Estimating Distance Between Solutions to ODE

1. Apr 26, 2005

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.

Last edited: Apr 26, 2005
2. Apr 26, 2005

Hippo

The third line is the contrapositive of the first.

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

3. Apr 26, 2005

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?

4. Apr 26, 2005

Hippo

Last edited by a moderator: Apr 26, 2005
5. Apr 26, 2005

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?