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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 $|f(x)| <M, \forall x >0$ ?

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

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  • $\begingroup$ 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 $C=H_m$. $\endgroup$ – Jack D'Aurizio Dec 10 '18 at 13:58
  • $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Dec 10 '18 at 14:55
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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.

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