# Lipschitz ODE problem

1. Sep 4, 2007

### wu_weidong

1. The problem statement, all variables and given/known data
Suppose the function f(t,x) is locally Lipschitz on the domain G in R^2, that is, |f(t,x_1)-f(t,x_2)| <= k(t) |x_1 - x_2| for all (t, x_1),(t,x_2) in G. Define I = (a,b) and phi_1(t) and phi_2(t) are 2 continuous functions on I. Assume that, if (t, phi_i(t)) is in G, then the function f(t, phi_i(t)) is an integrable function on I for i = 1, 2. Suppose that for i = 1,2 and t in I,

phi_i(t) = phi_i(t_0) + INTEGRATE (from t_0 to t){ f(s, phi_i(t)) }ds + E_i(t)

and (phi_1(t_0) - phi_2(t_0)| <= d

for some constant d. Show that for all t in (t_0, b) we have

|phi_1(t) - phi_2(t)| <= d exp(INTEGRATE(from t_0 to t){ k(s) }ds) + E(t) + INTEGRATE(from t_0 to t){ E(s)k(s)exp[INTEGRATE(from t_0 to s) { k(r) } dr] }ds

where E(t) = |E_1(t)| + |E_2(t)|

3. The attempt at a solution

I managed to get d exp(INTEGRATE(from t_0 to t) k(s) ds) using triangle inequality and Gronwall's inequality, but I cannot seem to get the last 2 terms in the inequality.

Here's what I did:

|phi_1(t) - phi_2(t)|
= |phi_1(t_0) + INTEGRATE (from t_0 to t){ f(s, phi_1(s)) }ds + E_1(t) - phi_2(t_0) - INTEGRATE (from t_0 to t){ f(s, phi_2(s)) }ds - E_2(t)|
= |phi_1(t_0) - phi_2(t_0) + INTEGRATE (from t_0 to t){ f(s, phi_1(s)) }ds - INTEGRATE (from t_0 to t){ f(s, phi_2(s)) }ds + E_1(t) - E_2(t)|
<= |phi_1(t_0) - phi_2(t_0)| + |INTEGRATE (from t_0 to t){ f(s, phi_1(s)) }ds - INTEGRATE (from t_0 to t){ f(s, phi_2(s)) }ds + E_1(t) - E_2(t)|
<= d + |E_1(t) - E_2(t)| + |INTEGRATE (from t_0 to t){ f(s, phi_1(s)) - f(s, phi_2(s)) }ds|
<= d + |E_1(t) - E_2(t)| + INTEGRATE (from t_0 to t){ |f(s, phi_1(s)) - f(s, phi_2(s))| }ds
<= d + |E_1(t) - E_2(t)| + INTEGRATE (from t_0 to t){ k(s)|phi_1(s) - phi_2(s)| } ds
<= (d + |E_1(t) - E_2(t)|) exp(INTEGRATE (from t_0 to t){ k(s) } ds)
= d exp(INTEGRATE (from t_0 to t){ k(s) } ds) + (|E_1(t) - E_2(t)|) exp(INTEGRATE (from t_0 to t){ k(s) } ds)

This is where I got stuck.

Thank you.

Regards,
Rayne

2. Sep 6, 2007

### EnumaElish

It would be much easier on the eye if you can re-type your post using [ t e x ] ... [ / t e x ] (single spaced).

If you have not used it before, click on the image to get a very basic starter course: $$\TeX$$

3. Sep 6, 2007

### wu_weidong

I rewrote the question in tex.

