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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 $y=0.$

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    @JuliánAguirre: Ah, of course, that's one thing that I missed; a non-periodic, non-stationary solution (connecting two equilibria).2012-03-01

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You have already proved two important facts: \frac12(y')^2=\frac12(y)^2-\frac14(y)^4+C,\tag1 and $y$ is bounded. Since $y\in L^2$ and $y$ is bounded, it follows that $y\in L^4$. Integrating $(1)$ on $\mathbb{R}$, it follows that $C=0$. It is now easy to conclude that $|y|\le\sqrt2$.

Edit

The solutions in $L^2$ with derivative in $L^2$are $ y=0,\quad y=\frac{2\sqrt2\,e^x}{1+e^{2x}}=\frac{\sqrt2}{\cosh x},\quad y=-\frac{2\sqrt2\,e^x}{1+e^{2x}}=-\frac{\sqrt2}{\cosh x} $ and their translates (since the equation is invariant uner translations).

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    @Hans Lundmark What is $\int_{-\infty}^{+\infty}1\,dx$?2012-03-01