Assume function $f:\mathbb R \to \mathbb R$ is bounded over the real line, and the bound is M. Prove that there is a unique bounded solution for the differential equation $y'-y = f(x)$. If f is periodic then the bounded solution is periodic.

My attempt:

Using the method of integrating factor, I arrived at: $$ y(x) = e^x (\int_{x_0}^x e^{-t}f(t)dt +C) \iff \lvert y(x) \rvert \le e^x (\int_{x_0}^x e^{-t} \lvert f(t) \rvert dt +C) \le e^xM (-e^{-x} + e^{-x_0} + C) = M(e^{x}C + e^{x - x_0} - 1)$$ which blows up as $x \to \infty$ unless I restrict my initial condition such that $C = -e^{-x_0}$, but that's impossible since I will get $\lvert y(x) \rvert \le -M$. Did I do anything wrong? Is there a single function satisfying this ODE that satisfies ANY initial conditions but is still bounded?

Please enlighten me.


First note that there exists the limit $$ D=\lim_{x\to+\infty}\int_{x_0}^x e^{-t}f(t)\,dt=\int_{x_0}^{+\infty} e^{-t}f(t)\,dt $$ exists (show for example that $c_n=\int_{x_0}^n e^{-t}f(t)\,dt$ is a Cauchy sequence). So, your solution is bounded for $x>0$ if and only if $C=-D$.

On the other hand, notice that any solution is bounded for $x<0$.

  • $\begingroup$ Ok I can just do the exhaustion method for computing the integral. Thank you. I guess the problem was not properly formulated then (?) $\endgroup$ – Nen Feb 14 '16 at 23:18
  • $\begingroup$ No, the problem is very well formulated. Only your solutions doesn't take all into account. $\endgroup$ – John B Feb 14 '16 at 23:19

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