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A Standard Integral Equation

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3

$begingroup$

Consider the integral equation

$$phi(x) = x + lambdaint_0^1 phi(s),ds$$

Integrating with respect to $x$ from $x=0$ to $x=1$:

$$int_0^1 phi(x),dx = int_0^1x,dx + lambda int_0^1Big[int_0^1phi(s),dsBig],dx$$

which is equivalent to

$$int_0^1 phi(x),dx = frac{1}{2} + lambda int_0^1phi(s),ds$$

How can I go from here in order to solve the problem for the homogeneous case and find the corresponding characteristic values and associated rank?

linear-algebra integration matrix-equations

share|cite|improve this question

edited 5 hours ago

LightningStrike

asked 5 hours ago

LightningStrikeLightningStrike

455

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What is $lambda$? What do you mean by "solve the problem"? I don't see what "the problem" is supposed to mean. Of which object do you want to find the characteristic values and ranks? Have you checked your definition of $phi$?
  $endgroup$
  – James
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My apologies, $lambda$ is an arbitrary constant. In essence I want to obtain an expression of $phi(x)$ which does not contain a function of $s$, which the initial integral equation has.
  $endgroup$
  – LightningStrike
  5 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Consider the integral equation

$$phi(x) = x + lambdaint_0^1 phi(s),ds$$

Integrating with respect to $x$ from $x=0$ to $x=1$:

$$int_0^1 phi(x),dx = int_0^1x,dx + lambda int_0^1Big[int_0^1phi(s),dsBig],dx$$

which is equivalent to

$$int_0^1 phi(x),dx = frac{1}{2} + lambda int_0^1phi(s),ds$$

How can I go from here in order to solve the problem for the homogeneous case and find the corresponding characteristic values and associated rank?

linear-algebra integration matrix-equations

share|cite|improve this question

edited 5 hours ago

LightningStrike

asked 5 hours ago

LightningStrikeLightningStrike

455

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What is $lambda$? What do you mean by "solve the problem"? I don't see what "the problem" is supposed to mean. Of which object do you want to find the characteristic values and ranks? Have you checked your definition of $phi$?
  $endgroup$
  – James
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My apologies, $lambda$ is an arbitrary constant. In essence I want to obtain an expression of $phi(x)$ which does not contain a function of $s$, which the initial integral equation has.
  $endgroup$
  – LightningStrike
  5 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Consider the integral equation

$$phi(x) = x + lambdaint_0^1 phi(s),ds$$

Integrating with respect to $x$ from $x=0$ to $x=1$:

$$int_0^1 phi(x),dx = int_0^1x,dx + lambda int_0^1Big[int_0^1phi(s),dsBig],dx$$

which is equivalent to

$$int_0^1 phi(x),dx = frac{1}{2} + lambda int_0^1phi(s),ds$$

How can I go from here in order to solve the problem for the homogeneous case and find the corresponding characteristic values and associated rank?

linear-algebra integration matrix-equations

share|cite|improve this question

edited 5 hours ago

LightningStrike

asked 5 hours ago

LightningStrikeLightningStrike

455

$endgroup$

share|cite|improve this question

edited 5 hours ago

LightningStrike

asked 5 hours ago

LightningStrikeLightningStrike

455

share|cite|improve this question

edited 5 hours ago

LightningStrike

asked 5 hours ago

LightningStrikeLightningStrike

455

asked 5 hours ago

LightningStrikeLightningStrike

455

asked 5 hours ago

LightningStrikeLightningStrike

455

4 Answers
4

active

oldest

votes

4

$begingroup$

Relabelling the dummy variable $xmapsto s$ on the LHS of your final equation, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac1{2(1-lambda)}$$

Thus $$phi(x)=x+fraclambda{2(1-lambda)}$$

share|cite|improve this answer

answered 5 hours ago

John DoeJohn Doe

11.3k11239

$endgroup$

add a comment | 

2

$begingroup$

Note $int_{0}^{1}{phi(s)ds}$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
Putting into FE yields:

$$x+alambda=x+lambdaint_{0}^{1}{(s+alambda )ds }iff alambda=lambdabig(frac{1}{2}+lambda abig)$$
If $lambda=0$ then $phi(x)=x$

if $lambdane 1$ $a=frac{1}{2}+lambda aiff ( 1-lambda)a=frac{1}{2}iff a=frac{1}{2-2lambda}$ and then $phi(x)=x+frac{lambda}{2-2lambda}$

If $lambda=1$ there won’t besuch $phi$.

