Solution of $y''+e^xy=0$ is unbounded as $xtoinfty$












1












$begingroup$


Consider the differential equation $y''+e^xy=0$. Can we say something about the behaviour of $y$ as $xtoinfty$? In particular, is it unbounded?



I think, to solve the equation, we need to use the power series method. By is there a way to understand the behaviour beforehand, say by using the sturm-picone or similar theorems?Thanks beforehand.










share|cite|improve this question









$endgroup$












  • $begingroup$
    When you ask about the boundedness of the solutions $y(x)$, what set is $x$ confined to?
    $endgroup$
    – user1337
    Dec 10 '18 at 8:53










  • $begingroup$
    @user1337 let us assume $xinmathbb{R}$
    $endgroup$
    – vidyarthi
    Dec 10 '18 at 9:00
















1












$begingroup$


Consider the differential equation $y''+e^xy=0$. Can we say something about the behaviour of $y$ as $xtoinfty$? In particular, is it unbounded?



I think, to solve the equation, we need to use the power series method. By is there a way to understand the behaviour beforehand, say by using the sturm-picone or similar theorems?Thanks beforehand.










share|cite|improve this question









$endgroup$












  • $begingroup$
    When you ask about the boundedness of the solutions $y(x)$, what set is $x$ confined to?
    $endgroup$
    – user1337
    Dec 10 '18 at 8:53










  • $begingroup$
    @user1337 let us assume $xinmathbb{R}$
    $endgroup$
    – vidyarthi
    Dec 10 '18 at 9:00














1












1








1


1



$begingroup$


Consider the differential equation $y''+e^xy=0$. Can we say something about the behaviour of $y$ as $xtoinfty$? In particular, is it unbounded?



I think, to solve the equation, we need to use the power series method. By is there a way to understand the behaviour beforehand, say by using the sturm-picone or similar theorems?Thanks beforehand.










share|cite|improve this question









$endgroup$




Consider the differential equation $y''+e^xy=0$. Can we say something about the behaviour of $y$ as $xtoinfty$? In particular, is it unbounded?



I think, to solve the equation, we need to use the power series method. By is there a way to understand the behaviour beforehand, say by using the sturm-picone or similar theorems?Thanks beforehand.







ordinary-differential-equations power-series






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asked Dec 10 '18 at 7:17









vidyarthividyarthi

2,9741832




2,9741832












  • $begingroup$
    When you ask about the boundedness of the solutions $y(x)$, what set is $x$ confined to?
    $endgroup$
    – user1337
    Dec 10 '18 at 8:53










  • $begingroup$
    @user1337 let us assume $xinmathbb{R}$
    $endgroup$
    – vidyarthi
    Dec 10 '18 at 9:00


















  • $begingroup$
    When you ask about the boundedness of the solutions $y(x)$, what set is $x$ confined to?
    $endgroup$
    – user1337
    Dec 10 '18 at 8:53










  • $begingroup$
    @user1337 let us assume $xinmathbb{R}$
    $endgroup$
    – vidyarthi
    Dec 10 '18 at 9:00
















$begingroup$
When you ask about the boundedness of the solutions $y(x)$, what set is $x$ confined to?
$endgroup$
– user1337
Dec 10 '18 at 8:53




$begingroup$
When you ask about the boundedness of the solutions $y(x)$, what set is $x$ confined to?
$endgroup$
– user1337
Dec 10 '18 at 8:53












$begingroup$
@user1337 let us assume $xinmathbb{R}$
$endgroup$
– vidyarthi
Dec 10 '18 at 9:00




$begingroup$
@user1337 let us assume $xinmathbb{R}$
$endgroup$
– vidyarthi
Dec 10 '18 at 9:00










