Characteristic Curve of a PDE












0












$begingroup$


Question from a Exam:




Consider the pde $xu_{xx}+2x^2u_{xy}=u_x-1$.
Find the characteristic curves of the above.




Can someone please tell me how are these types of problems handled?



I dont want exact solution.Please tell me how to solve these type of questions.






pde characteristics







share|cite|improve this question









$endgroup$












  • $begingroup$
    You can change the variable to $v=u_x$ and apply the characteristic curves method.
    $endgroup$
    – Dante Grevino
    Dec 3 '18 at 15:43










  • $begingroup$
    @DanteGrevino; i obtain $v=u_x,xv_x+2x^2v_y=v-1$
    $endgroup$
    – Join_PhD
    Dec 3 '18 at 15:50










  • $begingroup$
    then i have if $a=x,b=2x^2,c=v-1implies frac{dx}{a}=frac{dy}{2a^2}=frac{dv}{v-1}$
    $endgroup$
    – Join_PhD
    Dec 3 '18 at 15:51












  • $begingroup$
    @DanteGrevino;how to solve it now
    $endgroup$
    – Join_PhD
    Dec 3 '18 at 15:52










  • $begingroup$
    Do you have boundary conditions?
    $endgroup$
    – Dante Grevino
    Dec 3 '18 at 16:00
















0












$begingroup$


Question from a Exam:




Consider the pde $xu_{xx}+2x^2u_{xy}=u_x-1$.
Find the characteristic curves of the above.




Can someone please tell me how are these types of problems handled?



I dont want exact solution.Please tell me how to solve these type of questions.






pde characteristics







share|cite|improve this question









$endgroup$












  • $begingroup$
    You can change the variable to $v=u_x$ and apply the characteristic curves method.
    $endgroup$
    – Dante Grevino
    Dec 3 '18 at 15:43










  • $begingroup$
    @DanteGrevino; i obtain $v=u_x,xv_x+2x^2v_y=v-1$
    $endgroup$
    – Join_PhD
    Dec 3 '18 at 15:50










  • $begingroup$
    then i have if $a=x,b=2x^2,c=v-1implies frac{dx}{a}=frac{dy}{2a^2}=frac{dv}{v-1}$
    $endgroup$
    – Join_PhD
    Dec 3 '18 at 15:51












  • $begingroup$
    @DanteGrevino;how to solve it now
    $endgroup$
    – Join_PhD
    Dec 3 '18 at 15:52










  • $begingroup$
    Do you have boundary conditions?
    $endgroup$
    – Dante Grevino
    Dec 3 '18 at 16:00














0












0








0





$begingroup$


Question from a Exam:




Consider the pde $xu_{xx}+2x^2u_{xy}=u_x-1$.
Find the characteristic curves of the above.




Can someone please tell me how are these types of problems handled?



I dont want exact solution.Please tell me how to solve these type of questions.






pde characteristics







share|cite|improve this question









$endgroup$




Question from a Exam:




Consider the pde $xu_{xx}+2x^2u_{xy}=u_x-1$.
Find the characteristic curves of the above.




Can someone please tell me how are these types of problems handled?



I dont want exact solution.Please tell me how to solve these type of questions.






pde characteristics




pde characteristics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 15:24









Join_PhDJoin_PhD

3818




3818












  • $begingroup$
    You can change the variable to $v=u_x$ and apply the characteristic curves method.
    $endgroup$
    – Dante Grevino
    Dec 3 '18 at 15:43










  • $begingroup$
    @DanteGrevino; i obtain $v=u_x,xv_x+2x^2v_y=v-1$
    $endgroup$
    – Join_PhD
    Dec 3 '18 at 15:50










  • $begingroup$
    then i have if $a=x,b=2x^2,c=v-1implies frac{dx}{a}=frac{dy}{2a^2}=frac{dv}{v-1}$
    $endgroup$
    – Join_PhD
    Dec 3 '18 at 15:51












  • $begingroup$
    @DanteGrevino;how to solve it now
    $endgroup$
    – Join_PhD
    Dec 3 '18 at 15:52










  • $begingroup$
    Do you have boundary conditions?
    $endgroup$
    – Dante Grevino
    Dec 3 '18 at 16:00


















  • $begingroup$
    You can change the variable to $v=u_x$ and apply the characteristic curves method.
    $endgroup$
    – Dante Grevino
    Dec 3 '18 at 15:43










  • $begingroup$
    @DanteGrevino; i obtain $v=u_x,xv_x+2x^2v_y=v-1$
    $endgroup$
    – Join_PhD
    Dec 3 '18 at 15:50










