LOPEZ et al 2022a - Revision history

Gstinoco at 23:54, 23 December 2022

2022-12-23T23:54:22Z

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Gstinoco: Gstinoco moved page Review 244156969292 to LOPEZ et al 2022a

2022-11-16T02:57:39Z

Gstinoco moved page Review 244156969292 to LOPEZ et al 2022a

Loppital1 at 02:31, 15 November 2022

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Loppital1 at 02:03, 15 November 2022

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Loppital1 at 01:51, 15 November 2022

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Loppital1 at 01:32, 15 November 2022

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Loppital1 at 01:20, 15 November 2022

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Loppital1 at 00:58, 15 November 2022

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Loppital1 at 00:50, 15 November 2022

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Loppital1 at 00:40, 15 November 2022

2022-11-15T00:40:48Z

@@ Line 1,399: / Line 1,399: @@
 | style="width: 5px;text-align: right;white-space: nowrap;" | (77)
 |-
-| style="text-align: center;" | <math> \Gamma \frac{\mathbf{p}^{n}-\mathbf{p}^{n+1}}{\Delta t}+(K+C)\mathbf{p}^{n} =k_{1}\mathbf{y}^{n}\hbox{ ''', '''}n=N,...,1.   </math>
+| style="text-align: center;" | <math> \Gamma \frac{\mathbf{p}^{n}-\mathbf{p}^{n+1}}{\Delta t}+(K+C)\mathbf{p}^{n} =k_{1}\mathbf{y}^{n}\hbox{ , }n=N,...,1.   </math>
 | style="width: 5px;text-align: right;white-space: nowrap;" | (78)
 |}

@@ Line 1,741: / Line 1,741: @@
 *  We do <math display="inline">q = q + 1</math> and we repeat the process.
-The above time partitioning method has been applied with <math display="inline">\boldsymbol{\phi }_0=\boldsymbol{\theta }+\delta \boldsymbol{\theta }</math> and <math display="inline">\delta \boldsymbol{\theta }=[1e-2,1e-2,1e-2]</math>, the time interval under  consideration being <math display="inline">[0,20]</math>; we have used <math display="inline">\Delta T = 2.0</math>. After <math display="inline">t = 20</math>, we have taken <math display="inline">\mathbf{v} = \mathbf{0}}</math> in  ([[#eq-9|9]]) and in ([[#eq-5|5]]) to observe the evolution of the suddenly uncontrolled linear and nonlinear systems.  The results are reported in Figure [[#img-13|13]]. We observe that the system is practically stabilized for <math display="inline">1 \leq t \leq 20</math>, but if one stops controlling, the small residual perturbations of the system at <math display="inline">t = 20</math>, are sufficient to destabilize the linear and nonlinear systems and induces the nonlinear one to transition to a stable equilibrium in finite time.
+The above time partitioning method has been applied with <math display="inline">\boldsymbol{\phi }_0=\boldsymbol{\theta }+\delta \boldsymbol{\theta }</math> and <math display="inline">\delta \boldsymbol{\theta }=[1e-2,1e-2,1e-2]</math>, the time interval under  consideration being <math display="inline">[0,20]</math>; we have used <math display="inline">\Delta T = 2.0</math>. After <math display="inline">t = 20</math>, we have taken <math display="inline">\mathbf{v} = \mathbf{0}</math> in  ([[#eq-9|9]]) and in ([[#eq-5|5]]) to observe the evolution of the suddenly uncontrolled linear and nonlinear systems.  The results are reported in Figure [[#img-13|13]]. We observe that the system is practically stabilized for <math display="inline">1 \leq t \leq 20</math>, but if one stops controlling, the small residual perturbations of the system at <math display="inline">t = 20</math>, are sufficient to destabilize the linear and nonlinear systems and induces the nonlinear one to transition to a stable equilibrium in finite time.
 <div id='img-13'></div>

