Please help me solve this Eigenvalue problem
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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 Nov 18 at 23:18
gd1035
414210
asked Nov 18 at 23:07
Ray
62
add a comment |
up vote
1
down vote
favorite
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 Nov 18 at 23:18
gd1035
414210
asked Nov 18 at 23:07
Ray
62
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 |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 Nov 18 at 23:18
gd1035
414210
asked Nov 18 at 23:07
Ray
62
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 Nov 18 at 23:18
gd1035
414210
asked Nov 18 at 23:07
Ray
62
edited Nov 18 at 23:18
gd1035
414210
asked Nov 18 at 23:07
Ray
62
edited Nov 18 at 23:18
gd1035
414210
edited Nov 18 at 23:18
gd1035
414210
edited Nov 18 at 23:18
gd1035
414210
414210
asked Nov 18 at 23:07
Ray
62
asked Nov 18 at 23:07
Ray
62
asked Nov 18 at 23:07
Ray
62
62
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
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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