Dtjytyk

Is honey really a supersaturated solution? Does heating to un-crystalize redissolve it or melt it?

How are passwords stolen from companies if they only store hashes?

When to use a slotted vs. solid turner?

Professor being mistaken for a grad student

How to get the n-th line after a grepped one?

Relationship between sampajanna definitions in SN 47.2 and SN 47.35

Employee lack of ownership

Print a physical multiplication table

Is it true that good novels will automatically sell themselves on Amazon (and so on) and there is no need for one to waste time promoting?

Why do passenger jet manufacturers design their planes with stall prevention systems?

A diagram about partial derivatives of f(x,y)

Recruiter wants very extensive technical details about all of my previous work

What options are left, if Britain cannot decide?

Brexit - No Deal Rejection

Could the Saturn V actually have launched astronauts around Venus?

Do I need life insurance if I can cover my own funeral costs?

Why does overlay work only on the first tcolorbox?

How to write cleanly even if my character uses expletive language?

The German vowel “a” changes to the English “i”

Could this Scherzo by Beethoven be considered to be a fugue?

Simplify an interface for flexibly applying rules to periods of time

World War I as a war of liberals against authoritarians?

What is the relationship between relativity and the Doppler effect?

As a new Ubuntu desktop 18.04 LTS user, do I need to use ufw for a firewall or is iptables sufficient?

What functions can be represented as a series of eigenfunctions

Orthogonality of eigenfunctions of a linear operator.Finding eigenvalues and eigenfunctions for a BVPCan we construct Sturm Liouville problems from an orthogonal basis of functions?Cannot obtain a normalised eigenfunction for a boundary problem.Prove eigenfunctions corresponding to different eigenvalues are orthogonalInitial Value Problem EigenfunctionsEigenfunctions expansion in Sturm-Liouville Boundary Value ProblemGiven a differential equation, derive inner product for its eigenfunctions?The Proper Form of Eigenfunctions and Linear OperatorsHow to find eigenvalues and eigenfunctions of simple-looking differential operator

7

1

$begingroup$

Consider the differential equation:

$y'' = lambda y$

with the boundary conditions
$y(0) = y(2pi) = 0$.

This equation has eigenfunctions $mu_n(x) = sin(frac{nx}{2})$ with the corresponding eigenvalues $lambda_n = -frac{n^2}{4}$ for $n > 0$

• Am I right, that certain functions f(x) satisfying the same boundary conditions as above can be represented as an infinite series
  $f(x) = sum_1^infty c_n mu_n(x)$ with coefficients $c_n = frac{langle f,mu_n rangle}{langle mu_n, mu_n rangle}$? What conditions those certain functions need to satisfy?




• Can the previous claim be generalised for any set of eigenfunctions of some differential equation? I.E. suppose $Ly = lambda y$ is a differential equation ($L$ being the 2nd order differential operator) with boundary conditions $y(a) = y(b) = c$ . What functions can be represented as a weighted sum of the eigensolutions?

functional-analysis ordinary-differential-equations fourier-series eigenfunctions

share|cite|improve this question

edited Mar 11 at 17:02

mercury0114

asked Mar 11 at 11:25

mercury0114mercury0114

20118

$endgroup$

add a comment | 

7

1

$begingroup$

Consider the differential equation:

$y'' = lambda y$

with the boundary conditions
$y(0) = y(2pi) = 0$.

This equation has eigenfunctions $mu_n(x) = sin(frac{nx}{2})$ with the corresponding eigenvalues $lambda_n = -frac{n^2}{4}$ for $n > 0$

• Am I right, that certain functions f(x) satisfying the same boundary conditions as above can be represented as an infinite series
  $f(x) = sum_1^infty c_n mu_n(x)$ with coefficients $c_n = frac{langle f,mu_n rangle}{langle mu_n, mu_n rangle}$? What conditions those certain functions need to satisfy?




• Can the previous claim be generalised for any set of eigenfunctions of some differential equation? I.E. suppose $Ly = lambda y$ is a differential equation ($L$ being the 2nd order differential operator) with boundary conditions $y(a) = y(b) = c$ . What functions can be represented as a weighted sum of the eigensolutions?

functional-analysis ordinary-differential-equations fourier-series eigenfunctions

share|cite|improve this question

edited Mar 11 at 17:02

mercury0114

asked Mar 11 at 11:25

mercury0114mercury0114

20118

$endgroup$

add a comment | 

$begingroup$

Consider the differential equation:

$y'' = lambda y$

with the boundary conditions
$y(0) = y(2pi) = 0$.

