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A Standard Integral Equation

Linear versus non-linear integral equationsUnderstanding why the roots of homogeneous difference equation must be eigenvaluesIntegral equation solution: $y(x) = 1 + lambdaintlimits_0^2cos(x-t) y(t) mathrmdt$Integrating with respect to time a double derivative $ddotphi + frac bmdotphi = fracFmr$Integral with Bessel FunctionsEigenvalue problem for integrals in multiple dimensionsStuck on finding the $2times 2$ system of differential equationsConversion of second order ode into integral equationSolving a dual integral equation involving a zeroth-order Bessel functionHow to find a basis of eigenvectors??

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

$$int_0^1 phi(x),dx = frac12 + lambda int_0^1phi(s),ds$$

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 8 hours ago

LightningStrike

asked 8 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
  8 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
  8 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

$$int_0^1 phi(x),dx = frac12 + lambda int_0^1phi(s),ds$$

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 8 hours ago

LightningStrike

asked 8 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
  8 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
  8 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

$$int_0^1 phi(x),dx = frac12 + lambda int_0^1phi(s),ds$$

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 8 hours ago

LightningStrike

asked 8 hours ago

LightningStrikeLightningStrike

455

$endgroup$

share|cite|improve this question

edited 8 hours ago

LightningStrike

asked 8 hours ago

LightningStrikeLightningStrike

455

share|cite|improve this question

edited 8 hours ago

LightningStrike

asked 8 hours ago

LightningStrikeLightningStrike

455

asked 8 hours ago

LightningStrikeLightningStrike

455

asked 8 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, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac12(1-lambda)$$

Thus $$phi(x)=x+fraclambda2(1-lambda)$$

share|cite|improve this answer

answered 8 hours ago

John DoeJohn Doe

11.3k11239

$endgroup$

add a comment | 

2

$begingroup$

Note $int_0^1phi(s)ds$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
Putting into FE yields:

$$x+alambda=x+lambdaint_0^1(s+alambda )ds iff alambda=lambdabig(frac12+lambda abig)$$
If $lambda=0$ then $phi(x)=x$

if $lambdane 1$ $a=frac12+lambda aiff ( 1-lambda)a=frac12iff a=frac12-2lambda$ and then $phi(x)=x+fraclambda2-2lambda$

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

share|cite|improve this answer

answered 8 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:
$$
sum_n geq 0 a_n x^n = x + lambda sum_n geq 0a_n over n+1.
$$
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
$$
a_0 = lambda left(a_0 + a_1 over 2right).
$$
This gives a relationship between $a_0$ and $lambda$.

share|cite|improve this answer

answered 8 hours ago

avsavs

3,779514

$endgroup$

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 $$x+a=x+lambdaint _0^1x+adximplies\a=lambda(1over 2+a)implies\a=lambdaover 2-2lambda$$ and we obtain$$phi(x)=x+lambdaover 2-2lambdaquad,quad lambdane 1$$The case $lambda=1$ leads to no solution.

share|cite|improve this answer

answered 8 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, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac12(1-lambda)$$

Thus $$phi(x)=x+fraclambda2(1-lambda)$$

share|cite|improve this answer

answered 8 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, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac12(1-lambda)$$

Thus $$phi(x)=x+fraclambda2(1-lambda)$$

share|cite|improve this answer

answered 8 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, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac12(1-lambda)$$

Thus $$phi(x)=x+fraclambda2(1-lambda)$$

share|cite|improve this answer

answered 8 hours ago

John DoeJohn Doe

11.3k11239

$endgroup$

share|cite|improve this answer

answered 8 hours ago

John DoeJohn Doe

11.3k11239

answered 8 hours ago

John DoeJohn Doe

11.3k11239

answered 8 hours ago

John DoeJohn Doe

11.3k11239

2

$begingroup$

Note $int_0^1phi(s)ds$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
Putting into FE yields:

$$x+alambda=x+lambdaint_0^1(s+alambda )ds iff alambda=lambdabig(frac12+lambda abig)$$
If $lambda=0$ then $phi(x)=x$

if $lambdane 1$ $a=frac12+lambda aiff ( 1-lambda)a=frac12iff a=frac12-2lambda$ and then $phi(x)=x+fraclambda2-2lambda$

