Solving Fredholm Equation of the second kind
$begingroup$
Consider the Fredholm Equation of the second kind,
$$phi(x) = 3 + lambda int_{0}^{pi} text{cos}(x-s) , phi(s) ,ds$$
Where the analytical solution is found as,
$$phi(x) = 3 + frac{6lambda}{1 - lambda frac{pi}{2}},text{sin}(x)$$
How could one use Mathematica to find a numerical solution to the same integral equation by using the method of successive approximations (i.e. the Neumann series approach)?
numerical-integration integral-equations
edited Mar 2 at 20:25
J. M. is computer-less♦
97.3k10303463
asked Mar 1 at 19:28
LightningStrikeLightningStrike
735
$endgroup$
add a comment |
$begingroup$
Consider the Fredholm Equation of the second kind,
$$phi(x) = 3 + lambda int_{0}^{pi} text{cos}(x-s) , phi(s) ,ds$$
Where the analytical solution is found as,
$$phi(x) = 3 + frac{6lambda}{1 - lambda frac{pi}{2}},text{sin}(x)$$
How could one use Mathematica to find a numerical solution to the same integral equation by using the method of successive approximations (i.e. the Neumann series approach)?
numerical-integration integral-equations
edited Mar 2 at 20:25
J. M. is computer-less♦
97.3k10303463
asked Mar 1 at 19:28
LightningStrikeLightningStrike
735
$endgroup$
add a comment |
$begingroup$
Consider the Fredholm Equation of the second kind,
$$phi(x) = 3 + lambda int_{0}^{pi} text{cos}(x-s) , phi(s) ,ds$$
Where the analytical solution is found as,
$$phi(x) = 3 + frac{6lambda}{1 - lambda frac{pi}{2}},text{sin}(x)$$
How could one use Mathematica to find a numerical solution to the same integral equation by using the method of successive approximations (i.e. the Neumann series approach)?
numerical-integration integral-equations
edited Mar 2 at 20:25
J. M. is computer-less♦
97.3k10303463
asked Mar 1 at 19:28
LightningStrikeLightningStrike
735
$endgroup$
Consider the Fredholm Equation of the second kind,
$$phi(x) = 3 + lambda int_{0}^{pi} text{cos}(x-s) , phi(s) ,ds$$
Where the analytical solution is found as,
$$phi(x) = 3 + frac{6lambda}{1 - lambda frac{pi}{2}},text{sin}(x)$$
How could one use Mathematica to find a numerical solution to the same integral equation by using the method of successive approximations (i.e. the Neumann series approach)?
numerical-integration integral-equations
numerical-integration integral-equations
edited Mar 2 at 20:25
J. M. is computer-less♦
97.3k10303463
asked Mar 1 at 19:28
LightningStrikeLightningStrike
735
edited Mar 2 at 20:25
J. M. is computer-less♦
97.3k10303463
asked Mar 1 at 19:28
LightningStrikeLightningStrike
735
edited Mar 2 at 20:25
J. M. is computer-less♦
97.3k10303463
edited Mar 2 at 20:25
J. M. is computer-less♦
97.3k10303463
edited Mar 2 at 20:25
J. M. is computer-less♦
97.3k10303463
97.3k10303463
asked Mar 1 at 19:28
LightningStrikeLightningStrike
735
asked Mar 1 at 19:28
LightningStrikeLightningStrike
735
asked Mar 1 at 19:28
LightningStrikeLightningStrike
735
735
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Use DSolve:
PHI =
DSolveValue[ϕ[x] == 3 + λ Integrate[ Cos[x - s] ϕ[s], {s, 0, Pi}], ϕ, x]
(*Function[{x}, (3 (-2 + π λ - 4 λ Sin[x]))/(-2 + π λ)]*)
The solution can be further used in the form PHI[x].
edited Mar 2 at 23:50
m_goldberg
87.4k872198
answered Mar 1 at 19:34
Ulrich NeumannUlrich Neumann
9,513616
$endgroup$
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– LightningStrike
Mar 1 at 19:40
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
Mar 1 at 19:57
add a comment |
$begingroup$
Following Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:
n = 10; (* for example *)
ϕ[x_, 0] = 3;
Do[ϕ[x_, j_] = 3 + λ Integrate[Cos[x - p] ϕ[p, j - 1], {p, 0, π}], {j, n}]
The last term in the series ϕ[x,n] is the approximation to ϕ[x].
