charpits method to solve $u_x^2 + yu_y = u$












1














$u_x^2 + yu_y = u$ subject to $u(x,1) = 1 +x^2 /4$ for $-infty < x < infty $



Setting $p = du/dx$ and $q = du/dy$, I get $p^2 +yq = u$ and so I am able to write the diffeq as $F(p,q,y,u) = p^2 + qy - u = 0 $



Charpits equations then give me:
$dx/dt = 2p$,
$dy/dt = y$,
$dp/dt = p$,
$dq/dt = 0 $,
$du/dt = 2p^2 + qy$



How would I go about solving these equations in order to find u in explicit form and then find where it is uniquely defined by the data?






differential-equations pde partial-derivative characteristics







share|cite|improve this question





























    1














    $u_x^2 + yu_y = u$ subject to $u(x,1) = 1 +x^2 /4$ for $-infty < x < infty $



    Setting $p = du/dx$ and $q = du/dy$, I get $p^2 +yq = u$ and so I am able to write the diffeq as $F(p,q,y,u) = p^2 + qy - u = 0 $



    Charpits equations then give me:
    $dx/dt = 2p$,
    $dy/dt = y$,
    $dp/dt = p$,
    $dq/dt = 0 $,
    $du/dt = 2p^2 + qy$



    How would I go about solving these equations in order to find u in explicit form and then find where it is uniquely defined by the data?






    differential-equations pde partial-derivative characteristics







    share|cite|improve this question



























      1












      1








      1







      $u_x^2 + yu_y = u$ subject to $u(x,1) = 1 +x^2 /4$ for $-infty < x < infty $



      Setting $p = du/dx$ and $q = du/dy$, I get $p^2 +yq = u$ and so I am able to write the diffeq as $F(p,q,y,u) = p^2 + qy - u = 0 $



      Charpits equations then give me:
      $dx/dt = 2p$,
      $dy/dt = y$,
      $dp/dt = p$,
      $dq/dt = 0 $,
      $du/dt = 2p^2 + qy$



      How would I go about solving these equations in order to find u in explicit form and then find where it is uniquely defined by the data?






      differential-equations pde partial-derivative characteristics







      share|cite|improve this question















      $u_x^2 + yu_y = u$ subject to $u(x,1) = 1 +x^2 /4$ for $-infty < x < infty $



      Setting $p = du/dx$ and $q = du/dy$, I get $p^2 +yq = u$ and so I am able to write the diffeq as $F(p,q,y,u) = p^2 + qy - u = 0 $



      Charpits equations then give me:
      $dx/dt = 2p$,
      $dy/dt = y$,
      $dp/dt = p$,
      $dq/dt = 0 $,
      $du/dt = 2p^2 + qy$



      How would I go about solving these equations in order to find u in explicit form and then find where it is uniquely defined by the data?






      differential-equations pde partial-derivative characteristics




      differential-equations pde partial-derivative characteristics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 24 at 13:07

























      asked Nov 24 at 13:02









      pablo_mathscobar

      836




      836






















          1 Answer
          1






          active

          oldest

          votes


















          2














          You should be able to see that some equations can be solved directly, $y=y_0e^t$, $p=p_0e^t$, $q=q_0$, and then insert into the remaining equations, $$x=x_0+2p_0(e^t-1),$$ $$z=z_0+p_0^2(e^{2t}-1)+q_0y_0(e^t-1).$$



          Now apply $$dz=p,dx+q,dyimplies dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)$$ along the initial curve $x_0(s)=s$, $y_0(s)=1$, $z_0(s)=u(x_0(s),y_0(s))=1+frac{s^2}4$ to get $$frac{x_0}2=p_0cdot 1+q_0cdot 0.$$ Using the original equation $p_0^2+q_0=z_0$ gives $q_0=1$. Thus all collected,
          begin{align}
          x&=x_0y\
          z=u(x,y)&=1+frac{x_0^2}4+frac{x_0^2}4(y^2-1)+(y-1)=frac{x^2}4+y
          end{align}
          etc.






          share|cite|improve this answer























          • I think you assumed that $du/dt=2p2+qx$ and not $du/dt=2p2+qy$ Solving $du/dt=2p2+qy$ is $u = p_0^2 e^2t + q_0y_0e^t + u_0$ If $u_0=u(x,1)=1+x^2/4$, how do I find $p_0$ and $q_0$, is $p_0 = x/2$ and $q_0=0$
            – pablo_mathscobar
            Nov 24 at 13:23












