Acosta-Humánez et al 2017a - Revision history

Scipediacontent at 15:39, 16 March 2018

2018-03-16T15:39:09Z

Scipediacontent at 15:22, 16 March 2018

2018-03-16T15:22:56Z

Scipediacontent at 15:20, 16 March 2018

2018-03-16T15:20:28Z

Scipediacontent at 15:16, 16 March 2018

2018-03-16T15:16:48Z

Scipediacontent at 15:06, 16 March 2018

2018-03-16T15:06:09Z

Scipediacontent at 15:02, 16 March 2018

2018-03-16T15:02:45Z

Scipediacontent at 14:57, 16 March 2018

2018-03-16T14:57:39Z

Scipediacontent at 14:49, 16 March 2018

2018-03-16T14:49:19Z

Rimni at 14:20, 6 September 2017

2017-09-06T14:20:46Z

Rimni at 09:11, 21 July 2017

2017-07-21T09:11:54Z

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 {| style="text-align: left; margin:auto;width: 100%;"
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-| style="text-align: center;" | <math>\begin{array}{l}  \left\Vert A\left( \begin{array}{c}  \gamma _{1}^{\star }\\ \gamma _{2}^{\star }\\ \gamma _{3}^{\star } \end{array}  \right)  - A\left( \begin{array}{c}  \gamma _{1}\\ \gamma _{2}\\ \gamma _{3} \end{array}  \right) \right\Vert _{\infty }  & =\left\Vert A\left( \begin{array}{c}   \gamma _{1}^{\star }-\gamma _{1}\\ \gamma _{2}^{\star }-\gamma _{2}\\ \gamma _{3}^{\star }-\gamma _{3} \end{array}  \right) \right\Vert _{\infty }\\ & \leq \left\Vert A\right\Vert _{\infty }\left\Vert \left( \begin{array}{c}   \gamma _{1}^{\star }-\gamma _{1}\\ \gamma _{2}^{\star }-\gamma _{2}\\ \gamma _{3}^{\star }-\gamma _{3} \end{array}  \right) \right\Vert _{\infty } \end{array}</math>
+| style="text-align: center;" | <math>\left\Vert A\left( \begin{array}{c}  \gamma _{1}^{\star }\\ \gamma _{2}^{\star }\\ \gamma _{3}^{\star } \end{array}  \right)  - A\left( \begin{array}{c}  \gamma _{1}\\ \gamma _{2}\\ \gamma _{3} \end{array}  \right) \right\Vert _{\infty } =\left\Vert A\left( \begin{array}{c}   \gamma _{1}^{\star }-\gamma _{1}\\ \gamma _{2}^{\star }-\gamma _{2}\\ \gamma _{3}^{\star }-\gamma _{3} \end{array}  \right) \right\Vert _{\infty }</math>
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 {| style="text-align: left; margin:auto;width: 100%;"
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-| style="text-align: center;" | <math>\begin{array}{l}  \left\Vert A\right\Vert _{\infty }  & =\underset{1 \leq  i \leq  3}{\max }{\displaystyle \sum \limits _{j=1}^{3}}   \left\vert a_{ij}\right\vert  \wedge \\   {\displaystyle \sum \limits _{j=1}^{3}}   \left\vert a_{1j}\right\vert  & =\dfrac{2\sqrt{2}}{e^{x}+e^{-x}}, \\   {\displaystyle \sum \limits _{j=1}^{3}}   \left\vert a_{2j}\right\vert  & =\dfrac{\left\vert e^{x}-e^{-x}\right\vert  }{e^{x}+e^{-x}},\\    {\displaystyle \sum \limits _{j=1}^{3}}   \left\vert a_{3j}\right\vert  & =\dfrac{2\sqrt{2}}{e^{x}+e^{-x}} +\dfrac{\left\vert e^{x}-e^{-x}\right\vert }{e^{x}+e^{-x}}\\ & \Rightarrow \left\Vert A\right\Vert _{\infty }=\dfrac{2\sqrt{2}+\left\vert  e^{-x}-e^{x}\right\vert }{e^{x}+e^{-x}} \end{array}</math>