2 Answers
2






active

oldest

votes


















2












$begingroup$

In order to prove that the solutions of (the transformed equation through a change of variable)
$$ x u''(x) + u'(x) + x u(x) = 0 tag{1}$$
are bounded on $[1,+infty)$ we do not need to invoke the asymptotics of Bessel functions, we may directly exploit the structure of the differential equation. On short intervals $[a,b]$ it is reasonable to claim that the solution of $(1)$ with $u(a)=u_0, u'(a)=u_1$ is close to the solution of
$$ a v''(x) + v'(x) + a v(x) = 0 tag{2}$$
with boundary conditions $v(a)=u_0, v'(a)=u_1$. On the other hand $(2)$ is a ODE with constant coefficients and characteristic polynomial $az^2+z+a$, hence the solutions of $(2)$ are bounded by $sqrt{u_0^2+u_1^2+frac{u_0 u_1}{a}}$ on $[a,b]$. Let us consider the sequence of intervals $[H_1,H_2],[H_2,H_3],ldots$
By denoting as $sigma_m$ the following supremum
$$ sigma_m = sup_{xin[H_m,H_{m+1}]}left|u(x)-v(x)right| $$
(where the boundary conditions for $v$ on $[H_m,H_{m+1}]$ are given by the values of $u$ and $u'$ at the left endpoint) we have that
$$ {sigma_m}_{mgeq 1}in ell^1 tag{3} $$
allows to state that the boundedness of the solutions of $(1)$ is a consequence of the boundedness of the solutions of $(2)$. By Frobenius power series method both the solutions of $(1)$ and $(2)$ are analytic functions: by considering $d(x)=u(x)-v(x)$ on $[H_m,H_{m+1}]$ we have that both $d$ and $d'$ vanish at the left endpoint, so by considering the termwise difference between $(1)$ and $(2)$ it is not difficult to derive
$$ sigma_m ll |[H_m,H_{m+1}]|^2 H_m^2 ll frac{log^2 m}{m^2}$$
having ${sigma_m}_{mgeq 1}inell^1$ as a straightfoward consequence.





Now that the boundedness of the solutions of $(1)$ is proved, a "bootstrap" argument allows to prove that $u(x)^2+u'(x)^2 ll frac{1}{x}$ as $xto +infty$, hence the solutions of $(1)$ are not only bounded but convergent to zero and with a bounded derivative. Indeed, by defining the energy of a solution as $E(u)=u(x)^2+u'(x)^2$
and by multiplying both sides of $(1)$ by $2u'(x)$, we have



$$ 0 = x E'(u) + 2u'(x)^2 leq x E'(u) + 2 E(u)$$
and any non-trivial bound for the energy immediately gives a simultaneous bound for $u$ and $u'$.





Summarizing, Tricomi's approximation
$$ J_0(x) approx frac{sin(x)+cos(x)}{sqrt{pi x}} text{ for large values of }x$$
can be essentially derived by standard manipulations of Bessel's differential equation $(1)$.

A recommended book is Watson's A treatise on the theory of Bessel functions.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    great analysis! however, how did you transform the given equation to the form you first stated, did you take $e^x=u$?
    $endgroup$
    – vidyarthi
    Dec 11 '18 at 6:57






  • 1




    $begingroup$
    @vidyarthi: I mapped the variable $x$ of the original $DE$ into $2e^{x/2}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 11 '18 at 18:04



















3












$begingroup$

Your equation can be solved explicitly in terms of Bessel functions:
$$y(x)=C_1 J_0left(2 e^{x/2}right)+ C_2 Y_0left(2 e^{x/2}right). $$
Using the asymptotic behaviors of the Bessel functions $J_0(t),Y_0(t)$ as $t to 0^+$ and as $tto +infty$ we can deduce that the solution is bounded if and only if $C_2=0$.



Edit:



If you prefer the answer in terms of initial value problems, the solution defined by the initial conditions
$$y(x_0)=y_0, \
y'(x_0)=y_1, $$

is bounded if and only if
$$ y_1 J_0left(2 e^{x_0/2}right)+e^{x_0/2} y_0 J_1left(2 e^{x_0/2}right)=0. $$





Further Edit:



Changing the independent variable according to
$$t=2 e^{x/2}, quad y(x)=Y(t), $$
results in the Bessel equation of order $0$
$$ t^2 ddot{Y}+t dot{Y} + t^2 Y=0. $$
The solution is thus
$$ Y(t)=C_1 J_0(t)+C_2 Y_0(t), $$
and therefore $y(x)$ is given as above.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    By WKB approximation, $ysim e^{-x/4}[Acos(e^{x/2}/2)+Bsin(e^{x/2}/2)]$ which would be bounded. Could you give more detail on how to transform to the Bessel form?
    $endgroup$
    – LutzL
    Dec 10 '18 at 8:27










  • $begingroup$
    Note that for $xto 0^+$ the argument of the Bessel functions goes to $2$. Does this affect your claim on $C_2$?
    $endgroup$
    – LutzL
    Dec 10 '18 at 8:34










  • $begingroup$
    @LutzL I have added in the details. I am not interested in the limit $x to 0^+$. I am interested in the limits $x to -infty$ and $x to +infty$ which correspond to the limits $t to 0^+$ and $t to +infty$.
    $endgroup$
    – user1337
    Dec 10 '18 at 8:45