Here is what Mathematica returns for ϕ[x,10].
To investigate convergence, I guess we could look at the difference ϕ[x,n] - ϕ[x] as n gets large, since you know ϕ[x].
answered Mar 1 at 20:12
mjwmjw
5429
$endgroup$
$begingroup$
Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
$endgroup$
– LightningStrike
Mar 1 at 20:27
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:[Phi][x, n].
$endgroup$
– mjw
Mar 1 at 20:52
$begingroup$
Made some edits to my answer. Had a couple of typos. Within a function definition it isx_, otherwisex. Also,[Phi][x,j]needs two arguments, one forxand one for thejthapproximation. Hope its clear.
$endgroup$
– mjw
Mar 1 at 20:55
$begingroup$
@m_goldberg, how do you post symbols rather than the clutzy [Phi] type of notation here?
$endgroup$
– mjw
Mar 3 at 1:57
1
$begingroup$
I use halirutan's plug-in. You can learn more about it here
$endgroup$
– m_goldberg
Mar 3 at 2:53
|
show 1 more comment
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192434%2fsolving-fredholm-equation-of-the-second-kind%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use DSolve:
PHI =
DSolveValue[ϕ[x] == 3 + λ Integrate[ Cos[x - s] ϕ[s], {s, 0, Pi}], ϕ, x]
(*Function[{x}, (3 (-2 + π λ - 4 λ Sin[x]))/(-2 + π λ)]*)
The solution can be further used in the form PHI[x].
edited Mar 2 at 23:50
m_goldberg
87.4k872198
answered Mar 1 at 19:34
Ulrich NeumannUlrich Neumann
9,513616
$endgroup$
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– LightningStrike
Mar 1 at 19:40
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
Mar 1 at 19:57
add a comment |
$begingroup$
Use DSolve:
PHI =
DSolveValue[ϕ[x] == 3 + λ Integrate[ Cos[x - s] ϕ[s], {s, 0, Pi}], ϕ, x]
(*Function[{x}, (3 (-2 + π λ - 4 λ Sin[x]))/(-2 + π λ)]*)
The solution can be further used in the form PHI[x].
edited Mar 2 at 23:50
m_goldberg
87.4k872198
answered Mar 1 at 19:34
Ulrich NeumannUlrich Neumann
9,513616
$endgroup$
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– LightningStrike
Mar 1 at 19:40
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
Mar 1 at 19:57
add a comment |
$begingroup$
Use DSolve:
PHI =
DSolveValue[ϕ[x] == 3 + λ Integrate[ Cos[x - s] ϕ[s], {s, 0, Pi}], ϕ, x]
(*Function[{x}, (3 (-2 + π λ - 4 λ Sin[x]))/(-2 + π λ)]*)
The solution can be further used in the form PHI[x].
edited Mar 2 at 23:50
m_goldberg
87.4k872198
answered Mar 1 at 19:34
Ulrich NeumannUlrich Neumann
9,513616
$endgroup$
Use DSolve:
PHI =
DSolveValue[ϕ[x] == 3 + λ Integrate[ Cos[x - s] ϕ[s], {s, 0, Pi}], ϕ, x]
(*Function[{x}, (3 (-2 + π λ - 4 λ Sin[x]))/(-2 + π λ)]*)
The solution can be further used in the form PHI[x].
