$f''+pf'+qf=0$ where $qleq0$ and $f(0)=f(1)=0$ prove $f=0$ ($f$, $p$, $q$ defined on $[0,1]$)












2















For $f$ defined on $[0,1]$ twice differentiable, and two continuous function defined on $[0,1]$ named $p(x)$, $q(x)$, satisfying



$f''+pf'+qf=0$, where $qleq0$ and $f(0)=f(1)=0$,



prove $f=0$.




I haven't got an idea about the question. Any hint will be appreciated.



Edited: for $q(x_0)<0$ where $f'(x_0)=0$ the problem can easily be solved.
So now I'm mainly concerned about $q(x_0)=0$. It appears on my textbook, so i guess it's probably true. Counter examples are also welcomed.






real-analysis differential-equations







share|cite|improve this question




















  • 1




    Are $p,q$ constants or functions?
    – edm
    Nov 28 '18 at 5:33






  • 1




    For strictly negative $q$ one can use the same argument as in math.stackexchange.com/q/2998648/42969.
    – Martin R
    Nov 28 '18 at 6:12










  • @edm functions,sorry for my unclear statement
    – Oolong milk tea
    Nov 28 '18 at 9:30
















2















For $f$ defined on $[0,1]$ twice differentiable, and two continuous function defined on $[0,1]$ named $p(x)$, $q(x)$, satisfying



$f''+pf'+qf=0$, where $qleq0$ and $f(0)=f(1)=0$,



prove $f=0$.




I haven't got an idea about the question. Any hint will be appreciated.



Edited: for $q(x_0)<0$ where $f'(x_0)=0$ the problem can easily be solved.
So now I'm mainly concerned about $q(x_0)=0$. It appears on my textbook, so i guess it's probably true. Counter examples are also welcomed.






real-analysis differential-equations







share|cite|improve this question




















  • 1




    Are $p,q$ constants or functions?
    – edm
    Nov 28 '18 at 5:33






  • 1




    For strictly negative $q$ one can use the same argument as in math.stackexchange.com/q/2998648/42969.
    – Martin R
    Nov 28 '18 at 6:12










  • @edm functions,sorry for my unclear statement
    – Oolong milk tea
    Nov 28 '18 at 9:30














2












2








2


2






For $f$ defined on $[0,1]$ twice differentiable, and two continuous function defined on $[0,1]$ named $p(x)$, $q(x)$, satisfying



$f''+pf'+qf=0$, where $qleq0$ and $f(0)=f(1)=0$,



prove $f=0$.




I haven't got an idea about the question. Any hint will be appreciated.



Edited: for $q(x_0)<0$ where $f'(x_0)=0$ the problem can easily be solved.
So now I'm mainly concerned about $q(x_0)=0$. It appears on my textbook, so i guess it's probably true. Counter examples are also welcomed.






real-analysis differential-equations







share|cite|improve this question
















For $f$ defined on $[0,1]$ twice differentiable, and two continuous function defined on $[0,1]$ named $p(x)$, $q(x)$, satisfying



$f''+pf'+qf=0$, where $qleq0$ and $f(0)=f(1)=0$,



prove $f=0$.




I haven't got an idea about the question. Any hint will be appreciated.



Edited: for $q(x_0)<0$ where $f'(x_0)=0$ the problem can easily be solved.
So now I'm mainly concerned about $q(x_0)=0$. It appears on my textbook, so i guess it's probably true. Counter examples are also welcomed.






real-analysis differential-equations




real-analysis differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 18:44









LutzL

56.6k42054




56.6k42054










asked Nov 28 '18 at 3:57









Oolong milk teaOolong milk tea

787




787








  • 1




    Are $p,q$ constants or functions?
    – edm
    Nov 28 '18 at 5:33






  • 1




    For strictly negative $q$ one can use the same argument as in math.stackexchange.com/q/2998648/42969.
    – Martin R
    Nov 28 '18 at 6:12










  • @edm functions,sorry for my unclear statement
    – Oolong milk tea
    Nov 28 '18 at 9:30














  • 1




    Are $p,q$ constants or functions?
    – edm
    Nov 28 '18 at 5:33






  • 1




    For strictly negative $q$ one can use the same argument as in math.stackexchange.com/q/2998648/42969.
    – Martin R
    Nov 28 '18 at 6:12










  • @edm functions,sorry for my unclear statement
    – Oolong milk tea
    Nov 28 '18 at 9:30








