Was given this question as extra credit on an ODE exam. Didn't have time during the exam to consider it, but I have since then, and I'm stumped.

$$ f''(x) + f(x) = -f'(x)x^{2015}$$

$f(x)$ is twice differentiable and continuous everywhere. Show that $|f(x)| < M $ for some real $M$. Hint: compute the derivative of $f'(x)^2 + f(x)^2$.
I wrote this from memory, and I hope its correct or I've waiting quite a bit of time trying to figure this out. As per the hint, I compute the derivative and found that $$ \frac{d}{dx}[f'(x)^2 + f(x)^2] = 2f''(x)f'(x) + 2f'(x)f(x) = 2f'(x)[f(x) + f''(x)] $$ So, that would be $2f'(x)$ multiplied by the left hand side of the given differential equation. In other words, if we multiply through by $2f'(x)$, we have $$ \frac{d}{dx}[f'(x)^2 + f(x)^2] = -f'(x)x^{2015}*(2f'(x)) $$ $$ \frac{d}{dx}[f'(x)^2 + f(x)^2] = -2f'(x)^2x^{2015} $$

I didn't manage to produce anything fruitful from here. I tried to integrate both sides, but could not solve or draw any conclusions from the right hand side. I tried to isolate $x^{2015}$ and then integrate, but I couldn't solve the left hand integral. At this point, I was pretty stumped and tried a bunch of other ideas. I tried to solve the homogeneous system, to no avail. Then there was a fun attempt at creating a system of first order equations that left me with a, predictably, non-constant coefficient matrix. At that point, I decided to step away and ask for some help. Am I headed in the right direction with either of these attempts? If it helps, this is supposed to require only knowledge of calculus, so I'm certain the key lies above, especially considering the hint, but I'm not sure where to proceed. Solutions and hints equally appreciated!

EDIT: Forgot to mention, $f(x)$ is twice differentiable and continuous everywhere. Will include at top.

Update: Focused on studying for final exams but I will return to this! Still stumped.


What you have written down so far tells you that the non-negative, differentiable function $$ u(x) = f'(x)^2 + f(x)^2 $$

satisfies $u'(x) = -f'(x)^2x^{2015}$ so that $u'(0) = 0$, $u'(x) \leq 0$ for $x > 0$ and $u'(x) \geq 0$ for $x < 0$ (and further, it is clear that $f(x)^2 \leq u(x)$).

For some intuition, you can interpret this as a mass-spring system, change $x$ to $t$, and write this

$$ f'' = -f - f't^{2015}. $$

On the left-hand side, you have the total force, on the right, the opposite of the displacement (so this the usual restoring force in the spring) and then a force which, for $t > 0$, has the opposite sign of $f'$, so it is a damping force.

The quantity suggested in the hint, $u(x)$ is the total kinetic and potential energy in the system, and it makes sense that a damping force would decrease total energy for $t \geq 0$ (and thus the potential energy and displacement are bounded).

You can further check that this system is symmetric in time (i.e. $g(t) = f(-t)$ satisfies the same differential equation) so you get the same properties for $t \leq 0$.

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  • $\begingroup$ I apologize if I overlooked something relatively trivial, but I'm having a hard time reconciling your statements about u'. Given that we don't have that much information about f or f', how do you figure that u'(0) = 2f'(0)[f''(0) + f'(0)] = 0? $\endgroup$ – Gsp Dec 16 '14 at 5:51
  • $\begingroup$ @Zophir I have added more exposition $\endgroup$ – BaronVT Dec 16 '14 at 18:20

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