Legendre Eigenvalue problem
I have the eigenvalue problem,
$frac{d}{dx}big((1-x^2)frac{du}{dx}big)+lambda u=0$,
on $[-1,1]$ subject to single boundary condiction $u(-1) = u(1)$.
Assume that there is an eigenfunction of the form $u(x)=a_0+a_1x+a_2x^2$. Find the possible eigenvalues for such an eigenfunction.
To solve this problem, I plug the $u(x)$ and the 2nd derivative $u(x)$ into the $frac{d}{dx}big((1-x^2)frac{du}{dx}big)+lambda u=0$, solve for $x$.
But I couldn't get any eigenvalues. Any help will be appreciated.
eigenfunctions
edited Dec 26 at 0:56
Cosmas Zachos
1,518520
asked Nov 18 at 23:07
Ray
112
add a comment |
I have the eigenvalue problem,
$frac{d}{dx}big((1-x^2)frac{du}{dx}big)+lambda u=0$,
on $[-1,1]$ subject to single boundary condiction $u(-1) = u(1)$.
Assume that there is an eigenfunction of the form $u(x)=a_0+a_1x+a_2x^2$. Find the possible eigenvalues for such an eigenfunction.
To solve this problem, I plug the $u(x)$ and the 2nd derivative $u(x)$ into the $frac{d}{dx}big((1-x^2)frac{du}{dx}big)+lambda u=0$, solve for $x$.
But I couldn't get any eigenvalues. Any help will be appreciated.
eigenfunctions
edited Dec 26 at 0:56
Cosmas Zachos
1,518520
asked Nov 18 at 23:07
Ray
112
Please use Latex/MathJax to format your post.
– Stockfish
Nov 18 at 23:11
Include your calculations, then someone might help. And you shouldn't solve for $x$, but rather choose $a_0,a_1,a_2$ so that the equality $frac{d}{dx}big((1-x^2)frac{du}{dx}big)+lambda u=0$ is true for all $x$.
– Michał Miśkiewicz
Nov 18 at 23:32
@MichałMiśkiewicz Hi,sir. Could you please tell me
– Ray
Nov 18 at 23:44
@MichałMiśkiewicz why do I need to solve for a0,a2. Cause on my textbook, they used to solve for x, and get 3 different cases. like lambda greater than zero or less zero or equal zero.
– Ray
Nov 18 at 23:46
Well, the thing you're looking for is the function $u$, which is of the form $u(x)=a_0+a_1 x+a_2 x^2$. Thus, finding $u$ is equivalent to finding $a_0,a_1,a_2$. On the other hand, solving for $x$ makes no sense.
– Michał Miśkiewicz
Nov 19 at 9:24
add a comment |
I have the eigenvalue problem,
$frac{d}{dx}big((1-x^2)frac{du}{dx}big)+lambda u=0$,
on $[-1,1]$ subject to single boundary condiction $u(-1) = u(1)$.
Assume that there is an eigenfunction of the form $u(x)=a_0+a_1x+a_2x^2$. Find the possible eigenvalues for such an eigenfunction.
To solve this problem, I plug the $u(x)$ and the 2nd derivative $u(x)$ into the $frac{d}{dx}big((1-x^2)frac{du}{dx}big)+lambda u=0$, solve for $x$.
But I couldn't get any eigenvalues. Any help will be appreciated.
eigenfunctions
edited Dec 26 at 0:56
Cosmas Zachos
1,518520
asked Nov 18 at 23:07
Ray
112
I have the eigenvalue problem,
$frac{d}{dx}big((1-x^2)frac{du}{dx}big)+lambda u=0$,
on $[-1,1]$ subject to single boundary condiction $u(-1) = u(1)$.
Assume that there is an eigenfunction of the form $u(x)=a_0+a_1x+a_2x^2$. Find the possible eigenvalues for such an eigenfunction.
To solve this problem, I plug the $u(x)$ and the 2nd derivative $u(x)$ into the $frac{d}{dx}big((1-x^2)frac{du}{dx}big)+lambda u=0$, solve for $x$.
But I couldn't get any eigenvalues. Any help will be appreciated.
eigenfunctions
eigenfunctions
edited Dec 26 at 0:56
Cosmas Zachos
1,518520
asked Nov 18 at 23:07
Ray
112
edited Dec 26 at 0:56
Cosmas Zachos
1,518520
asked Nov 18 at 23:07
Ray
112
edited Dec 26 at 0:56
Cosmas Zachos
1,518520
edited Dec 26 at 0:56
Cosmas Zachos
1,518520
edited Dec 26 at 0:56
Cosmas Zachos
1,518520
1,518520
asked Nov 18 at 23:07
Ray
112
asked Nov 18 at 23:07
Ray
112
asked Nov 18 at 23:07
Ray
112
112
Please use Latex/MathJax to format your post.
