# Estimating Distance Between Solutions to ODE

• AKG
In summary, the author shows how to solve an ODE using the mean value theorem. They use the mean value theorem to prove that two solutions cannot cross each other. They also use the mean value theorem to prove that the integrand is positive. However, the middle line is unnecessary and incorrect anyways.
AKG
Homework Helper
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:
AKG said:
By Mean Value Theorem:

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

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

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

The third line is the contrapositive of the first.

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

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

Last edited by a moderator:
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?

## 1. How do you estimate the distance between solutions to ODE?

The distance between solutions to ODE can be estimated by calculating the difference between the solutions at different points in time. This can be done using various numerical methods such as Euler's method or Runge-Kutta method.

## 2. What is the importance of estimating the distance between solutions to ODE?

Estimating the distance between solutions to ODE is important in validating the accuracy of the numerical methods used to solve the ODE. It helps in determining the level of error in the solutions and can also provide insight into the stability of the numerical method.

## 3. Can the distance between solutions to ODE be calculated analytically?

In most cases, the distance between solutions to ODE cannot be calculated analytically as it involves solving the ODE itself. However, for simple ODEs, it is possible to calculate the distance between solutions using mathematical equations.

## 4. How does the choice of numerical method affect the estimated distance between solutions to ODE?

The choice of numerical method can significantly affect the estimated distance between solutions to ODE. Some methods are more accurate than others and can provide a closer estimation of the true distance. The stability of the numerical method can also impact the accuracy of the estimated distance.

## 5. Is there a standard measure for the distance between solutions to ODE?

There is no standard measure for the distance between solutions to ODE as it depends on the specific problem being solved and the numerical method used. However, commonly used measures include the Euclidean distance, the maximum difference between solutions, and the root mean square difference.

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