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

share|cite|improve this answer

answered 8 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,478211

$endgroup$

add a comment | 

2

$begingroup$

Note $int_0^1phi(s)ds$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
Putting into FE yields:

$$x+alambda=x+lambdaint_0^1(s+alambda )ds iff alambda=lambdabig(frac12+lambda abig)$$
If $lambda=0$ then $phi(x)=x$

if $lambdane 1$ $a=frac12+lambda aiff ( 1-lambda)a=frac12iff a=frac12-2lambda$ and then $phi(x)=x+fraclambda2-2lambda$

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

share|cite|improve this answer

answered 8 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,478211

$endgroup$

add a comment | 

$begingroup$

Note $int_0^1phi(s)ds$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
Putting into FE yields:

$$x+alambda=x+lambdaint_0^1(s+alambda )ds iff alambda=lambdabig(frac12+lambda abig)$$
If $lambda=0$ then $phi(x)=x$

if $lambdane 1$ $a=frac12+lambda aiff ( 1-lambda)a=frac12iff a=frac12-2lambda$ and then $phi(x)=x+fraclambda2-2lambda$

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

share|cite|improve this answer

answered 8 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,478211

$endgroup$

share|cite|improve this answer

answered 8 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,478211

answered 8 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,478211

answered 8 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:
$$
sum_n geq 0 a_n x^n = x + lambda sum_n geq 0a_n over n+1.
$$
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
$$
a_0 = lambda left(a_0 + a_1 over 2right).
$$
This gives a relationship between $a_0$ and $lambda$.

share|cite|improve this answer

answered 8 hours ago

avsavs

3,779514

$endgroup$

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:
$$
sum_n geq 0 a_n x^n = x + lambda sum_n geq 0a_n over n+1.
$$
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
$$
a_0 = lambda left(a_0 + a_1 over 2right).
$$
This gives a relationship between $a_0$ and $lambda$.

share|cite|improve this answer

answered 8 hours ago

avsavs

3,779514

$endgroup$

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:
$$
sum_n geq 0 a_n x^n = x + lambda sum_n geq 0a_n over n+1.
$$
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
$$
a_0 = lambda left(a_0 + a_1 over 2right).
$$
This gives a relationship between $a_0$ and $lambda$.

share|cite|improve this answer

answered 8 hours ago

avsavs

3,779514

$endgroup$

share|cite|improve this answer

answered 8 hours ago

avsavs

3,779514

answered 8 hours ago

avsavs

3,779514

answered 8 hours ago

avsavs

3,779514

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 $$x+a=x+lambdaint _0^1x+adximplies\a=lambda(1over 2+a)implies\a=lambdaover 2-2lambda$$ and we obtain$$phi(x)=x+lambdaover 2-2lambdaquad,quad lambdane 1$$The case $lambda=1$ leads to no solution.

share|cite|improve this answer

answered 8 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 $$x+a=x+lambdaint _0^1x+adximplies\a=lambda(1over 2+a)implies\a=lambdaover 2-2lambda$$ and we obtain$$phi(x)=x+lambdaover 2-2lambdaquad,quad lambdane 1$$The case $lambda=1$ leads to no solution.

share|cite|improve this answer

answered 8 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 $$x+a=x+lambdaint _0^1x+adximplies\a=lambda(1over 2+a)implies\a=lambdaover 2-2lambda$$ and we obtain$$phi(x)=x+lambdaover 2-2lambdaquad,quad lambdane 1$$The case $lambda=1$ leads to no solution.

share|cite|improve this answer

answered 8 hours ago

Mostafa AyazMostafa Ayaz

17.6k31039

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Mostafa AyazMostafa Ayaz

17.6k31039

answered 8 hours ago

Mostafa AyazMostafa Ayaz

17.6k31039

answered 8 hours ago

Mostafa AyazMostafa Ayaz

17.6k31039