+| style="text-align: center;" | <math>\left\Vert A\right\Vert _{\infty } =\underset{1 \leq  i \leq  3}{\max }{\displaystyle \sum \limits _{j=1}^{3}}   \left\vert a_{ij}\right\vert  \wedge</math>
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 {| style="text-align: left; margin:auto;width: 100%;"
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-| style="text-align: center;" | <math>\begin{array}{l}  \left\Vert A\right\Vert _{\infty }  & =\dfrac{2\sqrt{2}+\left\vert e^{x} -e^{-x}\right\vert }{e^{x}+e^{-x}}=\dfrac{e^{2x}+2\sqrt{2}e^{x}-1}{e^{2x} +1}=\\ & =\dfrac{e^{2x}-1}{e^{2x}+1}+\dfrac{2\sqrt{2}e^{x}}{e^{2x}+1} \end{array}</math>
+| style="text-align: center;" | <math>\left\Vert A\right\Vert _{\infty } =\dfrac{2\sqrt{2}+\left\vert e^{x} -e^{-x}\right\vert }{e^{x}+e^{-x}}=\dfrac{e^{2x}+2\sqrt{2}e^{x}-1}{e^{2x} +1}=</math>
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 {| style="text-align: left; margin:auto;width: 100%;"
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-| style="text-align: center;" | <math>\begin{array}{l}  \left\Vert A\right\Vert _{\infty }  & =\dfrac{2\sqrt{2}+\left\vert e^{x} -e^{-x}\right\vert }{e^{x}+e^{-x}}=\dfrac{2\sqrt{2}e^{x}+1-e^{2x}}{e^{2x} +1}=\\ & =\dfrac{2\sqrt{2}e^{x}}{e^{2x}+1}+\dfrac{1-e^{2x}}{e^{2x}+1}\leq \sqrt{2}+1  \end{array}</math>
+| style="text-align: center;" | <math>\left\Vert A\right\Vert _{\infty }=\dfrac{2\sqrt{2}+\left\vert e^{x} -e^{-x}\right\vert }{e^{x}+e^{-x}}=\dfrac{2\sqrt{2}e^{x}+1-e^{2x}}{e^{2x} +1}=</math>
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 {| style="text-align: left; margin:auto;width: 100%;"
 |-
-| style="text-align: center;" | <math>\begin{array}{l}  \left( \begin{array}{c}   \partial _{x}\gamma _{1}\\ \partial _{x}\gamma _{2}\\ \partial _{x}\gamma _{3} \end{array}  \right)  & =\left( \begin{array}{ccc} 0 & 0 & -\dfrac{2\sqrt{2}}{e^{x}+e^{-x}}\\ 0 & 0 & \dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\\ \dfrac{2\sqrt{2}}{e^{x}+e^{-x}} & -\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}} & 0  \end{array}  \right) \left( \begin{array}{c}   \gamma _{1}\\ \gamma _{2}\\ \gamma _{3} \end{array}  \right)\\ \left( 0,\mathbf{\gamma }_{0}\right) ^{T}  & =\left( 0,0,0.5,-0.5\right) ^{T} \end{array}</math>
+| style="text-align: center;" | <math>\left( \begin{array}{c}   \partial _{x}\gamma _{1}\\ \partial _{x}\gamma _{2}\\ \partial _{x}\gamma _{3} \end{array}  \right) =\left( \begin{array}{ccc} 0 & 0 & -\dfrac{2\sqrt{2}}{e^{x}+e^{-x}}\\ 0 & 0 & \dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\\ \dfrac{2\sqrt{2}}{e^{x}+e^{-x}} & -\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}} & 0  \end{array}  \right) \left( \begin{array}{c}   \gamma _{1}\\ \gamma _{2}\\ \gamma _{3} \end{array}  \right)</math>
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 |}

