T_p(t)$ and $\theta(t)$ are respectively the thrust and the flight bath angle of the rocket. We have

$$ \vec{T}_p(t) = u_{\max} \begin{pmatrix} \cos(\theta(t)) \\ \sin(\theta(t)) \end{pmatrix}, \tag{2.2} $$

<footnote n="2010">2010 Mathematics Subject Classification: 93C05, 53C35.
Submitted March 04, 2022. Published May 19, 2023</footnote>
<figure><img src="image_2.png" /></figure>

# Indirect Method for Solving Non-Linear Optimal Control of a Non-Rectilinear Motion of a Rocket With Variable Mass

Mohamed Aliane*, Nacima Moussouni, Kahina Louadj, Nicolas Boizot

**ABSTRACT:** In this paper, an optimal trajectory of the rocket angle with a variable mass will be calculated by considering the aerodynamic forces, the acceleration of gravity and moves with a non-rectilinear motion from an initial state to a final state with a known altitude. The aim is to optimize the lateral offset of the rocket. For this, we formulate an optimal control problem where the rocket angle is the control. In order to solve the problem, let applied Shooting method's based on the Pontryagin's maximum principle, and study the precision and a duration time. Finally, we validate the results by using MATLAB software.

**Key Words:** Optimal control, shooting method, aerodynamic forces, rocket.

## Contents

<table><tr><td>1 Introduction</td><td>1</td></tr><tr><td>2 Problem Statement</td><td>1</td></tr><tr><td>3 Solution of Problem by using Indirect Method</td><td>3</td></tr><tr><td>3.1 Necessary conditions of optimality</td><td>3</td></tr><tr><td>3.1.1 The trانversality conditions</td><td>5</td></tr><tr><td>3.2 Numerical solution by the Shooting method</td><td>5</td></tr><tr><td>4 Conclusion</td><td>9</td></tr></table>

## 1. Introduction

In this paper, we have presented a nonlinear optimal control model with free final time to maximize the lateral offset of a rocket moving from an initial position to a final position, where the control represents the flight bath angle of the rocket. In order to solve the considered problem, first the Pontryagin's maximum principle [15] is applied, then a numerical solution is found using the shooting method.

The problem of maximizing the lateral offset of a rocket which a non-rectilinear motion is formulated in Section 2. The resolution is detailed in Section 3. In particular, numerical results obtained with the help of the software MATLAB are given in Section 3.2. Finally, Section 4 concludes the article.

## 2. Problem Statement

Let consider $m(t)$, $t \in [0,T]$ the mass of the Rocket which moves from an initial position $M_0 = (x_{10}, x_{20})$ to a final position $M_f = (x_{1f}, x_{2f})$, where $x_{1f}$ is free.
$x(t) = (x_1(t), x_2(t))$ and $v(t) = (v_1(t), v_2(t))$ are respectively the position and speed of the rocket at the instant $t$. The motion equations are given by:

$$ \begin{cases} \dot{x}_1(t) = v_1(t), \\ \dot{x}_2(t) = v_2(t), \quad t \in [0, T]. \end{cases} \tag{2.1} $$

Let $T_p(t)$ and $\theta(t)$ are respectively the thrust and the flight bath angle of the rocket. We have

$$ \vec{T}_p(t) = u_{\max} \begin{pmatrix} \cos(\theta(t)) \\ \sin(\theta(t)) \end{pmatrix}, \tag{2.2} $$

<footnote n="2010">2010 Mathematics Subject Classification: 93C05, 53C35.
Submitted March 04, 2022. Published May 19, 2023</footnote>
where $u_{\max} = |\vec{T}_p(t)|$. The earth is supposed to be flat. Let consider aerodynamic forces $\vec{F}_a$ and the acceleration of the gravity $g$ is constant ($g = 9.80665 m.s^{-2}$). We obtain the following equations:

$$F_a(x_2(t), v_1(t), v_2(t)) = \alpha(v_1(t)^2 + v_2(t)^2)e^{-\beta x_2(t)} \begin{pmatrix} \cos(\phi(t)) \\ \sin(\phi(t)) \end{pmatrix}, \quad \vec{g} = \begin{pmatrix} 0 \\ -g \end{pmatrix}, \quad (2.3)$$

where $\alpha$, $\beta$ and $\phi$ the parameters of the aerodynamic forces with:

* $\alpha$: is the product of the drag coefficient: the surface of the machine and the atmospheric density.

* $\beta$: is the inverse of the altitude scale.

* $\phi$: is the angle of attack that the speed vector makes with the horizontal, hence:

$$\cos(\phi(t)) = \frac{v_1(t)}{\sqrt{v_1(t)^2 + v_2(t)^2}}, \quad (2.4)$$

and

$$\sin(\phi(t)) = \frac{v_2(t)}{\sqrt{v_1(t)^2 + v_2(t)^2}}, \quad (2.5)$$

<figure><img src="image_2.png" /><figcaption>Figure 1: Forces applied</figcaption></figure>

The dynamics are given by law of quantity of the movement theorem is given by the following formulas:

$$\frac{d\vec{p}}{dt} = \frac{(m + dm)(\vec{v} + d\vec{v}) - dm(\vec{v} + \vec{k}) - m\vec{v}}{dt} = \vec{R}. \quad (2.6)$$

By simplifying the formula (2.6), after eliminating the quadratic term to the differentials, we obtain:

$$\frac{d\vec{p}(t)}{dt} = m(t)\frac{d\vec{v}(t)}{dt} - \vec{k}\frac{dm(t)}{dt} = \vec{R}(t), \quad (2.7)$$

therefore

$$m(t)\frac{d\vec{v}}{dt} - \vec{k}\frac{dm(t)}{dt} = m(t)\vec{g} + \vec{F}_a(x_2(t), v_1(t), v_2(t)), \quad (2.8)$$

where $\vec{R} = m(t)\vec{g} + \vec{F}_a(x_2(t), v_1(t), v_2(t))$.

We obtain the following equation:

$$\frac{d\vec{v}(t)}{dt} = \vec{g} + \frac{\vec{T}_p(t)}{m(t)} + \frac{\vec{F}_a(x_2(t), v_1(t), v_2(t))}{m(t)}. \quad (2.9)$$
where $u_{\max} = |\vec{T}_p(t)|$. The earth is supposed to be flat. Let consider aerodynamic forces $\vec{F}_a$ and the acceleration of the gravity $g$ is constant ($g = 9.80665 m.s^{-2}$). We obtain the following equations:

$$F_a(x_2(t), v_1(t), v_2(t)) = \alpha(v_1(t)^2 + v_2(t)^2)e^{-\beta x_2(t)} \begin{pmatrix} \cos(\phi(t)) \\ \sin(\phi(t)) \end{pmatrix}, \quad \vec{g} = \begin{pmatrix} 0 \\ -g \end{pmatrix}, \quad (2.3)$$

where $\alpha$, $\beta$ and $\phi$ the parameters of the aerodynamic forces with:

* $\alpha$: is the product of the drag coefficient: the surface of the machine and the atmospheric density.

