I'm currently studying this non linear differential equation $$y''=y-y^3.$$ The assumption are that $y,y'$ are in $L^2(\mathbb R)$, however no boundary conditions are assigned. I am asked to prove that $\mid y(x)\mid\le \sqrt 2.$
My attempt goes as follows: multiply both sides by $y'$ and integrate to obtain:
$$(\clubsuit)\quad\mid y'(x)\mid=\sqrt{2\left(\frac{y^2}{2}-\frac{y^4}{4}+C_1\right)}.$$ Then what is inside the square root must be nonnegative, however I have no hypothesis on $C_1$ since no boundary conditions are provided.
Another thing which should be useful is that $y^2\in L^1(\mathbb R)$ and, since $$\int {(y^2)^'}=2\int \mid yy'\mid\leq \left(\int y^2\right)^{1/2}\left(\int y'^2\right)^{1/2}< +\infty, $$ then by noticing that $$\mid y^2(t)-y^2(0)\mid=\mid\int_0^t (y^2)'\mid< C.$$
Hence $y$ is bounded. But nowhere from here. Hints?
Finishing the exercise:
As Julian pointed out, from the fact that $y\in L^2(\mathbb R)$ and it is bounded we conclude that $$\int y^4\leq \|y\|_\infty^2\int y^2< \infty,$$ so that $y\in L^4(\mathbb R).$ Squaring and integrating $(\clubsuit)$ one must conclude $C_1=0$. The constant solutions are $$y=0.$$