How to solve the coupled differential equation












0












$begingroup$


I have the following differential equations.



$$ tau Vfrac{partial r}{partial x} + Wfrac{partial ^2r}{partial x^2} + [r - Cu(1-r^2)][1-r^2]=0,$$



$$ Vfrac{partial u}{partial x} + Dfrac{partial^2 u}{partial x^2} - frac{V}{2}frac{partial r}{partial x}=0,$$ with far field boundary condition as $u_{x=infty} = 0.6.$ and $u_{x=0} = 0$. Here, $tau, V, W, D, C$ are dimensionless constants.



Now, I would like to solve the second equation subject to boundary conditions. But the last term in the second equation confuses me on how to proceed. Can anyone help me out with this ?



thanks






ordinary-differential-equations numerical-methods







share|cite|improve this question











$endgroup$












  • $begingroup$
    I would rather say that the last term in the first equation is the most difficult to deal with. Is $x$ the only independent variable? Why are those partial differential equations instead of ordinary equations? There are a handful numerical methods that would help you to solve these equations if they are ordinary.
    $endgroup$
    – rafa11111
    Dec 18 '18 at 12:51










  • $begingroup$
    @rafa11111 Ok. let us say these are ordinary equations. I would like to solve it analytically.
    $endgroup$
    – newstudent
    Dec 18 '18 at 12:52










  • $begingroup$
    If you are sure these are ordinary equations, please edit the question to address so (change the equation, the tag, etc.). I don't think there is an analytical solution to these equations, mainly due to the nonlinear term in the first equation. If $C$ is a small number, you can use a perturbation method, that provides an approximate analytical solution.
    $endgroup$
    – rafa11111
    Dec 18 '18 at 13:01










  • $begingroup$
    You can integrate the second equation once then substitute for $r(x)$ in the first to give an equation in $u(x)$ alone. I think your journey will end there!
    $endgroup$
    – user121049
    Dec 19 '18 at 8:00










  • $begingroup$
    You need some more boundary conditions on $r$. The only hope I see is to assume $r$ stays small so you can set $1-r^2approx 1$ and linearize the first equation. The two equations are then easy to solve. Then check $r$ really stays small.
    $endgroup$
    – user121049
    Dec 20 '18 at 9:00
















0












$begingroup$


I have the following differential equations.



$$ tau Vfrac{partial r}{partial x} + Wfrac{partial ^2r}{partial x^2} + [r - Cu(1-r^2)][1-r^2]=0,$$



$$ Vfrac{partial u}{partial x} + Dfrac{partial^2 u}{partial x^2} - frac{V}{2}frac{partial r}{partial x}=0,$$ with far field boundary condition as $u_{x=infty} = 0.6.$ and $u_{x=0} = 0$. Here, $tau, V, W, D, C$ are dimensionless constants.



Now, I would like to solve the second equation subject to boundary conditions. But the last term in the second equation confuses me on how to proceed. Can anyone help me out with this ?



thanks






ordinary-differential-equations numerical-methods







share|cite|improve this question











$endgroup$












  • $begingroup$
    I would rather say that the last term in the first equation is the most difficult to deal with. Is $x$ the only independent variable? Why are those partial differential equations instead of ordinary equations? There are a handful numerical methods that would help you to solve these equations if they are ordinary.
    $endgroup$
    – rafa11111
    Dec 18 '18 at 12:51










  • $begingroup$
    @rafa11111 Ok. let us say these are ordinary equations. I would like to solve it analytically.
    $endgroup$
    – newstudent
    Dec 18 '18 at 12:52










  • $begingroup$
    If you are sure these are ordinary equations, please edit the question to address so (change the equation, the tag, etc.). I don't think there is an analytical solution to these equations, mainly due to the nonlinear term in the first equation. If $C$ is a small number, you can use a perturbation method, that provides an approximate analytical solution.
    $endgroup$
    – rafa11111
    Dec 18 '18 at 13:01










  • $begingroup$
    You can integrate the second equation once then substitute for $r(x)$ in the first to give an equation in $u(x)$ alone. I think your journey will end there!
    $endgroup$
    – user121049
    Dec 19 '18 at 8:00










  • $begingroup$
    You need some more boundary conditions on $r$. The only hope I see is to assume $r$ stays small so you can set $1-r^2approx 1$ and linearize the first equation. The two equations are then easy to solve. Then check $r$ really stays small.
    $endgroup$
    – user121049
    Dec 20 '18 at 9:00