* $\beta$: is the inverse of the altitude scale.

* $\phi$: is the angle of attack that the speed vector makes with the horizontal, hence:

$$\cos(\phi(t)) = \frac{v_1(t)}{\sqrt{v_1(t)^2 + v_2(t)^2}}, \quad (2.4)$$

and

$$\sin(\phi(t)) = \frac{v_2(t)}{\sqrt{v_1(t)^2 + v_2(t)^2}}, \quad (2.5)$$

<figure><img src="image_3.png" /><figcaption>Figure 1: Forces applied</figcaption></figure>

The dynamics are given by law of quantity of the movement theorem is given by the following formulas:

$$\frac{d\vec{p}}{dt} = \frac{(m + dm)(\vec{v} + d\vec{v}) - dm(\vec{v} + \vec{k}) - m\vec{v}}{dt} = \vec{R}. \quad (2.6)$$

By simplifying the formula (2.6), after eliminating the quadratic term to the differentials, we obtain:

$$\frac{d\vec{p}(t)}{dt} = m(t)\frac{d\vec{v}(t)}{dt} - \vec{k}\frac{dm(t)}{dt} = \vec{R}(t), \quad (2.7)$$

therefore

$$m(t)\frac{d\vec{v}}{dt} - \vec{k}\frac{dm(t)}{dt} = m(t)\vec{g} + \vec{F}_a(x_2(t), v_1(t), v_2(t)), \quad (2.8)$$

where $\vec{R} = m(t)\vec{g} + \vec{F}_a(x_2(t), v_1(t), v_2(t))$.

We obtain the following equation:

$$\frac{d\vec{v}(t)}{dt} = \vec{g} + \frac{\vec{T}_p(t)}{m(t)} + \frac{\vec{F}_a(x_2(t), v_1(t), v_2(t))}{m(t)}, \quad (2.9)$$
where

$$ \overrightarrow{T}_p = \overrightarrow{k} \frac{dm}{dt}, \frac{dm}{dt} = -bu_{\max} \quad \text{and} \quad \| \overrightarrow{k} \| = \frac{1}{b}. \qquad (2.10) $$

By projecting on the x-axis, we obtain:

$$ \frac{dv_1(t)}{dt} = \frac{u_{\max}}{m(t)} \cos(\theta(t)) - \frac{\alpha}{m(t)} v_1(t) \sqrt{v_1(t)^2 + v_2(t)^2} e^{-\beta x_2(t)}, \qquad (2.11) $$

and

$$ \frac{dv_2(t)}{dt} = \frac{u_{\max}}{m(t)} \sin(\theta(t)) - \frac{\alpha}{m(t)} v_2(t) \sqrt{v_1(t)^2 + v_2(t)^2} e^{-\beta x_2(t)} - g. \qquad (2.12) $$

We formulate our problem by the optimal control problem whose the control represents the flight bank angle and our goal is to bring back the orbit rocket of an altitude known a priori with a maximum lateral offset. The mathematical model is given as follows whose used the following data [19]:

$$ \left\{ \begin{array}{l} \text{Minimize } J(\theta, T) = -x_1(T), \\ \dot{x}_1(t) = v_1(t), \\ \dot{x}_2(t) = v_2(t), \\ \dot{v}_1(t) = \frac{u_{\max}}{m(t)} \cos(\theta(t)) - \frac{\alpha v_1(t) \sqrt{v_1(t)^2 + v_2(t)^2}}{m(t)} e^{-\beta x_2(t)}, \\ \dot{v}_2(t) = \frac{u_{\max}}{m(t)} \sin(\theta(t)) - \frac{\alpha v_2(t) \sqrt{v_1(t)^2 + v_2(t)^2}}{m(t)} e^{-\beta x_2(t)} - g, \\ \dot{m}(t) = -bu_{\max}, \\ x_1(0) = 0, x_2(0) = 0.005, v_1(0) = 0, v_2(0) = 0.01, m(0) = m_0 \\ x_2(T) = h, v_1(T) = v_c, v_2(T) = 0, \\ \theta(t) \in \mathbb{R}, t \in [0, T], T \text{ free,} \end{array} \right. \qquad (2.13) $$

where

$$ \alpha = 2.164, \beta = 0.113, v_c = 7.905 \, \text{km} \cdot \text{s}^{-1}, h = 180 \, \text{km}, m_0 = 122176.39 \, \text{kg}. $$

### 3. Solution of Problem by using Indirect Method

The analytical solution of the problem is very complicated. For this, the problem is solved numerically with the indirect shooting method based on the Pontryagin's maximum Principle.

#### 3.1. Necessary conditions of optimality

The theoretical solution of the problem is based on the Pontryagin's maximum principle, which gives a necessary condition of optimality. The Hamiltonian of the problem (2.13) is defined as follows:

$$ \begin{array}{l} H: \mathbb{R}^5 \times \mathbb{R}^5 \times \mathbb{R} \quad \rightarrow \quad \mathbb{R} \\ (x(t), p(t), \theta(t)) \quad \longmapsto \quad H(x(t), p(t), \theta(t)), \end{array} \qquad (3.1) $$

where

$$ x(t) = (x_1(t), x_2(t), v_1(t), v_2(t), m(t)) $$

and

$$ p(t) = (p_j(t), j = 1 \dots 5). $$

$$ \begin{aligned} H = & p_1(t)v_1(t) + p_2(t)v_2(t) \\ & + p_3(t) \left( \frac{u_{\max}}{m(t)} \cos(\theta(t)) - \frac{\alpha}{m(t)} v_1(t) \sqrt{v_1(t)^2 + v_2(t)^2} e^{-\beta x_2(t)} \right) \\ & + p_4(t) \left( \frac{u_{\max}}{m(t)} \sin(\theta(t)) - \frac{\alpha}{m(t)} v_2(t) \sqrt{v_1(t)^2 + v_2(t)^2} e^{-\beta x_2(t)} - g \right) \\ & - bp_5(t)u_{\max}. \end{aligned} \qquad (3.2) $$
where

$$ \vec{T}_p = \vec{k} \frac{dm}{dt}, \frac{dm}{dt} = -bu_{\max} \text{ and } \| \vec{k} \| = \frac{1}{b}. \qquad (2.10) $$