          • Corrected. The initial curve has $x_0(s)=s$, $y_0(s)=1$ and $z_0(s)=1+s^2/4$. Now insert into $$dot z_0(s)=p_0(s)dot x_0(s)+q(s)dot y_0(s)$$ to find $p_0(s)=s/2$. Then use the original equation to find $q_0(s)=1$.
            – LutzL
            Nov 24 at 13:39










          • when solving $dx/dt = 2p_0 e^t$ we get $x = 2p_0e^t + C_1$ since $x_0 = s$ and $p_0 = s/2$ surely this should give $C_1 = 0$ at t=0 and so the equation for x should be $x = 2p_0e^t $ and not $x=x0+2p_0e^t = s + 2p_0e^t$ as you claim
            – pablo_mathscobar
            Nov 24 at 13:44












          • Yes, if you integrate $e^t$ you get $(e^t-1)$ etc.
            – LutzL
            Nov 24 at 13:54










          • brilliant thanks
            – pablo_mathscobar
            Nov 24 at 14:11











          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%2f3011525%2fcharpits-method-to-solve-u-x2-yu-y-u%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          You should be able to see that some equations can be solved directly, $y=y_0e^t$, $p=p_0e^t$, $q=q_0$, and then insert into the remaining equations, $$x=x_0+2p_0(e^t-1),$$ $$z=z_0+p_0^2(e^{2t}-1)+q_0y_0(e^t-1).$$



          Now apply $$dz=p,dx+q,dyimplies dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)$$ along the initial curve $x_0(s)=s$, $y_0(s)=1$, $z_0(s)=u(x_0(s),y_0(s))=1+frac{s^2}4$ to get $$frac{x_0}2=p_0cdot 1+q_0cdot 0.$$ Using the original equation $p_0^2+q_0=z_0$ gives $q_0=1$. Thus all collected,
          begin{align}
          x&=x_0y\
          z=u(x,y)&=1+frac{x_0^2}4+frac{x_0^2}4(y^2-1)+(y-1)=frac{x^2}4+y
          end{align}
          etc.






          share|cite|improve this answer























          • I think you assumed that $du/dt=2p2+qx$ and not $du/dt=2p2+qy$ Solving $du/dt=2p2+qy$ is $u = p_0^2 e^2t + q_0y_0e^t + u_0$ If $u_0=u(x,1)=1+x^2/4$, how do I find $p_0$ and $q_0$, is $p_0 = x/2$ and $q_0=0$
            – pablo_mathscobar
            Nov 24 at 13:23












          • Corrected. The initial curve has $x_0(s)=s$, $y_0(s)=1$ and $z_0(s)=1+s^2/4$. Now insert into $$dot z_0(s)=p_0(s)dot x_0(s)+q(s)dot y_0(s)$$ to find $p_0(s)=s/2$. Then use the original equation to find $q_0(s)=1$.
            – LutzL
            Nov 24 at 13:39










          • when solving $dx/dt = 2p_0 e^t$ we get $x = 2p_0e^t + C_1$ since $x_0 = s$ and $p_0 = s/2$ surely this should give $C_1 = 0$ at t=0 and so the equation for x should be $x = 2p_0e^t $ and not $x=x0+2p_0e^t = s + 2p_0e^t$ as you claim
            – pablo_mathscobar
            Nov 24 at 13:44












          • Yes, if you integrate $e^t$ you get $(e^t-1)$ etc.
            – LutzL
            Nov 24 at 13:54










          • brilliant thanks
            – pablo_mathscobar
            Nov 24 at 14:11
















          2














          You should be able to see that some equations can be solved directly, $y=y_0e^t$, $p=p_0e^t$, $q=q_0$, and then insert into the remaining equations, $$x=x_0+2p_0(e^t-1),$$ $$z=z_0+p_0^2(e^{2t}-1)+q_0y_0(e^t-1).$$



          Now apply $$dz=p,dx+q,dyimplies dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)$$ along the initial curve $x_0(s)=s$, $y_0(s)=1$, $z_0(s)=u(x_0(s),y_0(s))=1+frac{s^2}4$ to get $$frac{x_0}2=p_0cdot 1+q_0cdot 0.$$ Using the original equation $p_0^2+q_0=z_0$ gives $q_0=1$. Thus all collected,
          begin{align}
          x&=x_0y\
          z=u(x,y)&=1+frac{x_0^2}4+frac{x_0^2}4(y^2-1)+(y-1)=frac{x^2}4+y
          end{align}
          etc.