0












0








0





$begingroup$


I have the following differential equations.



$$ tau Vfrac{partial r}{partial x} + Wfrac{partial ^2r}{partial x^2} + [r - Cu(1-r^2)][1-r^2]=0,$$



$$ Vfrac{partial u}{partial x} + Dfrac{partial^2 u}{partial x^2} - frac{V}{2}frac{partial r}{partial x}=0,$$ with far field boundary condition as $u_{x=infty} = 0.6.$ and $u_{x=0} = 0$. Here, $tau, V, W, D, C$ are dimensionless constants.



Now, I would like to solve the second equation subject to boundary conditions. But the last term in the second equation confuses me on how to proceed. Can anyone help me out with this ?



thanks






ordinary-differential-equations numerical-methods







share|cite|improve this question











$endgroup$




I have the following differential equations.



$$ tau Vfrac{partial r}{partial x} + Wfrac{partial ^2r}{partial x^2} + [r - Cu(1-r^2)][1-r^2]=0,$$



$$ Vfrac{partial u}{partial x} + Dfrac{partial^2 u}{partial x^2} - frac{V}{2}frac{partial r}{partial x}=0,$$ with far field boundary condition as $u_{x=infty} = 0.6.$ and $u_{x=0} = 0$. Here, $tau, V, W, D, C$ are dimensionless constants.



Now, I would like to solve the second equation subject to boundary conditions. But the last term in the second equation confuses me on how to proceed. Can anyone help me out with this ?



thanks






ordinary-differential-equations numerical-methods




ordinary-differential-equations numerical-methods






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 13:03







newstudent

















asked Dec 18 '18 at 12:38









newstudentnewstudent

366




366












  • $begingroup$
    I would rather say that the last term in the first equation is the most difficult to deal with. Is $x$ the only independent variable? Why are those partial differential equations instead of ordinary equations? There are a handful numerical methods that would help you to solve these equations if they are ordinary.
    $endgroup$
    – rafa11111
    Dec 18 '18 at 12:51










  • $begingroup$
    @rafa11111 Ok. let us say these are ordinary equations. I would like to solve it analytically.
    $endgroup$
    – newstudent
    Dec 18 '18 at 12:52










  • $begingroup$
    If you are sure these are ordinary equations, please edit the question to address so (change the equation, the tag, etc.). I don't think there is an analytical solution to these equations, mainly due to the nonlinear term in the first equation. If $C$ is a small number, you can use a perturbation method, that provides an approximate analytical solution.
    $endgroup$
    – rafa11111
    Dec 18 '18 at 13:01










  • $begingroup$
    You can integrate the second equation once then substitute for $r(x)$ in the first to give an equation in $u(x)$ alone. I think your journey will end there!
    $endgroup$
    – user121049
    Dec 19 '18 at 8:00










  • $begingroup$
    You need some more boundary conditions on $r$. The only hope I see is to assume $r$ stays small so you can set $1-r^2approx 1$ and linearize the first equation. The two equations are then easy to solve. Then check $r$ really stays small.
    $endgroup$
    – user121049
    Dec 20 '18 at 9:00


















  • $begingroup$
    I would rather say that the last term in the first equation is the most difficult to deal with. Is $x$ the only independent variable? Why are those partial differential equations instead of ordinary equations? There are a handful numerical methods that would help you to solve these equations if they are ordinary.
    $endgroup$
    – rafa11111
    Dec 18 '18 at 12:51










  • $begingroup$
    @rafa11111 Ok. let us say these are ordinary equations. I would like to solve it analytically.
    $endgroup$
    – newstudent
    Dec 18 '18 at 12:52










  • $begingroup$
    If you are sure these are ordinary equations, please edit the question to address so (change the equation, the tag, etc.). I don't think there is an analytical solution to these equations, mainly due to the nonlinear term in the first equation. If $C$ is a small number, you can use a perturbation method, that provides an approximate analytical solution.
    $endgroup$
    – rafa11111
    Dec 18 '18 at 13:01










  • $begingroup$
    You can integrate the second equation once then substitute for $r(x)$ in the first to give an equation in $u(x)$ alone. I think your journey will end there!
    $endgroup$
    – user121049
    Dec 19 '18 at 8:00










