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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:) –  Klara Oct 25 '12 at 2:31
    
Much appreciated :-). –  copper.hat Oct 25 '12 at 2:33
    
@Klara Is a $C^1$ function Lipschitz? –  Pragabhava Oct 25 '12 at 2:36
    
It is locally Lipschitz, but not globally Lipschitz. I'm slightly sceptical that the solution can be extended to the entire line... –  copper.hat Oct 25 '12 at 2:36
    
@ 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. –  Klara Oct 25 '12 at 2:41
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1 Answer 1

up vote 1 down vote accepted

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. –  copper.hat Oct 25 '12 at 3:20
    
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. –  Klara Oct 25 '12 at 3:38
    
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). –  copper.hat Oct 25 '12 at 4:42
    
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. –  copper.hat Oct 25 '12 at 6:47
    
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? –  Klara Oct 25 '12 at 17:28
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