By projecting on the x-axis, we obtain:

$$ \frac{dv_1(t)}{dt} = \frac{u_{\max}}{m(t)} \cos(\theta(t)) - \frac{\alpha}{m(t)} v_1(t) \sqrt{v_1(t)^2 + v_2(t)^2} e^{-\beta x_2(t)}, \qquad (2.11) $$

and

$$ \frac{dv_2(t)}{dt} = \frac{u_{\max}}{m(t)} \sin(\theta(t)) - \frac{\alpha}{m(t)} v_2(t) \sqrt{v_1(t)^2 + v_2(t)^2} e^{-\beta x_2(t)} - g. \qquad (2.12) $$

We formulate our problem by the optimal control problem whose the control represents the flight bank angle and our goal is to bring back the orbit rocket of an altitude known a priori with a maximum lateral offset. The mathematical model is given as follows whose used the following data [19]:

$$
\left\{
\begin{array}{l}
\text{Minimize } J(\theta, T) = -x_1(T), \\
\dot{x}_1(t) = v_1(t), \\
\dot{x}_2(t) = v_2(t), \\
\dot{v}_1(t) = \frac{u_{\max}}{m(t)} \cos(\theta(t)) - \frac{\alpha v_1(t) \sqrt{v_1(t)^2 + v_2(t)^2}}{m(t)} e^{-\beta x_2(t)}, \\
\dot{v}_2(t) = \frac{u_{\max}}{m(t)} \sin(\theta(t)) - \frac{\alpha v_2(t) \sqrt{v_1(t)^2 + v_2(t)^2}}{m(t)} e^{-\beta x_2(t)} - g, \\
\dot{m}(t) = -bu_{\max}, \\
x_1(0) = 0, x_2(0) = 0.005, v_1(0) = 0, v_2(0) = 0.01, m(0) = m_0 \\
x_2(T) = h, v_1(T) = v_c, v_2(T) = 0, \\
\theta(t) \in \mathbb{R}, t \in [0, T], T \text{ free},
\end{array}
\right.
\qquad (2.13)
$$

where

$$ \alpha = 2.164, \beta = 0.113, v_c = 7.905 \, \text{km} \cdot \text{s}^{-1}, h = 180 \, \text{km}, m_0 = 122176.39 \, \text{kg}. $$

### 3. Solution of Problem by using Indirect Method

The analytical solution of the problem is very complicated. For this, the problem is solved numerically with the indirect shooting method based on the Pontryagin's maximum Principle.

#### 3.1. Necessary conditions of optimality

The theoretical solution of the problem is based on the Pontryagin's maximum principle, which gives a necessary condition of optimality. The Hamiltonian of the problem (2.13) is defined as follows:

$$ H: \mathbb{R}^5 \times \mathbb{R}^5 \times \mathbb{R} \quad \rightarrow \mathbb{R} \\
(x(t), p(t), \theta(t)) \quad \longmapsto H(x(t), p(t), \theta(t)), \qquad (3.1) $$

where

$$ x(t) = (x_1(t), x_2(t), v_1(t), v_2(t), m(t)) $$

and

$$ p(t) = (p_j(t), j = 1 \dots 5). $$

$$ H = p_1(t)v_1(t) + p_2(t)v_2(t) \\ + p_3(t) \left( \frac{u_{\max}}{m(t)} \cos(\theta(t)) - \frac{\alpha}{m(t)} v_1(t) \sqrt{v_1(t)^2 + v_2(t)^2} e^{-\beta x_2(t)} \right) \\ + p_4(t) \left( \frac{u_{\max}}{m(t)} \sin(\theta(t)) - \frac{\alpha}{m(t)} v_2(t) \sqrt{v_1(t)^2 + v_2(t)^2} e^{-\beta x_2(t)} - g \right) \\ - bp_5(t)u_{\max}. \qquad (3.2) $$
where $p(t)$, is the adjoint vector, it is solution of the following system:

$$ egin{cases} 
\dot{p}_1(t) = 0, \\
p_2(t) = (p_3(t)v_1(t) + p_4(t)v_2(t)) \frac{-\alpha\beta\sqrt{v_1(t)^2 + v_2(t)^2}}{m(t)} e^{-\beta x_2(t)}, \\
\dot{p}_3(t) = -p_1(t) + \alpha \frac{p_3(t)(2v_1(t)^2 + v_2(t)^2) + p_4(t)v_1(t)v_2(t)}{m(t)\sqrt{v_1(t)^2 + v_2(t)^2}} e^{-\beta x_2(t)}, \\
\dot{p}_4(t) = -p_2(t) + \alpha \frac{p_3(t)v_1(t)v_2(t) + p_4(t)(v_1(t)^2 + 2v_2(t)^2)}{m(t)\sqrt{v_1(t)^2 + v_2(t)^2}} e^{-\beta x_2(t)}, \\
\dot{p}_5(t) = \frac{u_{\max}}{m(t)^2}(p_3(t)\cos(\theta(t)) + p_4(t)\sin(\theta(t))) \\
\qquad - \frac{\alpha\sqrt{v_1(t)^2 + v_2(t)^2}e^{-\beta x_2(t)}}{m(t)^2}(p_3(t)v_1(t) + p_4(t)v_2(t)), \quad t \in [0, T].
\end{cases} 
\quad (3.3) $$

The first order optimality condition gives:

$$ \frac{\partial H}{\partial \theta} = -\frac{u_{\max}}{m(t)} p_3(t) \sin(\theta) + \frac{u_{\max}}{m(t)} p_4(t) \cos(\theta) = 0, \quad (3.4) $$

Since $p_3(t) \neq 0, \forall t \in [0, T]$, the equation (3.4) is equivalent to:

$$ \tan(\theta(t)) = \frac{p_4(t)}{p_3(t)}, \quad \text{for all } t \in [0, T], \quad (3.5) $$

So

$$ \theta^*(t) = \arctan \left[ \frac{p_4(t)}{p_3(t)} \right], \quad t \in [0, T]. \quad (3.6) $$

The equation (3.5) is also equivalent to:

$$ \cos(\theta(t)) = \frac{p_3(t)}{\sqrt{p_3(t)^2 + p_4(t)^2}}, \quad (3.7) $$

$$ \sin(\theta(t)) = \frac{p_4(t)}{\sqrt{p_3(t)^2 + p_4(t)^2}}, \quad t \in [0, T]. \quad (3.8) $$

Furthermore, we have

$$ 
\begin{aligned}
\frac{\partial^2 H}{\partial \theta^2} &= -\frac{u_{\max}}{m(t)} p_3(t) \cos(\theta) - \frac{u_{\max}}{m(t)} p_4(t) \sin(\theta) \\
&= -\frac{u_{\max}}{m(t)} \left( p_3(t) \frac{p_3(t)}{\sqrt{p_3(t)^2 + p_4(t)^2}} \right) \\
&\quad -\frac{u_{\max}}{m(t)} \left( p_4(t) \frac{p_4(t)}{\sqrt{p_3(t)^2 + p_4(t)^2}} \right) \\
&= -\frac{u_{\max}}{m(t)} \sqrt{p_3(t)^2 + p_4(t)^2} < 0.
\end{aligned}
\quad (3.9) $$

Since $u_{\max}$ and $m(t), \forall t \in [0, T]$, are nonnegative, we get $\frac{\partial^2 H}{\partial \theta^2} < 0$.

