+Considérons le problème $ \mathcal{P} $ suivant :
+$$
+ \mathcal{P} \left \{
+ \begin{array}{l}
+ \displaystyle\min_{(x,y,z) \in \mathbb{R}^3} J(x,y,z) = x^2 + y^2 + z^2 -r^2 \\
+ g(x,y,z) = (g_1(x,y,z), g_2(x,y,z)) = (x^2 + y^2 - r_1^2, x^2 + z^2 -r_2^2) \leq 0 \\
+ \end{array}
+ \right .
+$$
+où $$ (r,r_1,r_2) \in \mathbb{R}_+^3. $$
+\textit{Entrées} : $ J $ et $ g $ de classe $ \mathcal{C}^2 $, $ \varepsilon = 0.01 $ la précision, $ (x_0,y_0,z_0) = $ point initial et $ (\lambda_{0_1},\lambda_{0_2}) = $ multiplicateur initial.
+\newline
+Le Lagrangien $ L $ de $ \mathcal{P} $ : $$ L((x,y,z),(\lambda_1,\lambda_2)) = x^2 + y^2 + z^2 -r^2 + \lambda_1(x^2 + y^2 - r_1^2) + \lambda_2(x^2 + z^2 -r_2^2). $$
+\newline
+Le gradient de $ J $ : $$ \nabla J(x,y,z) = (\frac{\partial J}{\partial x}(x,y,z),\frac{\partial J}{\partial y}(x,y,z),\frac{\partial J}{\partial z}(x,y,z)) = (2x,2y,2z). $$
+\newline
+Le gradient de $ g $ : $$ \nabla g(x,y,z) = (\nabla g_1(x,y,z),\nabla g_2(x,y,z)) $$
+$$ = ((\frac{\partial g_1}{\partial x}(x,y,z),\frac{\partial g_1}{\partial y}(x,y,z),\frac{\partial g_1}{\partial z}(x,y,z)),(\frac{\partial g_2}{\partial x}(x,y,z),\frac{\partial g_2}{\partial y}(x,y,z),\frac{\partial g_2}{\partial z}(x,y,z)) $$
+$$ = ((2x,2y,0),(2x,0,2z)). $$
+\newline
+Le gradient du Lagrangien $ L $ :
+$$ \nabla L((x,y,z),(\lambda_1,\lambda_2)) = \nabla J(x,y,z) + \lambda_1 \nabla g_1(x,y,z) + \lambda_2 \nabla g_2(x,y,z)) $$
+\newline
+La matrice hessienne de $ J $ : $$ H[J](x,y,z) =
+ \begin{pmatrix}
+ \frac{\partial^2 J}{\partial^2 x}(x,y,z) & \frac{\partial^2 J}{\partial x\partial y}(x,y,z) & \frac{\partial^2 J}{\partial x\partial z}(x,y,z) \\
+ \frac{\partial^2 J}{\partial y\partial x}(x,y,z) & \frac{\partial^2 J}{\partial^2 y}(x,y,z) & \frac{\partial^2 J}{\partial y\partial z}(x,y,z) \\
+ \frac{\partial^2 J}{\partial z\partial x}(x,y,z) & \frac{\partial^2 J}{\partial z\partial y}(x,y,z) & \frac{\partial^2 J}{\partial^2 z}(x,y,z) \\
+ \end{pmatrix} =
+ \begin{pmatrix}
+ 2 & 0 & 0 \\
+ 0 & 2 & 0 \\
+ 0 & 0 & 2 \\
+ \end{pmatrix} = 2Id_{\mathbb{R}^3} $$
+On en déduit que $ H[J](x,y,z) $ est inversible et que $ H[J](x,y,z)^{-1} = \frac{1}{2}Id_{\mathbb{R}^3} $.
+
+\hrulefill
+
+\subsection{Trace d'éxécution de PQS}
+
+Utilisons le problème $ \mathcal{P} $ précédent :
+
+$$
+ \mathcal{P} \left \{
+ \begin{array}{l}
+ \displaystyle\min_{(x,y,z) \in \mathbb{R}^3} J(x,y,z) = x^2 + y^2 + z^2 -r^2 \\
+ g(x,y,z) = (g_1(x,y,z), g_2(x,y,z)) = (x^2 + y^2 - r_1^2, x^2 + z^2 -r_2^2) \leq 0 \\
+ \end{array}
+ \right .
+$$
+où $$ (r,r_1,r_2) \in \mathbb{R}_+^3. $$
+\textit{Entrées} : $ J $ et $ g $ de classe $ \mathcal{C}^2 $, $ \varepsilon = (0.01,0.01,0.01) $, $ (x_0,y_0,z_0) = (80, 20 ,60)$ et $ (\lambda_{0_1},\lambda_{0_2}) = (1 , 1)$, les rayons : $r= 40$ et $r1= r2= 10$.
+\newline
+Le Lagrangien $ L $ de $ \mathcal{P} $ : $$ L((x,y,z),(\lambda_1,\lambda_2)) = x^2 + y^2 + z^2 -r^2 + \lambda_1(x^2 + y^2 - r_1^2) + \lambda_2(x^2 + z^2 -r_2^2). $$
+\newline
+Le Lagrangien $ L $ de $ \mathcal{P} $ avec les valeurs :
+ $ L((80,20,60),(1,1)) = 80^2 + 20^2 + 60^2 -60^2 + 1 * (80^2 +20y^2 - 30^2) + \lambda_2(80^2 + 60^2 -30^2). $
+ $ L((80,20,60),(1,1)) = 6400 + 400 + 3600 - 3600 + (6400 + 400 - 900) + (6400 + 3600 -900). $
+ $ L((80,20,60),(1,1)) = 21800. $
+
+ \begin{algorithm}
+ \caption {PQS du problème $ \mathcal{P} $}
+ \begin{algorithmic}
+ \REQUIRE $g(x,y,z)\leq 0$, $(x_0,y_0,z_0) = (80, 20 ,60)$
+ \ENSURE $\min_{(x,y,z) \in \mathbb{R}^3} J(x,y,z) = x^2 + y^2 + z^2 -r^2$ and \newline $g(x,y,z) = (g_1(x,y,z), g_2(x,y,z)) = (x^2 + y^2 - r_1^2, x^2 + z^2 -r_2^2) \leq 0 $
+ \STATE \textbf{Data :}
+ \STATE $k \leftarrow 0$
+ \STATE $x_k \leftarrow 80$
+ \STATE $y_k \leftarrow 20$
+ \STATE $z_k \leftarrow 60$
+ \STATE $x_a \leftarrow 30$
+ \STATE $y_a \leftarrow 10$
+ \STATE $z_a \leftarrow 40$
+ \STATE $r \leftarrow 40$
+ \STATE $r_1 \leftarrow 10$
+ \STATE $r_2 \leftarrow 10$
+ \STATE $\varepsilon \leftarrow 0.01$
+ \STATE $\lambda_1 = \lambda_2 = 1$
+ \STATE $ H[J](x,y,z)^{-1}\leftarrow \begin{pmatrix}
+ 0.5 & 0 & 0 \\
+ 0 & 0.5 & 0 \\
+ 0 & 0 & 0.5 \\ \end{pmatrix} $
+\newline