Boundary value problems: eigenvalue and eigenfunction
$begingroup$
I'm having trouble in understanding eigenvalues and eigenfunctions in BvP
the problem is:
$y''$ + $lambda$$y$ = $0$
$y(0)=0$
$y(2pi)$ = $0$.
Make characteristic polynomial
$r^2 + lambda = 0$
$r_1,_2 = pm sqrt{- lambda}$
the general solution is :
$$y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$$
Applying first boundary condition
$0=y(0)=c_1$
and applying the second boundary condition
$0=y(pi)=c_2 sinleft(2pisqrt{lambda}right)$.
I know the part how to solve BVP I just wanted to know how get
eigenvalue solution:
$$lambda_n = left(frac n2right)^2 = frac {n^2}{4},quad n=1,2,3...$$
and eigenfunction:
$$y_n= sin left(frac {nx}{2}right),quad n=1,2,3....$$
IF $lambda > 0$
ordinary-differential-equations eigenvalues-eigenvectors boundary-value-problem eigenfunctions
edited Jun 29 '16 at 12:00
Davide Giraudo
128k17154268
asked Jun 22 '16 at 8:09
clutchnicclutchnic
2416
$endgroup$
add a comment |
$begingroup$
I'm having trouble in understanding eigenvalues and eigenfunctions in BvP
the problem is:
$y''$ + $lambda$$y$ = $0$
$y(0)=0$
$y(2pi)$ = $0$.
Make characteristic polynomial
$r^2 + lambda = 0$
$r_1,_2 = pm sqrt{- lambda}$
the general solution is :
$$y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$$
Applying first boundary condition
$0=y(0)=c_1$
and applying the second boundary condition
$0=y(pi)=c_2 sinleft(2pisqrt{lambda}right)$.
I know the part how to solve BVP I just wanted to know how get
eigenvalue solution:
$$lambda_n = left(frac n2right)^2 = frac {n^2}{4},quad n=1,2,3...$$
and eigenfunction:
$$y_n= sin left(frac {nx}{2}right),quad n=1,2,3....$$
IF $lambda > 0$
ordinary-differential-equations eigenvalues-eigenvectors boundary-value-problem eigenfunctions
edited Jun 29 '16 at 12:00
Davide Giraudo
128k17154268
asked Jun 22 '16 at 8:09
clutchnicclutchnic
2416
$endgroup$
add a comment |
$begingroup$
I'm having trouble in understanding eigenvalues and eigenfunctions in BvP
the problem is:
$y''$ + $lambda$$y$ = $0$
$y(0)=0$
$y(2pi)$ = $0$.
Make characteristic polynomial
$r^2 + lambda = 0$
$r_1,_2 = pm sqrt{- lambda}$
the general solution is :
$$y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$$
Applying first boundary condition
$0=y(0)=c_1$
and applying the second boundary condition
$0=y(pi)=c_2 sinleft(2pisqrt{lambda}right)$.
I know the part how to solve BVP I just wanted to know how get
eigenvalue solution:
$$lambda_n = left(frac n2right)^2 = frac {n^2}{4},quad n=1,2,3...$$
and eigenfunction:
$$y_n= sin left(frac {nx}{2}right),quad n=1,2,3....$$
IF $lambda > 0$
ordinary-differential-equations eigenvalues-eigenvectors boundary-value-problem eigenfunctions
edited Jun 29 '16 at 12:00
Davide Giraudo
128k17154268
asked Jun 22 '16 at 8:09
clutchnicclutchnic
2416
$endgroup$
I'm having trouble in understanding eigenvalues and eigenfunctions in BvP
the problem is:
$y''$ + $lambda$$y$ = $0$
$y(0)=0$
$y(2pi)$ = $0$.
Make characteristic polynomial
$r^2 + lambda = 0$
$r_1,_2 = pm sqrt{- lambda}$
the general solution is :
$$y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$$
Applying first boundary condition
$0=y(0)=c_1$
and applying the second boundary condition
$0=y(pi)=c_2 sinleft(2pisqrt{lambda}right)$.
