What functions can be represented as a series of eigenfunctionsOrthogonality 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
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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
$begingroup$
Consider the differential equation:
$y'' = lambda y$
with the boundary conditions
$y(0) = y(2pi) = 0$.
This equation has eigenfunctions $mu_n(x) = sin(fracnx2)$ with the corresponding eigenvalues $lambda_n = -fracn^24$ 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 $c_n = fraclangle f,mu_n ranglelangle mu_n, mu_n rangle$? 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
$endgroup$
add a comment |
$begingroup$
Consider the differential equation:
$y'' = lambda y$
with the boundary conditions
$y(0) = y(2pi) = 0$.
This equation has eigenfunctions $mu_n(x) = sin(fracnx2)$ with the corresponding eigenvalues $lambda_n = -fracn^24$ 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 $c_n = fraclangle f,mu_n ranglelangle mu_n, mu_n rangle$? 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
$endgroup$
add a comment |
$begingroup$
Consider the differential equation:
$y'' = lambda y$
with the boundary conditions
$y(0) = y(2pi) = 0$.
This equation has eigenfunctions $mu_n(x) = sin(fracnx2)$ with the corresponding eigenvalues $lambda_n = -fracn^24$ 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 $c_n = fraclangle f,mu_n ranglelangle mu_n, mu_n rangle$? 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
$endgroup$
Consider the differential equation:
$y'' = lambda y$
with the boundary conditions
$y(0) = y(2pi) = 0$.
This equation has eigenfunctions $mu_n(x) = sin(fracnx2)$ with the corresponding eigenvalues $lambda_n = -fracn^24$ 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 $c_n = fraclangle f,mu_n ranglelangle mu_n, mu_n rangle$? 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
functional-analysis ordinary-differential-equations fourier-series eigenfunctions
edited 4 hours ago
mercury0114
asked 10 hours ago
mercury0114mercury0114
20118
20118
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 $sum_1^infty fraclangle mu_n,franglelangle mu_n,mu_nrangle mu_n(x)$ on a subset of $(0,2pi)$ of measure zero)
$endgroup$
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.
$endgroup$
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
$$
int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad textas nto infty.$$
If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $sum_1^infty fraclangle mu_n,franglelangle mu_n,mu_nrangle mu_n(x)$ on a subset of $(0,2pi)$ of measure zero)
$endgroup$
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 $sum_1^infty fraclangle mu_n,franglelangle mu_n,mu_nrangle mu_n(x)$ on a subset of $(0,2pi)$ of measure zero)
$endgroup$
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 $sum_1^infty fraclangle mu_n,franglelangle mu_n,mu_nrangle mu_n(x)$ on a subset of $(0,2pi)$ of measure zero)
$endgroup$
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 $sum_1^infty fraclangle mu_n,franglelangle mu_n,mu_nrangle mu_n(x)$ on a subset of $(0,2pi)$ of measure zero)
edited 4 hours ago
answered 8 hours ago
Thibaut DemaerelThibaut Demaerel
593312
593312
add a comment |
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
answered 7 hours ago
DisintegratingByPartsDisintegratingByParts
59.6k42581
59.6k42581
add a comment |
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
$$
int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad textas nto infty.$$
If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.
$endgroup$
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
$$
int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad textas nto infty.$$
If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.
$endgroup$
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
$$
int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad textas nto infty.$$
If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.
$endgroup$
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
$$
int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad textas nto infty.$$
If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.
answered 8 hours ago
Giuseppe NegroGiuseppe Negro
17.4k331126
17.4k331126
add a comment |
add a comment |
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-eigenfunctions, fourier-series, functional-analysis, ordinary-differential-equations