share|cite|improve this answer

answered 5 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,478211

$endgroup$

add a comment | 

0

$begingroup$

If you are after finding $phi(x)$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $[0, 1]$:
$$
phi(x) = sum_{n geq 0} a_{n} x^{n}.
$$
Substituting it into your equation, we get:
$$
sum_{n geq 0} a_{n} x^{n} = x + lambda sum_{n geq 0}{a_{n} over n+1}.
$$
Matching up the coefficients of the difference powers of $x$, we get:
$$
a_{n} = 0 quad mbox{ for } n geq 2,
$$
$$
a_{1} = 1,
$$
and
$$
a_{0} = lambda left(a_{0} + {a_{1} over 2}right).
$$
This gives a relationship between $a_{0}$ and $lambda$.

share|cite|improve this answer

answered 5 hours ago

avsavs

3,749514

$endgroup$

add a comment | 

0

$begingroup$

Note that since $lambdaint_0^1 phi(s),ds$ is a constant (with respect to $x$), then we can write$$phi(x)=x+a$$and by substitution we conclude that $$x+a=x+lambdaint _{0}^{1}x+adximplies\a=lambda({1over 2}+a)implies\a={lambdaover 2-2lambda}$$ and we obtain$$phi(x)=x+{lambdaover 2-2lambda}quad,quad lambdane 1$$The case $lambda=1$ leads to no solution.

share|cite|improve this answer

answered 5 hours ago

Mostafa AyazMostafa Ayaz

17.6k31039

$endgroup$

add a comment | 

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4

$begingroup$

Relabelling the dummy variable $xmapsto s$ on the LHS of your final equation, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac1{2(1-lambda)}$$

Thus $$phi(x)=x+fraclambda{2(1-lambda)}$$

share|cite|improve this answer

answered 5 hours ago

John DoeJohn Doe

11.3k11239

$endgroup$

add a comment | 

4

$begingroup$

Relabelling the dummy variable $xmapsto s$ on the LHS of your final equation, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac1{2(1-lambda)}$$

Thus $$phi(x)=x+fraclambda{2(1-lambda)}$$

share|cite|improve this answer

answered 5 hours ago

John DoeJohn Doe

11.3k11239

$endgroup$

add a comment | 

$begingroup$

Relabelling the dummy variable $xmapsto s$ on the LHS of your final equation, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac1{2(1-lambda)}$$

Thus $$phi(x)=x+fraclambda{2(1-lambda)}$$

share|cite|improve this answer

answered 5 hours ago

John DoeJohn Doe

11.3k11239

$endgroup$

share|cite|improve this answer

answered 5 hours ago

John DoeJohn Doe

11.3k11239

answered 5 hours ago

John DoeJohn Doe

11.3k11239

answered 5 hours ago

John DoeJohn Doe

11.3k11239

2

$begingroup$

Note $int_{0}^{1}{phi(s)ds}$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
Putting into FE yields:

$$x+alambda=x+lambdaint_{0}^{1}{(s+alambda )ds }iff alambda=lambdabig(frac{1}{2}+lambda abig)$$
If $lambda=0$ then $phi(x)=x$

if $lambdane 1$ $a=frac{1}{2}+lambda aiff ( 1-lambda)a=frac{1}{2}iff a=frac{1}{2-2lambda}$ and then $phi(x)=x+frac{lambda}{2-2lambda}$

If $lambda=1$ there won’t besuch $phi$.

share|cite|improve this answer

answered 5 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,478211

$endgroup$

add a comment | 

2

$begingroup$

Note $int_{0}^{1}{phi(s)ds}$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
Putting into FE yields:

$$x+alambda=x+lambdaint_{0}^{1}{(s+alambda )ds }iff alambda=lambdabig(frac{1}{2}+lambda abig)$$
If $lambda=0$ then $phi(x)=x$

if $lambdane 1$ $a=frac{1}{2}+lambda aiff ( 1-lambda)a=frac{1}{2}iff a=frac{1}{2-2lambda}$ and then $phi(x)=x+frac{lambda}{2-2lambda}$

If $lambda=1$ there won’t besuch $phi$.

share|cite|improve this answer

answered 5 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,478211

$endgroup$

add a comment | 

$begingroup$

Note $int_{0}^{1}{phi(s)ds}$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
Putting into FE yields:

$$x+alambda=x+lambdaint_{0}^{1}{(s+alambda )ds }iff alambda=lambdabig(frac{1}{2}+lambda abig)$$
If $lambda=0$ then $phi(x)=x$

if $lambdane 1$ $a=frac{1}{2}+lambda aiff ( 1-lambda)a=frac{1}{2}iff a=frac{1}{2-2lambda}$ and then $phi(x)=x+frac{lambda}{2-2lambda}$

If $lambda=1$ there won’t besuch $phi$.

