Consider the differential equation $y''+e^xy=0$. Can we say something about the behaviour of $y$ as $x\to\infty$? 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.

  • $\begingroup$ 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 $x\in\mathbb{R}$ $\endgroup$ – vidyarthi Dec 10 '18 at 9:00

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 $$ \sigma_m = \sup_{x\in[H_m,H_{m+1}]}\left|u(x)-v(x)\right| $$ (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 $$ \{\sigma_m\}_{m\geq 1}\in \ell^1 \tag{3} $$ 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 $\{\sigma_m\}_{m\geq 1}\in\ell^1$ 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 $x\to +\infty$, 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.

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

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


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 $$ y_1 J_0\left(2 e^{x_0/2}\right)+e^{x_0/2} y_0 J_1\left(2 e^{x_0/2}\right)=0. $$

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.

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    $\begingroup$ By WKB approximation, $y\sim e^{-x/4}[A\cos(e^{x/2}/2)+B\sin(e^{x/2}/2)]$ 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 $x\to 0^+$ 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 $x\to+\infty$. $\endgroup$ – 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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