  • $begingroup$
    then i have if $a=x,b=2x^2,c=v-1implies frac{dx}{a}=frac{dy}{2a^2}=frac{dv}{v-1}$
    $endgroup$
    – Join_PhD
    Dec 3 '18 at 15:51












  • $begingroup$
    @DanteGrevino;how to solve it now
    $endgroup$
    – Join_PhD
    Dec 3 '18 at 15:52










  • $begingroup$
    Do you have boundary conditions?
    $endgroup$
    – Dante Grevino
    Dec 3 '18 at 16:00
















$begingroup$
You can change the variable to $v=u_x$ and apply the characteristic curves method.
$endgroup$
– Dante Grevino
Dec 3 '18 at 15:43




$begingroup$
You can change the variable to $v=u_x$ and apply the characteristic curves method.
$endgroup$
– Dante Grevino
Dec 3 '18 at 15:43












$begingroup$
@DanteGrevino; i obtain $v=u_x,xv_x+2x^2v_y=v-1$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:50




$begingroup$
@DanteGrevino; i obtain $v=u_x,xv_x+2x^2v_y=v-1$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:50












$begingroup$
then i have if $a=x,b=2x^2,c=v-1implies frac{dx}{a}=frac{dy}{2a^2}=frac{dv}{v-1}$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:51






$begingroup$
then i have if $a=x,b=2x^2,c=v-1implies frac{dx}{a}=frac{dy}{2a^2}=frac{dv}{v-1}$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:51














$begingroup$
@DanteGrevino;how to solve it now
$endgroup$
– Join_PhD
Dec 3 '18 at 15:52




$begingroup$
@DanteGrevino;how to solve it now
$endgroup$
– Join_PhD
Dec 3 '18 at 15:52












$begingroup$
Do you have boundary conditions?
$endgroup$
– Dante Grevino
Dec 3 '18 at 16:00




$begingroup$
Do you have boundary conditions?
$endgroup$
– Dante Grevino
Dec 3 '18 at 16:00










2 Answers
2






active

oldest

votes


















0












$begingroup$

Your first step should be noting that part of the PDE has a $u_y$ or $u_yy$, so first define $v=u_x$. Your PDE translates to $xv_x+2x^2v_y=v-1$.



You have a constant term, and none of your coefficients are dependent on y. Therefore, $v=fy+g$ where f and g are functions of x. Substituting $v=fy+g, v_x=f'y+g', v_y=f$, you get $xf'y+xg'+2x^2f=fy+g-1$. Dividing this up into two equations (one with y and one without) gives that $xf'=f$ and that $xg'+2x^2f=g-1$. Solving the first of these equations gives that $f=Ax$. Then, solving the second equation gives:
$$xg'+2Ax^3=g-1$$
$$g-xg'=2Ax^3+1$$
$$g=-Ax^3+Bx+1$$
Together, this gives that $u_x=v=Axy-Ax^3+Bx+1$. Integrating with respect to x gives $u=frac{A}{2}x^2y-frac{A}{4}x^4+frac{B}{2}x^2+x+C$. Redefining our constants of integration gives $u=C_1(x^4-2x^2y)+C_2x^2+x+C_3$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $$xu_{xx}+2x^2u_{xy}=u_x-1 tag 1$$
    $v=u_x$



    $$xv_{x}+2x^2v_{y}=v-1 tag 2$$
    Charpit-Lagrange equations:
    $$frac{dx}{x}=frac{dy}{2x^2}=frac{dv}{v-1}$$
    First family of characteristic curves from $frac{dx}{x}=frac{dy}{2x^2}$ :
    $$y-x^2=c_1$$
    Second family of characteristic curves from $frac{dx}{x}=frac{dv}{v-1}$
    $$frac{v-1}{x}=c_2$$
    General solution of the PDE Eq.$(2)$:
    $$frac{v-1}{x}=F(y-x^2)$$
    where $F$ is an arbitrary function.
    $$v(x,y)=1+xF(y-x^2)$$
    $u=int left(1+xF(y-x^2)right)dx$
    $$u(x,y)=x+int xF(y-x^2) dx+G(y)$$
    $F$ and $G$ are arbitrary functions. They have to be determined according to some boundary conditions.



    Since no boundary condition is specified in the wording of the question it is not possible to fully solve the problem.






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024195%2fcharacteristic-curve-of-a-pde%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Your first step should be noting that part of the PDE has a $u_y$ or $u_yy$, so first define $v=u_x$. Your PDE translates to $xv_x+2x^2v_y=v-1$.