  • $begingroup$
    Ok, but the question only asks for boundedness under $xto+infty$.
    $endgroup$
    – LutzL
    Dec 10 '18 at 8:46










  • $begingroup$
    @LutzL Oh, I thought he meant both (directed) infinities. Let's hope that OP clarifies this point.
    $endgroup$
    – user1337
    Dec 10 '18 at 8:47











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2 Answers
2






active

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votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

In order to prove that the solutions of (the transformed equation through a change of variable)
$$ x u''(x) + u'(x) + x u(x) = 0 tag{1}$$
are bounded on $[1,+infty)$ we do not need to invoke the asymptotics of Bessel functions, we may directly exploit the structure of the differential equation. On short intervals $[a,b]$ it is reasonable to claim that the solution of $(1)$ with $u(a)=u_0, u'(a)=u_1$ is close to the solution of
$$ a v''(x) + v'(x) + a v(x) = 0 tag{2}$$
with boundary conditions $v(a)=u_0, v'(a)=u_1$. On the other hand $(2)$ is a ODE with constant coefficients and characteristic polynomial $az^2+z+a$, hence the solutions of $(2)$ are bounded by $sqrt{u_0^2+u_1^2+frac{u_0 u_1}{a}}$ on $[a,b]$. Let us consider the sequence of intervals $[H_1,H_2],[H_2,H_3],ldots$
By denoting as $sigma_m$ the following supremum
$$ sigma_m = sup_{xin[H_m,H_{m+1}]}left|u(x)-v(x)right| $$
(where the boundary conditions for $v$ on $[H_m,H_{m+1}]$ are given by the values of $u$ and $u'$ at the left endpoint) we have that
$$ {sigma_m}_{mgeq 1}in ell^1 tag{3} $$
allows to state that the boundedness of the solutions of $(1)$ is a consequence of the boundedness of the solutions of $(2)$. By Frobenius power series method both the solutions of $(1)$ and $(2)$ are analytic functions: by considering $d(x)=u(x)-v(x)$ on $[H_m,H_{m+1}]$ we have that both $d$ and $d'$ vanish at the left endpoint, so by considering the termwise difference between $(1)$ and $(2)$ it is not difficult to derive
$$ sigma_m ll |[H_m,H_{m+1}]|^2 H_m^2 ll frac{log^2 m}{m^2}$$
having ${sigma_m}_{mgeq 1}inell^1$ as a straightfoward consequence.





Now that the boundedness of the solutions of $(1)$ is proved, a "bootstrap" argument allows to prove that $u(x)^2+u'(x)^2 ll frac{1}{x}$ as $xto +infty$, hence the solutions of $(1)$ are not only bounded but convergent to zero and with a bounded derivative. Indeed, by defining the energy of a solution as $E(u)=u(x)^2+u'(x)^2$
and by multiplying both sides of $(1)$ by $2u'(x)$, we have



$$ 0 = x E'(u) + 2u'(x)^2 leq x E'(u) + 2 E(u)$$
and any non-trivial bound for the energy immediately gives a simultaneous bound for $u$ and $u'$.





Summarizing, Tricomi's approximation
$$ J_0(x) approx frac{sin(x)+cos(x)}{sqrt{pi x}} text{ for large values of }x$$
can be essentially derived by standard manipulations of Bessel's differential equation $(1)$.

A recommended book is Watson's A treatise on the theory of Bessel functions.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    great analysis! however, how did you transform the given equation to the form you first stated, did you take $e^x=u$?
    $endgroup$
    – vidyarthi
    Dec 11 '18 at 6:57






  • 1




    $begingroup$
    @vidyarthi: I mapped the variable $x$ of the original $DE$ into $2e^{x/2}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 11 '18 at 18:04
















