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The problem is:

Let $x(t)$ a solution of $$x'=-x(t^2-x^2),$$ so that $|x(t_0)|\lt |t_0|$. Show that for all $|t|\gt |t_0|$, $|x(t)|\lt |t|$.

Suppose that $t_0\geq 0$. Consider


Then $(t_0,x(t_0))$ is over the red line. I don't see why: $x$ solution of the above ODE implies that the graph of $x$ stay between the blue lines to the right of the red line and to the left of the orange line.


The attempt becomes an answer...

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The way the problem is currently stated, it is not true. I numerically solved the equation using $x(0)=.75$, and if we let $t_0=-1$, $|x(-1)|<1$. However, $|x(1)|>1$. – robjohn Apr 25 '12 at 23:19
@robjohn Thanks. I just copy the problem as is in the notes of the course. By now, I'll just skip this exercise. – leo Apr 25 '12 at 23:34
If you take the problem as stated and restrict to either $t\ge0$ or $t\le0$, then it is true. It is only the correlation between $x(t_0)$ and $x(-t_0)$ that is bad. Are you studying flow? – robjohn Apr 25 '12 at 23:41
@robjohn, I'm studying ordinary differential equations. If you refer to this kind of flow then I'm not. In my attempt to solve the exercise, I put the restriction $t_0\geq 0$. Is that a good approach to the problem? what do you think? – leo Apr 25 '12 at 23:50
that will give you a workable problem. However, you should also be able to solve the problem for $t_0<0$ as well. The problems are not exactly isomorphic. – robjohn Apr 26 '12 at 2:09
up vote 4 down vote accepted

For simplicity, multiply the equation by $2x$ and subtract $2t$ to get $$ (x^2-t^2)'=2x^2(x^2-t^2)-2t\tag{1} $$ First, let's deal with $t>0$.

$(1)$ says that if $x^2-t^2<0$, then we also have $(x^2-t^2)'<0$. Therefore, if $x(t_0)^2-t_0^2<0$, we have $x(t)^2-t^2<0$ for all $t\ge t_0$.

Next, let's deal with $t<0$.

$(1)$ says that if $x^2-t^2\ge0$, then we also have $(x^2-t^2)'\ge0$. Therefore, if $x(t)^2-t^2\ge0$, we have $x(t_0)^2-t_0^2\ge0$ for all $t_0\ge t$. The contrapositive says that for all $t\le t_0$, if $x(t_0)^2-t_0^2<0$, then we have $x(t)^2-t^2<0$.

A Graphical Explanation

Consider the direction field (aka slope field and flow diagram) for the solutions. That is, the vector field $\color{#0000c0}{\frac{\mathrm{d}}{\mathrm{d}t}(t,x)}$ which is tangent to all solutions $\color{#c00000}{(t,x)}$.


The important part to consider is the flow across the boundary $x^2=t^2$ between the regions where $\color{#00c000}{x^2< t^2}$ and $\color{#c000c0}{x^2>t^2}$. Notice that the on the boundary, the flow is horizontal since $x'=x(x^2-t^2)=0$.

$\hspace{4.2cm}$flow diagram

Thus, when $t<0$ (on the left), a solution can only move from the region where $\color{#00c000}{x^2< t^2}$ to the region where $\color{#c000c0}{x^2>t^2}$. When $t>0$ (on the right), a solution can only move from the region where $\color{#c000c0}{x^2>t^2}$ to the region where $\color{#00c000}{x^2< t^2}$.

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Neat! :)${}{}{}{}{}$ – t.b. Apr 27 '12 at 2:19

Consider first $t_0\geq 0$.

Suppose that the set $$C:=\{t\gt t_0:x(t)\geq t\}$$ is not empty, thus, since $C$ is bounded below by $t_0$, $$\tau=\inf C$$ is well defined. We can proof (by taking a sequence $C\ni t_n\to\tau$ and assuming continuity of $x$) that $\tau\in C$. In particular $\tau\gt t_0$ and then $$x(\tau)\geq\tau\gt 0,$$ then, there exist an $\alpha\gt 0$ such that $x$ is positive over $]\tau-\alpha,\tau+\alpha[\subset]t_0,\infty[$.

Then for all $t\in]\tau-\alpha,\tau[$, by the definition of $\tau$, $t\notin C$ (otherwise $t\in C$ with $t\lt\inf C$), so $0\lt x(t)\lt t$ and then $$x'(t)=x(t)(x(t)^2-t^2)\lt 0,$$ so $x$ is decreasing over $]\tau-\alpha,\tau[$.

Pick $t\in ]\tau-\alpha,\tau[$, then $$x(t)\geq x(\tau)\geq\tau\gt t,$$ thus $t\in C$, and that's contradictory.

Therefore $C=\emptyset$ and then for all $t\gt t_0$, $x(t)\lt t$.

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Hint: If $x(t_0) < |t_0|$ then, by the differential equation, $x'(t_0) = -x(t_0)(\textrm{something positive})$.

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If $x(t_0)=0$ then $x(t)=0$ for all $t\ge t_0$. Without loss of geneality, we assume $x(t_0)\ne0$.

If $0<x(t_0)<t_0$, you want to show that $0<x(t)\le t$ for all $t\ge t_0$. Now, $x'(t_0)=-x(t_0)(t_0^2-x(t_0)^2)<0$. By uniqueness of solutions $x(t)>0$ for all $t\>t_0$. And since the right hand side of the equation is negative on the region $\{(t,x):0<x<t,t>t_0\}$, it is easy to see that $x$ is decreasing.

If $x(t_0)=t_0$ then $x'(t_0)=0$. Taking derivatives in both sides of the equation we get $x''(t_0)=-2\,t_0\,x(t_0)<0$. Thus $x'(t)<0$ on some interval $(t_0,t_0+\delta)$, $\delta>0$, $x(t)<t$ on that same interval, and the argument in the previos paragraph carries through.

I leave to you the case $-t_0\le x(t_0)<0$.

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I don't see why $x(t_0)=0$ implies $x(t)=0$. – leo Apr 25 '12 at 18:52
@leo: note that $$ \frac{\mathrm{d}}{\mathrm{d}t}\log|x|=x^2-t^2 $$ When $|x|$ is close to $0$, $\log|x|$ is close to $-\infty$, yet the slope of $\log|x|$ is finite. Even if the slope were positive, which it isn't, it would take a long time to climb to something not close to $-\infty$. The closer $x$ gets to $0$, the longer it would take. Unfortunately, the slope of $\log|x|$ is not positive. – robjohn Apr 25 '12 at 22:48
$x(t)=0$ is certainly a solution. Uniqueness of solution (the right hand side is $C^\infty$, so it is locally Lipschitz,) implies that if the solution vanishes at some point, it is identically equal to $0$. – Julián Aguirre Apr 26 '12 at 8:22
I see. Thanks. ${}$ – leo Apr 27 '12 at 3:02

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