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 {| style="text-align: left; margin:auto;width: 100%;"
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-| style="text-align: center;" | <math>\begin{array}{l}  \dfrac{u+v}{u-v}  & =\widehat{\gamma }_{3},1-\widehat{\gamma }_{3}=\dfrac{-2v}{u-v},\dfrac{\widehat{\gamma }_{3}}{1-\widehat{\gamma }_{3}}=-\dfrac{u+v}{2v}\\ \dfrac{1-uv}{u-v}  & =\widehat{\gamma }_{1}, \dfrac{1+uv}{u-v} i=\widehat{\gamma }_{2} \end{array}</math>
+| style="text-align: center;" | <math>\dfrac{u+v}{u-v} = \widehat{\gamma }_{3},1-\widehat{\gamma }_{3}=\dfrac{-2v}{u-v},\dfrac{\widehat{\gamma }_{3}}{1-\widehat{\gamma }_{3}}=-\dfrac{u+v}{2v}</math>
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 |}

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 {| style="text-align: left; margin:auto;width: 100%;"
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-| style="text-align: center;" | <math>\begin{array}{l} \dfrac{u+v}{u-v}  & =\widehat{\gamma }_{3},1-\widehat{\gamma }_{3}=\dfrac{-2v}{u-v},\dfrac{\widehat{\gamma }_{3}}{1-\widehat{\gamma }_{3}}=-\dfrac{u+v}{2v}\end{array}</math>
+| style="text-align: center;" | <math> \dfrac{u+v}{u-v}=\widehat{\gamma }_{3},1-\widehat{\gamma }_{3}=\dfrac{-2v}{u-v},\dfrac{\widehat{\gamma }_{3}}{1-\widehat{\gamma }_{3}}=-\dfrac{u+v}{2v}</math>
 |-
-| style="text-align: center;" | <math>\begin{array}{l} \dfrac{1-uv}{u-v}  & =\widehat{\gamma }_{1}, \dfrac{1+uv}{u-v} i=\widehat{\gamma }_{2} \end{array}</math>
+| style="text-align: center;" | <math>\dfrac{1-uv}{u-v}=\widehat{\gamma }_{1}, \dfrac{1+uv}{u-v} i=\widehat{\gamma }_{2}</math>
 |}
 |}
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 {| style="text-align: left; margin:auto;width: 100%;"
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-| style="text-align: center;" | <math>\begin{array}{l}  \dfrac{\alpha ^{\prime \prime }\left( z\right) }{2\alpha \left( z\right) }-\left( \dfrac{\alpha ^{\prime }\left( z\right) }{2\alpha \left( z\right) }\right) ^{2}-\alpha \overline{\alpha }=r=\frac{s}{q}= \end{array}</math>
+| style="text-align: center;" | <math>\dfrac{\alpha ^{\prime \prime }\left( z\right) }{2\alpha \left( z\right) }-\left( \dfrac{\alpha ^{\prime }\left( z\right) }{2\alpha \left( z\right) }\right) ^{2}-\alpha \overline{\alpha }=r=\frac{s}{q}= </math>
 |-
 |style="text-align: center;" | <math>\begin{array}{l}=\dfrac{z^{12}+24z^{10}+63 z^{8}-128z^{6}-33z^{4}+8z^{2} +1}{2z^{2}\left( z^{2}+1\right) ^{2}\left( z^{4}+6z^{2}+1\right) ^{2}}+\end{array}</math>