I know the part how to solve BVP I just wanted to know how get
eigenvalue solution:
$$lambda_n = left(frac n2right)^2 = frac {n^2}{4},quad n=1,2,3...$$
and eigenfunction:
$$y_n= sin left(frac {nx}{2}right),quad n=1,2,3....$$
IF $lambda > 0$
ordinary-differential-equations eigenvalues-eigenvectors boundary-value-problem eigenfunctions
ordinary-differential-equations eigenvalues-eigenvectors boundary-value-problem eigenfunctions
edited Jun 29 '16 at 12:00
Davide Giraudo
128k17154268
asked Jun 22 '16 at 8:09
clutchnicclutchnic
2416
edited Jun 29 '16 at 12:00
Davide Giraudo
128k17154268
asked Jun 22 '16 at 8:09
clutchnicclutchnic
2416
edited Jun 29 '16 at 12:00
Davide Giraudo
128k17154268
edited Jun 29 '16 at 12:00
Davide Giraudo
128k17154268
edited Jun 29 '16 at 12:00
Davide Giraudo
128k17154268
128k17154268
asked Jun 22 '16 at 8:09
clutchnicclutchnic
2416
asked Jun 22 '16 at 8:09
clutchnicclutchnic
2416
asked Jun 22 '16 at 8:09
clutchnicclutchnic
2416
2416
add a comment |
add a comment |
1 Answer
1
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$begingroup$
As your said,
$y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$
and $0=y(0)=c_1$.
The key step is how to solve $0=y(2pi)=c_2 sinleft(2pisqrt{lambda}right)$
Indeed, we require
begin{align}2pisqrt{lambda}=npi, mbox{ where } n=1,2,cdotsend{align}
consequently, we deduce
begin{align}
&sqrtlambda=frac{n}{2}\
&lambda=left(frac{n}{2}right) ^2=frac{n^2}{4},mbox{ where } n=1,2,cdots
end{align}
Thusbegin{align}
y_n= sin left(frac {nx}{2}right),quad n=1,2,....end{align}
answered Jan 30 '17 at 17:50
Canrong TianCanrong Tian
165
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As your said,
$y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$
and $0=y(0)=c_1$.
The key step is how to solve $0=y(2pi)=c_2 sinleft(2pisqrt{lambda}right)$
Indeed, we require
begin{align}2pisqrt{lambda}=npi, mbox{ where } n=1,2,cdotsend{align}
consequently, we deduce
begin{align}
&sqrtlambda=frac{n}{2}\
&lambda=left(frac{n}{2}right) ^2=frac{n^2}{4},mbox{ where } n=1,2,cdots
end{align}
Thusbegin{align}
y_n= sin left(frac {nx}{2}right),quad n=1,2,....end{align}
answered Jan 30 '17 at 17:50
Canrong TianCanrong Tian
165
$endgroup$
add a comment |
$begingroup$
As your said,
$y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$
and $0=y(0)=c_1$.
The key step is how to solve $0=y(2pi)=c_2 sinleft(2pisqrt{lambda}right)$
Indeed, we require
begin{align}2pisqrt{lambda}=npi, mbox{ where } n=1,2,cdotsend{align}
consequently, we deduce
begin{align}
&sqrtlambda=frac{n}{2}\
&lambda=left(frac{n}{2}right) ^2=frac{n^2}{4},mbox{ where } n=1,2,cdots
end{align}
Thusbegin{align}
y_n= sin left(frac {nx}{2}right),quad n=1,2,....end{align}
answered Jan 30 '17 at 17:50
Canrong TianCanrong Tian
165
$endgroup$
add a comment |
$begingroup$
As your said,
$y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$
and $0=y(0)=c_1$.
The key step is how to solve $0=y(2pi)=c_2 sinleft(2pisqrt{lambda}right)$
Indeed, we require
begin{align}2pisqrt{lambda}=npi, mbox{ where } n=1,2,cdotsend{align}
consequently, we deduce
begin{align}
&sqrtlambda=frac{n}{2}\
&lambda=left(frac{n}{2}right) ^2=frac{n^2}{4},mbox{ where } n=1,2,cdots
end{align}
Thusbegin{align}
y_n= sin left(frac {nx}{2}right),quad n=1,2,....end{align}
answered Jan 30 '17 at 17:50
Canrong TianCanrong Tian
165
$endgroup$
As your said,
$y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$
and $0=y(0)=c_1$.
The key step is how to solve $0=y(2pi)=c_2 sinleft(2pisqrt{lambda}right)$
Indeed, we require
begin{align}2pisqrt{lambda}=npi, mbox{ where } n=1,2,cdotsend{align}
consequently, we deduce
begin{align}
&sqrtlambda=frac{n}{2}\
&lambda=left(frac{n}{2}right) ^2=frac{n^2}{4},mbox{ where } n=1,2,cdots
end{align}
Thusbegin{align}
y_n= sin left(frac {nx}{2}right),quad n=1,2,....end{align}
answered Jan 30 '17 at 17:50
Canrong TianCanrong Tian
165
answered Jan 30 '17 at 17:50
Canrong TianCanrong Tian
165
answered Jan 30 '17 at 17:50
Canrong TianCanrong Tian
165
answered Jan 30 '17 at 17:50
Canrong TianCanrong Tian
165
165
add a comment |
add a comment |
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