Solving Integral Equation by Converting to Differential Equations
$begingroup$
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 Mar 29 at 18:29
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 Mar 29 at 18:29
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 Mar 29 at 18:29
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 Mar 29 at 18:29
LightningStrikeLightningStrike
555
asked Mar 29 at 18:29
LightningStrikeLightningStrike
555
asked Mar 29 at 18:29
LightningStrikeLightningStrike
555
asked Mar 29 at 18:29
LightningStrikeLightningStrike
555
asked Mar 29 at 18:29
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 Mar 29 at 18:52
BotondBotond
6,54531034
$endgroup$
$begingroup$
In fact, the only valid solution for $phi(x)$ is $sin{(x)}$ because of the original equation.
$endgroup$
– Peter Foreman
Mar 29 at 19:03
$begingroup$
@PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
$endgroup$
– Botond
Mar 29 at 19:07
$begingroup$
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
Mar 29 at 19:17
$begingroup$
@LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
$endgroup$
– Botond
Mar 29 at 19:28
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 Mar 29 at 18:51
logologo
19610
$endgroup$
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
Mar 29 at 19:51
$begingroup$
may be you are right...but this is a common technique in an introductory course of integral equations.
$endgroup$
– logo
Mar 29 at 19:56
$begingroup$
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
Mar 29 at 20:05
add a comment |
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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 Mar 29 at 18:52
BotondBotond
6,54531034
$endgroup$
$begingroup$
In fact, the only valid solution for $phi(x)$ is $sin{(x)}$ because of the original equation.
$endgroup$
– Peter Foreman
Mar 29 at 19:03
$begingroup$
@PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
$endgroup$
– Botond
Mar 29 at 19:07
$begingroup$
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
Mar 29 at 19:17
$begingroup$
@LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
$endgroup$
– Botond
Mar 29 at 19:28
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 Mar 29 at 18:52
BotondBotond
6,54531034
$endgroup$
$begingroup$
In fact, the only valid solution for $phi(x)$ is $sin{(x)}$ because of the original equation.
$endgroup$
– Peter Foreman
Mar 29 at 19:03
$begingroup$
@PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
$endgroup$
– Botond
Mar 29 at 19:07
$begingroup$
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
Mar 29 at 19:17
$begingroup$
@LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
$endgroup$
– Botond
Mar 29 at 19:28
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 Mar 29 at 18:52
BotondBotond
6,54531034
$endgroup$
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 Mar 29 at 18:52
BotondBotond
6,54531034
edited Mar 29 at 19:00
answered Mar 29 at 18:52
BotondBotond
6,54531034
answered Mar 29 at 18:52
BotondBotond
6,54531034
answered Mar 29 at 18:52
BotondBotond
6,54531034
6,54531034
$begingroup$
In fact, the only valid solution for $phi(x)$ is $sin{(x)}$ because of the original equation.
$endgroup$
– Peter Foreman
Mar 29 at 19:03
$begingroup$
@PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
$endgroup$
– Botond
Mar 29 at 19:07
$begingroup$
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
Mar 29 at 19:17
$begingroup$
@LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
$endgroup$
– Botond
Mar 29 at 19:28
add a comment |
$begingroup$
In fact, the only valid solution for $phi(x)$ is $sin{(x)}$ because of the original equation.
$endgroup$
– Peter Foreman
Mar 29 at 19:03
$begingroup$
@PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
$endgroup$
– Botond
Mar 29 at 19:07
$begingroup$
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
Mar 29 at 19:17
$begingroup$
@LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
$endgroup$
– Botond
Mar 29 at 19:28
$begingroup$
In fact, the only valid solution for $phi(x)$ is $sin{(x)}$ because of the original equation.
$endgroup$
– Peter Foreman
Mar 29 at 19:03
$begingroup$
In fact, the only valid solution for $phi(x)$ is $sin{(x)}$ because of the original equation.
$endgroup$
– Peter Foreman
Mar 29 at 19:03
$begingroup$
@PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
$endgroup$
– Botond
Mar 29 at 19:07
$begingroup$
@PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
$endgroup$
– Botond
Mar 29 at 19:07
$begingroup$
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
Mar 29 at 19:17
$begingroup$
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
Mar 29 at 19:17
$begingroup$
@LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
$endgroup$
– Botond
Mar 29 at 19:28
$begingroup$
@LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
$endgroup$
– Botond
Mar 29 at 19:28
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 Mar 29 at 18:51
logologo
19610
$endgroup$
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
Mar 29 at 19:51
$begingroup$
may be you are right...but this is a common technique in an introductory course of integral equations.
$endgroup$
– logo
Mar 29 at 19:56
$begingroup$
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
Mar 29 at 20:05
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 Mar 29 at 18:51
logologo
19610
$endgroup$
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
Mar 29 at 19:51
$begingroup$
may be you are right...but this is a common technique in an introductory course of integral equations.
$endgroup$
– logo
Mar 29 at 19:56
$begingroup$
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
Mar 29 at 20:05
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 Mar 29 at 18:51
logologo
19610
$endgroup$
Differentiating both sides using Leibniz rule :
$${phi }'(x)=1-int_{0}^{x}{phi (s)ds}$$
Differentiate again:
$${phi }''(x)=-phi (x)$$
answered Mar 29 at 18:51
logologo
19610
edited Mar 29 at 18:56
answered Mar 29 at 18:51
logologo
19610
answered Mar 29 at 18:51
logologo
19610
answered Mar 29 at 18:51
logologo
19610
19610
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
Mar 29 at 19:51
$begingroup$
may be you are right...but this is a common technique in an introductory course of integral equations.
$endgroup$
– logo
Mar 29 at 19:56
$begingroup$
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
Mar 29 at 20:05
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
Mar 29 at 19:51
$begingroup$
may be you are right...but this is a common technique in an introductory course of integral equations.
$endgroup$
– logo
Mar 29 at 19:56
$begingroup$
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
Mar 29 at 20:05
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
Mar 29 at 19:51
$begingroup$
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
Mar 29 at 19:51
$begingroup$
may be you are right...but this is a common technique in an introductory course of integral equations.
$endgroup$
– logo
Mar 29 at 19:56
$begingroup$
may be you are right...but this is a common technique in an introductory course of integral equations.
$endgroup$
– logo
Mar 29 at 19:56
$begingroup$
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
Mar 29 at 20:05
$begingroup$
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
Mar 29 at 20:05
add a comment |
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