Therefore, the second order necessary optimality condition is checked. The dynamic system is autonomous, so the Hamiltonian is constant for all $t \in [0, T]$. Therefore, we will have

$$ H(t, x(t), p(t), \theta(t)) = c. \quad (3.10) $$
where $p(t)$, is the adjoint vector, it is solution of the following system:

$$ 
\left\{ 
\begin{aligned}
\dot{p}_1(t) &= 0, \\
p_2(t) &= (p_3(t)v_1(t) + p_4(t)v_2(t)) \frac{-\alpha\beta\sqrt{v_1(t)^2 + v_2(t)^2}}{m(t)} e^{-\beta x_2(t)}, \\
\dot{p}_3(t) &= -p_1(t) + \alpha \frac{p_3(t)(2v_1(t)^2 + v_2(t)^2) + p_4(t)v_1(t)v_2(t)}{m(t)\sqrt{v_1(t)^2 + v_2(t)^2}} e^{-\beta x_2(t)}, \\
\dot{p}_4(t) &= -p_2(t) + \alpha \frac{p_3(t)v_1(t)v_2(t) + p_4(t)(v_1(t)^2 + 2v_2(t)^2)}{m(t)\sqrt{v_1(t)^2 + v_2(t)^2}} e^{-\beta x_2(t)}, \\
\dot{p}_5(t) &= \frac{u_{\max}}{m(t)^2}(p_3(t)\cos(\theta(t)) + p_4(t)\sin(\theta(t))) \\
&\quad - \frac{\alpha\sqrt{v_1(t)^2 + v_2(t)^2}e^{-\beta x_2(t)}}{m(t)^2}(p_3(t)v_1(t) + p_4(t)v_2(t)), \quad t \in [0, T].
\end{aligned}
\right. 
\quad (3.3) 
$$

The first order optimality condition gives:

$$ \frac{\partial H}{\partial \theta} = -\frac{u_{\max}}{m(t)} p_3(t) \sin(\theta) + \frac{u_{\max}}{m(t)} p_4(t) \cos(\theta) = 0, \quad (3.4) $$

Since $p_3(t) \neq 0, \forall t \in [0, T]$, the equation (3.4) is equivalent to:

$$ \tan(\theta(t)) = \frac{p_4(t)}{p_3(t)}, \quad \text{for all } t \in [0, T], \quad (3.5) $$

So

$$ \theta^*(t) = \arctan \left[ \frac{p_4(t)}{p_3(t)} \right], \quad t \in [0, T]. \quad (3.6) $$

The equation (3.5) is also equivalent to:

$$ \cos(\theta(t)) = \frac{p_3(t)}{\sqrt{p_3(t)^2 + p_4(t)^2}}, \quad (3.7) $$

$$ \sin(\theta(t)) = \frac{p_4(t)}{\sqrt{p_3(t)^2 + p_4(t)^2}}, \quad t \in [0, T]. \quad (3.8) $$

Furthermore, we have

$$ 
\begin{aligned}
\frac{\partial^2 H}{\partial \theta^2} &= -\frac{u_{\max}}{m(t)} p_3(t) \cos(\theta) - \frac{u_{\max}}{m(t)} p_4(t) \sin(\theta) \\
&= -\frac{u_{\max}}{m(t)} \left( p_3(t) \frac{p_3(t)}{\sqrt{p_3(t)^2 + p_4(t)^2}} \right) \\
&\quad -\frac{u_{\max}}{m(t)} \left( p_4(t) \frac{p_4(t)}{\sqrt{p_3(t)^2 + p_4(t)^2}} \right) \\
&= -\frac{u_{\max}}{m(t)} \sqrt{p_3(t)^2 + p_4(t)^2} < 0.
\end{aligned} 
\quad (3.9) 
$$

Since $u_{\max}$ and $m(t), \forall t \in [0, T]$, are nonnegative, we get $\frac{\partial^2 H}{\partial \theta^2} < 0$.

Therefore, the second order necessary optimality condition is checked. The dynamic system is autonomous, so the Hamiltonian is constant for all $t \in [0, T]$. Therefore, we will have

$$ H(t, x(t), p(t), \theta(t)) = c. \quad (3.10) $$
3.1.1. The transversality conditions. Since T is free, according to the condition of transversality, we have:

$$H(T, x(T), p(T), \theta(T)) = 0. \tag{3.11}$$

From relationships (3.10) and (3.11), we get

$$H(t, x(t), p(t), \theta(t)) = 0, \quad \forall t \in [0, T]. \tag{3.12}$$

The transversality conditions can be used to calculate the first and the last components of $p(T)$.

## 3.2. Numerical solution by the Shooting method

The Shooting method is based on the Pontryagin's maximum principle. It consists to find a zero of the Shooting function by using Newton method's. The shooting method defined in the three following steps [1,2,12,17]:

* Step 1: Form a boundary value problem using the model equations and the adjoint vectors equations as well as the transversality conditions.

* Step 2: Determine the shooting function.

* Step 3: Solve a system of nonlinear equations.

Based on the Pontryagin's maximum principle, optimality equations are given as follows:

$$
\left\{
\begin{aligned}
\dot{x}_1(t) &= x_3(t), \\
\dot{x}_2(t) &= x_4(t), \\
\dot{x}_3(t) &= \frac{u_{\max}}{x_5(t)} \frac{p_3(t)}{\sqrt{p_3(t)^2 + p_4(t)^2}} - \frac{\alpha x_3(t) \sqrt{x_3(t)^2 + x_4(t)^2}}{x_5(t)} e^{-\beta x_2(t)}, \\
\dot{x}_4(t) &= \frac{u_{\max}}{x_5(t)} \frac{p_4(t)}{\sqrt{p_3(t)^2 + p_4(t)^2}} - \frac{\alpha x_4(t) \sqrt{x_3(t)^2 + x_4(t)^2}}{x_5(t)} e^{-\beta x_2(t)} - g, \\
\dot{x}_5(t) &= -b u_{\max}, \\
\dot{p}_1(t) &= 0, \\
\dot{p}_2(t) &= -(p_3(t)x_3(t) + p_4(t)x_4(t)) \frac{\alpha \beta \sqrt{x_3(t)^2 + x_4(t)^2}}{x_5(t)} e^{-\beta x_2(t)}, \\
\dot{p}_3(t) &= -p_1(t) + \alpha \frac{p_3(t)(2x_3(t)^2 + x_4(t)^2) + p_4(t)x_3(t)v_2(t)}{x_5(t) \sqrt{x_3(t)^2 + x_4(t)^2}} e^{-\beta x_2(t)}, \\
\dot{p}_4(t) &= -p_2(t) + \alpha \frac{p_3(t)x_3(t)x_4(t) + p_4(t)(x_3(t)^2 + 2x_4(t)^2)}{x_5(t) \sqrt{x_3(t)^2 + x_4(t)^2}} e^{-\beta x_2(t)}, \\
\dot{p}_5(t) &= \frac{u_{\max}}{x_5(t)^2} \sqrt{p_3(t) + p_4(t)} - \frac{\alpha \sqrt{x_3(t)^2 + x_4(t)^2} e^{-\beta x_2(t)}}{x_5(t)^2} (p_3(t)x_3(t) + p_4(t)x_4(t)), \\
x_1(0) &= 0, \quad x_2(0) = 0.005, \quad x_3(0) = 0, \quad x_4(0) = 0.01, \quad x_5(0) = m_0, \quad p_1(T) = 1, \quad p_5(T) = 0, \\
x_2(T) &= h, \quad x_4(T) = v_c, \quad x_4(T) = 0, \quad t \in [0, T],
\end{aligned}
\right.
\tag{3.13}
$$

where

$$x(t) = (x_j(t), j = 1 \dots 5) = (x_1(t), x_2(t), v_1(t), v_2(t), m(t)).$$

We construct the Shooting function:

$$G: \begin{array}{rcl} \mathbb{R}^{10} & \to & \mathbb{R}^{10} \\ (p(0), p(T)) & \to & G(p(0), p(T)), \end{array} \tag{3.14}$$
### 3.1.1. The transversality conditions.
Since $T$ is free, according to the condition of transversality, we have:

$$H(T, x(T), p(T), \theta(T)) = 0. \qquad (3.11)$$

From relationships (3.10) and (3.11), we get

$$H(t, x(t), p(t), \theta(t)) = 0, \quad \forall t \in [0, T]. \qquad (3.12)$$

The transversality conditions can be used to calculate the first and the last components of $p(T)$.

## 3.2. Numerical solution by the Shooting method

The Shooting method is based on the Pontryagin's maximum principle. It consists to find a zero of the Shooting function by using Newton method's. The shooting method defined in the three following steps [1,2,12,17]:

* Step 1: Form a boundary value problem using the model equations and the adjoint vectors equations as well as the transversality conditions.

* Step 2: Determine the shooting function.

* Step 3: Solve a system of nonlinear equations.

Based on the Pontryagin's maximum principle, optimality equations are given as follows:

$$ 
\left\{ 
\begin{aligned}
\dot{x}_1(t) &= x_3(t), \\
\dot{x}_2(t) &= x_4(t), \\
\dot{x}_3(t) &= \frac{u_{\max}}{x_5(t)} \frac{p_3(t)}{\sqrt{p_3(t)^2 + p_4(t)^2}} - \frac{\alpha x_3(t) \sqrt{x_3(t)^2 + x_4(t)^2}}{x_5(t)} e^{-\beta x_2(t)}, \\
\dot{x}_4(t) &= \frac{u_{\max}}{x_5(t)} \frac{p_4(t)}{\sqrt{p_3(t)^2 + p_4(t)^2}} - \frac{\alpha x_4(t) \sqrt{x_3(t)^2 + x_4(t)^2}}{x_5(t)} e^{-\beta x_2(t)} - g, \\
\dot{x}_5(t) &= -b u_{\max}, \\
\dot{p}_1(t) &= 0, \\
\dot{p}_2(t) &= -(p_3(t)x_3(t) + p_4(t)x_4(t)) \frac{\alpha \beta \sqrt{x_3(t)^2 + x_4(t)^2}}{x_5(t)} e^{-\beta x_2(t)}, \\
\dot{p}_3(t) &= -p_1(t) + \alpha \frac{p_3(t)(2x_3(t)^2 + x_4(t)^2) + p_4(t)x_3(t)v_2(t)}{x_5(t) \sqrt{x_3(t)^2 + x_4(t)^2}} e^{-\beta x_2(t)}, \\
\dot{p}_4(t) &= -p_2(t) + \alpha \frac{p_3(t)x_3(t)x_4(t) + p_4(t)(x_3(t)^2 + 2x_4(t)^2)}{x_5(t) \sqrt{x_3(t)^2 + x_4(t)^2}} e^{-\beta x_2(t)}, \\
\dot{p}_5(t) &= \frac{u_{\max}}{x_5(t)^2} \sqrt{p_3(t) + p_4(t)} - \frac{\alpha \sqrt{x_3(t)^2 + x_4(t)^2}}{x_5(t)^2} e^{-\beta x_2(t)} (p_3(t)x_3(t) + p_4(t)x_4(t)), \\
x_1(0) &= 0, \quad x_2(0) = 0.005, \quad x_3(0) = 0, \quad x_4(0) = 0.01, \quad x_5(0) = m_0, \quad p_1(T) = 1, \quad p_5(T) = 0, \\
x_2(T) &= h, \quad x_4(T) = v_c, \quad x_4(T) = 0, \quad t \in [0, T],
\end{aligned}
\right. 
\qquad (3.13) 
$$

where

$$x(t) = (x_j(t), j = 1 \dots 5) = (x_1(t), x_2(t), v_1(t), v_2(t), m(t)).$$

We construct the Shooting function:

$$ G: \begin{array}{rcl} \mathbb{R}^{10} & \to & \mathbb{R}^{10} \\ (p(0), p(T)) & \to & G(p(0), p(T)), \end{array} \qquad (3.14) $$
and

$$G(p(0), p(T)) = \begin{pmatrix} p_1(0) - 1 \\ p_2(0) - \lambda_1 \\ p_3(0) - \lambda_2 \\ p_4(0) - \lambda_3 \\ p_5(0) - \lambda_4 \\ p_1(T) - 1 \\ p_2(T) - \lambda_5 \\ p_3(T) - \lambda_6 \\ p_4(T) - \lambda_7 \\ p_5(T) \end{pmatrix} .$$