2












$begingroup$

In order to prove that the solutions of (the transformed equation through a change of variable)
$$ x u''(x) + u'(x) + x u(x) = 0 tag{1}$$
are bounded on $[1,+infty)$ we do not need to invoke the asymptotics of Bessel functions, we may directly exploit the structure of the differential equation. On short intervals $[a,b]$ it is reasonable to claim that the solution of $(1)$ with $u(a)=u_0, u'(a)=u_1$ is close to the solution of
$$ a v''(x) + v'(x) + a v(x) = 0 tag{2}$$
with boundary conditions $v(a)=u_0, v'(a)=u_1$. On the other hand $(2)$ is a ODE with constant coefficients and characteristic polynomial $az^2+z+a$, hence the solutions of $(2)$ are bounded by $sqrt{u_0^2+u_1^2+frac{u_0 u_1}{a}}$ on $[a,b]$. Let us consider the sequence of intervals $[H_1,H_2],[H_2,H_3],ldots$
By denoting as $sigma_m$ the following supremum
$$ sigma_m = sup_{xin[H_m,H_{m+1}]}left|u(x)-v(x)right| $$
(where the boundary conditions for $v$ on $[H_m,H_{m+1}]$ are given by the values of $u$ and $u'$ at the left endpoint) we have that
$$ {sigma_m}_{mgeq 1}in ell^1 tag{3} $$
allows to state that the boundedness of the solutions of $(1)$ is a consequence of the boundedness of the solutions of $(2)$. By Frobenius power series method both the solutions of $(1)$ and $(2)$ are analytic functions: by considering $d(x)=u(x)-v(x)$ on $[H_m,H_{m+1}]$ we have that both $d$ and $d'$ vanish at the left endpoint, so by considering the termwise difference between $(1)$ and $(2)$ it is not difficult to derive
$$ sigma_m ll |[H_m,H_{m+1}]|^2 H_m^2 ll frac{log^2 m}{m^2}$$
having ${sigma_m}_{mgeq 1}inell^1$ as a straightfoward consequence.





Now that the boundedness of the solutions of $(1)$ is proved, a "bootstrap" argument allows to prove that $u(x)^2+u'(x)^2 ll frac{1}{x}$ as $xto +infty$, hence the solutions of $(1)$ are not only bounded but convergent to zero and with a bounded derivative. Indeed, by defining the energy of a solution as $E(u)=u(x)^2+u'(x)^2$
and by multiplying both sides of $(1)$ by $2u'(x)$, we have



$$ 0 = x E'(u) + 2u'(x)^2 leq x E'(u) + 2 E(u)$$
and any non-trivial bound for the energy immediately gives a simultaneous bound for $u$ and $u'$.





Summarizing, Tricomi's approximation
$$ J_0(x) approx frac{sin(x)+cos(x)}{sqrt{pi x}} text{ for large values of }x$$
can be essentially derived by standard manipulations of Bessel's differential equation $(1)$.

A recommended book is Watson's A treatise on the theory of Bessel functions.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    great analysis! however, how did you transform the given equation to the form you first stated, did you take $e^x=u$?
    $endgroup$
    – vidyarthi
    Dec 11 '18 at 6:57






  • 1




    $begingroup$
    @vidyarthi: I mapped the variable $x$ of the original $DE$ into $2e^{x/2}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 11 '18 at 18:04














2












2








2





$begingroup$

In order to prove that the solutions of (the transformed equation through a change of variable)
$$ x u''(x) + u'(x) + x u(x) = 0 tag{1}$$
are bounded on $[1,+infty)$ we do not need to invoke the asymptotics of Bessel functions, we may directly exploit the structure of the differential equation. On short intervals $[a,b]$ it is reasonable to claim that the solution of $(1)$ with $u(a)=u_0, u'(a)=u_1$ is close to the solution of
$$ a v''(x) + v'(x) + a v(x) = 0 tag{2}$$
with boundary conditions $v(a)=u_0, v'(a)=u_1$. On the other hand $(2)$ is a ODE with constant coefficients and characteristic polynomial $az^2+z+a$, hence the solutions of $(2)$ are bounded by $sqrt{u_0^2+u_1^2+frac{u_0 u_1}{a}}$ on $[a,b]$. Let us consider the sequence of intervals $[H_1,H_2],[H_2,H_3],ldots$
By denoting as $sigma_m$ the following supremum
$$ sigma_m = sup_{xin[H_m,H_{m+1}]}left|u(x)-v(x)right| $$
(where the boundary conditions for $v$ on $[H_m,H_{m+1}]$ are given by the values of $u$ and $u'$ at the left endpoint) we have that
$$ {sigma_m}_{mgeq 1}in ell^1 tag{3} $$
allows to state that the boundedness of the solutions of $(1)$ is a consequence of the boundedness of the solutions of $(2)$. By Frobenius power series method both the solutions of $(1)$ and $(2)$ are analytic functions: by considering $d(x)=u(x)-v(x)$ on $[H_m,H_{m+1}]$ we have that both $d$ and $d'$ vanish at the left endpoint, so by considering the termwise difference between $(1)$ and $(2)$ it is not difficult to derive
$$ sigma_m ll |[H_m,H_{m+1}]|^2 H_m^2 ll frac{log^2 m}{m^2}$$
having ${sigma_m}_{mgeq 1}inell^1$ as a straightfoward consequence.