  • $begingroup$
    You need some more boundary conditions on $r$. The only hope I see is to assume $r$ stays small so you can set $1-r^2approx 1$ and linearize the first equation. The two equations are then easy to solve. Then check $r$ really stays small.
    $endgroup$
    – user121049
    Dec 20 '18 at 9:00
















$begingroup$
I would rather say that the last term in the first equation is the most difficult to deal with. Is $x$ the only independent variable? Why are those partial differential equations instead of ordinary equations? There are a handful numerical methods that would help you to solve these equations if they are ordinary.
$endgroup$
– rafa11111
Dec 18 '18 at 12:51




$begingroup$
I would rather say that the last term in the first equation is the most difficult to deal with. Is $x$ the only independent variable? Why are those partial differential equations instead of ordinary equations? There are a handful numerical methods that would help you to solve these equations if they are ordinary.
$endgroup$
– rafa11111
Dec 18 '18 at 12:51












$begingroup$
@rafa11111 Ok. let us say these are ordinary equations. I would like to solve it analytically.
$endgroup$
– newstudent
Dec 18 '18 at 12:52




$begingroup$
@rafa11111 Ok. let us say these are ordinary equations. I would like to solve it analytically.
$endgroup$
– newstudent
Dec 18 '18 at 12:52












$begingroup$
If you are sure these are ordinary equations, please edit the question to address so (change the equation, the tag, etc.). I don't think there is an analytical solution to these equations, mainly due to the nonlinear term in the first equation. If $C$ is a small number, you can use a perturbation method, that provides an approximate analytical solution.
$endgroup$
– rafa11111
Dec 18 '18 at 13:01




$begingroup$
If you are sure these are ordinary equations, please edit the question to address so (change the equation, the tag, etc.). I don't think there is an analytical solution to these equations, mainly due to the nonlinear term in the first equation. If $C$ is a small number, you can use a perturbation method, that provides an approximate analytical solution.
$endgroup$
– rafa11111
Dec 18 '18 at 13:01












$begingroup$
You can integrate the second equation once then substitute for $r(x)$ in the first to give an equation in $u(x)$ alone. I think your journey will end there!
$endgroup$
– user121049
Dec 19 '18 at 8:00




$begingroup$
You can integrate the second equation once then substitute for $r(x)$ in the first to give an equation in $u(x)$ alone. I think your journey will end there!
$endgroup$
– user121049
Dec 19 '18 at 8:00












$begingroup$
You need some more boundary conditions on $r$. The only hope I see is to assume $r$ stays small so you can set $1-r^2approx 1$ and linearize the first equation. The two equations are then easy to solve. Then check $r$ really stays small.
$endgroup$
– user121049
Dec 20 '18 at 9:00




$begingroup$
You need some more boundary conditions on $r$. The only hope I see is to assume $r$ stays small so you can set $1-r^2approx 1$ and linearize the first equation. The two equations are then easy to solve. Then check $r$ really stays small.
$endgroup$
– user121049
Dec 20 '18 at 9:00










1 Answer
1






active

oldest

votes


















1












$begingroup$

Disclaimer: I've just realized you can't actually solve for $r(x)$ in the first equation, which makes my answer kind of useless. I'll leave it up anyway, to learn from my mistakes.



Suppose you've already solved the first equation, and $r(x)$ is known, then the second equation is linear and non-homogeneous, and can easily be solved using variations of parameter



Let $a = frac{V}{D}$



$$ frac{d^2u}{dx^2} + afrac{du}{dx} = frac{a}{2}frac{dr}{dx} $$



The homogeneous equation has solution $u_h = c_1 + c_2e^{-ax}$, so guess a particular solution of the form



$$ u_p(x) = p(x) + q(x)e^{-ax} $$



Then



begin{align} p'(x) + q'(x)e^{-ax} &= 0 \ -aq'(x)e^{-ax} &= frac{a}{2}r'(x) end{align}



or $p'(x) = frac{1}{2}r'(x)$ and $q'(x) = -frac{1}{2}r'(x)e^{-ax}$



You can integrate to find



begin{align} p(x) &= frac{1}{2}r(x) \ q(x) &=-frac12int r'(x) e^{ax} = -frac12 r(x)e^{ax} dx + frac{a}{2}int r(x)e^{ax} dx end{align}



Therefore



$$ u_p(x) = frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$



Matching the boundary conditions, we find



$$ u(x) = u_{infty}(1-e^{-ax}) + frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$



where $u_infty$ is some constant depending on the behavior of $r(x)$ at ${xtoinfty}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    There is a $u(x)$ in the first equation so how do you solve it for $r(x)$?
    $endgroup$
    – user121049
    Dec 19 '18 at 20:37










  • $begingroup$
    @user121049 Wow I didn't even notice. Thanks for pointing it out.
    $endgroup$
    – Dylan
    Dec 20 '18 at 3:57











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

Disclaimer: I've just realized you can't actually solve for $r(x)$ in the first equation, which makes my answer kind of useless. I'll leave it up anyway, to learn from my mistakes.



