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Let $x'=(x-1)(x-4)(x+5)$ and $x(0)=3$ be an Initial Value Problem. Prove the existence and uniqueness of the solution and that the solution will be trapped in between $x=1$ and $x=4$ lines and that it can be extended in the entire line.

Any help is appreciated.

Thank you, Klara

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    @ copper I just did:)2012-10-25
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    Much appreciated :-).2012-10-25
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    @Klara Is a $C^1$ function Lipschitz?2012-10-25
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    It is locally Lipschitz, but not globally Lipschitz. I'm slightly sceptical that the solution can be extended to the entire line...2012-10-25
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    @ copper My professor blurted out things about this problem, he said to extend it on the entire line and I wrote bits and pieces of his intentions.2012-10-25
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    @clara Why don't you draw the _phase line_?2012-10-25
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    @ Pragabhava Yes, I think it is Locally Lischitz continuous.2012-10-25
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    @ Pragabhava How do you draw?2012-10-25
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    @Klara Well, if $-\infty < x < -5$ what is the sign of $x'$. What does that tell you of the solution? What if $-5 < x < 1$? And if $1 < x < 4$?2012-10-25
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    @ Pragabhava Well I know my solution starts in between lines $x=1$ and $x=4$ and there $x'$<0 while $x"$ is <0 close to x=1 and $x">0$ for x>=3.2012-10-25
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    So my solution starts at x(0)=3 and decreases and it is concave up till 2.646( inflection point) and then it decreases with concave down. But how do I show it never touches x=1 line2012-10-25
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    @Klara If it crosses $x=1$, what does that say about uniqueness?2012-10-25
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    @ Pragabhava that's what I'm trying to figure out.2012-10-25

1 Answers 1

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Note: I misinterpreted the question as looking for a solution for all initial conditions, whereas only $x(0) = 3$ is of interest, so this answer is a bit of overkill.

Let $f(x) = (x-1)(x-4)(x+5)$. A plot of $f$ is given below.

enter image description here

$f$ has zeros at $\{-5,1,4\}$, is strictly positive on $(-5,1)\cup(4,\infty)$ and strictly negative on $(-\infty, -5)\cup (1,4)$. $f$ is smooth, so is locally Lipschitz on bounded intervals. Hence solutions exist locally and are unique and smooth. We have $x(t) = x(0)$ is the unique solution for initial conditions $x(0) \in \{-5,1,4\}$.

The state space is $\mathbb{R}$, hence if $x(0)$ lies between two zeros of $f$, it must remain between the two zeros. To see this, suppose $z_1 \leq x(0) \leq z_2$, where $z_1,z_2$ are zeros of $f$ and suppose for the sake of contradiction that $x(t_1)$ lies outside the interval $[z_1,z_2]$. Continuity implies that for some $t_0 \in [0,t_1)$, $x(t_0) \in \{z_1,z_2\}$. However uniqueness of solution would imply that $x(t) \in \{z_1,z_2\}$ for all $t \geq t_0$, which is a contradiction. Hence $x(t)$ remains in the interval $[z_1,z_2]$. Since a uniform Lipschitz constant applies to $f$ for $x \in [z_1,z_2]$, we see that there exists a unique solution defined for all $t \geq 0$ (and $t<0$, but that is not a concern here).

This establishes the existence of a solution starting from $x(0) \in [-5,4]$.

However, it turns out that if $x(0)$ lies outside this interval, the solution 'blows up' in finite time. To see this, we again proceed by contradiction. Suppose a solution $x(t)$ exists for all $t \geq 0$, and $x(0) \notin [-5,4]$.

Note that $x(t)$ is increasing if $x(0) > 4$, and decreasing if $x(0) < -5$. So, if $x(0) < -5$, then $x(t) \leq x(0)$ for all $t \geq 0$ and if $x(0) > 4$, then $x(t) \geq x(0)$ for all $t\geq 0$.

Using partial fractions, we can rewrite the differential equation as $(\frac{1}{x-5} - \frac{3}{x-1}+\frac{2}{x-4}) \dot{x} = 54$. This can be written as (cf. separation of variables) $\frac{d}{dt} (\ln \phi(x)) = 54$ where $\phi(x) = \frac{(x+5)(x-4)^2}{(x-1)^3}$.

Note that $\phi(-5) = \phi(4) = 0$, and $\lim_{|x|\to \infty} \phi(x) = 1$. A tedious analysis shows that $\phi'(x) = \frac{54(x-4)}{(x-1)^4}$. It follows that $\phi(x) \in (0,1)$ on $(-\infty, -5) \cup (4,\infty)$. It also follows that $t \to \phi(x(t))$ is increasing.

We can integrate the differential equation to get $\phi(x(t)) = \phi(x(0)) e^{54 t}$, which contradicts the boundedness of $\phi$ on $(-\infty, -5) \cup (4,\infty)$.

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    Oops, sorry, got to go for a while, I will finish later.2012-10-25
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    Thank you copper. I am trying to understand the contradiction here. Could you please explain it in terms of mean value theorem. I never took analysis and now I'm taking advanced Calc.part I.2012-10-25
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    The contradiction comes from the intermediate value theorem which says that if $t \to x(t)$ is a continuous real-valued function on some interval $[a,b]$, then $x$ must take all values between $x(a)$ and $x(b)$ ($[x(a),x(b)]$, or $[x(b),x(a)]$, depending on sign).2012-10-25
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    Oops, I thought the question was a little broader, so the answer above is overkill. It may still be of interest, so I didn't prune it.2012-10-25
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    now what you stated is that if a function is continuous in a closed bounded interval [a,b] then the function f is bounded on [a,b] meaning between[f(a), f(b)], but this does not look like the MVT to me?2012-10-25
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    I was thinking, since x' is bounded locally then x is Lipschitz continuous and but if a and b are equilibrium points f(a)=f(b)=0 then where is my function bounded on?2012-10-25
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6240/discussion-between-klara-and-copper-hat)2012-10-25