@@ Line 567: / Line 567: @@
 {| style="text-align: left; margin:auto;width: 100%;"
 |-
-| style="text-align: center;" | <math>\mathbf{Au}=\mathbf{\beta },   </math>
+| style="text-align: center;" | <math>\mathbf{Au}=\boldsymbol{\beta },   </math>
 |}
 | style="width: 5px;text-align: right;white-space: nowrap;" | (26)
@@ Line 643: / Line 643: @@
 {| style="text-align: left; margin:auto;width: 100%;"
 |-
-| style="text-align: center;" | <math>\int \limits _{0}^{T}\mathbf{v}^{2}\cdot \mathbf{p}^{1}dt.=k_{2}\mathbf{y}^{1}(T)\mathbf{\cdot y}^{2}(T)+\int \limits _{0}^{T}k_{1}\mathbf{y}^{1}\cdot  \mathbf{y}^{2}dt.   </math>
+| style="text-align: center;" | <math>\int \limits _{0}^{T}\mathbf{v}^{2}\cdot \mathbf{p}^{1}dt=k_{2}\mathbf{y}^{1}(T)\mathbf{\cdot y}^{2}(T)+\int \limits _{0}^{T}k_{1}\mathbf{y}^{1}\cdot  \mathbf{y}^{2}dt.   </math>
 |}
 | style="width: 5px;text-align: right;white-space: nowrap;" | (29)
@@ Line 672: / Line 672: @@
 |}
-which implies the strong ellipticity of <math display="inline">\mathbf{A}</math> over <math display="inline">\mathcal{U}</math>. The linear operator <math display="inline">\mathbf{A}</math>, being continuous and strongly elliptic over <math display="inline">\mathcal{U}</math>, is an automorphism of <math display="inline">\mathcal{U}</math>. To identify the right hand side <math display="inline">\mathbf{\beta }</math> of equation ([[#eq-26|26]]), we introduce <math display="inline">Y_{0}</math> and <math display="inline">P_{0}</math> defined as the solutions of
+which implies the strong ellipticity of <math display="inline">\mathbf{A}</math> over <math display="inline">\mathcal{U}</math>. The linear operator <math display="inline">\mathbf{A}</math>, being continuous and strongly elliptic over <math display="inline">\mathcal{U}</math>, is an automorphism of <math display="inline">\mathcal{U}</math>. To identify the right hand side <math display="inline">\boldsymbol{\beta }</math> of equation ([[#eq-26|26]]), we introduce <math display="inline">\mathbf{Y}_{0}</math> and <math display="inline">\mathbf{P}_{0}</math> defined as the solutions of
 <span id="eq-31"></span>
@@ Line 748: / Line 748: @@
 |}
-the right hand side of ([[#eq-35|35]]) is the vector <math display="inline">\mathbf{\beta }</math> that we were looking for.
+the right hand side of ([[#eq-35|35]]) is the vector <math display="inline">\boldsymbol{\beta }</math> that we were looking for.
 From the properties of <math display="inline">\mathbf{A}</math>, problem ([[#eq-35|35]]) can be solved by a conjugate gradient algorithm operating in the Hilbert space <math display="inline">\mathcal{U} </math>. This algorithm will be described in the following subsubsection.
@@ Line 1,406: / Line 1,406: @@
 ====4.2.3 Functional equation for the discrete control solution of ([[#eq-60|60]])====
-Following the sketch for the continuous case we can show that the discrete version <math display="inline">\mathbf{A}^{\Delta t}</math> of operator <math display="inline">\mathbf{A}</math> and the discrete version <math display="inline">\mathbf{\beta }^{\Delta t}</math> of <math display="inline">\mathbf{\beta }</math> satisfies the equation
+Following the sketch for the continuous case we can show that the discrete version <math display="inline">\mathbf{A}^{\Delta t}</math> of operator <math display="inline">\mathbf{A}</math> and the discrete version <math display="inline">\boldsymbol{\beta }^{\Delta t}</math> of <math display="inline">\boldsymbol{\beta }</math> satisfies the equation
 <span id="eq-79"></span>
@@ Line 1,414: / Line 1,414: @@
 {| style="text-align: left; margin:auto;width: 100%;"
 |-
-| style="text-align: center;" | <math>\mathbf{A}^{\Delta t}\mathbf{u}^{\Delta t}=\mathbf{\beta }^{\Delta t},  </math>
+| style="text-align: center;" | <math>\mathbf{A}^{\Delta t}\mathbf{u}^{\Delta t}=\boldsymbol{\beta }^{\Delta t},  </math>
 |}
 | style="width: 5px;text-align: right;white-space: nowrap;" | (79)
@@ Line 1,423: / Line 1,423: @@
 ====4.2.4 Conjugate gradient solution of the discrete control problem ([[#eq-60|60]])====