This equation has eigenfunctions $mu_n(x) = sin(frac{nx}{2})$ with the corresponding eigenvalues $lambda_n = -frac{n^2}{4}$ for $n > 0$

• Am I right, that certain functions f(x) satisfying the same boundary conditions as above can be represented as an infinite series
  $f(x) = sum_1^infty c_n mu_n(x)$ with coefficients $c_n = frac{langle f,mu_n rangle}{langle mu_n, mu_n rangle}$? What conditions those certain functions need to satisfy?




• Can the previous claim be generalised for any set of eigenfunctions of some differential equation? I.E. suppose $Ly = lambda y$ is a differential equation ($L$ being the 2nd order differential operator) with boundary conditions $y(a) = y(b) = c$ . What functions can be represented as a weighted sum of the eigensolutions?

functional-analysis ordinary-differential-equations fourier-series eigenfunctions

share|cite|improve this question

edited Mar 11 at 17:02

mercury0114

asked Mar 11 at 11:25

mercury0114mercury0114

20118

$endgroup$

share|cite|improve this question

edited Mar 11 at 17:02

mercury0114

asked Mar 11 at 11:25

mercury0114mercury0114

20118

share|cite|improve this question

edited Mar 11 at 17:02

mercury0114

asked Mar 11 at 11:25

mercury0114mercury0114

20118

asked Mar 11 at 11:25

mercury0114mercury0114

20118

asked Mar 11 at 11:25

mercury0114mercury0114

20118

3 Answers
3

active

oldest

votes

5

$begingroup$

Your question situates within the realm of Sturm-Liouville theory and my subsequent answer applies to all differential operators (with associated eigenfunctions) that belong to this realm. You asked which continuous functions can be expanded in a series of eigenfunctions? The most natural answer turns out to be "functions that are square-integrable on the interval $(0,2pi)$". Also non-continuous, square-integrable functions $f$ can be expanded in this way (under the proviso that $f$ is allowed to differ from its series-expansion $sum_{1}^{infty} frac{langle mu_n,frangle}{langle mu_n,mu_nrangle} mu_n(x)$ on a subset of $(0,2pi)$ of measure zero)

share|cite|improve this answer

edited Mar 11 at 17:21

answered Mar 11 at 12:56

Thibaut DemaerelThibaut Demaerel

593312

$endgroup$

add a comment | 

1

$begingroup$

Consider the following problem on the interval $[a,b]$ for some $a < b$ and real angles $alpha,beta$:
$$
y''=lambda y,;;; a le x le b, \
cosalpha y(a)+sinalpha y'(a) = 0 \
cosbeta y(b)+sinbeta y'(b) = 0
$$
This gives rise to a discrete set of eigenvalues
$$
lambda_1 < lambda_2 < lambda_3 < cdots,
$$

and associated eigenfunctions $phi_n(x)$. For any function $fin L^2[a,b]$, the Fourier series for $f$ in these eigenfunctions converges to $f$ in $L^2[a,b]$. And you get pointwise convergence of the series at $xin(a,b)$ under the same type of Fourier conditions that you learned for the ordinary Fourier series. The endpoint conditions make the convergence at $x=a,b$ trickier, of course. Conditions of the type $y(a)=0=y(b)$, for example, forces any series in these eigenfunctions to converge to $0$ at the endpoints.

share|cite|improve this answer

answered Mar 11 at 14:00

DisintegratingByPartsDisintegratingByParts

59.8k42681

$endgroup$

add a comment | 

0

$begingroup$

You can expand all continuous functions, and even the non-continuous ones (but they must be square-integrable); however, the series will only converge in the $L^2(0, 1)$ sense. This means that, defining
$$
Sf_n:=sum_{k=1}^n c_n mu_n, $$
it holds that
$$
int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad text{as }nto infty.$$

If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.