– Stockfish
Nov 18 at 23:11
Include your calculations, then someone might help. And you shouldn't solve for $x$, but rather choose $a_0,a_1,a_2$ so that the equality $frac{d}{dx}big((1-x^2)frac{du}{dx}big)+lambda u=0$ is true for all $x$.
– Michał Miśkiewicz
Nov 18 at 23:32
@MichałMiśkiewicz Hi,sir. Could you please tell me
– Ray
Nov 18 at 23:44
@MichałMiśkiewicz why do I need to solve for a0,a2. Cause on my textbook, they used to solve for x, and get 3 different cases. like lambda greater than zero or less zero or equal zero.
– Ray
Nov 18 at 23:46
Well, the thing you're looking for is the function $u$, which is of the form $u(x)=a_0+a_1 x+a_2 x^2$. Thus, finding $u$ is equivalent to finding $a_0,a_1,a_2$. On the other hand, solving for $x$ makes no sense.
– Michał Miśkiewicz
Nov 19 at 9:24
add a comment |
Please use Latex/MathJax to format your post.
– Stockfish
Nov 18 at 23:11
Include your calculations, then someone might help. And you shouldn't solve for $x$, but rather choose $a_0,a_1,a_2$ so that the equality $frac{d}{dx}big((1-x^2)frac{du}{dx}big)+lambda u=0$ is true for all $x$.
– Michał Miśkiewicz
Nov 18 at 23:32
@MichałMiśkiewicz Hi,sir. Could you please tell me
– Ray
Nov 18 at 23:44
@MichałMiśkiewicz why do I need to solve for a0,a2. Cause on my textbook, they used to solve for x, and get 3 different cases. like lambda greater than zero or less zero or equal zero.
– Ray
Nov 18 at 23:46
Well, the thing you're looking for is the function $u$, which is of the form $u(x)=a_0+a_1 x+a_2 x^2$. Thus, finding $u$ is equivalent to finding $a_0,a_1,a_2$. On the other hand, solving for $x$ makes no sense.
– Michał Miśkiewicz
Nov 19 at 9:24
Please use Latex/MathJax to format your post.
– Stockfish
Nov 18 at 23:11
Please use Latex/MathJax to format your post.
– Stockfish
Nov 18 at 23:11
Include your calculations, then someone might help. And you shouldn't solve for $x$, but rather choose $a_0,a_1,a_2$ so that the equality $frac{d}{dx}big((1-x^2)frac{du}{dx}big)+lambda u=0$ is true for all $x$.
– Michał Miśkiewicz
Nov 18 at 23:32
Include your calculations, then someone might help. And you shouldn't solve for $x$, but rather choose $a_0,a_1,a_2$ so that the equality $frac{d}{dx}big((1-x^2)frac{du}{dx}big)+lambda u=0$ is true for all $x$.
– Michał Miśkiewicz
Nov 18 at 23:32
@MichałMiśkiewicz Hi,sir. Could you please tell me
– Ray
Nov 18 at 23:44
@MichałMiśkiewicz Hi,sir. Could you please tell me
– Ray
Nov 18 at 23:44
@MichałMiśkiewicz why do I need to solve for a0,a2. Cause on my textbook, they used to solve for x, and get 3 different cases. like lambda greater than zero or less zero or equal zero.
– Ray
Nov 18 at 23:46
@MichałMiśkiewicz why do I need to solve for a0,a2. Cause on my textbook, they used to solve for x, and get 3 different cases. like lambda greater than zero or less zero or equal zero.
– Ray
Nov 18 at 23:46
Well, the thing you're looking for is the function $u$, which is of the form $u(x)=a_0+a_1 x+a_2 x^2$. Thus, finding $u$ is equivalent to finding $a_0,a_1,a_2$. On the other hand, solving for $x$ makes no sense.
– Michał Miśkiewicz
Nov 19 at 9:24
Well, the thing you're looking for is the function $u$, which is of the form $u(x)=a_0+a_1 x+a_2 x^2$. Thus, finding $u$ is equivalent to finding $a_0,a_1,a_2$. On the other hand, solving for $x$ makes no sense.
– Michał Miśkiewicz
Nov 19 at 9:24
add a comment |
1 Answer
1
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Well, I assume it is sufficiently late I am not doing your homework for you. The boundary condition tells you that $a_1=0$, so it's missing from your Ansatz. Plugging the surviving piece into the equation yields
$$
partial ((1-x^2)2a_2)=-lambda(a_0+a_2x^2),
$$
so, comparing powers, $lambda=6$ and $a_2=-3a_0$.