      You have a constant term, and none of your coefficients are dependent on y. Therefore, $v=fy+g$ where f and g are functions of x. Substituting $v=fy+g, v_x=f'y+g', v_y=f$, you get $xf'y+xg'+2x^2f=fy+g-1$. Dividing this up into two equations (one with y and one without) gives that $xf'=f$ and that $xg'+2x^2f=g-1$. Solving the first of these equations gives that $f=Ax$. Then, solving the second equation gives:
      $$xg'+2Ax^3=g-1$$
      $$g-xg'=2Ax^3+1$$
      $$g=-Ax^3+Bx+1$$
      Together, this gives that $u_x=v=Axy-Ax^3+Bx+1$. Integrating with respect to x gives $u=frac{A}{2}x^2y-frac{A}{4}x^4+frac{B}{2}x^2+x+C$. Redefining our constants of integration gives $u=C_1(x^4-2x^2y)+C_2x^2+x+C_3$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Your first step should be noting that part of the PDE has a $u_y$ or $u_yy$, so first define $v=u_x$. Your PDE translates to $xv_x+2x^2v_y=v-1$.



        You have a constant term, and none of your coefficients are dependent on y. Therefore, $v=fy+g$ where f and g are functions of x. Substituting $v=fy+g, v_x=f'y+g', v_y=f$, you get $xf'y+xg'+2x^2f=fy+g-1$. Dividing this up into two equations (one with y and one without) gives that $xf'=f$ and that $xg'+2x^2f=g-1$. Solving the first of these equations gives that $f=Ax$. Then, solving the second equation gives:
        $$xg'+2Ax^3=g-1$$
        $$g-xg'=2Ax^3+1$$
        $$g=-Ax^3+Bx+1$$
        Together, this gives that $u_x=v=Axy-Ax^3+Bx+1$. Integrating with respect to x gives $u=frac{A}{2}x^2y-frac{A}{4}x^4+frac{B}{2}x^2+x+C$. Redefining our constants of integration gives $u=C_1(x^4-2x^2y)+C_2x^2+x+C_3$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Your first step should be noting that part of the PDE has a $u_y$ or $u_yy$, so first define $v=u_x$. Your PDE translates to $xv_x+2x^2v_y=v-1$.



          You have a constant term, and none of your coefficients are dependent on y. Therefore, $v=fy+g$ where f and g are functions of x. Substituting $v=fy+g, v_x=f'y+g', v_y=f$, you get $xf'y+xg'+2x^2f=fy+g-1$. Dividing this up into two equations (one with y and one without) gives that $xf'=f$ and that $xg'+2x^2f=g-1$. Solving the first of these equations gives that $f=Ax$. Then, solving the second equation gives:
          $$xg'+2Ax^3=g-1$$
          $$g-xg'=2Ax^3+1$$
          $$g=-Ax^3+Bx+1$$
          Together, this gives that $u_x=v=Axy-Ax^3+Bx+1$. Integrating with respect to x gives $u=frac{A}{2}x^2y-frac{A}{4}x^4+frac{B}{2}x^2+x+C$. Redefining our constants of integration gives $u=C_1(x^4-2x^2y)+C_2x^2+x+C_3$






          share|cite|improve this answer









          $endgroup$



          Your first step should be noting that part of the PDE has a $u_y$ or $u_yy$, so first define $v=u_x$. Your PDE translates to $xv_x+2x^2v_y=v-1$.



          You have a constant term, and none of your coefficients are dependent on y. Therefore, $v=fy+g$ where f and g are functions of x. Substituting $v=fy+g, v_x=f'y+g', v_y=f$, you get $xf'y+xg'+2x^2f=fy+g-1$. Dividing this up into two equations (one with y and one without) gives that $xf'=f$ and that $xg'+2x^2f=g-1$. Solving the first of these equations gives that $f=Ax$. Then, solving the second equation gives:
          $$xg'+2Ax^3=g-1$$
          $$g-xg'=2Ax^3+1$$
          $$g=-Ax^3+Bx+1$$
          Together, this gives that $u_x=v=Axy-Ax^3+Bx+1$. Integrating with respect to x gives $u=frac{A}{2}x^2y-frac{A}{4}x^4+frac{B}{2}x^2+x+C$. Redefining our constants of integration gives $u=C_1(x^4-2x^2y)+C_2x^2+x+C_3$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 16:07









          MoKo19MoKo19

          1914




          1914























              0












              $begingroup$

              $$xu_{xx}+2x^2u_{xy}=u_x-1 tag 1$$
              $v=u_x$



              $$xv_{x}+2x^2v_{y}=v-1 tag 2$$
              Charpit-Lagrange equations:
              $$frac{dx}{x}=frac{dy}{2x^2}=frac{dv}{v-1}$$
              First family of characteristic curves from $frac{dx}{x}=frac{dy}{2x^2}$ :
              $$y-x^2=c_1$$
              Second family of characteristic curves from $frac{dx}{x}=frac{dv}{v-1}$
              $$frac{v-1}{x}=c_2$$
              General solution of the PDE Eq.$(2)$:
              $$frac{v-1}{x}=F(y-x^2)$$
              where $F$ is an arbitrary function.
              $$v(x,y)=1+xF(y-x^2)$$
              $u=int left(1+xF(y-x^2)right)dx$
              $$u(x,y)=x+int xF(y-x^2) dx+G(y)$$
              $F$ and $G$ are arbitrary functions. They have to be determined according to some boundary conditions.