Suppose the function f(t,x) is locally Lipschitz on the domain $$G \subset \mathbb{R}^2$$, that is, $$|f(t,x_1)-f(t,x_2)| \leq k(t) |x_1 - x_2|$$ for all $$(t, x_1),(t,x_2) \in G$$. Define I = (a,b) and $$\phi_1(t)$$ and $$\phi_2(t)$$ are 2 continuous functions on I. Assume that, if $$(t, \phi_i(t)) \in G$$, then the function $$f(t, \phi_i(t))$$ is an integrable function on I for i = 1, 2. Suppose that for i = 1,2 and $$t \in I$$,

$$\phi_i(t) = \phi_i(t_0) + \int^t_{t_0} f(s, \phi_i(s))\,ds + E_i(t)$$

and $$|\phi_1(t_0) - \phi_2(t_0)| \leq \delta$$

for some constant $$\delta$$. Show that for all $$t \in (t_0, b)$$ we have

$$|\phi_1(t) - \phi_2(t)| \leq \delta e^{\int^t_{t_0} k(s) \,ds} + E(t) + \int^t_{t_0} E(s) k(s) e^{\int^s_{t_0} k(r) \,dr} \,ds$$

where $$E(t) = |E_1(t)| + |E_2(t)|$$

I managed to get $$\delta e^{\int^t_{t_0} k(s) \,ds}$$ using triangle inequality and Gronwall's inequality, but I cannot seem to get the last 2 terms in the inequality.

Here's what I did:

$$|\phi_1(t) - \phi_2(t)|$$
$$= |\phi_1(t_0) + \int^t_{t_0} f(s, \phi_1(s)) \,ds + E_1(t) - \phi_2(t_0) - \int^t_{t_0} f(s, \phi_2(s)) \,ds - E_2(t)|$$
$$= |\phi_1(t_0) - \phi_2(t_0) + \int^t_{t_0} f(s, \phi_1(s)) \,ds - \int^t_{t_0} f(s, \phi_2(s)) \,ds + E_1(t) - E_2(t)|$$
$$\leq |\phi_1(t_0) - \phi_2(t_0)| + |\int^t_{t_0} f(s, \phi_1(s)) }ds - \int^t_{t_0} f(s, \phi_2(s)) \,ds + E_1(t) - E_2(t)|$$
$$\leq \delta + |E_1(t) - E_2(t)| + |\int^t_{t_0} f(s, \phi_1(s)) - f(s, \phi_2(s)) \,ds|$$
$$\leq \delta + |E_1(t) + E_2(t)| + \int^t_{t_0} |f(s, \phi_1(s)) - f(s, \phi_2(s))| \,ds$$
$$\leq \delta + |E_1(t)| + |E_2(t)| + \int^t_{t_0} k(s)|\phi_1(s) - \phi_2(s)| \, ds$$
$$\leq (\delta + E(t)) e^{\int^t_{t_0} k(s)\, ds}$$
$$= \delta e^{\int^t_{t_0} k(s) \, ds} + E(t) e^{\int^t_{t_0} k(s) \, ds$$

This is where I got stuck.

4. Sep 8, 2007

### EnumaElish

How do you go from
$$\delta + |E_1(t)| + |E_2(t)| + \int^t_{t_0} k(s)|\phi_1(s) - \phi_2(s)| \, ds$$
to the next step?

5. Sep 8, 2007

### wu_weidong

$$|E_1(t)| + |E_2(t)| = E(t)$$

Taking Gronwall's inequality, that is,
$$\phi(t) \leq a \int^t_{t_0} \psi(s) \phi(s) \, ds + M,\, \, \, t_0 \leq t \leq t_0 + T$$

gives for $$t_0 \leq t \leq t_0 + T$$

$$\phi(t) \leq M e^{a \int^t_{t_0} \psi(s) \, ds}$$

Therefore, taking $$\delta + E(t)$$ as M, $$|\phi_1(s) - \phi_2(s)|$$ as $$\phi(t)$$, and $$k(s)$$ as $$\psi(s)$$, I get

$$(\delta + E(t)) e^{\int^t_{t_0} k(s)\, ds}$$

6. Sep 9, 2007

### EnumaElish

Is $a \int^t_{t_0} \psi(s) \phi(s) \, ds + M$ equal to $M e^{a \int^t_{t_0} \psi(s) \, ds$ ? My basics is a little rusty.

Last edited: Sep 9, 2007
7. Sep 9, 2007

### wu_weidong

Yes, in the Gronwall's inequality.