Now that the boundedness of the solutions of $(1)$ is proved, a "bootstrap" argument allows to prove that $u(x)^2+u'(x)^2 ll frac{1}{x}$ as $xto +infty$, hence the solutions of $(1)$ are not only bounded but convergent to zero and with a bounded derivative. Indeed, by defining the energy of a solution as $E(u)=u(x)^2+u'(x)^2$
and by multiplying both sides of $(1)$ by $2u'(x)$, we have



$$ 0 = x E'(u) + 2u'(x)^2 leq x E'(u) + 2 E(u)$$
and any non-trivial bound for the energy immediately gives a simultaneous bound for $u$ and $u'$.





Summarizing, Tricomi's approximation
$$ J_0(x) approx frac{sin(x)+cos(x)}{sqrt{pi x}} text{ for large values of }x$$
can be essentially derived by standard manipulations of Bessel's differential equation $(1)$.

A recommended book is Watson's A treatise on the theory of Bessel functions.






share|cite|improve this answer









$endgroup$



In order to prove that the solutions of (the transformed equation through a change of variable)
$$ x u''(x) + u'(x) + x u(x) = 0 tag{1}$$
are bounded on $[1,+infty)$ we do not need to invoke the asymptotics of Bessel functions, we may directly exploit the structure of the differential equation. On short intervals $[a,b]$ it is reasonable to claim that the solution of $(1)$ with $u(a)=u_0, u'(a)=u_1$ is close to the solution of
$$ a v''(x) + v'(x) + a v(x) = 0 tag{2}$$
with boundary conditions $v(a)=u_0, v'(a)=u_1$. On the other hand $(2)$ is a ODE with constant coefficients and characteristic polynomial $az^2+z+a$, hence the solutions of $(2)$ are bounded by $sqrt{u_0^2+u_1^2+frac{u_0 u_1}{a}}$ on $[a,b]$. Let us consider the sequence of intervals $[H_1,H_2],[H_2,H_3],ldots$
By denoting as $sigma_m$ the following supremum
$$ sigma_m = sup_{xin[H_m,H_{m+1}]}left|u(x)-v(x)right| $$
(where the boundary conditions for $v$ on $[H_m,H_{m+1}]$ are given by the values of $u$ and $u'$ at the left endpoint) we have that
$$ {sigma_m}_{mgeq 1}in ell^1 tag{3} $$
allows to state that the boundedness of the solutions of $(1)$ is a consequence of the boundedness of the solutions of $(2)$. By Frobenius power series method both the solutions of $(1)$ and $(2)$ are analytic functions: by considering $d(x)=u(x)-v(x)$ on $[H_m,H_{m+1}]$ we have that both $d$ and $d'$ vanish at the left endpoint, so by considering the termwise difference between $(1)$ and $(2)$ it is not difficult to derive
$$ sigma_m ll |[H_m,H_{m+1}]|^2 H_m^2 ll frac{log^2 m}{m^2}$$
having ${sigma_m}_{mgeq 1}inell^1$ as a straightfoward consequence.





Now that the boundedness of the solutions of $(1)$ is proved, a "bootstrap" argument allows to prove that $u(x)^2+u'(x)^2 ll frac{1}{x}$ as $xto +infty$, hence the solutions of $(1)$ are not only bounded but convergent to zero and with a bounded derivative. Indeed, by defining the energy of a solution as $E(u)=u(x)^2+u'(x)^2$
and by multiplying both sides of $(1)$ by $2u'(x)$, we have



$$ 0 = x E'(u) + 2u'(x)^2 leq x E'(u) + 2 E(u)$$
and any non-trivial bound for the energy immediately gives a simultaneous bound for $u$ and $u'$.





Summarizing, Tricomi's approximation
$$ J_0(x) approx frac{sin(x)+cos(x)}{sqrt{pi x}} text{ for large values of }x$$
can be essentially derived by standard manipulations of Bessel's differential equation $(1)$.

