# Twice diffrentiable function $f: \mathbb R \to \mathbb R$ such that $f''(x)+e^xf(x)=0, \forall x \in \mathbb R$ [closed]

Let $$f: \mathbb R \to \mathbb R$$ be a twice differentiable function such that $$f''(x)+e^xf(x)=0, \forall x \in \mathbb R$$. Then is it true that $$f(x)$$ is bounded in $$[0,\infty)$$ i.e. does there exist $$M >0$$ such that $$|f(x)| 0$$ ?

## closed as off-topic by Kavi Rama Murthy, Paramanand Singh, Gibbs, John B, BrahadeeshDec 10 '18 at 12:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Paramanand Singh, Gibbs, John B, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.

• Through a change of variable $(x=2 e^{z/2})$ the problem boils down to showing that the solutions of $$z u''(z) + u'(z) + z u(z)=0$$ are bounded on $z>2$. This can be done through a perturbative argument: if $C>2$ is a fixed constant, the solutions of the ordinary differential equation $C u''(z) + u'(z) + C u(z)=0$ are bounded since the roots of the characteristic polynomial $Ct^2+t+C$ have a non-positive real part. A working idea is to approximate the solution of the actual DE on the interval $(H_m,H_{m+1})$ with the solution of the ODE with $C=H_m$. – Jack D'Aurizio Dec 10 '18 at 13:58
• Actually you may prove that by defining an energy as $E(u)=u^2+u'^2$, the energy of the solutions of $z u''(z)+u'(z)+z u(z)=0$ converges to $0$ as $z\to +\infty$. The solutions are so not only bounded, but Lipschitz-continuous and convergent to zero. – Jack D'Aurizio Dec 10 '18 at 14:55

The explicit solution of $$f''(x) = -e^x f(x)$$ is given by $$f(x) = a_1 J_0(2e^{x/2}) + a_2 Y_0(2e^{x/2}),$$ where $$J_0$$ denote the Bessel function of the first kind and $$Y_0$$ the Bessel function of second kind. Since $$\lim_{x \rightarrow - \infty} J_0(x) =1$$ and $$\lim_{x \rightarrow \infty} J_0(x) =0$$, resp. $$\lim_{x \rightarrow \pm \infty} Y_0(x) =0$$, $$f(x)$$ is bounded.