Therefore, the problem (3.13) is equivalent to:

$$\left\{
\begin{aligned}
\dot{z}_1(t) &= z_3(t), \\
\dot{z}_2(t) &= z_4(t), \\
\dot{z}_3(t) &= \frac{u_{\max}}{z_5(t)} \frac{z_8(t)}{\sqrt{z_8(t)^2 + z_9(t)^2}} - \frac{\alpha z_3(t) \sqrt{z_3(t)^2 + z_4(t)^2}}{z_5(t)} e^{-\beta z_2(t)}, \\
\dot{z}_4(t) &= \frac{u_{\max}}{z_5(t)} \frac{z_9(t)}{\sqrt{z_8(t)^2 + z_9(t)^2}} - \frac{\alpha z_4(t) \sqrt{z_3(t)^2 + z_4(t)^2}}{z_5(t)} e^{-\beta z_2(t)} - g, \\
\dot{z}_5(t) &= -b u_{\max}, \\
\dot{z}_6(t) &= 0, \\
\dot{z}_7(t) &= -(z_8(t)z_3(t) + z_9(t)z_4(t)) \frac{\alpha \beta \sqrt{z_3(t)^2 + z_4(t)^2}}{z_5(t)} e^{-\beta z_2(t)}, \\
\dot{z}_8(t) &= -z_6(t) + \alpha \frac{z_8(t)(2z_3(t)^2 + z_4(t)^2) + z_9(t)z_3(t)z_4(t)}{x_5(t) \sqrt{z_3(t)^2 + z_4(t)^2}} e^{-\beta z_2(t)}, \\
\dot{z}_9(t) &= -z_7(t) + \alpha \frac{z_8(t)z_3(t)z_4(t) + z_9(t)(z_3(t)^2 + 2z_4(t)^2)}{z_5(t) \sqrt{z_3(t)^2 + z_4(t)^2}} e^{-\beta z_2(t)}, \\
\dot{z}_{10}(t) &= \frac{u_{\max}}{z_5(t)^2} \sqrt{z_8(t)^2 + z_9(t)^2} - \frac{\alpha \sqrt{z_3(t)^2 + z_4(t)^2} e^{-\beta z_2(t)}}{z_5(t)^2} (z_8(t)z_3(t) + z_9(t)z_4(t)), \\
z_1(0) &= 0, \quad z_2(0) = 0.005, \quad z_3(0) = 0, \quad z_4(0) = 0.01, \quad z_5(0) = m_0, \quad z_6(T) = 1, \quad z_{10}(T) = 0, \\
z_2(T) &= h, \quad z_4(T) = v_c, \quad z_4(T) = 0, \quad t \in [0, T],
\end{aligned}
\right.
\quad (3.15)$$

where $z(t) = (z_i(t), i = 1, \dots, 10) = (x(t), p(t))$.

The numerical solution of the nonlinear system $G(p(0), p(T)) = 0$ can be calculated using the Newton's method.
and

$$G(p(0), p(T)) = \begin{pmatrix} p_1(0) - 1 \\ p_2(0) - \lambda_1 \\ p_3(0) - \lambda_2 \\ p_4(0) - \lambda_3 \\ p_5(0) - \lambda_4 \\ p_1(T) - 1 \\ p_2(T) - \lambda_5 \\ p_3(T) - \lambda_6 \\ p_4(T) - \lambda_7 \\ p_5(T) \end{pmatrix} .$$

Therefore, the problem (3.13) is equivalent to:

$$\left\{
\begin{aligned}
\dot{z}_1(t) &= z_3(t), \\
\dot{z}_2(t) &= z_4(t), \\
\dot{z}_3(t) &= \frac{u_{\max}}{z_5(t)} \frac{z_8(t)}{\sqrt{z_8(t)^2 + z_9(t)^2}} - \frac{\alpha z_3(t) \sqrt{z_3(t)^2 + z_4(t)^2}}{z_5(t)} e^{-\beta z_2(t)}, \\
\dot{z}_4(t) &= \frac{u_{\max}}{z_5(t)} \frac{z_9(t)}{\sqrt{z_8(t)^2 + z_9(t)^2}} - \frac{\alpha z_4(t) \sqrt{z_3(t)^2 + z_4(t)^2}}{z_5(t)} e^{-\beta z_2(t)} - g, \\
\dot{z}_5(t) &= -b u_{\max}, \\
\dot{z}_6(t) &= 0, \\
\dot{z}_7(t) &= -(z_8(t)z_3(t) + z_9(t)z_4(t)) \frac{\alpha \beta \sqrt{z_3(t)^2 + z_4(t)^2}}{z_5(t)} e^{-\beta z_2(t)}, \\
\dot{z}_8(t) &= -z_6(t) + \alpha \frac{z_8(t)(2z_3(t)^2 + z_4(t)^2) + z_9(t)z_3(t)z_4(t)}{x_5(t) \sqrt{z_3(t)^2 + z_4(t)^2}} e^{-\beta z_2(t)}, \\
\dot{z}_9(t) &= -z_7(t) + \alpha \frac{z_8(t)z_3(t)z_4(t) + z_9(t)(z_3(t)^2 + 2z_4(t)^2)}{z_5(t) \sqrt{z_3(t)^2 + z_4(t)^2}} e^{-\beta z_2(t)}, \\
\dot{z}_{10}(t) &= \frac{u_{\max}}{z_5(t)^2} \sqrt{z_8(t)^2 + z_9(t)^2} - \frac{\alpha \sqrt{z_3(t)^2 + z_4(t)^2} e^{-\beta z_2(t)}}{z_5(t)^2} (z_8(t)z_3(t) + z_9(t)z_4(t)), \\
z_1(0) &= 0, \quad z_2(0) = 0.005, \quad z_3(0) = 0, \quad z_4(0) = 0.01, \quad z_5(0) = m_0, \quad z_6(T) = 1, \quad z_{10}(T) = 0, \\
z_2(T) &= h, \quad z_4(T) = v_c, \quad z_4(T) = 0, \quad t \in [0, T],
\end{aligned}
\right.
\quad (3.15)$$

where $z(t) = (z_i(t), i = 1, \dots, 10) = (x(t), p(t))$.

The numerical solution of the nonlinear system $G(p(0), p(T)) = 0$ can be calculated using the Newton's method.
<figure><img src="image_3.png" /><figcaption>Figure 2: Variation of the flight bank angle according to the time</figcaption></figure>

• From Figure 2, we note that the angle $\theta(t)$ decreases as a function of time from $\theta(0) = 0.35\pi$ to $\theta(t_c) = -1.57\pi$, where $t_c = 136.1s$ ($t_c$ is the commutation time), then increase quickly in the interval $[t_c, t_c + \epsilon]$, where $\epsilon = 0.5$, until it get to the angle $\theta(t_c + \epsilon) = 1.57\pi$, then it settles at this time in the interval $[t_c + \epsilon, T]$. In the reality of these results, the rocket change the direction at time $t_c = 136.1s$.

