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Solving Integral Equation by Converting to Differential Equations

2

$begingroup$

Consider the problem

$$phi(x) = x - int_0^x(x-s)phi(s),ds$$

How can we solve this by converting to a differential equation?

ordinary-differential-equations integral-equations integro-differential-equations

share|cite|improve this question

asked 5 hours ago

LightningStrikeLightningStrike

555

$endgroup$

add a comment | 

2

$begingroup$

Consider the problem

$$phi(x) = x - int_0^x(x-s)phi(s),ds$$

How can we solve this by converting to a differential equation?

ordinary-differential-equations integral-equations integro-differential-equations

share|cite|improve this question

asked 5 hours ago

LightningStrikeLightningStrike

555

$endgroup$

add a comment | 

$begingroup$

Consider the problem

$$phi(x) = x - int_0^x(x-s)phi(s),ds$$

How can we solve this by converting to a differential equation?

ordinary-differential-equations integral-equations integro-differential-equations

share|cite|improve this question

asked 5 hours ago

LightningStrikeLightningStrike

555

$endgroup$

share|cite|improve this question

asked 5 hours ago

LightningStrikeLightningStrike

555

share|cite|improve this question

asked 5 hours ago

LightningStrikeLightningStrike

555

asked 5 hours ago

LightningStrikeLightningStrike

555

asked 5 hours ago

LightningStrikeLightningStrike

555

2 Answers
2

active

oldest

votes

4

$begingroup$

We have that
$$phi(x)=x-xint_0^x phi(s) mathrmd s + int_0^x s phi(s)mathrmds$$
From this, we can see that $phi(0)=0$.
We can differentiate both sides and use the product rule and the FTC1 to get:
$$phi'(x)=1-int_0^x phi(s) mathrmds -x phi(x)+xphi(x)$$
$$phi'(x)=1-int_0^x phi(s) mathrmd s$$
From this, we can see that $phi'(0)=1$. We can differentiate it again:
$$phi''(x)=-phi(x)$$
Which is an alternative definition of the $sin$ function.

share|cite|improve this answer

edited 4 hours ago

answered 5 hours ago

BotondBotond

6,49331034

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In fact, the only valid solution for $phi(x)$ is $sin(x)$ because of the original equation.
  $endgroup$
  – Peter Foreman
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
  $endgroup$
  – Botond
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
  $endgroup$
  – LightningStrike
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrmds$?
  $endgroup$
  – Botond
  4 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Differentiating both sides using Leibniz rule :

$$phi '(x)=1-int_0^xphi (s)ds$$

Differentiate again:

$$phi ''(x)=-phi (x)$$

share|cite|improve this answer

edited 4 hours ago

answered 5 hours ago

logologo

1048

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
  $endgroup$
  – Botond
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  may be you are right...but this is a common technique in an introductory course of integral equations.
  $endgroup$
  – logo
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
  $endgroup$
  – Botond
  3 hours ago
  
  

  
  
  
  

add a comment | 

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4

$begingroup$

We have that
$$phi(x)=x-xint_0^x phi(s) mathrmd s + int_0^x s phi(s)mathrmds$$
From this, we can see that $phi(0)=0$.
We can differentiate both sides and use the product rule and the FTC1 to get:
$$phi'(x)=1-int_0^x phi(s) mathrmds -x phi(x)+xphi(x)$$
$$phi'(x)=1-int_0^x phi(s) mathrmd s$$
From this, we can see that $phi'(0)=1$. We can differentiate it again:
$$phi''(x)=-phi(x)$$
Which is an alternative definition of the $sin$ function.

share|cite|improve this answer

edited 4 hours ago

answered 5 hours ago

BotondBotond

6,49331034

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In fact, the only valid solution for $phi(x)$ is $sin(x)$ because of the original equation.
  $endgroup$
  – Peter Foreman
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
  $endgroup$
  – Botond
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
  $endgroup$
  – LightningStrike
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrmds$?
  $endgroup$
  – Botond
  4 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

We have that
$$phi(x)=x-xint_0^x phi(s) mathrmd s + int_0^x s phi(s)mathrmds$$
From this, we can see that $phi(0)=0$.
We can differentiate both sides and use the product rule and the FTC1 to get:
$$phi'(x)=1-int_0^x phi(s) mathrmds -x phi(x)+xphi(x)$$
$$phi'(x)=1-int_0^x phi(s) mathrmd s$$
From this, we can see that $phi'(0)=1$. We can differentiate it again:
$$phi''(x)=-phi(x)$$
Which is an alternative definition of the $sin$ function.

share|cite|improve this answer

edited 4 hours ago

answered 5 hours ago

BotondBotond

6,49331034

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In fact, the only valid solution for $phi(x)$ is $sin(x)$ because of the original equation.
  $endgroup$
  – Peter Foreman
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
  $endgroup$
  – Botond
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
  $endgroup$
  – LightningStrike
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrmds$?
  $endgroup$
  – Botond
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

We have that
$$phi(x)=x-xint_0^x phi(s) mathrmd s + int_0^x s phi(s)mathrmds$$
From this, we can see that $phi(0)=0$.
We can differentiate both sides and use the product rule and the FTC1 to get:
$$phi'(x)=1-int_0^x phi(s) mathrmds -x phi(x)+xphi(x)$$
$$phi'(x)=1-int_0^x phi(s) mathrmd s$$
From this, we can see that $phi'(0)=1$. We can differentiate it again:
$$phi''(x)=-phi(x)$$
Which is an alternative definition of the $sin$ function.

share|cite|improve this answer

edited 4 hours ago

answered 5 hours ago

BotondBotond

6,49331034

$endgroup$

share|cite|improve this answer

edited 4 hours ago

answered 5 hours ago

BotondBotond

6,49331034

answered 5 hours ago

BotondBotond

6,49331034

answered 5 hours ago

BotondBotond

6,49331034

1

$begingroup$

Differentiating both sides using Leibniz rule :

$$phi '(x)=1-int_0^xphi (s)ds$$

Differentiate again:

$$phi ''(x)=-phi (x)$$

share|cite|improve this answer

edited 4 hours ago

answered 5 hours ago

logologo

1048

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
  $endgroup$
  – Botond
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  may be you are right...but this is a common technique in an introductory course of integral equations.
  $endgroup$
  – logo
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
  $endgroup$
  – Botond
  3 hours ago
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Differentiating both sides using Leibniz rule :

$$phi '(x)=1-int_0^xphi (s)ds$$

Differentiate again:

$$phi ''(x)=-phi (x)$$

share|cite|improve this answer

edited 4 hours ago

answered 5 hours ago

logologo

1048

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
  $endgroup$
  – Botond
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  may be you are right...but this is a common technique in an introductory course of integral equations.
  $endgroup$
  – logo
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
  $endgroup$
  – Botond
  3 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Differentiating both sides using Leibniz rule :

$$phi '(x)=1-int_0^xphi (s)ds$$

Differentiate again:

$$phi ''(x)=-phi (x)$$

share|cite|improve this answer

edited 4 hours ago

answered 5 hours ago

logologo

1048

$endgroup$

share|cite|improve this answer

edited 4 hours ago

answered 5 hours ago

logologo

1048

answered 5 hours ago

logologo

1048

answered 5 hours ago

logologo

1048