Solving Integral Equation by Converting to Differential Equations The Next CEO of Stack...
Unreliable Magic - Is it worth it?
Is there an analogue of projective spaces for proper schemes?
Does it take more energy to get to Venus or to Mars?
Is there a difference between "Fahrstuhl" and "Aufzug"
How do I go from 300 unfinished/half written blog posts, to published posts?
Skipping indices in a product
What flight has the highest ratio of time difference to flight time?
Are there any unintended negative consequences to allowing PCs to gain multiple levels at once in a short milestone-XP game?
calculus parametric curve length
How do scammers retract money, while you can’t?
How to start emacs in "nothing" mode (`fundamental-mode`)
What happens if you roll doubles 3 times then land on "Go to jail?"
How to transpose the 1st and -1th levels of arbitrarily nested array?
Extending anchors in TikZ
WOW air has ceased operation, can I get my tickets refunded?
Preparing Indesign booklet with .psd graphics for print
What is ( CFMCC ) on ILS approach chart?
Elegant way to replace substring in a regex with optional groups in Python?
Solving Integral Equation by Converting to Differential Equations
How fast would a person need to move to trick the eye?
Bold, vivid family
Is it possible to search for a directory/file combination?
How to avoid supervisors with prejudiced views?
Anatomically Correct Strange Women In Ponds Distributing Swords
Solving Integral Equation by Converting to Differential Equations
The Next CEO of Stack OverflowAre there methods to solve coupled integral and integro-differential equations?Voltera equationSolve integral equation by converting to differential equationHow can I solve this integral equation by converting it to a differential equationConverting a integral equation to differential equationSolving integro-differential equation - numericallySolution of Differential equation as an integral equationConverting Differential Operator to Integral Equationreference for converting an integro-differential equation to a differential algebraic equationSolving second order ordinary differential equation with variable constants
$begingroup$
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
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
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
asked 5 hours ago
LightningStrikeLightningStrike
555
$endgroup$
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
ordinary-differential-equations integral-equations integro-differential-equations
asked 5 hours ago
LightningStrikeLightningStrike
555
asked 5 hours ago
LightningStrikeLightningStrike
555
asked 5 hours ago
LightningStrikeLightningStrike
555
asked 5 hours ago
LightningStrikeLightningStrike
555
asked 5 hours ago
LightningStrikeLightningStrike
555
555
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We have that
$$phi(x)=x-xint_0^x phi(s) mathrm{d} s + int_0^x s phi(s)mathrm{d}s$$
From this, we can see that $phi(0)=0$.
We can differentiate both sides and use the product rule and the FTC1 to get:
$$phi'(x)=1-int_0^x phi(s) mathrm{d}s -x phi(x)+xphi(x)$$
$$phi'(x)=1-int_0^x phi(s) mathrm{d} s$$
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.
answered 4 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 $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
$endgroup$
– Botond
4 hours ago
add a comment |
$begingroup$
Differentiating both sides using Leibniz rule :
$${phi }'(x)=1-int_{0}^{x}{phi (s)ds}$$
Differentiate again:
$${phi }''(x)=-phi (x)$$
answered 4 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
3 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 |
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: "69"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
},
noCode: 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%2fmath.stackexchange.com%2fquestions%2f3167442%2fsolving-integral-equation-by-converting-to-differential-equations%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
$begingroup$
We have that
$$phi(x)=x-xint_0^x phi(s) mathrm{d} s + int_0^x s phi(s)mathrm{d}s$$
From this, we can see that $phi(0)=0$.
We can differentiate both sides and use the product rule and the FTC1 to get:
$$phi'(x)=1-int_0^x phi(s) mathrm{d}s -x phi(x)+xphi(x)$$
$$phi'(x)=1-int_0^x phi(s) mathrm{d} s$$
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.
answered 4 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 $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
$endgroup$
– Botond
4 hours ago
add a comment |
$begingroup$
We have that
$$phi(x)=x-xint_0^x phi(s) mathrm{d} s + int_0^x s phi(s)mathrm{d}s$$
From this, we can see that $phi(0)=0$.
We can differentiate both sides and use the product rule and the FTC1 to get:
$$phi'(x)=1-int_0^x phi(s) mathrm{d}s -x phi(x)+xphi(x)$$
$$phi'(x)=1-int_0^x phi(s) mathrm{d} s$$
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.
answered 4 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 $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
$endgroup$
– Botond
4 hours ago
add a comment |
$begingroup$
We have that
$$phi(x)=x-xint_0^x phi(s) mathrm{d} s + int_0^x s phi(s)mathrm{d}s$$
From this, we can see that $phi(0)=0$.
We can differentiate both sides and use the product rule and the FTC1 to get:
$$phi'(x)=1-int_0^x phi(s) mathrm{d}s -x phi(x)+xphi(x)$$
$$phi'(x)=1-int_0^x phi(s) mathrm{d} s$$
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.
answered 4 hours ago
BotondBotond
6,49331034
$endgroup$
We have that
$$phi(x)=x-xint_0^x phi(s) mathrm{d} s + int_0^x s phi(s)mathrm{d}s$$
From this, we can see that $phi(0)=0$.
We can differentiate both sides and use the product rule and the FTC1 to get:
$$phi'(x)=1-int_0^x phi(s) mathrm{d}s -x phi(x)+xphi(x)$$
$$phi'(x)=1-int_0^x phi(s) mathrm{d} s$$
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.
answered 4 hours ago
BotondBotond
6,49331034
edited 4 hours ago
answered 4 hours ago
BotondBotond
6,49331034
answered 4 hours ago
BotondBotond
6,49331034
answered 4 hours ago
BotondBotond
6,49331034
6,49331034
$begingroup$
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 $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
$endgroup$
– Botond
4 hours ago
add a comment |
$begingroup$
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 $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
$endgroup$
– Botond
4 hours ago
$begingroup$
In fact, the only valid solution for $phi(x)$ is $sin{(x)}$ because of the original equation.
$endgroup$
– Peter Foreman
4 hours ago
$begingroup$
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
$begingroup$
@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
$begingroup$
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 $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
$endgroup$
– Botond
4 hours ago
$begingroup$
@LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
$endgroup$
– Botond
4 hours ago
add a comment |
$begingroup$
Differentiating both sides using Leibniz rule :
$${phi }'(x)=1-int_{0}^{x}{phi (s)ds}$$
Differentiate again:
$${phi }''(x)=-phi (x)$$
answered 4 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
3 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 :
$${phi }'(x)=1-int_{0}^{x}{phi (s)ds}$$
Differentiate again:
$${phi }''(x)=-phi (x)$$
answered 4 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
3 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 :
$${phi }'(x)=1-int_{0}^{x}{phi (s)ds}$$
Differentiate again:
$${phi }''(x)=-phi (x)$$
answered 4 hours ago
logologo
1048
$endgroup$
Differentiating both sides using Leibniz rule :
$${phi }'(x)=1-int_{0}^{x}{phi (s)ds}$$
Differentiate again:
$${phi }''(x)=-phi (x)$$
answered 4 hours ago
logologo
1048
edited 4 hours ago
answered 4 hours ago
logologo
1048
answered 4 hours ago
logologo
1048
answered 4 hours ago
logologo
1048
1048
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
3 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
$begingroup$
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
3 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
1
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
3 hours ago
$begingroup$
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
3 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
$begingroup$
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
$begingroup$
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 |
Thanks for contributing an answer to Mathematics 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%2fmath.stackexchange.com%2fquestions%2f3167442%2fsolving-integral-equation-by-converting-to-differential-equations%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