1




1




Are $p,q$ constants or functions?
– edm
Nov 28 '18 at 5:33




Are $p,q$ constants or functions?
– edm
Nov 28 '18 at 5:33




1




1




For strictly negative $q$ one can use the same argument as in math.stackexchange.com/q/2998648/42969.
– Martin R
Nov 28 '18 at 6:12




For strictly negative $q$ one can use the same argument as in math.stackexchange.com/q/2998648/42969.
– Martin R
Nov 28 '18 at 6:12












@edm functions,sorry for my unclear statement
– Oolong milk tea
Nov 28 '18 at 9:30




@edm functions,sorry for my unclear statement
– Oolong milk tea
Nov 28 '18 at 9:30










2 Answers
2






active

oldest

votes


















4














As $f'(0)ne 0$ by the uniqueness theorem, we can set $f'(0)=1$, as the linearity allows arbitrary rescaling.



Then let $$a=min{sin(0,1]colon f(s)le0}$$ be the smallest positive root of $f$, such that $f(x)>0$ for $xin (0,a)$.



On this interval one has then,
begin{align}
f''(x)+p(x)f'(x)=(-q(x))f(x)ge 0,
end{align}
and with the anti-derivative $P(x)=int_0^xp(s)ds$,
begin{align}
(e^{P(x)}f'(x))'ge 0.
end{align}
Integrating twice results in
begin{align}
&e^{P(x)}f'(x)ge e^{P(0)}f'(0)=1\[1em]
text{ and }~~ &f(x)ge int_0^xe^{-P(s)} dsge e^{-max_{sin[0,1]} P(s)}x
end{align}
This is a contradiction, as $f(a)=0$ and $f(a)ge e^{-max_{sin[0,1]} P(s)}a>0$ can not be true at the same time.




This result can easily extended to the more general claim that:




Any solution of the given differential equation can have at most one root.







share|cite|improve this answer























  • A little problem: how do you make the conclusion $f'(0)>0$ by uniqueness theorem? Maybe we have fuction like $f＝e^{-frac{1}{x}}$ around $0$.
    – Zero
    Nov 29 '18 at 11:41










  • @Zero : Because then the initial values are $f(0)=0$ and $f'(0)=0$, and $f=0$ is a solution. The ODE is linear, satisfies thus the Lipschitz condition, so that uniqueness follows. Now take the assumed solution and divide by $f'(0)$ to get another solution with $f'(0)=1$.
    – LutzL
    Nov 29 '18 at 13:02












  • Thank you, I get what you mean now..
    – Zero
    Nov 29 '18 at 13:09





















0














Your differential equation is a homogeneous second order ODE (I take it the coefficients are constant, it's not clear). That means that the characteristic polynomial should give us the possible solutions. The polynomial would be:
$$ r^{2} + p r + q = 0 $$
And the roots are, then:
$$ r_{1} = frac{-p + sqrt{p^{2} - 4q}}{2} $$
$$ r_{2} = frac{-p - sqrt{p^{2} - 4q}}{2} $$
Now, there are three different cases for the solutions of homogenous second order ODEs with constant coefficients, either the roots are both real and different, they are both real and the same, or they are both complex. But notice that $ p in mathbb{R} $ implies that $ p^{2} geq 0 $, and $q leq 0 $ of course implies that $-4q geq 0$. Then, the square root expression can only be a non-negative number, and there are your 2 remaining possible cases, so you'll just need to evaluate the general solution for either of them.



On one case, you have that both roots are equal. Think about what conditions are required for $p,q$ for that to be possible. And in the case that the roots are different, you will need to evaluate the values you are given in the general solution, of the form:
$$y(t) = c_{1} e^{r_{1}t} + c_{2} e^{r_{2}t}$$






share|cite|improve this answer



















  • 5




    The $p$ and $q$ are functions, not constants.
    – RRL
    Nov 28 '18 at 5:40











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2 Answers
2






active

oldest

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4














As $f'(0)ne 0$ by the uniqueness theorem, we can set $f'(0)=1$, as the linearity allows arbitrary rescaling.



Then let $$a=min{sin(0,1]colon f(s)le0}$$ be the smallest positive root of $f$, such that $f(x)>0$ for $xin (0,a)$.



On this interval one has then,
begin{align}
f''(x)+p(x)f'(x)=(-q(x))f(x)ge 0,
end{align}
and with the anti-derivative $P(x)=int_0^xp(s)ds$,
begin{align}
(e^{P(x)}f'(x))'ge 0.
end{align}
Integrating twice results in
begin{align}
&e^{P(x)}f'(x)ge e^{P(0)}f'(0)=1\[1em]
text{ and }~~ &f(x)ge int_0^xe^{-P(s)} dsge e^{-max_{sin[0,1]} P(s)}x
end{align}
This is a contradiction, as $f(a)=0$ and $f(a)ge e^{-max_{sin[0,1]} P(s)}a>0$ can not be true at the same time.