A recommended book is Watson's A treatise on the theory of Bessel functions.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 21:27









Jack D'AurizioJack D'Aurizio

290k33282662




290k33282662












  • $begingroup$
    great analysis! however, how did you transform the given equation to the form you first stated, did you take $e^x=u$?
    $endgroup$
    – vidyarthi
    Dec 11 '18 at 6:57






  • 1




    $begingroup$
    @vidyarthi: I mapped the variable $x$ of the original $DE$ into $2e^{x/2}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 11 '18 at 18:04


















  • $begingroup$
    great analysis! however, how did you transform the given equation to the form you first stated, did you take $e^x=u$?
    $endgroup$
    – vidyarthi
    Dec 11 '18 at 6:57






  • 1




    $begingroup$
    @vidyarthi: I mapped the variable $x$ of the original $DE$ into $2e^{x/2}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 11 '18 at 18:04
















$begingroup$
great analysis! however, how did you transform the given equation to the form you first stated, did you take $e^x=u$?
$endgroup$
– vidyarthi
Dec 11 '18 at 6:57




$begingroup$
great analysis! however, how did you transform the given equation to the form you first stated, did you take $e^x=u$?
$endgroup$
– vidyarthi
Dec 11 '18 at 6:57




1




1




$begingroup$
@vidyarthi: I mapped the variable $x$ of the original $DE$ into $2e^{x/2}$.
$endgroup$
– Jack D'Aurizio
Dec 11 '18 at 18:04




$begingroup$
@vidyarthi: I mapped the variable $x$ of the original $DE$ into $2e^{x/2}$.
$endgroup$
– Jack D'Aurizio
Dec 11 '18 at 18:04











3












$begingroup$

Your equation can be solved explicitly in terms of Bessel functions:
$$y(x)=C_1 J_0left(2 e^{x/2}right)+ C_2 Y_0left(2 e^{x/2}right). $$
Using the asymptotic behaviors of the Bessel functions $J_0(t),Y_0(t)$ as $t to 0^+$ and as $tto +infty$ we can deduce that the solution is bounded if and only if $C_2=0$.



Edit:



If you prefer the answer in terms of initial value problems, the solution defined by the initial conditions
$$y(x_0)=y_0, \
y'(x_0)=y_1, $$

is bounded if and only if
$$ y_1 J_0left(2 e^{x_0/2}right)+e^{x_0/2} y_0 J_1left(2 e^{x_0/2}right)=0. $$





Further Edit:



Changing the independent variable according to
$$t=2 e^{x/2}, quad y(x)=Y(t), $$
results in the Bessel equation of order $0$
$$ t^2 ddot{Y}+t dot{Y} + t^2 Y=0. $$
The solution is thus
$$ Y(t)=C_1 J_0(t)+C_2 Y_0(t), $$
and therefore $y(x)$ is given as above.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    By WKB approximation, $ysim e^{-x/4}[Acos(e^{x/2}/2)+Bsin(e^{x/2}/2)]$ which would be bounded. Could you give more detail on how to transform to the Bessel form?
    $endgroup$
    – LutzL
    Dec 10 '18 at 8:27










  • $begingroup$
    Note that for $xto 0^+$ the argument of the Bessel functions goes to $2$. Does this affect your claim on $C_2$?
    $endgroup$
    – LutzL
    Dec 10 '18 at 8:34










  • $begingroup$
    @LutzL I have added in the details. I am not interested in the limit $x to 0^+$. I am interested in the limits $x to -infty$ and $x to +infty$ which correspond to the limits $t to 0^+$ and $t to +infty$.
    $endgroup$
    – user1337
    Dec 10 '18 at 8:45












  • $begingroup$
    Ok, but the question only asks for boundedness under $xto+infty$.
    $endgroup$
    – LutzL
    Dec 10 '18 at 8:46










  • $begingroup$
    @LutzL Oh, I thought he meant both (directed) infinities. Let's hope that OP clarifies this point.
    $endgroup$
    – user1337
    Dec 10 '18 at 8:47
















3












$begingroup$

Your equation can be solved explicitly in terms of Bessel functions:
$$y(x)=C_1 J_0left(2 e^{x/2}right)+ C_2 Y_0left(2 e^{x/2}right). $$
Using the asymptotic behaviors of the Bessel functions $J_0(t),Y_0(t)$ as $t to 0^+$ and as $tto +infty$ we can deduce that the solution is bounded if and only if $C_2=0$.



