# Solution of $y''+e^xy=0$ is unbounded as $x\to\infty$

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.

• When you ask about the boundedness of the solutions $y(x)$, what set is $x$ confined to? – user1337 Dec 10 '18 at 8:53
• @user1337 let us assume $x\in\mathbb{R}$ – 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.

• great analysis! however, how did you transform the given equation to the form you first stated, did you take $e^x=u$? – vidyarthi Dec 11 '18 at 6:57
• @vidyarthi: I mapped the variable $x$ of the original $DE$ into $2e^{x/2}$. – 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 $$C_2=0$$.

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

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.

• 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? – LutzL Dec 10 '18 at 8:27
• Note that for $x\to 0^+$ the argument of the Bessel functions goes to $2$. Does this affect your claim on $C_2$? – LutzL Dec 10 '18 at 8:34
• @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 $t \to +\infty$. – user1337 Dec 10 '18 at 8:45
• Ok, but the question only asks for boundedness under $x\to+\infty$. – LutzL Dec 10 '18 at 8:46
• @LutzL Oh, I thought he meant both (directed) infinities. Let's hope that OP clarifies this point. – user1337 Dec 10 '18 at 8:47