Suppose |f(t)|< c/t^(1+a) for t≥ t_0 , with constants c,a>0…

Suppose $|f(t)|\lt c/t^{1+a}$ for $t\geq t_0$, with constants $c,a>0$. Prove that $u''-\bigl[1-f(t)\bigr]u=0$ has two solutions $u_1(t)=e^{t}\bigl[1+O(1/t^a)\bigr]$ and $u_2(t)=e^{-t}\bigl[1+O(1/t^a)\bigr]$ as $t\to\infty$.

Step 1: Strategy: a perturbed constant-coefficient equation

Rewrite the equation as $u''-u=-f(t)u=:g(t)$, a perturbation of $u''-u=0$, whose two independent solutions are $e^{t}$ and $e^{-t}$. The idea is to convert this into an integral equation for $u$ using the Green's function of the unperturbed operator, then show the perturbation only contributes a controllable $O(t^{-a})$ correction — but, as will become clear, the growing and decaying solutions require genuinely different constructions.

Step 2: A Green's-function identity

For any (sufficiently well-behaved) function $g$, define $v(t)=\displaystyle\int_t^\infty \sinh(s-t)\,g(s)\,ds$. Writing $\sinh(s-t)=\tfrac12\bigl(e^{s}e^{-t}-e^{-s}e^{t}\bigr)$,

\[ v(t) = \tfrac12 e^{-t}\!\int_t^\infty\! e^{s}g(s)\,ds \;-\; \tfrac12 e^{t}\!\int_t^\infty\! e^{-s}g(s)\,ds. \]

Differentiating under the integral sign (Leibniz's rule): the boundary term at $s=t$ vanishes each time because $\sinh(0)=0$, so

\[ v'(t) = -\int_t^\infty \cosh(s-t)\,g(s)\,ds, \qquad v''(t) = g(t) + \int_t^\infty \sinh(s-t)\,g(s)\,ds = g(t)+v(t). \]

Hence $v''-v=g$ — this is exactly the variation-of-parameters / Green's-function representation for $u''-u=g$ built from the fundamental system $\{e^t,e^{-t}\}$ (Wronskian $-2$).

Step 3: The decaying solution $u_2$: a single Volterra equation

Taking $g=-f\cdot u_2$ in the identity above and adding the homogeneous term $e^{-t}$ gives a natural candidate: $u_2(t) = e^{-t} + \int_t^\infty \sinh(t-s)\,f(s)\,u_2(s)\,ds.$

This integral converges: for large $s$, $\sinh(t-s)\sim -\tfrac12 e^{s-t}$ and $u_2(s)\sim e^{-s}$, so the integrand behaves like $-\tfrac12 f(s)e^{-t}$, which is absolutely integrable since $f(s)=O(s^{-(1+a)})$. Normalizing $w(t)=u_2(t)e^{t}$ and using $\sinh(t-s)e^{t-s}=\tfrac12\bigl(e^{-2(s-t)}-1\bigr)$ (with $s\geq t$):

\[ w(t) = 1 + \int_t^\infty \frac{e^{-2(s-t)}-1}{2}\,f(s)\,w(s)\,ds. \]

Since $s\geq t$, $e^{-2(s-t)}\in(0,1]$, so the kernel is bounded: $\bigl|e^{-2(s-t)}-1\bigr|\leq 1$. Hence

\[ |w(t)-1| \;\leq\; \frac12\int_t^\infty |f(s)|\,|w(s)|\,ds. \]

Step 4: The bootstrap for $u_2$

Let $\Phi(t)=\displaystyle\int_t^\infty |f(s)|\,ds \leq \int_t^\infty \frac{c}{s^{1+a}}\,ds = \frac{c}{a\,t^a} = O(t^{-a})$. Choosing $t_0$ large enough that $\tfrac12\Phi(t_0)<1$, a Picard-iteration / contraction argument on the integral equation first yields a coarse bound $|w(t)|\leq M$ for $t\geq t_0$; feeding this back into the inequality above gives the sharp estimate

\[ |w(t)-1| \leq \frac{M}{2}\Phi(t) = O(t^{-a}). \] $u_2(t) = e^{-t$.

Step 5: Why $u_1$ needs a different construction

The naive analogue $u_1(t)=e^{t}+\displaystyle\int_t^\infty \sinh(t-s)f(s)u_1(s)\,ds$ does not converge: for large $s$, $\sinh(t-s)\sim -\tfrac12 e^{s-t}$ and now $u_1(s)\sim e^{s}$, so the integrand behaves like $-\tfrac12 f(s)e^{2s-t}$. Since $f$ decays only polynomially, this exponential growth is never integrable — a single forward Volterra equation cannot represent the growing solution. Instead, split the particular-solution formula into two pieces carrying different integration ranges:

\[ u_1(t) = e^{t}\left[1+\frac12\int_t^\infty e^{-s}f(s)u_1(s)\,ds\right] + \frac12 e^{-t}\int_{t_0}^{t} e^{s}f(s)u_1(s)\,ds. \]

Direct differentiation (Leibniz's rule, the same Green's-function mechanism as in Step 2) confirms this satisfies $u_1''-u_1=-f\,u_1$. The two pieces are each well-behaved for a different reason: the forward integral ($t$ to $\infty$) has integrand $e^{-s}f(s)u_1(s)\sim f(s)$ since $e^{-s}u_1(s)$ stays bounded — integrable by the same argument as for $u_2$; the fixed-basepoint integral ($t_0$ to $t$) is a proper, finite integral for every finite $t$, so it never faces a convergence-at-infinity issue at all.

Step 6: Estimating both pieces of $u_1$

Normalize $v(t)=u_1(t)e^{-t}$. The forward-integral term becomes

\[ \frac12\int_t^\infty f(s)\,v(s)\,ds, \qquad \text{bounded (as for $u_2$) by} \qquad \frac{M}{2}\Phi(t) = O(t^{-a}). \]

The fixed-basepoint term, after dividing the whole expression for $u_1$ by $e^t$ (which turns its prefactor $e^{-t}$ into $e^{-2t}$), contributes

\[ \frac12 e^{-2t}\int_{t_0}^{t} e^{2s}f(s)v(s)\,ds. \]

Since $e^{2s}$ grows while $f$ decays only polynomially, this integral is dominated by its value near the upper endpoint $s=t$ (a Laplace-type / integration-by-parts estimate): $\displaystyle\int_{t_0}^t e^{2s}f(s)\,ds = O\bigl(e^{2t}f(t)\bigr)=O\bigl(e^{2t}t^{-(1+a)}\bigr)$. So this whole term is

\[ O\bigl(e^{-2t}\cdot e^{2t}t^{-(1+a)}\bigr) = O\bigl(t^{-(1+a)}\bigr), \]

which is strictly smaller than the forward term's $O(t^{-a})$. Combining both contributions, $v(t) = 1 + O(t^{-a}) + O\bigl(t^{-(1+a)}\bigr) = 1+O(t^{-a}).$ $u_1(t) = e^{t$.

Step 7: Linear independence

To leading order, differentiating the small correction in each asymptotic form produces an even smaller term, so $u_1'\sim e^t\bigl[1+O(t^{-a})\bigr]$ and $u_2'\sim -e^{-t}\bigl[1+O(t^{-a})\bigr]$. The Wronskian is then

\[ W(u_1,u_2) = u_1u_2'-u_1'u_2 \sim e^{t}(-e^{-t})-e^{t}e^{-t} = -2+O(t^{-a}) \neq 0 \]

for $t$ sufficiently large, so $\{u_1,u_2\}$ form a genuine fundamental system as $t\to\infty$.

$u_1(t) = e^{t$.