What functions can be represented as a series of eigenfunctions
$begingroup$
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
$endgroup$
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
$endgroup$
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
$endgroup$
Consider the differential equation:
$y'' = lambda y$
with the boundary conditions
$y(0) = y(2pi) = 0$.
This equation has eigenfunctions $mu_n(x) = sin(frac{nx}{2})$ 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
functional-analysis ordinary-differential-equations fourier-series eigenfunctions
edited 6 hours ago
mercury0114
asked 11 hours ago
mercury0114mercury0114
20118
20118
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3 Answers
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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 $sum_{1}^{infty} frac{langle mu_n,frangle}{langle mu_n,mu_nrangle} mu_n(x)$ 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
$$
int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad text{as }nto infty.$$
If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.
$endgroup$
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 $sum_{1}^{infty} frac{langle mu_n,frangle}{langle mu_n,mu_nrangle} mu_n(x)$ 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 $sum_{1}^{infty} frac{langle mu_n,frangle}{langle mu_n,mu_nrangle} mu_n(x)$ 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 $sum_{1}^{infty} frac{langle mu_n,frangle}{langle mu_n,mu_nrangle} mu_n(x)$ 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 $sum_{1}^{infty} frac{langle mu_n,frangle}{langle mu_n,mu_nrangle} mu_n(x)$ on a subset of $(0,2pi)$ of measure zero)
edited 5 hours ago
answered 10 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 9 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
$$
int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad text{as }nto infty.$$
If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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
$$
int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad text{as }nto infty.$$
If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.
answered 10 hours ago
Giuseppe NegroGiuseppe Negro
17.4k331126
17.4k331126
add a comment |
add a comment |
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