<figure><img src="image_4.png" /><figcaption>Figure 3: Variation of the lateral offset according to the time</figcaption></figure>

• From Figure 3, we remark that $x_1(t)$ is rapidly increases with time, it reaches the optimal lateral offset $x_1^*(T) = 218.9km$. The angle is proportional than lateral offset, for this last, lateral increases.
<figure><img src="image_4.png" /><figcaption>Figure 2: Variation of the flight bank angle according to the time</figcaption></figure>

• From Figure 2, we note that the angle $\theta(t)$ decreases as a function of time from $\theta(0) = 0.35\pi$ to $\theta(t_c) = -1.57\pi$, where $t_c = 136.1s$ ($t_c$ is the commutation time), then increase quickly in the interval $[t_c, t_c + \epsilon]$, where $\epsilon = 0.5$, until it get to the angle $\theta(t_c + \epsilon) = 1.57\pi$, then it settles at this time in the interval $[t_c + \epsilon, T]$. In the reality of these results, the rocket change the direction at time $t_c = 136.1s$.

<figure><img src="image_5.png" /><figcaption>Figure 3: Variation of the lateral offset according to the time</figcaption></figure>

• From Figure 3, we remark that $x_1(t)$ is rapidly increases with time, it reaches the optimal lateral offset $x_1^*(T) = 218.9km$. The angle is proportional than lateral offset, for this last, lateral increases.
<figure><img src="image_5.png" /><figcaption>Figure 4: Variation of the altitude according to the time</figcaption></figure>

• Furthermore, we see that $x_2(t)$ reaches the altitude target 180km. While the rocket is in movement, the altitude increase (See Figure 4).

<figure><img src="image_6.png" /><figcaption>Figure 5: Variation of the speed components according to the time</figcaption></figure>

• The velocity $v_1(t)$ is also increasing with time, it varies from $v_1(0) = 0$ to $v_1(T) = v_c = 7.904\,\text{km}\cdot\text{s}^{-1}$, $v_1(t)$ is the speed which used to follows trajectory and to prepare to enter orbit, $v_2(t)$ grows from $v_2(0) = 0.005\,\text{km}\cdot\text{s}^{-1}$ to $v_2(t_c + \epsilon) = 2.359\,\text{km}\cdot\text{s}^{-1}$ where $t_c + \epsilon = 136.6\,\text{s}$, then decreases from $v_2(t_c + \epsilon)$ to $v_2(T) = 0$, $v_2(t)$ is the speed which guarantees its arrival in orbit (See Figure 5).
<figure><img src="image_6.png" /><figcaption>Figure 4: Variation of the altitude according to the time</figcaption></figure>

• Furthermore, we see that $x_2(t)$ reaches the altitude target 180km. While the rocket is in movement, the altitude increase (See Figure 4).

<figure><img src="image_7.png" /><figcaption>Figure 5: Variation of the speed components according to the time</figcaption></figure>

• The velocity $v_1(t)$ is also increasing with time, it varies from $v_1(0) = 0$ to $v_1(T) = v_c = 7.904\,\text{km}\cdot\text{s}^{-1}$, $v_1(t)$ is the speed which used to follows trajectory and to prepare to enter orbit, $v_2(t)$ grows from $v_2(0) = 0.005\,\text{km}\cdot\text{s}^{-1}$ to $v_2(t_c + \epsilon) = 2.359\,\text{km}\cdot\text{s}^{-1}$ where $t_c + \epsilon = 136.6\,\text{s}$, then decreases from $v_2(t_c + \epsilon)$ to $v_2(T) = 0$, $v_2(t)$ is the speed which guarantees its arrival in orbit (See Figure 5).
<figure><img src="image_7.png" /><figcaption>Figure 6: Mass of the rocket</figcaption></figure>

• From Figure 6, we remark that $m(t)$ of the vehicle aerospatial decreases in the form of a linear function from $m(0) = 122176.39\text{kg}$ to $m(T) = 3600\text{kg}$ within a period $T = 158.3\text{s}$, more the rocket in motion, the mass decreases.

• Finally, we note that the execution time of the Shooting method necessary to find the optimal solution is 1.87s.

## 4. Conclusion

In this work, we have modelled a practical problem arising in aerospace field as a nonlinear optimal control problem. First, the Pontryagin's maximum principle has been used which gives a necessary condition of optimality. Then the considered problem is solved numerically with the shooting method. The choice is carried the shooting method in the numerical resolution for its precision and its speed. The results found coincide with our reality: when the time $t$ increases, $x_2(t)$ increases and $v_2(t)$ then decreases at time $t_c$ until it vanishes, to avoid the explosion of gear at the orbital entrance.

## References

1. Aliane M, Moussouni N, Bentobache M., *Optimal control of a rectilinear motion of a rocket*. Statistics, Optimization & Information Computing, 8(1): 281-295, (2020)

2. Aliane M, Moussouni N, Bentobache M., *Nonlinear optimal control of the heel angle of a rocket*, 6th International Conference on Control, Decision and Information Technologies (CODIT'19); Paris, France, April 23-26.(2019)

3. Aronna M.S, Bonnans J.F, Martinon P., *A shooting algorithm for optimal control problems with singular arcs*. J.Optim. Theory Appl, 158(2): 419-459, (2013)

4. Ascher UM, Petzold LR., *Computer methods for ordinary differential equations and differentiel-algebraic equations*, Society for industrial and Applied Mathematics, Philadelphia, PA. (1998)

5. Avanci G, Matteis G., *Two-timescale-integration method for inverse simulation*. Journal og guidance, Control, and Dynamics, 22, 395-401, (1993)

6. Betts J.T., *Practical methods for optimal control and estimation using nonlinear programming*. SIAM, Philadelphia, PA, second edition.(2010)

7. Chinchuluun A, Enkhbat R, Pardalos P.M., *A novel approach for nonconvex optimal control problems*, Optimization: A Journal of Mathematical Programming and Operations Research, 58(7): 781-789, (2009)

8. Cox A., *Optimal control of a two-stage rocket*. AAE508 project proposal, Aeronautical and Astronautical Engineering, Purdue University. (2014)

9. Garrett R.R., *Numerical Methods For Solving Optimal Control Problems*. Master's Thesis, university of Tennessee, (2015)

10. Gergand J, Haberkoun T., *Orbital transfer: some links between the low-thrust and the impulse cases*, Acta Astronautica 60, 6(9):649-657, (2007)

11. Lertmann G., *On class of variational problems in rocket flight*, Journal Aerospace Sci, 26, n9, 586-591, (1959)
<figure><img src="image_8.png" /><figcaption>Figure 6: Mass of the rocket</figcaption></figure>

• From Figure 6, we remark that $m(t)$ of the vehicle aerospatial decreases in the form of a linear function from $m(0) = 122176.39\text{kg}$ to $m(T) = 3600\text{kg}$ within a period $T = 158.3\text{s}$, more the rocket in motion, the mass decreases.