share|cite|improve this answer

answered 5 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,478211

$endgroup$

share|cite|improve this answer

answered 5 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,478211

answered 5 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,478211

answered 5 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,478211

0

$begingroup$

If you are after finding $phi(x)$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $[0, 1]$:
$$
phi(x) = sum_{n geq 0} a_{n} x^{n}.
$$
Substituting it into your equation, we get:
$$
sum_{n geq 0} a_{n} x^{n} = x + lambda sum_{n geq 0}{a_{n} over n+1}.
$$
Matching up the coefficients of the difference powers of $x$, we get:
$$
a_{n} = 0 quad mbox{ for } n geq 2,
$$
$$
a_{1} = 1,
$$
and
$$
a_{0} = lambda left(a_{0} + {a_{1} over 2}right).
$$
This gives a relationship between $a_{0}$ and $lambda$.

share|cite|improve this answer

answered 5 hours ago

avsavs

3,749514

$endgroup$

add a comment | 

0

$begingroup$

If you are after finding $phi(x)$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $[0, 1]$:
$$
phi(x) = sum_{n geq 0} a_{n} x^{n}.
$$
Substituting it into your equation, we get:
$$
sum_{n geq 0} a_{n} x^{n} = x + lambda sum_{n geq 0}{a_{n} over n+1}.
$$
Matching up the coefficients of the difference powers of $x$, we get:
$$
a_{n} = 0 quad mbox{ for } n geq 2,
$$
$$
a_{1} = 1,
$$
and
$$
a_{0} = lambda left(a_{0} + {a_{1} over 2}right).
$$
This gives a relationship between $a_{0}$ and $lambda$.

share|cite|improve this answer

answered 5 hours ago

avsavs

3,749514

$endgroup$

add a comment | 

$begingroup$

If you are after finding $phi(x)$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $[0, 1]$:
$$
phi(x) = sum_{n geq 0} a_{n} x^{n}.
$$
Substituting it into your equation, we get:
$$
sum_{n geq 0} a_{n} x^{n} = x + lambda sum_{n geq 0}{a_{n} over n+1}.
$$
Matching up the coefficients of the difference powers of $x$, we get:
$$
a_{n} = 0 quad mbox{ for } n geq 2,
$$
$$
a_{1} = 1,
$$
and
$$
a_{0} = lambda left(a_{0} + {a_{1} over 2}right).
$$
This gives a relationship between $a_{0}$ and $lambda$.

share|cite|improve this answer

answered 5 hours ago

avsavs

3,749514

$endgroup$

share|cite|improve this answer

answered 5 hours ago

avsavs

3,749514

answered 5 hours ago

avsavs

3,749514

answered 5 hours ago

avsavs

3,749514

0

$begingroup$

Note that since $lambdaint_0^1 phi(s),ds$ is a constant (with respect to $x$), then we can write$$phi(x)=x+a$$and by substitution we conclude that $$x+a=x+lambdaint _{0}^{1}x+adximplies\a=lambda({1over 2}+a)implies\a={lambdaover 2-2lambda}$$ and we obtain$$phi(x)=x+{lambdaover 2-2lambda}quad,quad lambdane 1$$The case $lambda=1$ leads to no solution.

share|cite|improve this answer

answered 5 hours ago

Mostafa AyazMostafa Ayaz

17.6k31039

$endgroup$

add a comment | 

0

$begingroup$

Note that since $lambdaint_0^1 phi(s),ds$ is a constant (with respect to $x$), then we can write$$phi(x)=x+a$$and by substitution we conclude that $$x+a=x+lambdaint _{0}^{1}x+adximplies\a=lambda({1over 2}+a)implies\a={lambdaover 2-2lambda}$$ and we obtain$$phi(x)=x+{lambdaover 2-2lambda}quad,quad lambdane 1$$The case $lambda=1$ leads to no solution.

share|cite|improve this answer

answered 5 hours ago

Mostafa AyazMostafa Ayaz

17.6k31039

$endgroup$

add a comment | 

$begingroup$

Note that since $lambdaint_0^1 phi(s),ds$ is a constant (with respect to $x$), then we can write$$phi(x)=x+a$$and by substitution we conclude that $$x+a=x+lambdaint _{0}^{1}x+adximplies\a=lambda({1over 2}+a)implies\a={lambdaover 2-2lambda}$$ and we obtain$$phi(x)=x+{lambdaover 2-2lambda}quad,quad lambdane 1$$The case $lambda=1$ leads to no solution.

share|cite|improve this answer

answered 5 hours ago

Mostafa AyazMostafa Ayaz

17.6k31039

$endgroup$

share|cite|improve this answer

answered 5 hours ago

Mostafa AyazMostafa Ayaz

17.6k31039

answered 5 hours ago

Mostafa AyazMostafa Ayaz

17.6k31039

answered 5 hours ago

Mostafa AyazMostafa Ayaz

17.6k31039