Finding eigenfunctions and eigenvalues from a differential equation
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1
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Consider the differential equation
$$X''(x)+lambda X=0$$
on $0 leq x leq 1$with boundary conditions
$$X'(0)+X(0)=0 text{and} X(1)=0.$$
I have a few problems here that I think I figured out, but I would appreciate another look or some hints as to what I can fix. Or, maybe I am totally wrong!
$textbf{My first goal:}$
Find an eigenfunction associated with eigenvalue $lambda=0.$
An eigenvalue $lambda =0$ would mean that $X''(x)=0$. This means that the solution takes the form
$$X(x)=Ax+B.$$
Since $X'(0)=A$ and $X(0)=B$,
$$X'(0)+X(0)=0 iff A=-B.$$
Therefore an eigenfunction that works would be $boxed{X_{0}(x)=-2x+2}.$
$textbf{My second goal:}$
Find an expression for all eigenvalues $lambda = beta ^2>0.$
This one requires a little more work. The solution to $X'(x)+beta ^2 X=0$ takes the form
$$X(x)=Acosbeta x +Bsin beta x.$$
Taking derivatives, one easily finds that $X'(0)=Bbeta$ and $X(0)=A.$ Thus we obtain
$$Bbeta + A=0 implies beta= frac{-A}{B}.$$
Finally, this gives $boxed{beta=frac{A^2}{B^2}}$
What are your thoughts? Thanks in advance!
Edit: I just realized that my last problem did not necessarily satisfy $X(1)=0.$
differential-equations pde eigenfunctions
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up vote
1
down vote
favorite
Consider the differential equation
$$X''(x)+lambda X=0$$
on $0 leq x leq 1$with boundary conditions
$$X'(0)+X(0)=0 text{and} X(1)=0.$$
I have a few problems here that I think I figured out, but I would appreciate another look or some hints as to what I can fix. Or, maybe I am totally wrong!
$textbf{My first goal:}$
Find an eigenfunction associated with eigenvalue $lambda=0.$
An eigenvalue $lambda =0$ would mean that $X''(x)=0$. This means that the solution takes the form
$$X(x)=Ax+B.$$
Since $X'(0)=A$ and $X(0)=B$,
$$X'(0)+X(0)=0 iff A=-B.$$
Therefore an eigenfunction that works would be $boxed{X_{0}(x)=-2x+2}.$
$textbf{My second goal:}$
Find an expression for all eigenvalues $lambda = beta ^2>0.$
This one requires a little more work. The solution to $X'(x)+beta ^2 X=0$ takes the form
$$X(x)=Acosbeta x +Bsin beta x.$$
Taking derivatives, one easily finds that $X'(0)=Bbeta$ and $X(0)=A.$ Thus we obtain
$$Bbeta + A=0 implies beta= frac{-A}{B}.$$
Finally, this gives $boxed{beta=frac{A^2}{B^2}}$
What are your thoughts? Thanks in advance!
Edit: I just realized that my last problem did not necessarily satisfy $X(1)=0.$
differential-equations pde eigenfunctions
1
First goal, a better eigenfunction would be $$X(x) = Ax - A, quad forall A in mathbb{R}$$ Always try to keep it as general as possible. Second goal, applying both boundary conditions yields begin{align} beta &= -frac{A}{B} \ &= tan beta end{align} and using the oddness of $tan( cdot)$, we get $$lambda = tan^{2}(sqrt{lambda})$$
– Mattos
2 days ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the differential equation
$$X''(x)+lambda X=0$$
on $0 leq x leq 1$with boundary conditions
$$X'(0)+X(0)=0 text{and} X(1)=0.$$
I have a few problems here that I think I figured out, but I would appreciate another look or some hints as to what I can fix. Or, maybe I am totally wrong!
$textbf{My first goal:}$
Find an eigenfunction associated with eigenvalue $lambda=0.$
An eigenvalue $lambda =0$ would mean that $X''(x)=0$. This means that the solution takes the form
$$X(x)=Ax+B.$$
Since $X'(0)=A$ and $X(0)=B$,
$$X'(0)+X(0)=0 iff A=-B.$$
Therefore an eigenfunction that works would be $boxed{X_{0}(x)=-2x+2}.$
$textbf{My second goal:}$
Find an expression for all eigenvalues $lambda = beta ^2>0.$
This one requires a little more work. The solution to $X'(x)+beta ^2 X=0$ takes the form
$$X(x)=Acosbeta x +Bsin beta x.$$
Taking derivatives, one easily finds that $X'(0)=Bbeta$ and $X(0)=A.$ Thus we obtain
$$Bbeta + A=0 implies beta= frac{-A}{B}.$$
Finally, this gives $boxed{beta=frac{A^2}{B^2}}$
What are your thoughts? Thanks in advance!
