Ordinary Differential Equations - Existence/Uniqueness Proof

  • #1
215
11
I'm having some difficulty with a problem from Boyce & DiPrima's Elementary Differential Equations and Boundary Value Problems, 9th Edition. The problem comes from Section 2.8: The Existence and Uniqueness Theorem and is part of a collection of problems intended to show that the sequence ##{\phi_n(t)}## converges, where

$$\phi_{n+1}(t) = \int_0^t f[s,\phi_n(s)] ds$$

with ##\phi_0(t) = 0##. The problem requires the following two results:

1) If ##\phi_{n-1}(t)## and ##\phi_n(t)## are members of the sequence ##{\phi_n(t)}## then
$$|f[t,\phi_n(t)]-f[t,\phi_{n-1}(t)| \leq K|\phi_n(t) - \phi_{n-1}(t)|$$
where ##K## is chosen to be the maximum value of ##\frac{∂f}{∂y}## in the region ##D##.

2) If ##|t|<h##, then
$$|\phi_1(t)| \leq M|t|$$
where ##M## is chosen so that ##|f(t,y)|\leq M## for ##(t,y)## in ##D##.

The problem is as follows: "Use the results of Problem 16 and part (a) of Problem 17 [Results 1 and 2 above, respectively] to show that ##|\phi_2(t) - \phi_1(t)| \leq \frac{MK|t|^2}{2}##."

(The problem and results are found in Boyce & DiPrima's Elementary Differential Equations and Boundary Value Problems, 9th Edition; Problem 2.8.17(b); pg. 120.)


I've been struggling with this problem even though it seems like it should be fairly straightforward. From now on I'm going to write ##\phi_n(t) = \phi_n## for the sake of laziness. I've been playing around with a few different approaches - I tried writing out the left-hand side of my desired result explicitly using the definition of ##\phi_n## which gives me

$$|\phi_2 - \phi_1| = | \int_0^t f(s,\phi_1) ds - \int_0^t f(s,\phi_0) ds | = | \int_0^t f(s,\phi_1) - f(s,\phi_0) ds | $$

I got stuck here. I can almost apply Result 1 now, but not quite since I have my integrand inside the absolute value...and even if I could it didn't yield promising results. Then I thought of using the inequality ##|x-y| \leq | |x| - |y| |## which let me express ##|\phi_1| ## using Result 2 but I was again stuck with no way to write ##|\phi_2|##.

I noticed that the right-hand side of the first result looks very similar to the left-hand side of what I wish to show when you take ##n=2## but that didn't get me very far either:

$$|f[t,\phi_n(t)]-f[t,\phi_{n-1}(t)| \leq K|\phi_n(t) - \phi_{n-1}(t)|$$

So I could try to show

$$|f[t,\phi_n(t)]-f[t,\phi_{n-1}(t)| \leq \frac{MK^2|t|^2}{2}$$

But then I realized that this is only true if I assume that what I want to show is true and so this shouldn't be the approach I take.

Any guidance would be appreciated. I feel like this should be pretty straightforward, but I'm lost for ideas right now. Thanks in advance for any help!
 

Answers and Replies

  • #2
RUber
Homework Helper
1,687
344
Starting with what you have:
## | \phi_2 - \phi_1 | \leq \left| \int _0^t f(s, \phi_2) - f(s, \phi_1) ds \right| ##
You can move the absolute value inside the integral to get:
## \left| \int _0^t f(s, \phi_2) - f(s, \phi_1) ds \right| \leq \int _0^t \left| f(s, \phi_2) - f(s, \phi_1) \right| ds##
Then you can apply rule 1.
Expanding that out and doing the same steps should get you to something that looks like :
## \int_0^t | \phi_1 | ds ##
Which should look like the result you need.
 
  • #3
RUber
Homework Helper
1,687
344
Sorry, I used the wrong subscripts on the phis inside the integral...after you apply rule 1, use the fact that ##\phi_0= 0##, and then apply rule 2.
Just to be clear...when I see ## MK|t|^2 / 2 ##, I am looking for something that looks like ##\int_0^t MK|s| ds##.
 
  • #4
STEMucator
Homework Helper
2,075
140
I got stuck here. I can almost apply Result 1 now, but not quite since I have my integrand inside the absolute value...and even if I could it didn't yield promising results. Then I thought of using the inequality |x−y|≤||x|−|y|| which let me express |ϕ1| using Result 2 but I was again stuck with no way to write |ϕ2|.
I just want to mention the inequality ##|x - y| \leq \left| |x| - |y| \right|## is not true. The correct inequality would be ##|x - y| \geq \left| |x| - |y| \right|##.

You have correctly expanded the left hand side of what you want to show:

$$| \phi_2(t) - \phi_1(t) | = \left| \int_0^t f(s, \phi_1(s)) \space ds - \int_0^t f(s, \phi_0(s)) \space ds \right| = \left| \int_0^t f(s, \phi_1(s)) - f(s, \phi_0(s)) \space ds \right|$$

As RUber has stated, you can use the property ##\left| \int_a^b f(x) \space dx \right| \leq \int_a^b |f(x)| \space dx## to write:

$$\left| \int_0^t f(s, \phi_1(s)) - f(s, \phi_0(s)) \space ds \right| \leq \int_0^t \left| f(s, \phi_1(s)) - f(s, \phi_0(s)) \right| \space ds$$

After applying rule #1 and the triangle inequality, you are going to wind up with something like:

$$K \int_0^t \left| \phi_1(s) - \phi_0(s) \right| \space ds \leq K \int_0^t |\phi_1(s)| + |\phi_0(s)| \space ds$$

Then you need to apply rule #2 in terms of ##s##. The condition ##\phi_0(t) = 0## will also be important to use.
 
Last edited:

Related Threads on Ordinary Differential Equations - Existence/Uniqueness Proof

Replies
13
Views
4K
Replies
14
Views
4K
Replies
0
Views
7K
Replies
0
Views
3K
Replies
0
Views
1K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
3
Views
4K
Top