Consider the ode $$ f''(t)-e^{-2t} f(t)=0. $$ What is the general behaviour of $|f|$ for large $t$s?
Is it true that there exists an $A\in \mathbb{R}$ and there exists a positive $B$ such that $|f(t)|\sim A+Bt $ for all large enough $t$?
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$\begingroup$ It looks like it is approximately linear, all right, but why do you expect $A$ to be positive? $\endgroup$– fedjaCommented Jul 14, 2019 at 14:49
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$\begingroup$ Thanks. I think I was mislead by the modulus on f, but I agree now that A could be possibly negative. $\endgroup$– LuciaCommented Jul 14, 2019 at 14:54
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4$\begingroup$ It is good idea sometimes to check on a computer (maple, mathematica) to see if this equation has any "explicit" solutions. For example, the code dsolve(diff(f(x),x,x)-exp(-2x)*f(x)=0,f(x)); immediately gives $f(x)=C_1 BesselI(0,\exp(-x))+C_2 BesselK(0, \exp(-x))$, and the asymptotics of the Bessel functions are known. Now it suggests to do the change of variables $\exp(-t)=x$ and see that your ode reduces to the standard Bessel ode for the function $g(v)=f(-\ln v)$. $\endgroup$– Paata IvanishviliCommented Jul 14, 2019 at 15:11
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5$\begingroup$ Or just write the general solution in the form $\sum_{k\ge 0}L_k(t)e^{-2kt}$ where $L_k$ are linear functions, $L_0$ is arbitrary, and the rest of $L_k$ are determined from the obvious recurrence relations. That often works even if there is no explicit form. $\endgroup$– fedjaCommented Jul 14, 2019 at 15:21
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2 Answers
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The general solution of this differential equation is $$ f \left( t \right) =a\,{{ I}_{0}\left({{\rm e}^{-t}}\right) }+b\,{{ K}_{0}\left({{\rm e}^{-t}}\right)} $$ where $I_0$ and $K_0$ are modified Bessel functions. As $t \to +\infty$, $$I_0({\rm e}^{-t}) = 1 + O\left({\rm e}^{-2t}\right)$$ while $$K_0({\rm e}^{-t}) = \ln(2) - \gamma + t + O\left({\rm e}^{-2t}\right) $$
answered Jul 14, 2019 at 15:00
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The Mathematica 12.0 code
AsymptoticDSolveValue[ f''[t] - Exp[-2 t]*f[t] == 0, f, {t, Infinity, 1}]
outputs $$c_1 \left(\frac{e^{-4 t}}{64}+\frac{e^{-2 t}}{4}+1\right)+c_2 e^{-2 t} \left(\frac{t}{2}-\frac{\gamma }{2}+\frac{1}{2}+\frac{\log (2)}{2}\right)+c_2 (2 t-2 \gamma +2 \log (2)).$$
