Learn Integral Equations for Self-Teaching | ODE Graduate at 19

  • Context: Graduate 
  • Thread starter Thread starter epr2008
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary

Discussion Overview

The discussion revolves around self-teaching integral equations, particularly focusing on a lemma related to continuous functions and their integrals. Participants explore the connections between integral equations and differential equations, as well as the challenges of understanding the material presented in a textbook.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant shares their experience of self-studying integral equations and expresses difficulty with the explanations in the textbook by Peter Collins, particularly regarding a lemma involving continuous functions.
  • The participant attempts to prove the lemma using integration by parts, leading to a differential equation with variable coefficients, and questions whether this approach is valid.
  • Another participant suggests that the original integral's domain in the (x',t) plane should be considered, indicating that a change in the order of integration could clarify the problem.
  • A different participant points out that the original statement does not imply any derivatives of the function f, questioning the relevance of the first participant's approach.
  • The first participant acknowledges their initial misunderstanding but seeks validation for their reasoning regarding the relationship between integral and differential equations.
  • There is a discussion about the properties of integrable functions, with one participant asserting that being integrable does not imply differentiability on the same interval.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the first participant's approach to the lemma and the relationship between integrability and differentiability. There is no consensus on these points, indicating ongoing debate and exploration of the concepts.

Contextual Notes

Participants highlight limitations in understanding the material, particularly regarding assumptions about the properties of functions involved in the discussion. The exploration of the lemma and its implications remains unresolved.

Who May Find This Useful

This discussion may be useful for individuals interested in self-teaching integral equations, exploring the connections between integral and differential equations, or those seeking to understand the nuances of mathematical proofs and reasoning in this area.

epr2008
Messages
44
Reaction score
0
I'm 19 and i just finished ode at a community college. there are no more math classes for me to take there so I've been planning teaching myself this semester. I was good at calc and ode's so i read about integral equations and figured i would start. I am using a book by Peter Collins but he's not very good at explaining the in between steps. he always says "i'll leave this for the reader to prove." well some of these are really not easy to see... so, I started off trying to prove a lemma he introduced in the first chapter and out of nowhere i ended up with something that i think is much more enlightening i just don't know if its right, so i was looking for some input...

The lemma to be proved:
Suppose that f:[a,b][tex]\rightarrow[/tex][tex]\Re[/tex] is continuous. Then
[tex]\int\limits_a^x {\int\limits_a^{x'} {f(t)dtdx' = \int\limits_a^x {(x - t)f(t)dt} } }[/tex]

Ok so at first before i figured out what he was talking about i started out like the book did with the rhs of the equation. and i thought about it and i said well what is the integral of that? ok integration by parts...

[tex]\begin{array}{l}<br /> F(x) = \int\limits_a^x {(x - t)f(t)dt} \\ <br /> u = f(t) \\ <br /> du = f'(t)dt \\ <br /> dv = (x - t)dt \\ <br /> v = xt - {\textstyle{1 \over 2}}{t^2} \\ <br /> F(x) = (xt - {\textstyle{1 \over 2}}{t^2})f(t) - \int {(xt - {\textstyle{1 \over 2}}{t^2})f'(t)dt} ]\nolimits_a^x \\ <br /> \end{array}[/tex]

Now if f is a continuously differentiable function in t then integration by parts will eventually yield a differential equation with variable coefficients of the form
[tex]F(x) = \sum\limits_0^\infty {{{( - 1)}^n}} {f^n}(t)\frac{{(x + t)n + 2x + t}}{{(n + 1)(n + 2)}}{t^{n + 1}} ]\nolimits_a^x[/tex]

Now obviously had i chosen u=(x-t) and dv=f(t)dt then i would have come up with another integral but with no derivatives in the equation right? Is this somewhat along the lines of the basic theory of differential and integral equations or am i just wrong?
 
Physics news on Phys.org
You are completely on the wrong track. What you need to do is look at the domain of integration in the (x',t) plane. The original integral is t first then x'. Figure out what the domain would look like if you did x' first then t. Once you do that, the answer you need immediately pops out.
 
Oh i know that for that particular theorem i was on the wrong track. I figured that out problem out. i was just wondering if that was a reasonable argument?
 
There was nothing in the original statement which said that f had any derivatives. Even if it was analytic, I don't see how your approach gets you anywhere.
 
i understand it doesn't really get me anywhere but what i was trying to do is get a feel for where integral equations come from. so i tried to relate it to a differential equation.

and yes it doesn't say anything about derivatives i was just looking at it from a different perspective... if a function is integrable on an interval shouldn't it also be differentiable on that same interval?
 
if a function is integrable on an interval shouldn't it also be differentiable on that same interval?
absolutely not!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K