About positive eigenvectors of $-Delta$ on $mathbb R^N$
$begingroup$
Im interested whether there exists positive solutions of
$$
(*)begin{cases}
-Delta u=lambda u\ uin H^1(mathbb R^N)
end{cases}
$$
for some $lambdainmathbb R$. Here, we hve $-Delta=-sumpartial_i^2$.
On the one hand, it es well known that in case $N=1$ the equation $-u''=lambda u$ has the general solution
$$
begin{cases} asin(sqrt{lambda}x)+bcos(sqrt{lambda}x) & text{ if }lambda>0\
at+b & text{ if }lambda=0\
ae^{sqrt{|lambda|}x}+be^{-sqrt{|lambda|}x} & text{ if }lambda<0
end{cases}
$$
But obviously non of these are square integrable. So I claimed
There is no nontrivial and positive solution of $(*)$ for $Nleq 4$.
I would appreciate if you can verify if my proof it correct or point out if I missed a point.
Proof:
Let be $lambdageq 0$ and $uin H^1(mathbb R^N)$ a positive solution of $(*)$. Then we get $-Delta u=lambda ugeq 0$. In this case, there are theorems which claims for $Ngeq 4$ that $u$ has to be constantly zero.
So, we have to consider the case $lambda<0$. A solution of $(*)$ is a critical point of the function $I(u):=frac12|nabla u|_2^2-frac{lambda}2|u|_2^2$, where $|~cdot~|_2$ denotes the $L^2(mathbb R^N)$-norm. But then we get
$$
0=nabla I(u)[u]=|nabla u|_2^2-lambda|u|_2^2=|nabla u|_2^2+|lambda|cdot|u|_2^2.
$$
But this yields $uequiv 0$.
End of the proof
Ok, how about the eigenvectors of $-Delta$ for $Ngeq 2$? The conclusion is, that the eigenvectors of $-Delta$ have either sign changes or they are not in $L^2(mathbb R^N)$. Are there some eigenvectors explicitly known?
functional-analysis pde
asked Jan 23 at 19:55
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
add a comment |
$begingroup$
Im interested whether there exists positive solutions of
$$
(*)begin{cases}
-Delta u=lambda u\ uin H^1(mathbb R^N)
end{cases}
$$
for some $lambdainmathbb R$. Here, we hve $-Delta=-sumpartial_i^2$.
On the one hand, it es well known that in case $N=1$ the equation $-u''=lambda u$ has the general solution
$$
begin{cases} asin(sqrt{lambda}x)+bcos(sqrt{lambda}x) & text{ if }lambda>0\
at+b & text{ if }lambda=0\
ae^{sqrt{|lambda|}x}+be^{-sqrt{|lambda|}x} & text{ if }lambda<0
end{cases}
$$
But obviously non of these are square integrable. So I claimed
There is no nontrivial and positive solution of $(*)$ for $Nleq 4$.
I would appreciate if you can verify if my proof it correct or point out if I missed a point.
Proof:
Let be $lambdageq 0$ and $uin H^1(mathbb R^N)$ a positive solution of $(*)$. Then we get $-Delta u=lambda ugeq 0$. In this case, there are theorems which claims for $Ngeq 4$ that $u$ has to be constantly zero.
So, we have to consider the case $lambda<0$. A solution of $(*)$ is a critical point of the function $I(u):=frac12|nabla u|_2^2-frac{lambda}2|u|_2^2$, where $|~cdot~|_2$ denotes the $L^2(mathbb R^N)$-norm. But then we get
$$
0=nabla I(u)[u]=|nabla u|_2^2-lambda|u|_2^2=|nabla u|_2^2+|lambda|cdot|u|_2^2.
$$
But this yields $uequiv 0$.
End of the proof
Ok, how about the eigenvectors of $-Delta$ for $Ngeq 2$? The conclusion is, that the eigenvectors of $-Delta$ have either sign changes or they are not in $L^2(mathbb R^N)$. Are there some eigenvectors explicitly known?
functional-analysis pde
asked Jan 23 at 19:55
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
$begingroup$
The Laplacian has no eigenfunctions in $L^2$. This can be easily verified using Fourier transform. See also this closely related question: math.stackexchange.com/questions/790401/… and probably many others on this site.