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 {| style="text-align: left; margin:auto;width: 100%;"
 |-
-| style="text-align: center;" | <math>\begin{array}{l}  \dfrac{u+v}{u-v}  & =\widehat{\gamma }_{3},1-\widehat{\gamma }_{3}=\dfrac{-2v}{u-v},\dfrac{\widehat{\gamma }_{3}}{1-\widehat{\gamma }_{3}}=-\dfrac{u+v}{2v}\\ \dfrac{1-uv}{u-v}  & =\widehat{\gamma }_{1}, \dfrac{1+uv}{u-v} i=\widehat{\gamma }_{2} \end{array}</math>
+| style="text-align: center;" | <math>\begin{array}{l} \dfrac{u+v}{u-v}  & =\widehat{\gamma }_{3},1-\widehat{\gamma }_{3}=\dfrac{-2v}{u-v},\dfrac{\widehat{\gamma }_{3}}{1-\widehat{\gamma }_{3}}=-\dfrac{u+v}{2v}\end{array}</math>
 |}
 |}
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 {| style="text-align: left; margin:auto;width: 100%;"
 |-
-| style="text-align: center;" | <math>\begin{array}{l}  & \dfrac{\alpha ^{\prime \prime }\left( z\right) }{2\alpha \left( z\right) }-\left( \dfrac{\alpha ^{\prime }\left( z\right) }{2\alpha \left( z\right) }\right) ^{2}-\alpha \overline{\alpha }=r=\frac{s}{q}= \end{array}</math>
+| style="text-align: center;" | <math>\begin{array}{l}  \dfrac{\alpha ^{\prime \prime }\left( z\right) }{2\alpha \left( z\right) }-\left( \dfrac{\alpha ^{\prime }\left( z\right) }{2\alpha \left( z\right) }\right) ^{2}-\alpha \overline{\alpha }=r=\frac{s}{q}= \end{array}</math>
 |-
 |style="text-align: center;" | <math>\begin{array}{l}=\dfrac{z^{12}+24z^{10}+63 z^{8}-128z^{6}-33z^{4}+8z^{2} +1}{2z^{2}\left( z^{2}+1\right) ^{2}\left( z^{4}+6z^{2}+1\right) ^{2}}+\end{array}</math>

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 {| style="text-align: left; margin:auto;width: 100%;"
 |-
-| style="text-align: center;" | <math>\begin{array}{l}  & \dfrac{\alpha ^{\prime \prime }\left( z\right) }{2\alpha \left( z\right) }-\left( \dfrac{\alpha ^{\prime }\left( z\right) }{2\alpha \left( z\right) }\right) ^{2}-\alpha \overline{\alpha }=r=\frac{s}{q}= \\ & =\dfrac{z^{12}+24z^{10}+63 z^{8}-128z^{6}-33z^{4}+8z^{2} +1}{2z^{2}\left( z^{2}+1\right) ^{2}\left( z^{4}+6z^{2}+1\right) ^{2}}+ \\ & +\dfrac{-2\sqrt{2}\left( 3z^{11}+21z^{9}+46z^{7}+42z^{5}+15z^{3} + z\right) i}{2z^{2}\left( z^{2}+1\right) ^{2}\left( z^{4}+6z^{2}+1\right) ^{2}} \end{array}</math>
+| style="text-align: center;" | <math>\begin{array}{l}  & \dfrac{\alpha ^{\prime \prime }\left( z\right) }{2\alpha \left( z\right) }-\left( \dfrac{\alpha ^{\prime }\left( z\right) }{2\alpha \left( z\right) }\right) ^{2}-\alpha \overline{\alpha }=r=\frac{s}{q}= \end{array}</math>
 |}
 |}

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 {| style="text-align: left; margin:auto;width: 100%;"
 |-
-| style="text-align: center;" | <math>\begin{array}{l}  f  & =e^{\frac{1}{2}\int \frac{\partial _{z}\alpha }{\alpha }}=e^{\frac{1}{2} \ln \alpha }\Rightarrow f=\alpha ^{\frac{1}{2}}, t=y\alpha ^{\frac{1}{2}}\\ \partial _{z}^{2}t  & =\left (\dfrac{1}{2}\frac{\partial _{z}^{2}\alpha }{} {\alpha }-\frac{1}{4}\left ( \frac{\partial _{z}\alpha }{\alpha }\right ) ^{2}-\alpha \overline{\alpha }\right ) t  \end{array}</math>
+| style="text-align: center;" | <math>\begin{array}{l}  f  & =e^{\frac{1}{2}\int \frac{\partial _{z}\alpha }{\alpha }}=e^{\frac{1}{2} \ln \alpha }\Rightarrow f=\alpha ^{\frac{1}{2}}, t=y\alpha ^{\frac{1}{2}}\end{array}</math>
 |}
 |}
 El proceso de algebrización del sistema diferencial lineal ha permitido  expresar el coeficiente de <math>t</math>, <math>\dfrac{1}{2}\frac{\partial _{z}^{2}\alpha  }{\alpha }-\frac{1}{4}\left( \frac{\partial _{z}\alpha }{\alpha }\right) ^{2}-\alpha \overline{\alpha }</math>, como una función racional, en efecto