share|cite|improve this answer

answered Mar 11 at 13:04

Giuseppe NegroGiuseppe Negro

17.5k332126

$endgroup$

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
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3143573%2fwhat-functions-can-be-represented-as-a-series-of-eigenfunctions%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

5

$begingroup$

Your question situates within the realm of Sturm-Liouville theory and my subsequent answer applies to all differential operators (with associated eigenfunctions) that belong to this realm. You asked which continuous functions can be expanded in a series of eigenfunctions? The most natural answer turns out to be "functions that are square-integrable on the interval $(0,2pi)$". Also non-continuous, square-integrable functions $f$ can be expanded in this way (under the proviso that $f$ is allowed to differ from its series-expansion $sum_{1}^{infty} frac{langle mu_n,frangle}{langle mu_n,mu_nrangle} mu_n(x)$ on a subset of $(0,2pi)$ of measure zero)

share|cite|improve this answer

edited Mar 11 at 17:21

answered Mar 11 at 12:56

Thibaut DemaerelThibaut Demaerel

593312

$endgroup$

add a comment | 

5

$begingroup$

Your question situates within the realm of Sturm-Liouville theory and my subsequent answer applies to all differential operators (with associated eigenfunctions) that belong to this realm. You asked which continuous functions can be expanded in a series of eigenfunctions? The most natural answer turns out to be "functions that are square-integrable on the interval $(0,2pi)$". Also non-continuous, square-integrable functions $f$ can be expanded in this way (under the proviso that $f$ is allowed to differ from its series-expansion $sum_{1}^{infty} frac{langle mu_n,frangle}{langle mu_n,mu_nrangle} mu_n(x)$ on a subset of $(0,2pi)$ of measure zero)

share|cite|improve this answer

edited Mar 11 at 17:21

answered Mar 11 at 12:56

Thibaut DemaerelThibaut Demaerel

593312

$endgroup$

add a comment | 

$begingroup$

Your question situates within the realm of Sturm-Liouville theory and my subsequent answer applies to all differential operators (with associated eigenfunctions) that belong to this realm. You asked which continuous functions can be expanded in a series of eigenfunctions? The most natural answer turns out to be "functions that are square-integrable on the interval $(0,2pi)$". Also non-continuous, square-integrable functions $f$ can be expanded in this way (under the proviso that $f$ is allowed to differ from its series-expansion $sum_{1}^{infty} frac{langle mu_n,frangle}{langle mu_n,mu_nrangle} mu_n(x)$ on a subset of $(0,2pi)$ of measure zero)

share|cite|improve this answer

edited Mar 11 at 17:21

answered Mar 11 at 12:56

Thibaut DemaerelThibaut Demaerel

593312

$endgroup$

share|cite|improve this answer

edited Mar 11 at 17:21

answered Mar 11 at 12:56

Thibaut DemaerelThibaut Demaerel

593312

answered Mar 11 at 12:56

Thibaut DemaerelThibaut Demaerel

593312

answered Mar 11 at 12:56

Thibaut DemaerelThibaut Demaerel

593312

1

$begingroup$

Consider the following problem on the interval $[a,b]$ for some $a < b$ and real angles $alpha,beta$:
$$
y''=lambda y,;;; a le x le b, \
cosalpha y(a)+sinalpha y'(a) = 0 \
cosbeta y(b)+sinbeta y'(b) = 0
$$
This gives rise to a discrete set of eigenvalues
$$
lambda_1 < lambda_2 < lambda_3 < cdots,
$$

and associated eigenfunctions $phi_n(x)$. For any function $fin L^2[a,b]$, the Fourier series for $f$ in these eigenfunctions converges to $f$ in $L^2[a,b]$. And you get pointwise convergence of the series at $xin(a,b)$ under the same type of Fourier conditions that you learned for the ordinary Fourier series. The endpoint conditions make the convergence at $x=a,b$ trickier, of course. Conditions of the type $y(a)=0=y(b)$, for example, forces any series in these eigenfunctions to converge to $0$ at the endpoints.