Edit:



If you prefer the answer in terms of initial value problems, the solution defined by the initial conditions
$$y(x_0)=y_0, \
y'(x_0)=y_1, $$

is bounded if and only if
$$ y_1 J_0left(2 e^{x_0/2}right)+e^{x_0/2} y_0 J_1left(2 e^{x_0/2}right)=0. $$





Further Edit:



Changing the independent variable according to
$$t=2 e^{x/2}, quad y(x)=Y(t), $$
results in the Bessel equation of order $0$
$$ t^2 ddot{Y}+t dot{Y} + t^2 Y=0. $$
The solution is thus
$$ Y(t)=C_1 J_0(t)+C_2 Y_0(t), $$
and therefore $y(x)$ is given as above.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    By WKB approximation, $ysim e^{-x/4}[Acos(e^{x/2}/2)+Bsin(e^{x/2}/2)]$ which would be bounded. Could you give more detail on how to transform to the Bessel form?
    $endgroup$
    – LutzL
    Dec 10 '18 at 8:27










  • $begingroup$
    Note that for $xto 0^+$ the argument of the Bessel functions goes to $2$. Does this affect your claim on $C_2$?
    $endgroup$
    – LutzL
    Dec 10 '18 at 8:34










  • $begingroup$
    @LutzL I have added in the details. I am not interested in the limit $x to 0^+$. I am interested in the limits $x to -infty$ and $x to +infty$ which correspond to the limits $t to 0^+$ and $t to +infty$.
    $endgroup$
    – user1337
    Dec 10 '18 at 8:45












  • $begingroup$
    Ok, but the question only asks for boundedness under $xto+infty$.
    $endgroup$
    – LutzL
    Dec 10 '18 at 8:46










  • $begingroup$
    @LutzL Oh, I thought he meant both (directed) infinities. Let's hope that OP clarifies this point.
    $endgroup$
    – user1337
    Dec 10 '18 at 8:47














3












3








3





$begingroup$

Your equation can be solved explicitly in terms of Bessel functions:
$$y(x)=C_1 J_0left(2 e^{x/2}right)+ C_2 Y_0left(2 e^{x/2}right). $$
Using the asymptotic behaviors of the Bessel functions $J_0(t),Y_0(t)$ as $t to 0^+$ and as $tto +infty$ we can deduce that the solution is bounded if and only if $C_2=0$.



Edit:



If you prefer the answer in terms of initial value problems, the solution defined by the initial conditions
$$y(x_0)=y_0, \
y'(x_0)=y_1, $$

is bounded if and only if
$$ y_1 J_0left(2 e^{x_0/2}right)+e^{x_0/2} y_0 J_1left(2 e^{x_0/2}right)=0. $$





Further Edit:



Changing the independent variable according to
$$t=2 e^{x/2}, quad y(x)=Y(t), $$
results in the Bessel equation of order $0$
$$ t^2 ddot{Y}+t dot{Y} + t^2 Y=0. $$
The solution is thus
$$ Y(t)=C_1 J_0(t)+C_2 Y_0(t), $$
and therefore $y(x)$ is given as above.






share|cite|improve this answer











$endgroup$



Your equation can be solved explicitly in terms of Bessel functions:
$$y(x)=C_1 J_0left(2 e^{x/2}right)+ C_2 Y_0left(2 e^{x/2}right). $$
Using the asymptotic behaviors of the Bessel functions $J_0(t),Y_0(t)$ as $t to 0^+$ and as $tto +infty$ we can deduce that the solution is bounded if and only if $C_2=0$.



Edit:



If you prefer the answer in terms of initial value problems, the solution defined by the initial conditions
$$y(x_0)=y_0, \
y'(x_0)=y_1, $$

is bounded if and only if
$$ y_1 J_0left(2 e^{x_0/2}right)+e^{x_0/2} y_0 J_1left(2 e^{x_0/2}right)=0. $$





Further Edit:



Changing the independent variable according to
$$t=2 e^{x/2}, quad y(x)=Y(t), $$
results in the Bessel equation of order $0$
$$ t^2 ddot{Y}+t dot{Y} + t^2 Y=0. $$
The solution is thus
$$ Y(t)=C_1 J_0(t)+C_2 Y_0(t), $$
and therefore $y(x)$ is given as above.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 8:43

























answered Dec 10 '18 at 7:25









user1337user1337

16.7k43491




16.7k43491








  • 1




    $begingroup$
    By WKB approximation, $ysim e^{-x/4}[Acos(e^{x/2}/2)+Bsin(e^{x/2}/2)]$ which would be bounded. Could you give more detail on how to transform to the Bessel form?
    $endgroup$
    – LutzL
    Dec 10 '18 at 8:27










