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Consider the ODE:

[tex]x'(t) = f(t, x(t)) = \cos (t) - 2x(t) - x(t)^{2005}[/tex]

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

[tex]|x_a(t) - x_b(t)| < |a - b|\exp (-2t)[/tex]

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

[tex]\frac{x_a(t) - x_b(t)}{a - b} < \exp (-2t)[/tex]

Now, consider the time-t mapping [itex]\phi _t[/itex] defined by:

[tex]\phi _t (x_0) = x_{x_0}(t) = \phi (t, x_0)[/tex]

where [itex]\phi[/itex] is the flow associated with the ODE. The above inequality can be re-written:

[tex]\frac{\phi _t (a) - \phi _t (b)}{a - b} < \exp (-2t)[/tex]

By Mean Value Theorem:

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

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

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

[tex](\forall c)(\phi _t'(c) < \exp(-2t))[/tex]*

[tex]\phi _t'(c) = \frac{\partial \phi}{\partial \x_0}(t, c)[/tex]

My book gives that:

[tex]\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 )[/tex]

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

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

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

[tex]\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[/tex]

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

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

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

[tex]\frac{1}{\frac{\partial \phi}{\partial x_0}(s, c)} < 1 + \frac{2005}{2}\phi (s, c)^{2004}[/tex]

And it would suffice to show that:

[tex]\frac{1}{\frac{\partial \phi}{\partial x_0}(s, c)} < 1[/tex]*

[tex]\frac{\partial \phi}{\partial x_0}(s, c) > 1[/tex]

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

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

[tex]\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[/tex]

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

But 1 is positive, anything to the exponent 2004 is positive, and the parital derivative of [itex]\phi[/itex] 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

[tex]x'(t) = f(t, x(t)) = \cos (t) - 2x(t) - x(t)^{2005}[/tex]

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

[tex]|x_a(t) - x_b(t)| < |a - b|\exp (-2t)[/tex]

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

[tex]\frac{x_a(t) - x_b(t)}{a - b} < \exp (-2t)[/tex]

Now, consider the time-t mapping [itex]\phi _t[/itex] defined by:

[tex]\phi _t (x_0) = x_{x_0}(t) = \phi (t, x_0)[/tex]

where [itex]\phi[/itex] is the flow associated with the ODE. The above inequality can be re-written:

[tex]\frac{\phi _t (a) - \phi _t (b)}{a - b} < \exp (-2t)[/tex]

By Mean Value Theorem:

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

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

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

[tex](\forall c)(\phi _t'(c) < \exp(-2t))[/tex]*

[tex]\phi _t'(c) = \frac{\partial \phi}{\partial \x_0}(t, c)[/tex]

My book gives that:

[tex]\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 )[/tex]

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

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

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

[tex]\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[/tex]

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

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

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

[tex]\frac{1}{\frac{\partial \phi}{\partial x_0}(s, c)} < 1 + \frac{2005}{2}\phi (s, c)^{2004}[/tex]

And it would suffice to show that:

[tex]\frac{1}{\frac{\partial \phi}{\partial x_0}(s, c)} < 1[/tex]*

[tex]\frac{\partial \phi}{\partial x_0}(s, c) > 1[/tex]

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

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

[tex]\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[/tex]

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

But 1 is positive, anything to the exponent 2004 is positive, and the parital derivative of [itex]\phi[/itex] 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.
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