share|cite|improve this answer

answered Mar 11 at 14:00

DisintegratingByPartsDisintegratingByParts

59.8k42681

$endgroup$

add a comment | 

1

$begingroup$

Consider the following problem on the interval $[a,b]$ for some $a < b$ and real angles $alpha,beta$:
$$
y''=lambda y,;;; a le x le b, \
cosalpha y(a)+sinalpha y'(a) = 0 \
cosbeta y(b)+sinbeta y'(b) = 0
$$
This gives rise to a discrete set of eigenvalues
$$
lambda_1 < lambda_2 < lambda_3 < cdots,
$$

and associated eigenfunctions $phi_n(x)$. For any function $fin L^2[a,b]$, the Fourier series for $f$ in these eigenfunctions converges to $f$ in $L^2[a,b]$. And you get pointwise convergence of the series at $xin(a,b)$ under the same type of Fourier conditions that you learned for the ordinary Fourier series. The endpoint conditions make the convergence at $x=a,b$ trickier, of course. Conditions of the type $y(a)=0=y(b)$, for example, forces any series in these eigenfunctions to converge to $0$ at the endpoints.

share|cite|improve this answer

answered Mar 11 at 14:00

DisintegratingByPartsDisintegratingByParts

59.8k42681

$endgroup$

add a comment | 

$begingroup$

Consider the following problem on the interval $[a,b]$ for some $a < b$ and real angles $alpha,beta$:
$$
y''=lambda y,;;; a le x le b, \
cosalpha y(a)+sinalpha y'(a) = 0 \
cosbeta y(b)+sinbeta y'(b) = 0
$$
This gives rise to a discrete set of eigenvalues
$$
lambda_1 < lambda_2 < lambda_3 < cdots,
$$

and associated eigenfunctions $phi_n(x)$. For any function $fin L^2[a,b]$, the Fourier series for $f$ in these eigenfunctions converges to $f$ in $L^2[a,b]$. And you get pointwise convergence of the series at $xin(a,b)$ under the same type of Fourier conditions that you learned for the ordinary Fourier series. The endpoint conditions make the convergence at $x=a,b$ trickier, of course. Conditions of the type $y(a)=0=y(b)$, for example, forces any series in these eigenfunctions to converge to $0$ at the endpoints.

share|cite|improve this answer

answered Mar 11 at 14:00

DisintegratingByPartsDisintegratingByParts

59.8k42681

$endgroup$

share|cite|improve this answer

answered Mar 11 at 14:00

DisintegratingByPartsDisintegratingByParts

59.8k42681

answered Mar 11 at 14:00

DisintegratingByPartsDisintegratingByParts

59.8k42681

answered Mar 11 at 14:00

DisintegratingByPartsDisintegratingByParts

59.8k42681

0

$begingroup$

You can expand all continuous functions, and even the non-continuous ones (but they must be square-integrable); however, the series will only converge in the $L^2(0, 1)$ sense. This means that, defining
$$
Sf_n:=sum_{k=1}^n c_n mu_n, $$
it holds that
$$
int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad text{as }nto infty.$$

If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.

share|cite|improve this answer

answered Mar 11 at 13:04

Giuseppe NegroGiuseppe Negro

17.5k332126

$endgroup$

add a comment | 

0

$begingroup$

You can expand all continuous functions, and even the non-continuous ones (but they must be square-integrable); however, the series will only converge in the $L^2(0, 1)$ sense. This means that, defining
$$
Sf_n:=sum_{k=1}^n c_n mu_n, $$
it holds that
$$
int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad text{as }nto infty.$$

If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.

share|cite|improve this answer

answered Mar 11 at 13:04

Giuseppe NegroGiuseppe Negro

17.5k332126

$endgroup$

add a comment | 

$begingroup$

You can expand all continuous functions, and even the non-continuous ones (but they must be square-integrable); however, the series will only converge in the $L^2(0, 1)$ sense. This means that, defining
$$
Sf_n:=sum_{k=1}^n c_n mu_n, $$
it holds that
$$
int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad text{as }nto infty.$$

If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.

share|cite|improve this answer

answered Mar 11 at 13:04

Giuseppe NegroGiuseppe Negro

17.5k332126

$endgroup$

share|cite|improve this answer

answered Mar 11 at 13:04

Giuseppe NegroGiuseppe Negro

17.5k332126

answered Mar 11 at 13:04

Giuseppe NegroGiuseppe Negro

17.5k332126

answered Mar 11 at 13:04

Giuseppe NegroGiuseppe Negro

17.5k332126