-Using <math display="inline">\mathbf{y}_{q}^{n}=\{ \mathbf{y}_{iq}^{n}\} _{i=1}^{3}</math> to denote the discrete value of the vector-valued function <math display="inline">\mathbf{y}</math> at time <math display="inline">n\Delta t</math> and iteration <math display="inline">q</math>; similarly, <math display="inline">\mathbf{u}_{q}^{n}</math> will denote the discrete value of the control <math display="inline">\mathbf{u}</math> at time <math display="inline">n\Delta t</math> and iteration <math display="inline">q</math>, the conjugate gradient algorithm ([[#eq-38|38]])-([[#eq-44|44]]) to solve the finite dimensional problem ([[#eq-60|60]]) reads as follow:
+Using <math display="inline">\mathbf{y}_{q}^{n}=\{ y_{iq}^{n}\} _{i=1}^{3}</math> to denote the discrete value of the vector-valued function <math display="inline">\mathbf{y}</math> at time <math display="inline">n\Delta t</math> and iteration <math display="inline">q</math>; similarly, <math display="inline">\mathbf{u}_{q}^{n}</math> will denote the discrete value of the control <math display="inline">\mathbf{u}</math> at time <math display="inline">n\Delta t</math> and iteration <math display="inline">q</math>, the conjugate gradient algorithm ([[#eq-38|38]])-([[#eq-44|44]]) to solve the finite dimensional problem ([[#eq-60|60]]) reads as follow:
 * Suppose
@@ Line 1,440: / Line 1,440: @@
 |}
-* Compute <math display="inline">\{ \mathbf{y}_{0}^{n}\} _{n=0}^{N}=\{ \{ \mathbf{y}_{i0}^{n}\} _{i=1}^{3}\} _{n=0}^{N}</math> and <math display="inline">\{ \mathbf{p}_{0}^{n}\} _{n=1}^{N+1}=\{ \{ \mathbf{p}_{i0}^{n}\} _{i=1}^{3}\} _{n=1}^{N+1}</math> via the solution of
+* Compute <math display="inline">\{ \mathbf{y}_{0}^{n}\} _{n=0}^{N}=\{ \{ y_{i0}^{n}\} _{i=1}^{3}\} _{n=0}^{N}</math> and <math display="inline">\{ \mathbf{p}_{0}^{n}\} _{n=1}^{N+1}=\{ \{ p_{i0}^{n}\} _{i=1}^{3}\} _{n=1}^{N+1}</math> via the solution of
 <span id="eq-81"></span>
@@ Line 1,741: / Line 1,741: @@
 *  We do <math display="inline">q = q + 1</math> and we repeat the process.
-The above time partitioning method has been applied with <math display="inline">\boldsymbol{\phi }_0=\boldsymbol{\theta }+\delta \boldsymbol{\theta }</math> and <math display="inline">\delta \boldsymbol{\theta }=[1e-2,1e-2,1e-2]</math>, the time interval under  consideration being <math display="inline">[0,20]</math>; we have used <math display="inline">\Delta T = 2.0</math>. After <math display="inline">t = 20</math>, we have taken <math display="inline">\mathbf{v = 0}</math> in  ([[#eq-9|9]]) and in ([[#eq-5|5]]) to observe the evolution of the suddenly uncontrolled linear and nonlinear systems.  The results are reported in Figure [[#img-13|13]]. We observe that the system is practically stabilized for <math display="inline">1 \leq t \leq 20</math>, but if one stops controlling, the small residual perturbations of the system at <math display="inline">t = 20</math>, are sufficient to destabilize the linear and nonlinear systems and induces the nonlinear one to transition to a stable equilibrium in finite time.
+The above time partitioning method has been applied with <math display="inline">\boldsymbol{\phi }_0=\boldsymbol{\theta }+\delta \boldsymbol{\theta }</math> and <math display="inline">\delta \boldsymbol{\theta }=[1e-2,1e-2,1e-2]</math>, the time interval under  consideration being <math display="inline">[0,20]</math>; we have used <math display="inline">\Delta T = 2.0</math>. After <math display="inline">t = 20</math>, we have taken <math display="inline">\mathbf{v} = \mathbf{0}}</math> in  ([[#eq-9|9]]) and in ([[#eq-5|5]]) to observe the evolution of the suddenly uncontrolled linear and nonlinear systems.  The results are reported in Figure [[#img-13|13]]. We observe that the system is practically stabilized for <math display="inline">1 \leq t \leq 20</math>, but if one stops controlling, the small residual perturbations of the system at <math display="inline">t = 20</math>, are sufficient to destabilize the linear and nonlinear systems and induces the nonlinear one to transition to a stable equilibrium in finite time.
 <div id='img-13'></div>