              Since no boundary condition is specified in the wording of the question it is not possible to fully solve the problem.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $$xu_{xx}+2x^2u_{xy}=u_x-1 tag 1$$
                $v=u_x$



                $$xv_{x}+2x^2v_{y}=v-1 tag 2$$
                Charpit-Lagrange equations:
                $$frac{dx}{x}=frac{dy}{2x^2}=frac{dv}{v-1}$$
                First family of characteristic curves from $frac{dx}{x}=frac{dy}{2x^2}$ :
                $$y-x^2=c_1$$
                Second family of characteristic curves from $frac{dx}{x}=frac{dv}{v-1}$
                $$frac{v-1}{x}=c_2$$
                General solution of the PDE Eq.$(2)$:
                $$frac{v-1}{x}=F(y-x^2)$$
                where $F$ is an arbitrary function.
                $$v(x,y)=1+xF(y-x^2)$$
                $u=int left(1+xF(y-x^2)right)dx$
                $$u(x,y)=x+int xF(y-x^2) dx+G(y)$$
                $F$ and $G$ are arbitrary functions. They have to be determined according to some boundary conditions.



                Since no boundary condition is specified in the wording of the question it is not possible to fully solve the problem.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$xu_{xx}+2x^2u_{xy}=u_x-1 tag 1$$
                  $v=u_x$



                  $$xv_{x}+2x^2v_{y}=v-1 tag 2$$
                  Charpit-Lagrange equations:
                  $$frac{dx}{x}=frac{dy}{2x^2}=frac{dv}{v-1}$$
                  First family of characteristic curves from $frac{dx}{x}=frac{dy}{2x^2}$ :
                  $$y-x^2=c_1$$
                  Second family of characteristic curves from $frac{dx}{x}=frac{dv}{v-1}$
                  $$frac{v-1}{x}=c_2$$
                  General solution of the PDE Eq.$(2)$:
                  $$frac{v-1}{x}=F(y-x^2)$$
                  where $F$ is an arbitrary function.
                  $$v(x,y)=1+xF(y-x^2)$$
                  $u=int left(1+xF(y-x^2)right)dx$
                  $$u(x,y)=x+int xF(y-x^2) dx+G(y)$$
                  $F$ and $G$ are arbitrary functions. They have to be determined according to some boundary conditions.



                  Since no boundary condition is specified in the wording of the question it is not possible to fully solve the problem.






                  share|cite|improve this answer









                  $endgroup$



                  $$xu_{xx}+2x^2u_{xy}=u_x-1 tag 1$$
                  $v=u_x$



                  $$xv_{x}+2x^2v_{y}=v-1 tag 2$$
                  Charpit-Lagrange equations:
                  $$frac{dx}{x}=frac{dy}{2x^2}=frac{dv}{v-1}$$
                  First family of characteristic curves from $frac{dx}{x}=frac{dy}{2x^2}$ :
                  $$y-x^2=c_1$$
                  Second family of characteristic curves from $frac{dx}{x}=frac{dv}{v-1}$
                  $$frac{v-1}{x}=c_2$$
                  General solution of the PDE Eq.$(2)$:
                  $$frac{v-1}{x}=F(y-x^2)$$
                  where $F$ is an arbitrary function.
                  $$v(x,y)=1+xF(y-x^2)$$
                  $u=int left(1+xF(y-x^2)right)dx$
                  $$u(x,y)=x+int xF(y-x^2) dx+G(y)$$
                  $F$ and $G$ are arbitrary functions. They have to be determined according to some boundary conditions.



                  Since no boundary condition is specified in the wording of the question it is not possible to fully solve the problem.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 8 '18 at 8:50









                  JJacquelinJJacquelin

                  43.3k21853




                  43.3k21853






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024195%2fcharacteristic-curve-of-a-pde%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Plaza Victoria

                      Image
                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                      Read more

                      Puebla de Zaragoza

                      Image
                      Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                      Read more

                      Musa

                      Image
                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                      Read more