Suppose you've already solved the first equation, and $r(x)$ is known, then the second equation is linear and non-homogeneous, and can easily be solved using variations of parameter



Let $a = frac{V}{D}$



$$ frac{d^2u}{dx^2} + afrac{du}{dx} = frac{a}{2}frac{dr}{dx} $$



The homogeneous equation has solution $u_h = c_1 + c_2e^{-ax}$, so guess a particular solution of the form



$$ u_p(x) = p(x) + q(x)e^{-ax} $$



Then



begin{align} p'(x) + q'(x)e^{-ax} &= 0 \ -aq'(x)e^{-ax} &= frac{a}{2}r'(x) end{align}



or $p'(x) = frac{1}{2}r'(x)$ and $q'(x) = -frac{1}{2}r'(x)e^{-ax}$



You can integrate to find



begin{align} p(x) &= frac{1}{2}r(x) \ q(x) &=-frac12int r'(x) e^{ax} = -frac12 r(x)e^{ax} dx + frac{a}{2}int r(x)e^{ax} dx end{align}



Therefore



$$ u_p(x) = frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$



Matching the boundary conditions, we find



$$ u(x) = u_{infty}(1-e^{-ax}) + frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$



where $u_infty$ is some constant depending on the behavior of $r(x)$ at ${xtoinfty}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    There is a $u(x)$ in the first equation so how do you solve it for $r(x)$?
    $endgroup$
    – user121049
    Dec 19 '18 at 20:37










  • $begingroup$
    @user121049 Wow I didn't even notice. Thanks for pointing it out.
    $endgroup$
    – Dylan
    Dec 20 '18 at 3:57
















1












$begingroup$

Disclaimer: I've just realized you can't actually solve for $r(x)$ in the first equation, which makes my answer kind of useless. I'll leave it up anyway, to learn from my mistakes.



Suppose you've already solved the first equation, and $r(x)$ is known, then the second equation is linear and non-homogeneous, and can easily be solved using variations of parameter



Let $a = frac{V}{D}$



$$ frac{d^2u}{dx^2} + afrac{du}{dx} = frac{a}{2}frac{dr}{dx} $$



The homogeneous equation has solution $u_h = c_1 + c_2e^{-ax}$, so guess a particular solution of the form



$$ u_p(x) = p(x) + q(x)e^{-ax} $$



Then



begin{align} p'(x) + q'(x)e^{-ax} &= 0 \ -aq'(x)e^{-ax} &= frac{a}{2}r'(x) end{align}



or $p'(x) = frac{1}{2}r'(x)$ and $q'(x) = -frac{1}{2}r'(x)e^{-ax}$



You can integrate to find



begin{align} p(x) &= frac{1}{2}r(x) \ q(x) &=-frac12int r'(x) e^{ax} = -frac12 r(x)e^{ax} dx + frac{a}{2}int r(x)e^{ax} dx end{align}



Therefore



$$ u_p(x) = frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$



Matching the boundary conditions, we find



$$ u(x) = u_{infty}(1-e^{-ax}) + frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$



where $u_infty$ is some constant depending on the behavior of $r(x)$ at ${xtoinfty}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    There is a $u(x)$ in the first equation so how do you solve it for $r(x)$?
    $endgroup$
    – user121049
    Dec 19 '18 at 20:37










  • $begingroup$
    @user121049 Wow I didn't even notice. Thanks for pointing it out.
    $endgroup$
    – Dylan
    Dec 20 '18 at 3:57














1












1








1





$begingroup$

Disclaimer: I've just realized you can't actually solve for $r(x)$ in the first equation, which makes my answer kind of useless. I'll leave it up anyway, to learn from my mistakes.