@@ Line 572: / Line 572: @@
 |}
-where the linear operator <math display="inline">\mathbf{A}</math> is a strongly elliptic and symmetric isomorphism from <math display="inline">\mathcal{U}</math> into itself (an automorphism of <math display="inline">\mathcal{U)}</math> and where <math display="inline">\mathbf{\beta }\in \mathcal{U}</math>. A candidate for <math display="inline">\mathbf{A}</math> is the linear operator from <math display="inline">\mathcal{U}</math> into itself defined by
+where the linear operator <math display="inline">\mathbf{A}</math> is a strongly elliptic and symmetric isomorphism from <math display="inline">\mathcal{U}</math> into itself (an automorphism of <math display="inline">\mathcal{U)}</math> and where <math display="inline">\boldsymbol{\beta }\in \mathcal{U}</math>. A candidate for <math display="inline">\mathbf{A}</math> is the linear operator from <math display="inline">\mathcal{U}</math> into itself defined by
 {| class="formulaSCP" style="width: 100%; text-align: left;"

@@ Line 1,679: / Line 1,679: @@
 |[[Image:Draft_LOPEZ_262416069-Ufigure7J23L16.png|276px|u<sup>∆t</sup> (top) and \| y<sup>∆t</sup>(⋅)\|  (bottom) for several values of k₁ and k₂. The unstable equilibrium θ is given by \{ n₁,n₂,n₃\} =\{ 1,0,0(u)\} . The junctions used to control are 2 and 3.]]
 |- style="text-align: center; font-size: 75%;"
-| colspan="2" | '''Figure 7:''' <math>\mathbf{u}^{\Delta t}</math> (top) and <math>\| \mathbf{y}^{\Delta t}(\cdot )\| </math> (bottom) for several values of <math>k_1</math> and <math>k_2</math>. The unstable equilibrium <math>\boldsymbol{\theta }</math> is given by <math>\{ n_{1},n_{2},n_{3}\} =\{ 1,0,0(u)\} </math>. The junctions used to control are 2 and 3.
+| colspan="2" | '''Figure 7:''' <math>\mathbf{u}^{\Delta t}</math> (left) and <math>\| \mathbf{y}^{\Delta t}(\cdot )\| </math> (right) for several values of <math>k_1</math> and <math>k_2</math>. The unstable equilibrium <math>\boldsymbol{\theta }</math> is given by <math>\{ n_{1},n_{2},n_{3}\} =\{ 1,0,0(u)\} </math>. The junctions used to control are 2 and 3.
 |}
@@ Line 1,687: / Line 1,687: @@
 |[[Image:Draft_LOPEZ_262416069-Ufigure9J23L16.png|384px|ln\| g<sub>q</sub><sup>∆t</sup>\|  for several values of k₁ and k₂. The unstable equilibrium θ is given by \{ n₁,n₂,n₃\} =\{ 1,0,0(u)\} . The junctions used to control are 2 and 3.]]
 |- style="text-align: center; font-size: 75%;"
-| colspan="1" | '''Figure 8:''' <math>ln\| \mathbf{g}_q^{\Delta t}\| </math> for several values of <math>k_1</math> and <math>k_2</math>. The unstable equilibrium <math>\boldsymbol{\theta }</math> is given by <math>\{ n_{1},n_{2},n_{3}\} =\{ 1,0,0(u)\} </math>. The junctions used to control are 2 and 3.
+| colspan="1" | '''Figure 8:''' <math>Ln\| \mathbf{g}_q^{\Delta t}\| </math> for several values of <math>k_1</math> and <math>k_2</math>. The unstable equilibrium <math>\boldsymbol{\theta }</math> is given by <math>\{ n_{1},n_{2},n_{3}\} =\{ 1,0,0(u)\} </math>. The junctions used to control are 2 and 3.
 |}
@@ Line 1,700: / Line 1,700: @@
 |[[Image:Draft_LOPEZ_262416069-Ufigure10J123L16.png|360px|u<sup>∆t</sup> (top) and \| y<sup>∆t</sup>(⋅)\|  (bottom) for several values of k₁ and k₂. The unstable equilibrium θ is given by \{ n₁,n₂,n₃\} =\{ 1,0,0(u)\} . The junctions used to control are 1, 2 and 3.]]
 |- style="text-align: center; font-size: 75%;"
-| colspan="2" | '''Figure 9:''' <math>\mathbf{u}^{\Delta t}</math> (top) and <math>\| \mathbf{y}^{\Delta t}(\cdot )\| </math> (bottom) for several values of <math>k_1</math> and <math>k_2</math>. The unstable equilibrium <math>\boldsymbol{\theta }</math> is given by <math>\{ n_{1},n_{2},n_{3}\} =\{ 1,0,0(u)\} </math>. The junctions used to control are 1, 2 and 3.