          share|cite|improve this answer























          • I think you assumed that $du/dt=2p2+qx$ and not $du/dt=2p2+qy$ Solving $du/dt=2p2+qy$ is $u = p_0^2 e^2t + q_0y_0e^t + u_0$ If $u_0=u(x,1)=1+x^2/4$, how do I find $p_0$ and $q_0$, is $p_0 = x/2$ and $q_0=0$
            – pablo_mathscobar
            Nov 24 at 13:23












          • Corrected. The initial curve has $x_0(s)=s$, $y_0(s)=1$ and $z_0(s)=1+s^2/4$. Now insert into $$dot z_0(s)=p_0(s)dot x_0(s)+q(s)dot y_0(s)$$ to find $p_0(s)=s/2$. Then use the original equation to find $q_0(s)=1$.
            – LutzL
            Nov 24 at 13:39










          • when solving $dx/dt = 2p_0 e^t$ we get $x = 2p_0e^t + C_1$ since $x_0 = s$ and $p_0 = s/2$ surely this should give $C_1 = 0$ at t=0 and so the equation for x should be $x = 2p_0e^t $ and not $x=x0+2p_0e^t = s + 2p_0e^t$ as you claim
            – pablo_mathscobar
            Nov 24 at 13:44












          • Yes, if you integrate $e^t$ you get $(e^t-1)$ etc.
            – LutzL
            Nov 24 at 13:54










          • brilliant thanks
            – pablo_mathscobar
            Nov 24 at 14:11














          2












          2








          2






          You should be able to see that some equations can be solved directly, $y=y_0e^t$, $p=p_0e^t$, $q=q_0$, and then insert into the remaining equations, $$x=x_0+2p_0(e^t-1),$$ $$z=z_0+p_0^2(e^{2t}-1)+q_0y_0(e^t-1).$$



          Now apply $$dz=p,dx+q,dyimplies dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)$$ along the initial curve $x_0(s)=s$, $y_0(s)=1$, $z_0(s)=u(x_0(s),y_0(s))=1+frac{s^2}4$ to get $$frac{x_0}2=p_0cdot 1+q_0cdot 0.$$ Using the original equation $p_0^2+q_0=z_0$ gives $q_0=1$. Thus all collected,
          begin{align}
          x&=x_0y\
          z=u(x,y)&=1+frac{x_0^2}4+frac{x_0^2}4(y^2-1)+(y-1)=frac{x^2}4+y
          end{align}
          etc.






          share|cite|improve this answer














          You should be able to see that some equations can be solved directly, $y=y_0e^t$, $p=p_0e^t$, $q=q_0$, and then insert into the remaining equations, $$x=x_0+2p_0(e^t-1),$$ $$z=z_0+p_0^2(e^{2t}-1)+q_0y_0(e^t-1).$$



          Now apply $$dz=p,dx+q,dyimplies dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)$$ along the initial curve $x_0(s)=s$, $y_0(s)=1$, $z_0(s)=u(x_0(s),y_0(s))=1+frac{s^2}4$ to get $$frac{x_0}2=p_0cdot 1+q_0cdot 0.$$ Using the original equation $p_0^2+q_0=z_0$ gives $q_0=1$. Thus all collected,
          begin{align}
          x&=x_0y\
          z=u(x,y)&=1+frac{x_0^2}4+frac{x_0^2}4(y^2-1)+(y-1)=frac{x^2}4+y
          end{align}
          etc.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 24 at 13:53

























          answered Nov 24 at 13:08









          LutzL

          55.8k42054




          55.8k42054












          • I think you assumed that $du/dt=2p2+qx$ and not $du/dt=2p2+qy$ Solving $du/dt=2p2+qy$ is $u = p_0^2 e^2t + q_0y_0e^t + u_0$ If $u_0=u(x,1)=1+x^2/4$, how do I find $p_0$ and $q_0$, is $p_0 = x/2$ and $q_0=0$
            – pablo_mathscobar
            Nov 24 at 13:23












          • Corrected. The initial curve has $x_0(s)=s$, $y_0(s)=1$ and $z_0(s)=1+s^2/4$. Now insert into $$dot z_0(s)=p_0(s)dot x_0(s)+q(s)dot y_0(s)$$ to find $p_0(s)=s/2$. Then use the original equation to find $q_0(s)=1$.
            – LutzL
            Nov 24 at 13:39










          • when solving $dx/dt = 2p_0 e^t$ we get $x = 2p_0e^t + C_1$ since $x_0 = s$ and $p_0 = s/2$ surely this should give $C_1 = 0$ at t=0 and so the equation for x should be $x = 2p_0e^t $ and not $x=x0+2p_0e^t = s + 2p_0e^t$ as you claim
            – pablo_mathscobar
            Nov 24 at 13:44












