Estimating Distance Between Solutions to ODE

[AKG]

Consider the ODE:

$$x'(t) = f(t, x(t)) = \cos (t) - 2x(t) - x(t)^{2005}$$

For any two solutions $x_a,\ x_b$ with initial values $x_a(0) = a,\ x_b(0) = b$, prove that for any t > 0, the following inequality holds:

$$|x_a(t) - x_b(t)| < |a - b|\exp (-2t)$$

Since the solutions can't cross each other, we know that if a > b, then $x_a(t) > x_b(t)$ for all t, so we can stipulate that a > b and remove the absolute value bars. We have:

$$\frac{x_a(t) - x_b(t)}{a - b} < \exp (-2t)$$

Now, consider the time-t mapping $\phi _t$ defined by:

$$\phi _t (x_0) = x_{x_0}(t) = \phi (t, x_0)$$

where $\phi$ is the flow associated with the ODE. The above inequality can be re-written:

$$\frac{\phi _t (a) - \phi _t (b)}{a - b} < \exp (-2t)$$

By Mean Value Theorem:

$$\frac{\phi _t (a) - \phi _t (b)}{a - b} \geq \exp (-2t) \Rightarrow (\exists c \in (a, b))(\phi _t '(c) \geq \exp (-2t))$$

$$(\forall c \in (a,b))(\phi _t '(c) < \exp(-2t)) \Rightarrow \frac{\phi _t (a) - \phi _t (b)}{a - b} < \exp (-2t)$$

Since this must hold for arbitrary a, b, we it suffices to show that:

$$(\forall c)(\phi _t'(c) < \exp(-2t))$$*

$$\phi _t'(c) = \frac{\partial \phi}{\partial \x_0}(t, c)$$

My book gives that:

$$\frac{\partial \phi}{\partial x_0}(t, c) = \exp \left (\int _0 ^t \frac{\partial f}{\partial x_0}(s, \phi (s, c))\, ds\right )$$

I want to show that this is less than exp(-2t), which means:

$$\int _0 ^t \frac{\partial f}{\partial x_0}(s, \phi (s, c))\, ds < -2t$$

$$\int _0 ^t \frac{\partial}{\partial x_0}\left (\cos (s) - 2\phi (s, c) - \phi (s, c)^{2005}\right )\, ds < -2t$$

$$\int _0 ^t -2\frac{\partial \phi}{\partial x_0} (s, c) - 2005\phi (s, c)^{2004}\frac{\partial \phi}{\partial x_0}(s, c)\, ds < \int _0 ^t -2\, ds$$

$$\int _0 ^t \frac{\partial \phi}{\partial x_0}(s, c)\left (1 + \frac{2005}{2}\phi (s, c)^{2004}\right ) - 1\, ds > 0$$

It would be enough to show that the integrand is positive:

$$\frac{\partial \phi}{\partial x_0}(s, c)\left (1 + \frac{2005}{2}\phi (s, c)^{2004}\right ) - 1 > 0$$*

$$\frac{1}{\frac{\partial \phi}{\partial x_0}(s, c)} < 1 + \frac{2005}{2}\phi (s, c)^{2004}$$

And it would suffice to show that:

$$\frac{1}{\frac{\partial \phi}{\partial x_0}(s, c)} < 1$$*

$$\frac{\partial \phi}{\partial x_0}(s, c) > 1$$

$$\exp \left (\int _0 ^s \frac{\partial f}{\partial x_0}(r, \phi (r, c))\, dr\right ) > \exp (0)$$

$$\int _0 ^s \frac{\partial f}{\partial x_0}(r, \phi (r, c))\, dr > 0$$

$$\int _0 ^s -2\frac{\partial \phi}{\partial x_0} (r, c) - 2005\phi (r, c)^{2004}\frac{\partial \phi}{\partial x_0}(r, c)\, dr > 0$$

$$\int _0 ^s \frac{\partial \phi}{\partial x_0} (r, c)\left (1 + \frac{2005}{2}\phi (r, c)^{2004}\right )\, dr < 0$$

But 1 is positive, anything to the exponent 2004 is positive, and the parital derivative of $\phi$ is positive, otherwise two solutions would cross. So I cannot show this last inequality, but the steps indicated by * were not necessary step, just sufficient, so those steps are where I went wrong. But how do I correct it? I can't seem to figure the problem out. My guess is that the whole approach above is wrong, so how do I do this problem, especially in general? Thanks.

 

[Hippo]

By Mean Value Theorem:

$$\frac{\phi _t (a) - \phi _t (b)}{a - b} \geq \exp (-2t) \Rightarrow (\exists c \in (a, b))(\phi _t '(c) \geq \exp (-2t))$$

(([/color] $$~(\exists c \in (a, b))(\phi _t '(c) \geq \exp (-2t)) \Rightarrow \frac{\phi _t (a) - \phi _t (b)}{a - b} < \exp (-2t)$$ ))[/color]

$$(\forall c \in (a,b))(\phi _t '(c) < \exp(-2t)) \Rightarrow \frac{\phi _t (a) - \phi _t (b)}{a - b} < \exp (-2t)$$

The third line is the contrapositive of the first.

However, the middle line (in big red quotation marks) is unnecessary and incorrect anyways.

 

[AKG]

The problem with the second line was that the "~" symbol didn't show up. Anyways, any ideas as to how to do the prolbem?

 

[Hippo]

I'm thinking about it.

 

[AKG]

When I asked someone else this, they told me that, by mean value theorem:

$$\exists !u\ \mbox{such that} \ |x_1(a, t) - x_2(b, t)| = \frac{dx}{da} (u(t), t) |a - b|$$

Although I didn't find out what $x_1,\ x_2,\ x$ were. They said that, using this, estimate u to get the answer. This doesn't make any sense to me, but maybe this gives you an idea as to how to do this?