+| colspan="2" | '''Figure 9:''' <math>\mathbf{u}^{\Delta t}</math> (left) and <math>\| \mathbf{y}^{\Delta t}(\cdot )\| </math> (right) for several values of <math>k_1</math> and <math>k_2</math>. The unstable equilibrium <math>\boldsymbol{\theta }</math> is given by <math>\{ n_{1},n_{2},n_{3}\} =\{ 1,0,0(u)\} </math>. The junctions used to control are 1, 2 and 3.
 |}
@@ Line 1,708: / Line 1,708: @@
 |[[Image:Draft_LOPEZ_262416069-Ufigure12J123L16.png|384px|ln\| g<sub>q</sub><sup>∆t</sup>\|  for several values of k₁ and k₂. The unstable equilibrium θ is given by \{ n₁,n₂,n₃\} =\{ 1,0,0(u)\} . The junctions used to control are 1, 2 and 3.]]
 |- style="text-align: center; font-size: 75%;"
-| colspan="1" | '''Figure 10:''' <math>ln\| \mathbf{g}_q^{\Delta t}\| </math> for several values of <math>k_1</math> and <math>k_2</math>. The unstable equilibrium <math>\boldsymbol{\theta }</math> is given by <math>\{ n_{1},n_{2},n_{3}\} =\{ 1,0,0(u)\} </math>. The junctions used to control are 1, 2 and 3.
+| colspan="1" | '''Figure 10:''' <math>Ln\| \mathbf{g}_q^{\Delta t}\| </math> for several values of <math>k_1</math> and <math>k_2</math>. The unstable equilibrium <math>\boldsymbol{\theta }</math> is given by <math>\{ n_{1},n_{2},n_{3}\} =\{ 1,0,0(u)\} </math>. The junctions used to control are 1, 2 and 3.
 |}
@@ Line 1,719: / Line 1,719: @@
 |[[Image:Draft_LOPEZ_262416069-Ufigure17t12L16.png|240px|Optimal controls for the linear system (top), and Euclidean norm of the controlled solution ϕ<sup>∆t</sup> of the nonlinear system  (bottom).]]
 |- style="text-align: center; font-size: 75%;"
-| colspan="2" | '''Figure 11:''' Optimal controls for the linear system (top), and Euclidean norm of the controlled solution <math>\boldsymbol{\phi }^{\Delta t}</math> of the nonlinear system  (bottom).
+| colspan="2" | '''Figure 11:''' Optimal controls (left), and Euclidean norm of the controlled solution <math>\boldsymbol{\phi }^{\Delta t}</math> of the nonlinear system  (right).
 |}
@@ Line 1,730: / Line 1,730: @@
 |[[Image:Draft_LOPEZ_262416069-Ufigure14J123L16.png|360px|Extended controlled solution y<sub>i</sub><sup>∆t</sup> of the linear perturbation model (top), and extended controlled solution ϕ<sub>i</sub><sup>∆t</sup> of the nonlinear system (bottom). After t=2 the controls are extended as cero.]]
 |- style="text-align: center; font-size: 75%;"
-| colspan="2" | '''Figure 12:''' Extended controlled solution <math>y_i^{\Delta t}</math> of the linear perturbation model (top), and extended controlled solution <math>\phi _i^{\Delta t}</math> of the nonlinear system (bottom). After <math>t=2</math> the controls are extended as cero.
+| colspan="2" | '''Figure 12:''' Extended controlled solution <math>y_i^{\Delta t}</math> of the linear perturbation model (left), and extended controlled solution <math>\phi _i^{\Delta t}</math> of the nonlinear system (right). After <math>t=2</math> the controls are extended as cero.
 |}
@@ Line 1,751: / Line 1,751: @@
 | colspan="2"|[[Image:Draft_LOPEZ_262416069-Ufigure20t28L16.png|300px|Controls calculated each two seconds (top); Euclidean norm of the controlled solution y<sup>∆t</sup> using the controls in the top (middle); Euclidean norm of the controlled solution ϕ<sup>∆t</sup> using the controls in the top (bottom).]]
 |- style="text-align: center; font-size: 75%;"
-| colspan="2" | '''Figure 13:''' Controls calculated each two seconds (top); Euclidean norm of the controlled solution <math>\mathbf{y}^{\Delta t}</math> using the controls in the top (middle); Euclidean norm of the controlled solution <math>\boldsymbol{\phi }^{\Delta t}</math> using the controls in the top (bottom).
+| colspan="2" | '''Figure 13:''' Controls calculated each two seconds (top-left); Euclidean norm of the controlled solution <math>\mathbf{y}^{\Delta t}</math> using the controls in the top (top-right); Euclidean norm of the controlled solution <math>\boldsymbol{\phi }^{\Delta t}</math> using the controls in the top-left (bottom).
 |}