In fact, do you now see how your particular case (n=2) falls into the general order-n polynomial solution
$$
P_npropto partial^n (x^2-1)^n
$$
of
$$
P_n(x)= partial ((1-x^2)partial P_n) +n(n+1)P_n=0 ~~?
$$
answered Dec 25 at 22:12
Cosmas Zachos
1,518520
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
votes
Well, I assume it is sufficiently late I am not doing your homework for you. The boundary condition tells you that $a_1=0$, so it's missing from your Ansatz. Plugging the surviving piece into the equation yields
$$
partial ((1-x^2)2a_2)=-lambda(a_0+a_2x^2),
$$
so, comparing powers, $lambda=6$ and $a_2=-3a_0$.
In fact, do you now see how your particular case (n=2) falls into the general order-n polynomial solution
$$
P_npropto partial^n (x^2-1)^n
$$
of
$$
P_n(x)= partial ((1-x^2)partial P_n) +n(n+1)P_n=0 ~~?
$$
answered Dec 25 at 22:12
Cosmas Zachos
1,518520
add a comment |
Well, I assume it is sufficiently late I am not doing your homework for you. The boundary condition tells you that $a_1=0$, so it's missing from your Ansatz. Plugging the surviving piece into the equation yields
$$
partial ((1-x^2)2a_2)=-lambda(a_0+a_2x^2),
$$
so, comparing powers, $lambda=6$ and $a_2=-3a_0$.
In fact, do you now see how your particular case (n=2) falls into the general order-n polynomial solution
$$
P_npropto partial^n (x^2-1)^n
$$
of
$$
P_n(x)= partial ((1-x^2)partial P_n) +n(n+1)P_n=0 ~~?
$$
answered Dec 25 at 22:12
Cosmas Zachos
1,518520
add a comment |
Well, I assume it is sufficiently late I am not doing your homework for you. The boundary condition tells you that $a_1=0$, so it's missing from your Ansatz. Plugging the surviving piece into the equation yields
$$
partial ((1-x^2)2a_2)=-lambda(a_0+a_2x^2),
$$
so, comparing powers, $lambda=6$ and $a_2=-3a_0$.
In fact, do you now see how your particular case (n=2) falls into the general order-n polynomial solution
$$
P_npropto partial^n (x^2-1)^n
$$
of
$$
P_n(x)= partial ((1-x^2)partial P_n) +n(n+1)P_n=0 ~~?
$$
answered Dec 25 at 22:12
Cosmas Zachos
1,518520
Well, I assume it is sufficiently late I am not doing your homework for you. The boundary condition tells you that $a_1=0$, so it's missing from your Ansatz. Plugging the surviving piece into the equation yields
$$
partial ((1-x^2)2a_2)=-lambda(a_0+a_2x^2),
$$
so, comparing powers, $lambda=6$ and $a_2=-3a_0$.
In fact, do you now see how your particular case (n=2) falls into the general order-n polynomial solution
$$
P_npropto partial^n (x^2-1)^n
$$
of
$$
P_n(x)= partial ((1-x^2)partial P_n) +n(n+1)P_n=0 ~~?
$$
answered Dec 25 at 22:12
Cosmas Zachos
1,518520
answered Dec 25 at 22:12
Cosmas Zachos
1,518520
answered Dec 25 at 22:12
Cosmas Zachos
1,518520
answered Dec 25 at 22:12
Cosmas Zachos
1,518520
1,518520
add a comment |
add a comment |
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– Stockfish
Nov 18 at 23:11
Include your calculations, then someone might help. And you shouldn't solve for $x$, but rather choose $a_0,a_1,a_2$ so that the equality $frac{d}{dx}big((1-x^2)frac{du}{dx}big)+lambda u=0$ is true for all $x$.
– Michał Miśkiewicz
Nov 18 at 23:32
@MichałMiśkiewicz Hi,sir. Could you please tell me
– Ray
Nov 18 at 23:44
@MichałMiśkiewicz why do I need to solve for a0,a2. Cause on my textbook, they used to solve for x, and get 3 different cases. like lambda greater than zero or less zero or equal zero.
– Ray
Nov 18 at 23:46
Well, the thing you're looking for is the function $u$, which is of the form $u(x)=a_0+a_1 x+a_2 x^2$. Thus, finding $u$ is equivalent to finding $a_0,a_1,a_2$. On the other hand, solving for $x$ makes no sense.
– Michał Miśkiewicz
Nov 19 at 9:24