          • Yes, if you integrate $e^t$ you get $(e^t-1)$ etc.
            – LutzL
            Nov 24 at 13:54










          • brilliant thanks
            – pablo_mathscobar
            Nov 24 at 14:11


















          • I think you assumed that $du/dt=2p2+qx$ and not $du/dt=2p2+qy$ Solving $du/dt=2p2+qy$ is $u = p_0^2 e^2t + q_0y_0e^t + u_0$ If $u_0=u(x,1)=1+x^2/4$, how do I find $p_0$ and $q_0$, is $p_0 = x/2$ and $q_0=0$
            – pablo_mathscobar
            Nov 24 at 13:23












          • Corrected. The initial curve has $x_0(s)=s$, $y_0(s)=1$ and $z_0(s)=1+s^2/4$. Now insert into $$dot z_0(s)=p_0(s)dot x_0(s)+q(s)dot y_0(s)$$ to find $p_0(s)=s/2$. Then use the original equation to find $q_0(s)=1$.
            – LutzL
            Nov 24 at 13:39










          • when solving $dx/dt = 2p_0 e^t$ we get $x = 2p_0e^t + C_1$ since $x_0 = s$ and $p_0 = s/2$ surely this should give $C_1 = 0$ at t=0 and so the equation for x should be $x = 2p_0e^t $ and not $x=x0+2p_0e^t = s + 2p_0e^t$ as you claim
            – pablo_mathscobar
            Nov 24 at 13:44












          • Yes, if you integrate $e^t$ you get $(e^t-1)$ etc.
            – LutzL
            Nov 24 at 13:54










          • brilliant thanks
            – pablo_mathscobar
            Nov 24 at 14:11
















          I think you assumed that $du/dt=2p2+qx$ and not $du/dt=2p2+qy$ Solving $du/dt=2p2+qy$ is $u = p_0^2 e^2t + q_0y_0e^t + u_0$ If $u_0=u(x,1)=1+x^2/4$, how do I find $p_0$ and $q_0$, is $p_0 = x/2$ and $q_0=0$
          – pablo_mathscobar
          Nov 24 at 13:23






          I think you assumed that $du/dt=2p2+qx$ and not $du/dt=2p2+qy$ Solving $du/dt=2p2+qy$ is $u = p_0^2 e^2t + q_0y_0e^t + u_0$ If $u_0=u(x,1)=1+x^2/4$, how do I find $p_0$ and $q_0$, is $p_0 = x/2$ and $q_0=0$
          – pablo_mathscobar
          Nov 24 at 13:23














          Corrected. The initial curve has $x_0(s)=s$, $y_0(s)=1$ and $z_0(s)=1+s^2/4$. Now insert into $$dot z_0(s)=p_0(s)dot x_0(s)+q(s)dot y_0(s)$$ to find $p_0(s)=s/2$. Then use the original equation to find $q_0(s)=1$.
          – LutzL
          Nov 24 at 13:39




          Corrected. The initial curve has $x_0(s)=s$, $y_0(s)=1$ and $z_0(s)=1+s^2/4$. Now insert into $$dot z_0(s)=p_0(s)dot x_0(s)+q(s)dot y_0(s)$$ to find $p_0(s)=s/2$. Then use the original equation to find $q_0(s)=1$.
          – LutzL
          Nov 24 at 13:39












          when solving $dx/dt = 2p_0 e^t$ we get $x = 2p_0e^t + C_1$ since $x_0 = s$ and $p_0 = s/2$ surely this should give $C_1 = 0$ at t=0 and so the equation for x should be $x = 2p_0e^t $ and not $x=x0+2p_0e^t = s + 2p_0e^t$ as you claim
          – pablo_mathscobar
          Nov 24 at 13:44






          when solving $dx/dt = 2p_0 e^t$ we get $x = 2p_0e^t + C_1$ since $x_0 = s$ and $p_0 = s/2$ surely this should give $C_1 = 0$ at t=0 and so the equation for x should be $x = 2p_0e^t $ and not $x=x0+2p_0e^t = s + 2p_0e^t$ as you claim
          – pablo_mathscobar
          Nov 24 at 13:44














          Yes, if you integrate $e^t$ you get $(e^t-1)$ etc.
          – LutzL
          Nov 24 at 13:54




          Yes, if you integrate $e^t$ you get $(e^t-1)$ etc.
          – LutzL
          Nov 24 at 13:54












          brilliant thanks
          – pablo_mathscobar
          Nov 24 at 14:11




          brilliant thanks
          – pablo_mathscobar
          Nov 24 at 14:11


















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3011525%2fcharpits-method-to-solve-u-x2-yu-y-u%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