← Older revision	Revision as of 02:57, 16 November 2022
(No difference)

@@ Line 526: / Line 526: @@
 ====3.3.3 Optimality conditions for problem ([[#eq-10|10]])====
-Let <math display="inline">\mathbf{u}</math> be the solution of problem ([[#eq-10|10]]) and let us denote by <math display="inline">\mathbf{y}</math> (respectively, <math display="inline">\mathbf{p}</math>) the corresponding solution of the state system ([[#eq-9|9]]) (respectively, of the adjoint system ([[#eq-17|17]])). It follows from Subsubsection 2.3.2 that <math display="inline">DJ(\mathbf{u})=\mathbf{0}</math> is equivalent to the following (optimality) system:
+Let <math display="inline">\mathbf{u}</math> be the solution of problem ([[#eq-10|10]]) and let us denote by <math display="inline">\mathbf{y}</math> (respectively, <math display="inline">\mathbf{p}</math>) the corresponding solution of the state system ([[#eq-9|9]]) (respectively, of the adjoint system ([[#eq-17|17]])). It follows from Subsubsection 3.3.2 that <math display="inline">DJ(\mathbf{u})=\mathbf{0}</math> is equivalent to the following (optimality) system:
 <span id="eq-21"></span>

@@ Line 230: / Line 230: @@
 |[[Image:Draft_LOPEZ_262416069-cfigure4.png|300px|Evolution to state \{ 1,1,0(s)\} (top), \{ 2,1,0(s)\} (bottom) from an approximation to the unstable equilibrium \{ 1,0,0(u)\} (top), \{ 2,1,0(u)\} (bottom).]]
 |- style="text-align: center; font-size: 75%;"
-| colspan="2" | '''Figure 2:''' Evolution to state <math>\{ 1,1,0(s)\} </math>(top), <math>\{ 2,1,0(s)\} </math>(bottom) from an approximation to the unstable equilibrium <math>\{ 1,0,0(u)\} </math>(top), <math>\{ 2,1,0(u)\} </math>(bottom).
+| colspan="2" | '''Figure 2:''' Evolution to state <math>\{ 1,1,0(s)\} </math>(left), <math>\{ 2,1,0(s)\} </math>(right) from an approximation to the unstable equilibrium <math>\{ 1,0,0(u)\} </math>(left), <math>\{ 2,1,0(u)\} </math>(right).
 |}
@@ Line 255: / Line 255: @@
 (c) Apply the above control to the nonlinear system.
-Let us consider an unstable state <math display="inline">\boldsymbol{\theta } = \{ \theta _1,\theta _2,\theta _3  \} </math> of ([[#eq-5|5]]) and a small initial variation <math display="inline">\delta \boldsymbol{\theta } </math> of <math display="inline">\boldsymbol{\theta } </math>. The perturbation <math display="inline">\delta \boldsymbol{\phi } </math> of the steady-state <math display="inline">\theta </math> in <math display="inline">(0,T)</math> satisfies approximately the following linear model