Edit: I just realized that my last problem did not necessarily satisfy $X(1)=0.$
differential-equations pde eigenfunctions
Consider the differential equation
$$X''(x)+lambda X=0$$
on $0 leq x leq 1$with boundary conditions
$$X'(0)+X(0)=0 text{and} X(1)=0.$$
I have a few problems here that I think I figured out, but I would appreciate another look or some hints as to what I can fix. Or, maybe I am totally wrong!
$textbf{My first goal:}$
Find an eigenfunction associated with eigenvalue $lambda=0.$
An eigenvalue $lambda =0$ would mean that $X''(x)=0$. This means that the solution takes the form
$$X(x)=Ax+B.$$
Since $X'(0)=A$ and $X(0)=B$,
$$X'(0)+X(0)=0 iff A=-B.$$
Therefore an eigenfunction that works would be $boxed{X_{0}(x)=-2x+2}.$
$textbf{My second goal:}$
Find an expression for all eigenvalues $lambda = beta ^2>0.$
This one requires a little more work. The solution to $X'(x)+beta ^2 X=0$ takes the form
$$X(x)=Acosbeta x +Bsin beta x.$$
Taking derivatives, one easily finds that $X'(0)=Bbeta$ and $X(0)=A.$ Thus we obtain
$$Bbeta + A=0 implies beta= frac{-A}{B}.$$
Finally, this gives $boxed{beta=frac{A^2}{B^2}}$
What are your thoughts? Thanks in advance!
Edit: I just realized that my last problem did not necessarily satisfy $X(1)=0.$
differential-equations pde eigenfunctions
differential-equations pde eigenfunctions
edited 2 days ago
asked 2 days ago
StatGuy
530110
530110
1
First goal, a better eigenfunction would be $$X(x) = Ax - A, quad forall A in mathbb{R}$$ Always try to keep it as general as possible. Second goal, applying both boundary conditions yields begin{align} beta &= -frac{A}{B} \ &= tan beta end{align} and using the oddness of $tan( cdot)$, we get $$lambda = tan^{2}(sqrt{lambda})$$
– Mattos
2 days ago
add a comment |
1
First goal, a better eigenfunction would be $$X(x) = Ax - A, quad forall A in mathbb{R}$$ Always try to keep it as general as possible. Second goal, applying both boundary conditions yields begin{align} beta &= -frac{A}{B} \ &= tan beta end{align} and using the oddness of $tan( cdot)$, we get $$lambda = tan^{2}(sqrt{lambda})$$
– Mattos
2 days ago
1
1
First goal, a better eigenfunction would be $$X(x) = Ax - A, quad forall A in mathbb{R}$$ Always try to keep it as general as possible. Second goal, applying both boundary conditions yields begin{align} beta &= -frac{A}{B} \ &= tan beta end{align} and using the oddness of $tan( cdot)$, we get $$lambda = tan^{2}(sqrt{lambda})$$
– Mattos
2 days ago
First goal, a better eigenfunction would be $$X(x) = Ax - A, quad forall A in mathbb{R}$$ Always try to keep it as general as possible. Second goal, applying both boundary conditions yields begin{align} beta &= -frac{A}{B} \ &= tan beta end{align} and using the oddness of $tan( cdot)$, we get $$lambda = tan^{2}(sqrt{lambda})$$
– Mattos
2 days ago
add a comment |
1 Answer
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Since you want $X(1)=0$, you cat set $X'(1)$ to any non-zero constant ($0$ leads to $Xequiv 0$.) So I'd pick a value of $1$. Then
$$
X_{lambda}(x) = frac{sin(sqrt{lambda}(x-1))}{sqrt{lambda}}
$$
Choosing a constant value for $X'(1)$ forces the limiting case as $lambdarightarrow 0$ to be the correct solution for $lambda=0$ as well, which is
$$
X_{0} = x-1.
$$
You can check that this is an eigenfunction. So $lambda=0$ is an eigenvalue, with corresponding eigenfucntion $x-1$.