  • $begingroup$
    Note that for $xto 0^+$ the argument of the Bessel functions goes to $2$. Does this affect your claim on $C_2$?
    $endgroup$
    – LutzL
    Dec 10 '18 at 8:34










  • $begingroup$
    @LutzL I have added in the details. I am not interested in the limit $x to 0^+$. I am interested in the limits $x to -infty$ and $x to +infty$ which correspond to the limits $t to 0^+$ and $t to +infty$.
    $endgroup$
    – user1337
    Dec 10 '18 at 8:45












  • $begingroup$
    Ok, but the question only asks for boundedness under $xto+infty$.
    $endgroup$
    – LutzL
    Dec 10 '18 at 8:46










  • $begingroup$
    @LutzL Oh, I thought he meant both (directed) infinities. Let's hope that OP clarifies this point.
    $endgroup$
    – user1337
    Dec 10 '18 at 8:47














  • 1




    $begingroup$
    By WKB approximation, $ysim e^{-x/4}[Acos(e^{x/2}/2)+Bsin(e^{x/2}/2)]$ which would be bounded. Could you give more detail on how to transform to the Bessel form?
    $endgroup$
    – LutzL
    Dec 10 '18 at 8:27










  • $begingroup$
    Note that for $xto 0^+$ the argument of the Bessel functions goes to $2$. Does this affect your claim on $C_2$?
    $endgroup$
    – LutzL
    Dec 10 '18 at 8:34










  • $begingroup$
    @LutzL I have added in the details. I am not interested in the limit $x to 0^+$. I am interested in the limits $x to -infty$ and $x to +infty$ which correspond to the limits $t to 0^+$ and $t to +infty$.
    $endgroup$
    – user1337
    Dec 10 '18 at 8:45












  • $begingroup$
    Ok, but the question only asks for boundedness under $xto+infty$.
    $endgroup$
    – LutzL
    Dec 10 '18 at 8:46










  • $begingroup$
    @LutzL Oh, I thought he meant both (directed) infinities. Let's hope that OP clarifies this point.
    $endgroup$
    – user1337
    Dec 10 '18 at 8:47








1




1




$begingroup$
By WKB approximation, $ysim e^{-x/4}[Acos(e^{x/2}/2)+Bsin(e^{x/2}/2)]$ which would be bounded. Could you give more detail on how to transform to the Bessel form?
$endgroup$
– LutzL
Dec 10 '18 at 8:27




$begingroup$
By WKB approximation, $ysim e^{-x/4}[Acos(e^{x/2}/2)+Bsin(e^{x/2}/2)]$ which would be bounded. Could you give more detail on how to transform to the Bessel form?
$endgroup$
– LutzL
Dec 10 '18 at 8:27












$begingroup$
Note that for $xto 0^+$ the argument of the Bessel functions goes to $2$. Does this affect your claim on $C_2$?
$endgroup$
– LutzL
Dec 10 '18 at 8:34




$begingroup$
Note that for $xto 0^+$ the argument of the Bessel functions goes to $2$. Does this affect your claim on $C_2$?
$endgroup$
– LutzL
Dec 10 '18 at 8:34












$begingroup$
@LutzL I have added in the details. I am not interested in the limit $x to 0^+$. I am interested in the limits $x to -infty$ and $x to +infty$ which correspond to the limits $t to 0^+$ and $t to +infty$.
$endgroup$
– user1337
Dec 10 '18 at 8:45






$begingroup$
@LutzL I have added in the details. I am not interested in the limit $x to 0^+$. I am interested in the limits $x to -infty$ and $x to +infty$ which correspond to the limits $t to 0^+$ and $t to +infty$.
$endgroup$
– user1337
Dec 10 '18 at 8:45














$begingroup$
Ok, but the question only asks for boundedness under $xto+infty$.
$endgroup$
– LutzL
Dec 10 '18 at 8:46




$begingroup$
Ok, but the question only asks for boundedness under $xto+infty$.
$endgroup$
– LutzL
Dec 10 '18 at 8:46












$begingroup$
@LutzL Oh, I thought he meant both (directed) infinities. Let's hope that OP clarifies this point.
$endgroup$
– user1337
Dec 10 '18 at 8:47




$begingroup$
@LutzL Oh, I thought he meant both (directed) infinities. Let's hope that OP clarifies this point.
$endgroup$
– user1337
Dec 10 '18 at 8:47


















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