• Finally, we note that the execution time of the Shooting method necessary to find the optimal solution is 1.87s.

## 4. Conclusion

In this work, we have modelled a practical problem arising in aerospace field as a nonlinear optimal control problem. First, the Pontryagin's maximum principle has been used which gives a necessary condition of optimality. Then the considered problem is solved numerically with the shooting method. The choice is carried the shooting method in the numerical resolution for its precision and its speed. The results found coincide with our reality: when the time $t$ increases, $x_2(t)$ increases and $v_2(t)$ then decreases at time $t_c$ until it vanishes, to avoid the explosion of gear at the orbital entrance.

## References

1. Aliane M, Moussouni N, Bentobache M., *Optimal control of a rectilinear motion of a rocket*. Statistics, Optimization & Information Computing, 8(1): 281-295, (2020)

2. Aliane M, Moussouni N, Bentobache M., *Nonlinear optimal control of the heel angle of a rocket*, 6th International Conference on Control, Decision and Information Technologies (CODIT'19); Paris, France, April 23-26.(2019)

3. Aronna M.S, Bonnans J.F, Martinon P., *A shooting algorithm for optimal control problems with singular arcs*. J.Optim. Theory Appl, 158(2): 419-459, (2013)

4. Ascher UM, Petzold LR., *Computer methods for ordinary differential equations and differentiel-algebraic equations*, Society for industrial and Applied Mathematics, Philadelphia, PA. (1998)

5. Avanci G, Matteis G., *Two-timescale-integration method for inverse simulation*. Journal og guidance, Control, and Dynamics, 22, 395-401, (1993)

6. Betts J.T., *Practical methods for optimal control and estimation using nonlinear programming*. SIAM, Philadelphia, PA, second edition.(2010)

7. Chinchuluun A, Enkhbat R, Pardalos P.M., *A novel approach for nonconvex optimal control problems*, Optimization: A Journal of Mathematical Programming and Operations Research, 58(7): 781-789, (2009)

8. Cox A., *Optimal control of a two-stage rocket*. AAE508 project proposal, Aeronautical and Astronautical Engineering, Purdue University. (2014)

9. Garrett R.R., *Numerical Methods For Solving Optimal Control Problems*. Master's Thesis, university of Tennessee, (2015)

10. Gergand J, Haberkoun T., *Orbital transfer: some links between the low-thrust and the impulse cases*, Acta Astronautica 60, 6(9):649-657, (2007)

11. Lertmann G., *On class of variational problems in rocket flight*, Journal Aerospace Sci, 26, n9, 586-591, (1959)
12. Louadj K, Spiteri P, Demim F, Aidene M, Nemra A, Messine F., *Application Optimal Control for a Problem Aircraft Flight*, Engineering Science and Technology Review, 1(1):156-164, (2018)

13. Moussouni N, Aidene M., *Optimization of cereal output in presence of locusts*. An International Journal of Optimization and Control: Theories & Applications 6:1-10, (2016)

14. Marec J.P., *Optimal space trajectories*. Elsevier scientific publishing company, New York. (1979)

15. Pontryagin L.S, Boltyanskii V.G, Gamkrelidze R.V, Mishchenko E.F., *The mathematical theory of optimal processes*. Intersciences Publisher, New York. (1962)

16. Prussing J.E and B.A. Conway B.A., *Orbital Mechanics*. Oxford University Press, New York, second edition. (2013)

17. Trelat E., *Optimal control theory and applications*. Vuibert, Concrete mathematics collection, Paris. (in french). (2005)

18. Prussing J.E and Sandrik S.L., *Second-Order Necessary Conditions and sufficient conditions applied to continuous thrust trajectories*. Journal of guidance, Control, and Dynamics, Engineering Note, 28(4): 812-816, (2005)

19. Tsai, Hung-i Bruce., *Optimal trajectory designs and systems engineering analyses of reusable launch vehicles*. 15 Retrospective Theses and Dissertations, (2003)

Mohamed Aliane,
Department of Mathematics and Computer Science,
University of Medea, Algeria.
E-mail address: aliane.mohamed@univ-medea.dz

and

Nacima Moussouni,
Laboratory L2CSP,
University of Tizi-Ouzou, Algeria.
E-mail address: nmoussouni@yahoo.fr

and

Kahina Louadj,
Laboratory L2CSP,
University of Tizi-Ouzou, Algeria.
E-mail address: louadj-kahina@yahoo.fr

and

Nicolas Boizot,
Laboratoire des Sciences de l'Information et systèmes, UMR,
Université de Toulon, la Garde, France.
E-mail address: nicolas.boizot@univ-tln.fr
12. Louadj K, Spiteri P, Demim F, Aidene M, Nemra A, Messine F., *Application Optimal Control for a Problem Aircraft Flight*, Engineering Science and Technology Review, 1(1):156-164, (2018)

13. Moussouni N, Aidene M., *Optimization of cereal output in presence of locusts*. An International Journal of Optimization and Control: Theories & Applications 6:1-10, (2016)

14. Marec J.P., *Optimal space trajectories*. Elsevier scientific publishing company, New York. (1979)

15. Pontryagin L.S, Boltyanskii V.G, Gamkrelidze R.V, Mishchenko E.F., *The mathematical theory of optimal processes*. Intersciences Publisher, New York. (1962)

16. Prussing J.E and B.A. Conway B.A., *Orbital Mechanics*. Oxford University Press, New York, second edition. (2013)

17. Trelat E., *Optimal control theory and applications*. Vuibert, Concrete mathematics collection, Paris. (in french). (2005)

18. Prussing J.E and Sandrik S.L., *Second-Order Necessary Conditions and sufficient conditions applied to continuous thrust trajectories*. Journal of guidance, Control, and Dynamics, Engineering Note, 28(4): 812-816, (2005)

19. Tsai, Hung-i Bruce., *Optimal trajectory designs and systems engineering analyses of reusable launch vehicles*. 15 Retrospective Theses and Dissertations, (2003)

Mohamed Aliane,
Department of Mathematics and Computer Science,
University of Medea, Algeria.
E-mail address: aliane.mohamed@univ-medea.dz

and

Nacima Moussouni,
Laboratory L2CSP,
University of Tizi-Ouzou, Algeria.
E-mail address: nmoussouni@yahoo.fr

and

Kahina Louadj,
Laboratory L2CSP,
University of Tizi-Ouzou, Algeria.
E-mail address: louadj-kahina@yahoo.fr

and

Nicolas Boizot,
Laboratoire des Sciences de l'Information et systèmes, UMR,
Université de Toulon, la Garde, France.
E-mail address: nicolas.boizot@univ-tln.fr