The general eigenvalue equation becomes
$$
X_{lambda}(0)+X_{lambda}'(0)=0 \
-frac{sin(sqrt{lambda})}{sqrt{lambda}}+cos(sqrt{lambda})=0
$$
The limit at $lambdarightarrow 0$ is $0$. So $lambda_0=0$ is an eigenvalue with $X_{0}=x-1$. For $lambdane 0$, the solutions are zeros of
$$
tan(sqrt{lambda})=sqrt{lambda}.
$$
This is a transcendental equation. You can plot the graphs of $y=tan(x)$ and $y=x$, and check the intersections of the graphs for $x ge 0$. The negative values of $sqrt{lambda}$ can be ignored because they lead to duplicate values of $lambda$.
The non-negative solutions are ordered as follows:
$$
sqrt{lambda_0} = 0 < sqrt{lambda_1} < frac{pi}{2} < sqrt{lambda_2} < frac{3pi}{2} < sqrt{lambda_3} < frac{5pi}{2} < cdots
$$
Asymptotically, $sqrt{lambda_n} approx frac{(2n-1)pi}{2}$ or $lambda_n approx frac{(2n-1)^2pi^2}{4}$ as $nrightarrowinfty$.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Since you want $X(1)=0$, you cat set $X'(1)$ to any non-zero constant ($0$ leads to $Xequiv 0$.) So I'd pick a value of $1$. Then
$$
X_{lambda}(x) = frac{sin(sqrt{lambda}(x-1))}{sqrt{lambda}}
$$
Choosing a constant value for $X'(1)$ forces the limiting case as $lambdarightarrow 0$ to be the correct solution for $lambda=0$ as well, which is
$$
X_{0} = x-1.
$$
You can check that this is an eigenfunction. So $lambda=0$ is an eigenvalue, with corresponding eigenfucntion $x-1$.
The general eigenvalue equation becomes
$$
X_{lambda}(0)+X_{lambda}'(0)=0 \
-frac{sin(sqrt{lambda})}{sqrt{lambda}}+cos(sqrt{lambda})=0
$$
The limit at $lambdarightarrow 0$ is $0$. So $lambda_0=0$ is an eigenvalue with $X_{0}=x-1$. For $lambdane 0$, the solutions are zeros of
$$
tan(sqrt{lambda})=sqrt{lambda}.
$$
This is a transcendental equation. You can plot the graphs of $y=tan(x)$ and $y=x$, and check the intersections of the graphs for $x ge 0$. The negative values of $sqrt{lambda}$ can be ignored because they lead to duplicate values of $lambda$.
The non-negative solutions are ordered as follows:
$$
sqrt{lambda_0} = 0 < sqrt{lambda_1} < frac{pi}{2} < sqrt{lambda_2} < frac{3pi}{2} < sqrt{lambda_3} < frac{5pi}{2} < cdots
$$
Asymptotically, $sqrt{lambda_n} approx frac{(2n-1)pi}{2}$ or $lambda_n approx frac{(2n-1)^2pi^2}{4}$ as $nrightarrowinfty$.
add a comment |
up vote
0
down vote
accepted
Since you want $X(1)=0$, you cat set $X'(1)$ to any non-zero constant ($0$ leads to $Xequiv 0$.) So I'd pick a value of $1$. Then
$$
X_{lambda}(x) = frac{sin(sqrt{lambda}(x-1))}{sqrt{lambda}}
$$
Choosing a constant value for $X'(1)$ forces the limiting case as $lambdarightarrow 0$ to be the correct solution for $lambda=0$ as well, which is
$$
X_{0} = x-1.
$$
You can check that this is an eigenfunction. So $lambda=0$ is an eigenvalue, with corresponding eigenfucntion $x-1$.
The general eigenvalue equation becomes
$$
X_{lambda}(0)+X_{lambda}'(0)=0 \
-frac{sin(sqrt{lambda})}{sqrt{lambda}}+cos(sqrt{lambda})=0
$$
The limit at $lambdarightarrow 0$ is $0$. So $lambda_0=0$ is an eigenvalue with $X_{0}=x-1$. For $lambdane 0$, the solutions are zeros of
$$
tan(sqrt{lambda})=sqrt{lambda}.
$$
This is a transcendental equation. You can plot the graphs of $y=tan(x)$ and $y=x$, and check the intersections of the graphs for $x ge 0$. The negative values of $sqrt{lambda}$ can be ignored because they lead to duplicate values of $lambda$.
