A Standard Integral EquationLinear 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??
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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??
$begingroup$
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
asked 11 hours ago
LightningStrikeLightningStrike
455
$endgroup$
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
asked 11 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
11 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
11 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
asked 11 hours ago
LightningStrikeLightningStrike
455
$endgroup$
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
linear-algebra integration matrix-equations
asked 11 hours ago
LightningStrikeLightningStrike
455
asked 11 hours ago
LightningStrikeLightningStrike
455
edited 11 hours ago
LightningStrike
asked 11 hours ago
LightningStrikeLightningStrike
455
asked 11 hours ago
LightningStrikeLightningStrike
455
asked 11 hours ago
LightningStrikeLightningStrike
455
455
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
11 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
11 hours ago
add a comment |
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
11 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
11 hours ago
1
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
11 hours ago
$begingroup$
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
11 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
11 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
11 hours ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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 $$phi(x)=x+fraclambda2(1-lambda)$$
answered 11 hours ago
John DoeJohn Doe
11.3k11239
$endgroup$
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 $phi$.
answered 11 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,468211
$endgroup$
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 $lambda$.
answered 11 hours ago
avsavs
3,789514
$endgroup$
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.
answered 11 hours ago
Mostafa AyazMostafa Ayaz
17.6k31039
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $$phi(x)=x+fraclambda2(1-lambda)$$
answered 11 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 $$phi(x)=x+fraclambda2(1-lambda)$$
answered 11 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 $$phi(x)=x+fraclambda2(1-lambda)$$
answered 11 hours ago
John DoeJohn Doe
11.3k11239
$endgroup$
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 $$phi(x)=x+fraclambda2(1-lambda)$$
answered 11 hours ago
John DoeJohn Doe
11.3k11239
answered 11 hours ago
John DoeJohn Doe
11.3k11239
answered 11 hours ago
John DoeJohn Doe
11.3k11239
answered 11 hours ago
John DoeJohn Doe
11.3k11239
11.3k11239
add a comment |
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 $phi$.
answered 11 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,468211
$endgroup$
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 $phi$.
answered 11 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,468211
$endgroup$
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 $phi$.
answered 11 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,468211
$endgroup$
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 $phi$.
answered 11 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,468211
answered 11 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,468211
answered 11 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,468211
answered 11 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,468211
1,468211
add a comment |
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 $lambda$.
answered 11 hours ago
avsavs
3,789514
$endgroup$
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 $lambda$.
answered 11 hours ago
avsavs
3,789514
$endgroup$
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 $lambda$.
answered 11 hours ago
avsavs
3,789514
$endgroup$
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 $lambda$.
answered 11 hours ago
avsavs
3,789514
answered 11 hours ago
avsavs
3,789514
answered 11 hours ago
avsavs
3,789514
answered 11 hours ago
avsavs
3,789514
3,789514
add a comment |
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.
answered 11 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.
answered 11 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.
answered 11 hours ago
Mostafa AyazMostafa Ayaz
17.6k31039
$endgroup$
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.
answered 11 hours ago
Mostafa AyazMostafa Ayaz
17.6k31039
answered 11 hours ago
Mostafa AyazMostafa Ayaz
17.6k31039
answered 11 hours ago
Mostafa AyazMostafa Ayaz
17.6k31039
answered 11 hours ago
Mostafa AyazMostafa Ayaz
17.6k31039
17.6k31039
add a comment |
add a comment |
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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
11 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
11 hours ago