edited Mar 2 at 23:50
m_goldberg
87.4k872198
answered Mar 1 at 19:34
Ulrich NeumannUlrich Neumann
9,513616
edited Mar 2 at 23:50
m_goldberg
87.4k872198
edited Mar 2 at 23:50
m_goldberg
87.4k872198
edited Mar 2 at 23:50
m_goldberg
87.4k872198
87.4k872198
answered Mar 1 at 19:34
Ulrich NeumannUlrich Neumann
9,513616
answered Mar 1 at 19:34
Ulrich NeumannUlrich Neumann
9,513616
answered Mar 1 at 19:34
Ulrich NeumannUlrich Neumann
9,513616
9,513616
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– LightningStrike
Mar 1 at 19:40
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
Mar 1 at 19:57
add a comment |
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– LightningStrike
Mar 1 at 19:40
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
Mar 1 at 19:57
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– LightningStrike
Mar 1 at 19:40
$begingroup$
Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
$endgroup$
– LightningStrike
Mar 1 at 19:40
1
1
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
Mar 1 at 19:57
$begingroup$
@ user57401 I modified my answer!
$endgroup$
– Ulrich Neumann
Mar 1 at 19:57
add a comment |
$begingroup$
Following Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:
n = 10; (* for example *)
ϕ[x_, 0] = 3;
Do[ϕ[x_, j_] = 3 + λ Integrate[Cos[x - p] ϕ[p, j - 1], {p, 0, π}], {j, n}]
The last term in the series ϕ[x,n] is the approximation to ϕ[x].
Here is what Mathematica returns for ϕ[x,10].
To investigate convergence, I guess we could look at the difference ϕ[x,n] - ϕ[x] as n gets large, since you know ϕ[x].
answered Mar 1 at 20:12
mjwmjw
5429
$endgroup$
$begingroup$
Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
$endgroup$
– LightningStrike
Mar 1 at 20:27
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:[Phi][x, n].
$endgroup$
– mjw
Mar 1 at 20:52
$begingroup$
Made some edits to my answer. Had a couple of typos. Within a function definition it isx_, otherwisex. Also,[Phi][x,j]needs two arguments, one forxand one for thejthapproximation. Hope its clear.
$endgroup$
– mjw
Mar 1 at 20:55
$begingroup$
@m_goldberg, how do you post symbols rather than the clutzy [Phi] type of notation here?
$endgroup$
– mjw
Mar 3 at 1:57
1
$begingroup$
I use halirutan's plug-in. You can learn more about it here
$endgroup$
– m_goldberg
Mar 3 at 2:53
|
show 1 more comment
$begingroup$
Following Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:
n = 10; (* for example *)
ϕ[x_, 0] = 3;
Do[ϕ[x_, j_] = 3 + λ Integrate[Cos[x - p] ϕ[p, j - 1], {p, 0, π}], {j, n}]
The last term in the series ϕ[x,n] is the approximation to ϕ[x].
Here is what Mathematica returns for ϕ[x,10].
To investigate convergence, I guess we could look at the difference ϕ[x,n] - ϕ[x] as n gets large, since you know ϕ[x].
answered Mar 1 at 20:12
mjwmjw
5429
$endgroup$
$begingroup$
Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
$endgroup$
– LightningStrike
Mar 1 at 20:27
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:[Phi][x, n].
$endgroup$
– mjw
Mar 1 at 20:52
$begingroup$
Made some edits to my answer. Had a couple of typos. Within a function definition it isx_, otherwisex. Also,[Phi][x,j]needs two arguments, one forxand one for thejthapproximation. Hope its clear.
$endgroup$
– mjw
Mar 1 at 20:55
$begingroup$
@m_goldberg, how do you post symbols rather than the clutzy [Phi] type of notation here?
$endgroup$
– mjw
Mar 3 at 1:57
1
$begingroup$
I use halirutan's plug-in. You can learn more about it here
$endgroup$
– m_goldberg
Mar 3 at 2:53
|
show 1 more comment
$begingroup$
Following Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:
n = 10; (* for example *)
ϕ[x_, 0] = 3;
Do[ϕ[x_, j_] = 3 + λ Integrate[Cos[x - p] ϕ[p, j - 1], {p, 0, π}], {j, n}]
The last term in the series ϕ[x,n] is the approximation to ϕ[x].
Here is what Mathematica returns for ϕ[x,10].