Suppose you've already solved the first equation, and $r(x)$ is known, then the second equation is linear and non-homogeneous, and can easily be solved using variations of parameter



Let $a = frac{V}{D}$



$$ frac{d^2u}{dx^2} + afrac{du}{dx} = frac{a}{2}frac{dr}{dx} $$



The homogeneous equation has solution $u_h = c_1 + c_2e^{-ax}$, so guess a particular solution of the form



$$ u_p(x) = p(x) + q(x)e^{-ax} $$



Then



begin{align} p'(x) + q'(x)e^{-ax} &= 0 \ -aq'(x)e^{-ax} &= frac{a}{2}r'(x) end{align}



or $p'(x) = frac{1}{2}r'(x)$ and $q'(x) = -frac{1}{2}r'(x)e^{-ax}$



You can integrate to find



begin{align} p(x) &= frac{1}{2}r(x) \ q(x) &=-frac12int r'(x) e^{ax} = -frac12 r(x)e^{ax} dx + frac{a}{2}int r(x)e^{ax} dx end{align}



Therefore



$$ u_p(x) = frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$



Matching the boundary conditions, we find



$$ u(x) = u_{infty}(1-e^{-ax}) + frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$



where $u_infty$ is some constant depending on the behavior of $r(x)$ at ${xtoinfty}$






share|cite|improve this answer











$endgroup$



Disclaimer: I've just realized you can't actually solve for $r(x)$ in the first equation, which makes my answer kind of useless. I'll leave it up anyway, to learn from my mistakes.



Suppose you've already solved the first equation, and $r(x)$ is known, then the second equation is linear and non-homogeneous, and can easily be solved using variations of parameter



Let $a = frac{V}{D}$



$$ frac{d^2u}{dx^2} + afrac{du}{dx} = frac{a}{2}frac{dr}{dx} $$



The homogeneous equation has solution $u_h = c_1 + c_2e^{-ax}$, so guess a particular solution of the form



$$ u_p(x) = p(x) + q(x)e^{-ax} $$



Then



begin{align} p'(x) + q'(x)e^{-ax} &= 0 \ -aq'(x)e^{-ax} &= frac{a}{2}r'(x) end{align}



or $p'(x) = frac{1}{2}r'(x)$ and $q'(x) = -frac{1}{2}r'(x)e^{-ax}$



You can integrate to find



begin{align} p(x) &= frac{1}{2}r(x) \ q(x) &=-frac12int r'(x) e^{ax} = -frac12 r(x)e^{ax} dx + frac{a}{2}int r(x)e^{ax} dx end{align}



Therefore



$$ u_p(x) = frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$



Matching the boundary conditions, we find



$$ u(x) = u_{infty}(1-e^{-ax}) + frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$



where $u_infty$ is some constant depending on the behavior of $r(x)$ at ${xtoinfty}$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 4:03

























answered Dec 19 '18 at 7:48









DylanDylan

14.1k31127




14.1k31127












  • $begingroup$
    There is a $u(x)$ in the first equation so how do you solve it for $r(x)$?
    $endgroup$
    – user121049
    Dec 19 '18 at 20:37










  • $begingroup$
    @user121049 Wow I didn't even notice. Thanks for pointing it out.
    $endgroup$
    – Dylan
    Dec 20 '18 at 3:57


















  • $begingroup$
    There is a $u(x)$ in the first equation so how do you solve it for $r(x)$?
    $endgroup$
    – user121049
    Dec 19 '18 at 20:37










  • $begingroup$
    @user121049 Wow I didn't even notice. Thanks for pointing it out.
    $endgroup$
    – Dylan
    Dec 20 '18 at 3:57
















$begingroup$
There is a $u(x)$ in the first equation so how do you solve it for $r(x)$?
$endgroup$
– user121049
Dec 19 '18 at 20:37




$begingroup$
There is a $u(x)$ in the first equation so how do you solve it for $r(x)$?
$endgroup$
– user121049
Dec 19 '18 at 20:37












$begingroup$
@user121049 Wow I didn't even notice. Thanks for pointing it out.
$endgroup$
– Dylan
Dec 20 '18 at 3:57




$begingroup$
@user121049 Wow I didn't even notice. Thanks for pointing it out.
$endgroup$
– Dylan
Dec 20 '18 at 3:57


















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Plaza Victoria

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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
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Puebla de Zaragoza

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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
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Musa

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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
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