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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

img

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.

Attempt.

The attempt becomes an answer...

  • 0
    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.2012-04-26

4 Answers 4

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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 $\color{#c00000}{(t,x)}$.

$\hspace{3cm}$solution

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! :)${}{}{}{}{}$2012-04-27
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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, you want to show that 0 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):0t_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) 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 see. Thanks. ${}$2012-04-27