$endgroup$
– MaoWao
Jan 24 at 10:28
$begingroup$
@MaoWao Nice, I see. So, assume $f$ is an $L^2$ eigenfunction to an eigenvalue $lambdainmathbb C$, then using Fourier transformation gives $(|xi|^2-lambda)hat f=0$, which implies $hat f=0$ and then $f=0$. Is it correct?
$endgroup$
– Mundron Schmidt
Jan 25 at 21:58
$begingroup$
Yes, that is correct.
$endgroup$
– MaoWao
Jan 31 at 10:11
add a comment |
$begingroup$
Im interested whether there exists positive solutions of
$$
(*)begin{cases}
-Delta u=lambda u\ uin H^1(mathbb R^N)
end{cases}
$$
for some $lambdainmathbb R$. Here, we hve $-Delta=-sumpartial_i^2$.
On the one hand, it es well known that in case $N=1$ the equation $-u''=lambda u$ has the general solution
$$
begin{cases} asin(sqrt{lambda}x)+bcos(sqrt{lambda}x) & text{ if }lambda>0\
at+b & text{ if }lambda=0\
ae^{sqrt{|lambda|}x}+be^{-sqrt{|lambda|}x} & text{ if }lambda<0
end{cases}
$$
But obviously non of these are square integrable. So I claimed
There is no nontrivial and positive solution of $(*)$ for $Nleq 4$.
I would appreciate if you can verify if my proof it correct or point out if I missed a point.
Proof:
Let be $lambdageq 0$ and $uin H^1(mathbb R^N)$ a positive solution of $(*)$. Then we get $-Delta u=lambda ugeq 0$. In this case, there are theorems which claims for $Ngeq 4$ that $u$ has to be constantly zero.
So, we have to consider the case $lambda<0$. A solution of $(*)$ is a critical point of the function $I(u):=frac12|nabla u|_2^2-frac{lambda}2|u|_2^2$, where $|~cdot~|_2$ denotes the $L^2(mathbb R^N)$-norm. But then we get
$$
0=nabla I(u)[u]=|nabla u|_2^2-lambda|u|_2^2=|nabla u|_2^2+|lambda|cdot|u|_2^2.
$$
But this yields $uequiv 0$.
End of the proof
Ok, how about the eigenvectors of $-Delta$ for $Ngeq 2$? The conclusion is, that the eigenvectors of $-Delta$ have either sign changes or they are not in $L^2(mathbb R^N)$. Are there some eigenvectors explicitly known?
functional-analysis pde
asked Jan 23 at 19:55
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
Im interested whether there exists positive solutions of
$$
(*)begin{cases}
-Delta u=lambda u\ uin H^1(mathbb R^N)
end{cases}
$$
for some $lambdainmathbb R$. Here, we hve $-Delta=-sumpartial_i^2$.
On the one hand, it es well known that in case $N=1$ the equation $-u''=lambda u$ has the general solution
$$
begin{cases} asin(sqrt{lambda}x)+bcos(sqrt{lambda}x) & text{ if }lambda>0\
at+b & text{ if }lambda=0\
ae^{sqrt{|lambda|}x}+be^{-sqrt{|lambda|}x} & text{ if }lambda<0
end{cases}
$$
But obviously non of these are square integrable. So I claimed
There is no nontrivial and positive solution of $(*)$ for $Nleq 4$.
I would appreciate if you can verify if my proof it correct or point out if I missed a point.
Proof:
Let be $lambdageq 0$ and $uin H^1(mathbb R^N)$ a positive solution of $(*)$. Then we get $-Delta u=lambda ugeq 0$. In this case, there are theorems which claims for $Ngeq 4$ that $u$ has to be constantly zero.
So, we have to consider the case $lambda<0$. A solution of $(*)$ is a critical point of the function $I(u):=frac12|nabla u|_2^2-frac{lambda}2|u|_2^2$, where $|~cdot~|_2$ denotes the $L^2(mathbb R^N)$-norm. But then we get
$$
0=nabla I(u)[u]=|nabla u|_2^2-lambda|u|_2^2=|nabla u|_2^2+|lambda|cdot|u|_2^2.
$$
But this yields $uequiv 0$.