This result can easily extended to the more general claim that:




Any solution of the given differential equation can have at most one root.







share|cite|improve this answer























  • A little problem: how do you make the conclusion $f'(0)>0$ by uniqueness theorem? Maybe we have fuction like $f＝e^{-frac{1}{x}}$ around $0$.
    – Zero
    Nov 29 '18 at 11:41










  • @Zero : Because then the initial values are $f(0)=0$ and $f'(0)=0$, and $f=0$ is a solution. The ODE is linear, satisfies thus the Lipschitz condition, so that uniqueness follows. Now take the assumed solution and divide by $f'(0)$ to get another solution with $f'(0)=1$.
    – LutzL
    Nov 29 '18 at 13:02












  • Thank you, I get what you mean now..
    – Zero
    Nov 29 '18 at 13:09


















4














As $f'(0)ne 0$ by the uniqueness theorem, we can set $f'(0)=1$, as the linearity allows arbitrary rescaling.



Then let $$a=min{sin(0,1]colon f(s)le0}$$ be the smallest positive root of $f$, such that $f(x)>0$ for $xin (0,a)$.



On this interval one has then,
begin{align}
f''(x)+p(x)f'(x)=(-q(x))f(x)ge 0,
end{align}
and with the anti-derivative $P(x)=int_0^xp(s)ds$,
begin{align}
(e^{P(x)}f'(x))'ge 0.
end{align}
Integrating twice results in
begin{align}
&e^{P(x)}f'(x)ge e^{P(0)}f'(0)=1\[1em]
text{ and }~~ &f(x)ge int_0^xe^{-P(s)} dsge e^{-max_{sin[0,1]} P(s)}x
end{align}
This is a contradiction, as $f(a)=0$ and $f(a)ge e^{-max_{sin[0,1]} P(s)}a>0$ can not be true at the same time.




This result can easily extended to the more general claim that:




Any solution of the given differential equation can have at most one root.







share|cite|improve this answer























  • A little problem: how do you make the conclusion $f'(0)>0$ by uniqueness theorem? Maybe we have fuction like $f＝e^{-frac{1}{x}}$ around $0$.
    – Zero
    Nov 29 '18 at 11:41










  • @Zero : Because then the initial values are $f(0)=0$ and $f'(0)=0$, and $f=0$ is a solution. The ODE is linear, satisfies thus the Lipschitz condition, so that uniqueness follows. Now take the assumed solution and divide by $f'(0)$ to get another solution with $f'(0)=1$.
    – LutzL
    Nov 29 '18 at 13:02












  • Thank you, I get what you mean now..
    – Zero
    Nov 29 '18 at 13:09
















4












4








4






As $f'(0)ne 0$ by the uniqueness theorem, we can set $f'(0)=1$, as the linearity allows arbitrary rescaling.



Then let $$a=min{sin(0,1]colon f(s)le0}$$ be the smallest positive root of $f$, such that $f(x)>0$ for $xin (0,a)$.



On this interval one has then,
begin{align}
f''(x)+p(x)f'(x)=(-q(x))f(x)ge 0,
end{align}
and with the anti-derivative $P(x)=int_0^xp(s)ds$,
begin{align}
(e^{P(x)}f'(x))'ge 0.
end{align}
Integrating twice results in
begin{align}
&e^{P(x)}f'(x)ge e^{P(0)}f'(0)=1\[1em]
text{ and }~~ &f(x)ge int_0^xe^{-P(s)} dsge e^{-max_{sin[0,1]} P(s)}x
end{align}
This is a contradiction, as $f(a)=0$ and $f(a)ge e^{-max_{sin[0,1]} P(s)}a>0$ can not be true at the same time.




This result can easily extended to the more general claim that:




Any solution of the given differential equation can have at most one root.







share|cite|improve this answer














As $f'(0)ne 0$ by the uniqueness theorem, we can set $f'(0)=1$, as the linearity allows arbitrary rescaling.



Then let $$a=min{sin(0,1]colon f(s)le0}$$ be the smallest positive root of $f$, such that $f(x)>0$ for $xin (0,a)$.



