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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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    I'm a bit confused by all this. To me it seems simpler. The constant solutions are $y=0$ and $y=\pm 1$ (not $\pm \sqrt{2}$). Since this is a conservative system with a "well-type" potential $V(y)=-\frac12 y^2 + \frac14 y^4$, any non-constant solution will be periodic, and therefore not in $L^2$. Changing sign is no problem; the uniqueness theorem applies to first-order systems, and so we should write $y'=p$, $p'=\dots$, and then $y$ can pass through zero (with $p$ nonzero).2012-02-29
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    (cont.) But if the energy of the system is negative, so that the particle is in one of the two dips of the potential, then $|y| < \sqrt{2}$ and $y$ cannot change sign. Have I missed something?2012-02-29
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    How stupid from me. It was not a first order differential equation, due to the module. I erase the edit. Thank you, your solution does convince me. Thanks2012-02-29
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    Phase plane analysis shows that the unique solution $z$ with $z(0)=\sqrt2$, $z'(0)=0$ is such that $z(x)>0$ for all $x\in\mathbb{R}$, $z$ is even, $z$ is strictly decreasing on $(0,+\infty)$ and $\lim_{x\to+\infty}z(x)=0$. I strongly suspect that $z$ is in $L^2$, but I have not proved it.2012-02-29
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    I have edited my answer to include the three solutions in $L^2$ with derivative in $L^2$.2012-02-29
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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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    My god... i can't believe it. I was so focused on using some theorems on local unicity that i missed the obvious part. Lol. Btw thak you very much Julian2012-02-29
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    @Hans Lundmark What is $\int_{-\infty}^{+\infty}1\,dx$?2012-03-01