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 ==Resumen==
-Este trabajo contrasta métodos numéricos con métodos algebraicos  aplicados ambos a la resolución de un sistema de ecuaciones  diferenciales lineales 3-dimensionales con matrices antisimétricas  definidas en un cuerpo diferencial no constante. Al mismo sistema se aplican métodos y algorítmos propios de la Teoría de Galois Diferencial, lo que permite resolverlo algebraicamente y métodos numé ricos, en particular métodos de la familia de Runge - Kutta. Por último, se calculan los errores absolutos y relativos entre las soluciones Liouvillianas, obtenidas mediante la resolución algebraica y  las soluciones obtenidas aplicando métodos numéricos.
+Este trabajo contrasta métodos numéricos con métodos algebraicos aplicados ambos a la resolución de un sistema de ecuaciones diferenciales lineales 3-dimensionales con matrices antisimétricas definidas en un cuerpo diferencial no constante. Al mismo sistema se aplican métodos y algorítmos propios de la Teoría de Galois Diferencial, lo que permite resolverlo algebraicamente y métodos numéricos, en particular métodos de la familia de Runge - Kutta. Por último, se calculan los errores absolutos y relativos entre las soluciones Liouvillianas, obtenidas mediante la resolución algebraica y  las soluciones obtenidas aplicando métodos numéricos.
-''Palabras claves'': Teoría de Galois diferencial, métodos de la familia de Runge -  Kutta, soluciones Liouvillianas, sistemas de ecuaciones diferenciales,  matrices antisimétricas, cuerpo diferencial no constante.
+''Palabras claves'': Teoría de Galois diferencial, métodos de la familia de Runge - Kutta, soluciones Liouvillianas, sistemas de ecuaciones diferenciales,  matrices antisimétricas, cuerpo diferencial no constante.
 ==Abstract==

← Older revision		Revision as of 14:20, 6 September 2017
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	This work contrasts numerical methods with algebraic methods. These methods are applied to solve a three dimensional linear differential system with skew symmetric matrices defined in a non- constant differential field. Algorithms and methods of Differential Galois Theory, are used to provide an algebraic solution, while numerical methods, in particular, methods from Runge - Kutta family, are applied to the same system. Finally, the absolute and relative errors between Liouvillians solution are calculated comparing the solutions obtained by means of algebraic methods and by means of numerical methods.		This work contrasts numerical methods with algebraic methods. These methods are applied to solve a three dimensional linear differential system with skew symmetric matrices defined in a non- constant differential field. Algorithms and methods of Differential Galois Theory, are used to provide an algebraic solution, while numerical methods, in particular, methods from Runge - Kutta family, are applied to the same system. Finally, the absolute and relative errors between Liouvillians solution are calculated comparing the solutions obtained by means of algebraic methods and by means of numerical methods.
−

	==1 Introducción==		==1 Introducción==

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 ==Agradecimientos==
-Los autores agradecen a la Vicerrectoría de Investigaciones de la Universidad Simón Bolívar, por el apoyo brindado para la publicación de este artículo. Los autores agradecen también a la Universidad del Norte, en donde se realizó la mayor parte del trabajo. Finalmente los autores agradecen a los árbitros anónimos por su ayuda para mejorar este documento.
+Los autores agradecen a la oficina de la DIDI de la Universidad del Norte, por el apoyo brindado para la publicación de este artículo. Los autores agradecen también a la Universidad Simón Bolívar. Finalmente los autores agradecen a los árbitros anónimos por su ayuda para mejorar este documento.
 ==Bibliografía==