On this interval one has then,
begin{align}
f''(x)+p(x)f'(x)=(-q(x))f(x)ge 0,
end{align}
and with the anti-derivative $P(x)=int_0^xp(s)ds$,
begin{align}
(e^{P(x)}f'(x))'ge 0.
end{align}
Integrating twice results in
begin{align}
&e^{P(x)}f'(x)ge e^{P(0)}f'(0)=1\[1em]
text{ and }~~ &f(x)ge int_0^xe^{-P(s)} dsge e^{-max_{sin[0,1]} P(s)}x
end{align}
This is a contradiction, as $f(a)=0$ and $f(a)ge e^{-max_{sin[0,1]} P(s)}a>0$ can not be true at the same time.




This result can easily extended to the more general claim that:




Any solution of the given differential equation can have at most one root.








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 28 '18 at 19:10

























answered Nov 28 '18 at 13:54









LutzLLutzL

56.6k42054




56.6k42054












  • A little problem: how do you make the conclusion $f'(0)>0$ by uniqueness theorem? Maybe we have fuction like $f＝e^{-frac{1}{x}}$ around $0$.
    – Zero
    Nov 29 '18 at 11:41










  • @Zero : Because then the initial values are $f(0)=0$ and $f'(0)=0$, and $f=0$ is a solution. The ODE is linear, satisfies thus the Lipschitz condition, so that uniqueness follows. Now take the assumed solution and divide by $f'(0)$ to get another solution with $f'(0)=1$.
    – LutzL
    Nov 29 '18 at 13:02












  • Thank you, I get what you mean now..
    – Zero
    Nov 29 '18 at 13:09




















  • A little problem: how do you make the conclusion $f'(0)>0$ by uniqueness theorem? Maybe we have fuction like $f＝e^{-frac{1}{x}}$ around $0$.
    – Zero
    Nov 29 '18 at 11:41










  • @Zero : Because then the initial values are $f(0)=0$ and $f'(0)=0$, and $f=0$ is a solution. The ODE is linear, satisfies thus the Lipschitz condition, so that uniqueness follows. Now take the assumed solution and divide by $f'(0)$ to get another solution with $f'(0)=1$.
    – LutzL
    Nov 29 '18 at 13:02












  • Thank you, I get what you mean now..
    – Zero
    Nov 29 '18 at 13:09


















A little problem: how do you make the conclusion $f'(0)>0$ by uniqueness theorem? Maybe we have fuction like $f＝e^{-frac{1}{x}}$ around $0$.
– Zero
Nov 29 '18 at 11:41




A little problem: how do you make the conclusion $f'(0)>0$ by uniqueness theorem? Maybe we have fuction like $f＝e^{-frac{1}{x}}$ around $0$.
– Zero
Nov 29 '18 at 11:41












@Zero : Because then the initial values are $f(0)=0$ and $f'(0)=0$, and $f=0$ is a solution. The ODE is linear, satisfies thus the Lipschitz condition, so that uniqueness follows. Now take the assumed solution and divide by $f'(0)$ to get another solution with $f'(0)=1$.
– LutzL
Nov 29 '18 at 13:02






@Zero : Because then the initial values are $f(0)=0$ and $f'(0)=0$, and $f=0$ is a solution. The ODE is linear, satisfies thus the Lipschitz condition, so that uniqueness follows. Now take the assumed solution and divide by $f'(0)$ to get another solution with $f'(0)=1$.
– LutzL
Nov 29 '18 at 13:02














Thank you, I get what you mean now..
– Zero
Nov 29 '18 at 13:09






Thank you, I get what you mean now..
– Zero
Nov 29 '18 at 13:09













0














Your differential equation is a homogeneous second order ODE (I take it the coefficients are constant, it's not clear). That means that the characteristic polynomial should give us the possible solutions. The polynomial would be:
$$ r^{2} + p r + q = 0 $$
And the roots are, then:
$$ r_{1} = frac{-p + sqrt{p^{2} - 4q}}{2} $$
$$ r_{2} = frac{-p - sqrt{p^{2} - 4q}}{2} $$
Now, there are three different cases for the solutions of homogenous second order ODEs with constant coefficients, either the roots are both real and different, they are both real and the same, or they are both complex. But notice that $ p in mathbb{R} $ implies that $ p^{2} geq 0 $, and $q leq 0 $ of course implies that $-4q geq 0$. Then, the square root expression can only be a non-negative number, and there are your 2 remaining possible cases, so you'll just need to evaluate the general solution for either of them.