To investigate convergence, I guess we could look at the difference ϕ[x,n] - ϕ[x] as n gets large, since you know ϕ[x].
answered Mar 1 at 20:12
mjwmjw
5429
$endgroup$
Following Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:
n = 10; (* for example *)
ϕ[x_, 0] = 3;
Do[ϕ[x_, j_] = 3 + λ Integrate[Cos[x - p] ϕ[p, j - 1], {p, 0, π}], {j, n}]
The last term in the series ϕ[x,n] is the approximation to ϕ[x].
Here is what Mathematica returns for ϕ[x,10].
To investigate convergence, I guess we could look at the difference ϕ[x,n] - ϕ[x] as n gets large, since you know ϕ[x].
answered Mar 1 at 20:12
mjwmjw
5429
edited yesterday
answered Mar 1 at 20:12
mjwmjw
5429
answered Mar 1 at 20:12
mjwmjw
5429
answered Mar 1 at 20:12
mjwmjw
5429
5429
$begingroup$
Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
$endgroup$
– LightningStrike
Mar 1 at 20:27
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:[Phi][x, n].
$endgroup$
– mjw
Mar 1 at 20:52
$begingroup$
Made some edits to my answer. Had a couple of typos. Within a function definition it isx_, otherwisex. Also,[Phi][x,j]needs two arguments, one forxand one for thejthapproximation. Hope its clear.
$endgroup$
– mjw
Mar 1 at 20:55
$begingroup$
@m_goldberg, how do you post symbols rather than the clutzy [Phi] type of notation here?
$endgroup$
– mjw
Mar 3 at 1:57
1
$begingroup$
I use halirutan's plug-in. You can learn more about it here
$endgroup$
– m_goldberg
Mar 3 at 2:53
|
show 1 more comment
$begingroup$
Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
$endgroup$
– LightningStrike
Mar 1 at 20:27
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:[Phi][x, n].
$endgroup$
– mjw
Mar 1 at 20:52
$begingroup$
Made some edits to my answer. Had a couple of typos. Within a function definition it isx_, otherwisex. Also,[Phi][x,j]needs two arguments, one forxand one for thejthapproximation. Hope its clear.
$endgroup$
– mjw
Mar 1 at 20:55
$begingroup$
@m_goldberg, how do you post symbols rather than the clutzy [Phi] type of notation here?
$endgroup$
– mjw
Mar 3 at 1:57
1
$begingroup$
I use halirutan's plug-in. You can learn more about it here
$endgroup$
– m_goldberg
Mar 3 at 2:53
$begingroup$
Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
$endgroup$
– LightningStrike
Mar 1 at 20:27
$begingroup$
Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
$endgroup$
– LightningStrike
Mar 1 at 20:27
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:
[Phi][x, n].$endgroup$
– mjw
Mar 1 at 20:52
$begingroup$
Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value:
[Phi][x, n].$endgroup$
– mjw
Mar 1 at 20:52
$begingroup$
Made some edits to my answer. Had a couple of typos. Within a function definition it is
x_, otherwise x. Also, [Phi][x,j] needs two arguments, one for x and one for the jth approximation. Hope its clear.$endgroup$
– mjw
Mar 1 at 20:55
$begingroup$
Made some edits to my answer. Had a couple of typos. Within a function definition it is
x_, otherwise x. Also, [Phi][x,j] needs two arguments, one for x and one for the jth approximation. Hope its clear.$endgroup$
– mjw
Mar 1 at 20:55
$begingroup$
@m_goldberg, how do you post symbols rather than the clutzy [Phi] type of notation here?
$endgroup$
– mjw
Mar 3 at 1:57
$begingroup$
@m_goldberg, how do you post symbols rather than the clutzy [Phi] type of notation here?
$endgroup$
– mjw
Mar 3 at 1:57
1
1
$begingroup$
I use halirutan's plug-in. You can learn more about it here
$endgroup$
– m_goldberg
Mar 3 at 2:53
$begingroup$
I use halirutan's plug-in. You can learn more about it here
$endgroup$
– m_goldberg
Mar 3 at 2:53
|
show 1 more comment
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192434%2fsolving-fredholm-equation-of-the-second-kind%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