End of the proof
Ok, how about the eigenvectors of $-Delta$ for $Ngeq 2$? The conclusion is, that the eigenvectors of $-Delta$ have either sign changes or they are not in $L^2(mathbb R^N)$. Are there some eigenvectors explicitly known?
functional-analysis pde
functional-analysis pde
asked Jan 23 at 19:55
Mundron SchmidtMundron Schmidt
7,5042729
asked Jan 23 at 19:55
Mundron SchmidtMundron Schmidt
7,5042729
asked Jan 23 at 19:55
Mundron SchmidtMundron Schmidt
7,5042729
asked Jan 23 at 19:55
Mundron SchmidtMundron Schmidt
7,5042729
asked Jan 23 at 19:55
Mundron SchmidtMundron Schmidt
7,5042729
7,5042729
$begingroup$
The Laplacian has no eigenfunctions in $L^2$. This can be easily verified using Fourier transform. See also this closely related question: math.stackexchange.com/questions/790401/… and probably many others on this site.
$endgroup$
– MaoWao
Jan 24 at 10:28
$begingroup$
@MaoWao Nice, I see. So, assume $f$ is an $L^2$ eigenfunction to an eigenvalue $lambdainmathbb C$, then using Fourier transformation gives $(|xi|^2-lambda)hat f=0$, which implies $hat f=0$ and then $f=0$. Is it correct?
$endgroup$
– Mundron Schmidt
Jan 25 at 21:58
$begingroup$
Yes, that is correct.
$endgroup$
– MaoWao
Jan 31 at 10:11
add a comment |
$begingroup$
The Laplacian has no eigenfunctions in $L^2$. This can be easily verified using Fourier transform. See also this closely related question: math.stackexchange.com/questions/790401/… and probably many others on this site.
$endgroup$
– MaoWao
Jan 24 at 10:28
$begingroup$
@MaoWao Nice, I see. So, assume $f$ is an $L^2$ eigenfunction to an eigenvalue $lambdainmathbb C$, then using Fourier transformation gives $(|xi|^2-lambda)hat f=0$, which implies $hat f=0$ and then $f=0$. Is it correct?
$endgroup$
– Mundron Schmidt
Jan 25 at 21:58
$begingroup$
Yes, that is correct.
$endgroup$
– MaoWao
Jan 31 at 10:11
$begingroup$
The Laplacian has no eigenfunctions in $L^2$. This can be easily verified using Fourier transform. See also this closely related question: math.stackexchange.com/questions/790401/… and probably many others on this site.
$endgroup$
– MaoWao
Jan 24 at 10:28
$begingroup$
The Laplacian has no eigenfunctions in $L^2$. This can be easily verified using Fourier transform. See also this closely related question: math.stackexchange.com/questions/790401/… and probably many others on this site.
$endgroup$
– MaoWao
Jan 24 at 10:28
$begingroup$
@MaoWao Nice, I see. So, assume $f$ is an $L^2$ eigenfunction to an eigenvalue $lambdainmathbb C$, then using Fourier transformation gives $(|xi|^2-lambda)hat f=0$, which implies $hat f=0$ and then $f=0$. Is it correct?
$endgroup$
– Mundron Schmidt
Jan 25 at 21:58
$begingroup$
@MaoWao Nice, I see. So, assume $f$ is an $L^2$ eigenfunction to an eigenvalue $lambdainmathbb C$, then using Fourier transformation gives $(|xi|^2-lambda)hat f=0$, which implies $hat f=0$ and then $f=0$. Is it correct?
$endgroup$
– Mundron Schmidt
Jan 25 at 21:58
$begingroup$
Yes, that is correct.
$endgroup$
– MaoWao
Jan 31 at 10:11
$begingroup$
Yes, that is correct.
$endgroup$
– MaoWao
Jan 31 at 10:11
add a comment |
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$begingroup$
The Laplacian has no eigenfunctions in $L^2$. This can be easily verified using Fourier transform. See also this closely related question: math.stackexchange.com/questions/790401/… and probably many others on this site.
$endgroup$
– MaoWao
Jan 24 at 10:28
$begingroup$
@MaoWao Nice, I see. So, assume $f$ is an $L^2$ eigenfunction to an eigenvalue $lambdainmathbb C$, then using Fourier transformation gives $(|xi|^2-lambda)hat f=0$, which implies $hat f=0$ and then $f=0$. Is it correct?
$endgroup$
– Mundron Schmidt
Jan 25 at 21:58
$begingroup$
Yes, that is correct.
$endgroup$
– MaoWao
Jan 31 at 10:11