+Let us consider an unstable state <math display="inline">\boldsymbol{\theta } = \{ \theta _1,\theta _2,\theta _3  \} </math> of ([[#eq-5|5]]) and a small initial variation <math display="inline">\delta \boldsymbol{\theta } </math> of <math display="inline">\boldsymbol{\theta } </math>. The perturbation <math display="inline">\delta \boldsymbol{\phi } </math> of the steady-state <math display="inline">\boldsymbol{\theta } </math> in <math display="inline">(0,T)</math> satisfies approximately the following linear model
 <span id="eq-7"></span>

@@ Line 243: / Line 243: @@
 The organization of this paper is as follows.  In Section 2 we focus on a methodology to stabilize the JJAM system around an unstable equilibrium configuration, via a controllability approach. Section 3 is concerning the practical aspects of this methodology. In Section 4 we show the numerical results and in Section 5 we give the conclusions.
-==3 Formulation of the optimal control problem for the stabiliza-tion of the JJAM and the conjugate gradient algorithm==
+==3 Formulation of the optimal control problem for the stabilization of the JJAM and the conjugate gradient algorithm==
 ===3.1 The approach===
@@ Line 384: / Line 384: @@
 {| style="text-align: left; margin:auto;width: 100%;"
 |-
-| style="text-align: center;" | <math>DJ(\mathbf{u})=\mathbf{0.}   </math>
+| style="text-align: center;" | <math>DJ(\mathbf{u})=\mathbf{0}.   </math>
 |}
 | style="width: 5px;text-align: right;white-space: nowrap;" | (12)
@@ Line 400: / Line 400: @@
 |}
-and from the properties (need to be shown) of the operator <math display="inline">\mathbf{v}\rightarrow DJ(\mathbf{v})-DJ(\mathbf{0})</math>, problem <math display="inline">DJ(\mathbf{u})=0</math> could be solved by a (quadratic case)-conjugate gradient algorithm operating in the space <math display="inline">\mathcal{U}</math>. The practical implementation of the above algorithm would requires the explicit knowledge of <math display="inline">DJ(\mathbf{v})</math>.
+and from the properties (need to be shown) of the operator <math display="inline">\mathbf{v}\rightarrow DJ(\mathbf{v})-DJ(\mathbf{0})</math>, problem <math display="inline">DJ(\mathbf{u})=\mathbf{0}</math> could be solved by a (quadratic case)-conjugate gradient algorithm operating in the space <math display="inline">\mathcal{U}</math>. The practical implementation of the above algorithm would requires the explicit knowledge of <math display="inline">DJ(\mathbf{v})</math>.
 ====3.3.2 <span id='lb-3.3.2'></span>Computing DJ(v)====