On one case, you have that both roots are equal. Think about what conditions are required for $p,q$ for that to be possible. And in the case that the roots are different, you will need to evaluate the values you are given in the general solution, of the form:
$$y(t) = c_{1} e^{r_{1}t} + c_{2} e^{r_{2}t}$$






share|cite|improve this answer



















  • 5




    The $p$ and $q$ are functions, not constants.
    – RRL
    Nov 28 '18 at 5:40
















0














Your differential equation is a homogeneous second order ODE (I take it the coefficients are constant, it's not clear). That means that the characteristic polynomial should give us the possible solutions. The polynomial would be:
$$ r^{2} + p r + q = 0 $$
And the roots are, then:
$$ r_{1} = frac{-p + sqrt{p^{2} - 4q}}{2} $$
$$ r_{2} = frac{-p - sqrt{p^{2} - 4q}}{2} $$
Now, there are three different cases for the solutions of homogenous second order ODEs with constant coefficients, either the roots are both real and different, they are both real and the same, or they are both complex. But notice that $ p in mathbb{R} $ implies that $ p^{2} geq 0 $, and $q leq 0 $ of course implies that $-4q geq 0$. Then, the square root expression can only be a non-negative number, and there are your 2 remaining possible cases, so you'll just need to evaluate the general solution for either of them.



On one case, you have that both roots are equal. Think about what conditions are required for $p,q$ for that to be possible. And in the case that the roots are different, you will need to evaluate the values you are given in the general solution, of the form:
$$y(t) = c_{1} e^{r_{1}t} + c_{2} e^{r_{2}t}$$






share|cite|improve this answer



















  • 5




    The $p$ and $q$ are functions, not constants.
    – RRL
    Nov 28 '18 at 5:40














0












0








0






Your differential equation is a homogeneous second order ODE (I take it the coefficients are constant, it's not clear). That means that the characteristic polynomial should give us the possible solutions. The polynomial would be:
$$ r^{2} + p r + q = 0 $$
And the roots are, then:
$$ r_{1} = frac{-p + sqrt{p^{2} - 4q}}{2} $$
$$ r_{2} = frac{-p - sqrt{p^{2} - 4q}}{2} $$
Now, there are three different cases for the solutions of homogenous second order ODEs with constant coefficients, either the roots are both real and different, they are both real and the same, or they are both complex. But notice that $ p in mathbb{R} $ implies that $ p^{2} geq 0 $, and $q leq 0 $ of course implies that $-4q geq 0$. Then, the square root expression can only be a non-negative number, and there are your 2 remaining possible cases, so you'll just need to evaluate the general solution for either of them.



On one case, you have that both roots are equal. Think about what conditions are required for $p,q$ for that to be possible. And in the case that the roots are different, you will need to evaluate the values you are given in the general solution, of the form:
$$y(t) = c_{1} e^{r_{1}t} + c_{2} e^{r_{2}t}$$






share|cite|improve this answer














Your differential equation is a homogeneous second order ODE (I take it the coefficients are constant, it's not clear). That means that the characteristic polynomial should give us the possible solutions. The polynomial would be:
$$ r^{2} + p r + q = 0 $$
And the roots are, then:
$$ r_{1} = frac{-p + sqrt{p^{2} - 4q}}{2} $$
$$ r_{2} = frac{-p - sqrt{p^{2} - 4q}}{2} $$
Now, there are three different cases for the solutions of homogenous second order ODEs with constant coefficients, either the roots are both real and different, they are both real and the same, or they are both complex. But notice that $ p in mathbb{R} $ implies that $ p^{2} geq 0 $, and $q leq 0 $ of course implies that $-4q geq 0$. Then, the square root expression can only be a non-negative number, and there are your 2 remaining possible cases, so you'll just need to evaluate the general solution for either of them.



On one case, you have that both roots are equal. Think about what conditions are required for $p,q$ for that to be possible. And in the case that the roots are different, you will need to evaluate the values you are given in the general solution, of the form:
$$y(t) = c_{1} e^{r_{1}t} + c_{2} e^{r_{2}t}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 28 '18 at 5:37

























answered Nov 28 '18 at 5:30









Joaquin C.Joaquin C.

154




154








  • 5




    The $p$ and $q$ are functions, not constants.
    – RRL
    Nov 28 '18 at 5:40














  • 5




    The $p$ and $q$ are functions, not constants.
    – RRL
    Nov 28 '18 at 5:40








5




5




The $p$ and $q$ are functions, not constants.
– RRL
Nov 28 '18 at 5:40




The $p$ and $q$ are functions, not constants.
– RRL
Nov 28 '18 at 5:40


















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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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