differential equation 2 Help me please to resolve this question
Let the second order differential equation $$u''+a(x)u=0..........(1)$$ where $a \in \mathcal{C}^1([0,+\infty[)$
How we 
1-prouve that if $a(x) \rightarrow +\infty$ when $x \rightarrow + \infty$, then all solutions of this equation are bounded on $[0,+\infty[$
2- Prouve that if all solutions of the precedent equation are bounded in $[0,+\infty[$ and if $b(x) \rightarrow 0$ when  $x \rightarrow +\infty$ or $\int_0^{\infty} |b(s)|ds < \infty$, then, all the solutions of the equation $$u'' + (a(x) + b(x)) u = 0............(2)$$ are bounded on $[0,+\infty[$.
 A: What you're trying to prove is only true under the additional assumption that $a$ is non-decreasing. In that case the following approach works

I'm assuming that $a$ is always positive (which actually loses no generality).
Physically the equation represents a harmonic oscillator where the spring constant changes dynamically as specified by $a(x)$. This suggests that looking at the total "energy" might be useful:
  $$ E = au^2 + (\frac{du}{dx})^2 $$
  How does this change with time? We find
  $$ \frac{dE}{dx} =
\frac{da}{dx} u^2 + 2au\frac{du}{dx} + 2\frac{du}{dx} \frac{d^2u}{dx^2} =
\frac{da}{dx} u^2 \le \frac{da}{dx} \frac{E}{a} $$
(The assumption $\frac{da}{dx}\ge 0$ was used for the last inequality).
Therefore $\frac{d}{dx}\log E \le \frac{d}{dx}\log a$, so $E/a$ is non-increasing, and since $u^2$ is at most $E/a$, $u$ must be bounded.

However, if $a$ is allowed to decrease temporarily (but still tend to infinity eventually), it is possible for the adversary to find an $a$ that will make the amplitude of $u$ grow without bounds, by following the following procedure:


*

*At time $x=0$ we set $n=2$ and start with $u(0)=1$ and $u'(0)=0$.

*Whenever $u'=0$, increase $n$ by one if $|u|>n$. Then, no matter whether $n$ was increased or not, set $a=n$.

*When $u$ crosses the $x$ axis, decrease $a$ to $n-1$.

*Wait for the next point where $u'=0$. Then $|u|$ will be larger than the $|u|$ at the previous stationary point, by a factor of $\sqrt{\frac{n}{n-1}}$. Go back to step $2$.
This procedure produces an $a$ that eventually tends to infinity, but such that the extremal values of $u$ diverge just as fast as $a$ does.
Strictly speaking the $a$ thus constructed is a step function and not $\mathcal C^1$, but it's not a problem to bridge the jumps of $a$ smoothly (say, with appropriately scaled and compressed copies of $x^3-3x$ on $[-1,1]$) without changing the essential features of the construction.