The non-negative solutions are ordered as follows:
$$
sqrt{lambda_0} = 0 < sqrt{lambda_1} < frac{pi}{2} < sqrt{lambda_2} < frac{3pi}{2} < sqrt{lambda_3} < frac{5pi}{2} < cdots
$$
Asymptotically, $sqrt{lambda_n} approx frac{(2n-1)pi}{2}$ or $lambda_n approx frac{(2n-1)^2pi^2}{4}$ as $nrightarrowinfty$.
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Since you want $X(1)=0$, you cat set $X'(1)$ to any non-zero constant ($0$ leads to $Xequiv 0$.) So I'd pick a value of $1$. Then
$$
X_{lambda}(x) = frac{sin(sqrt{lambda}(x-1))}{sqrt{lambda}}
$$
Choosing a constant value for $X'(1)$ forces the limiting case as $lambdarightarrow 0$ to be the correct solution for $lambda=0$ as well, which is
$$
X_{0} = x-1.
$$
You can check that this is an eigenfunction. So $lambda=0$ is an eigenvalue, with corresponding eigenfucntion $x-1$.
The general eigenvalue equation becomes
$$
X_{lambda}(0)+X_{lambda}'(0)=0 \
-frac{sin(sqrt{lambda})}{sqrt{lambda}}+cos(sqrt{lambda})=0
$$
The limit at $lambdarightarrow 0$ is $0$. So $lambda_0=0$ is an eigenvalue with $X_{0}=x-1$. For $lambdane 0$, the solutions are zeros of
$$
tan(sqrt{lambda})=sqrt{lambda}.
$$
This is a transcendental equation. You can plot the graphs of $y=tan(x)$ and $y=x$, and check the intersections of the graphs for $x ge 0$. The negative values of $sqrt{lambda}$ can be ignored because they lead to duplicate values of $lambda$.
The non-negative solutions are ordered as follows:
$$
sqrt{lambda_0} = 0 < sqrt{lambda_1} < frac{pi}{2} < sqrt{lambda_2} < frac{3pi}{2} < sqrt{lambda_3} < frac{5pi}{2} < cdots
$$
Asymptotically, $sqrt{lambda_n} approx frac{(2n-1)pi}{2}$ or $lambda_n approx frac{(2n-1)^2pi^2}{4}$ as $nrightarrowinfty$.
Since you want $X(1)=0$, you cat set $X'(1)$ to any non-zero constant ($0$ leads to $Xequiv 0$.) So I'd pick a value of $1$. Then
$$
X_{lambda}(x) = frac{sin(sqrt{lambda}(x-1))}{sqrt{lambda}}
$$
Choosing a constant value for $X'(1)$ forces the limiting case as $lambdarightarrow 0$ to be the correct solution for $lambda=0$ as well, which is
$$
X_{0} = x-1.
$$
You can check that this is an eigenfunction. So $lambda=0$ is an eigenvalue, with corresponding eigenfucntion $x-1$.
The general eigenvalue equation becomes
$$
X_{lambda}(0)+X_{lambda}'(0)=0 \
-frac{sin(sqrt{lambda})}{sqrt{lambda}}+cos(sqrt{lambda})=0
$$
The limit at $lambdarightarrow 0$ is $0$. So $lambda_0=0$ is an eigenvalue with $X_{0}=x-1$. For $lambdane 0$, the solutions are zeros of
$$
tan(sqrt{lambda})=sqrt{lambda}.
$$
This is a transcendental equation. You can plot the graphs of $y=tan(x)$ and $y=x$, and check the intersections of the graphs for $x ge 0$. The negative values of $sqrt{lambda}$ can be ignored because they lead to duplicate values of $lambda$.
The non-negative solutions are ordered as follows:
$$
sqrt{lambda_0} = 0 < sqrt{lambda_1} < frac{pi}{2} < sqrt{lambda_2} < frac{3pi}{2} < sqrt{lambda_3} < frac{5pi}{2} < cdots
$$
Asymptotically, $sqrt{lambda_n} approx frac{(2n-1)pi}{2}$ or $lambda_n approx frac{(2n-1)^2pi^2}{4}$ as $nrightarrowinfty$.
edited 4 hours ago
answered 7 hours ago
DisintegratingByParts
57.5k42375
57.5k42375
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First goal, a better eigenfunction would be $$X(x) = Ax - A, quad forall A in mathbb{R}$$ Always try to keep it as general as possible. Second goal, applying both boundary conditions yields begin{align} beta &= -frac{A}{B} \ &= tan beta end{align} and using the oddness of $tan( cdot)$, we get $$lambda = tan^{2}(sqrt{lambda})$$
– Mattos
2 days ago