Mixing
Eigenvalue of a given operator
up vote
2
down vote
favorite
If $u_0$ is a positive radial symmetric nontrival solution of
$$
-frac{1}{2}frac{d^2u}{dx^2}+lambda u -u^3=0
$$
Then how to show $-3lambda$ is a eigenvalue of
$$
Lu=-frac{1}{2}frac{d^2u}{dx^2}+lambda u -3u_0^2 u
$$
and the corresponding eigenfunction is $u_0^2$ ?
What I try:
begin{align}
Lu_0^2
&=-frac{1}{2}frac{d^2u_0^2}{dx^2}+lambda u_0^2 -3u_0^4 \
&=-(frac{du_0}{dx})^2 -(u_0 frac{d^2 u_0}{dx^2}) + lambda u_0^2 -3 u_0^4 \
&=-(frac{du_0}{dx})^2-frac{1}{2}(u_0 frac{d^2 u_0}{dx^2})-2u_0^4
+ [-frac{1}{2}(u_0 frac{d^2 u_0}{dx^2})+ lambda u_0^2 - u_0^4] \
&=-(frac{du_0}{dx})^2-frac{1}{2}(u_0 frac{d^2 u_0}{dx^2})-2u_0^4
end{align}
then I don't how to deal the $-(frac{du_0}{dx})^2$, what should I do ?
differential-equations eigenvalues-eigenvectors eigenfunctions elliptic-equations
asked Nov 15 at 13:57
lanse7pty
1,7651823
add a comment |
up vote
2
down vote
favorite
If $u_0$ is a positive radial symmetric nontrival solution of
$$
-frac{1}{2}frac{d^2u}{dx^2}+lambda u -u^3=0
$$
Then how to show $-3lambda$ is a eigenvalue of
$$
Lu=-frac{1}{2}frac{d^2u}{dx^2}+lambda u -3u_0^2 u
$$
and the corresponding eigenfunction is $u_0^2$ ?
What I try:
begin{align}
Lu_0^2
&=-frac{1}{2}frac{d^2u_0^2}{dx^2}+lambda u_0^2 -3u_0^4 \
&=-(frac{du_0}{dx})^2 -(u_0 frac{d^2 u_0}{dx^2}) + lambda u_0^2 -3 u_0^4 \
&=-(frac{du_0}{dx})^2-frac{1}{2}(u_0 frac{d^2 u_0}{dx^2})-2u_0^4
+ [-frac{1}{2}(u_0 frac{d^2 u_0}{dx^2})+ lambda u_0^2 - u_0^4] \
&=-(frac{du_0}{dx})^2-frac{1}{2}(u_0 frac{d^2 u_0}{dx^2})-2u_0^4
end{align}
then I don't how to deal the $-(frac{du_0}{dx})^2$, what should I do ?
differential-equations eigenvalues-eigenvectors eigenfunctions elliptic-equations
asked Nov 15 at 13:57
lanse7pty
1,7651823
You could replace the second derivative term with the help of the original equation and then use and "integral of motion" of the same equation.
– minimax
Nov 15 at 14:33
1
Snookie is right. Something is missing. If $u_0=sqrt{lambda}$ (provided that $lambda>0$), then $u_0$ is positive and satisfies the first DE. But $Lu_0^2=-2lambda u_0^2$. So, $u_0^2$ is an eigenfunction of $L$ with eigenvalue $-2lambda$.
– Zvi
Nov 15 at 15:10
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If $u_0$ is a positive radial symmetric nontrival solution of
$$
-frac{1}{2}frac{d^2u}{dx^2}+lambda u -u^3=0
$$
Then how to show $-3lambda$ is a eigenvalue of
$$
Lu=-frac{1}{2}frac{d^2u}{dx^2}+lambda u -3u_0^2 u
$$
and the corresponding eigenfunction is $u_0^2$ ?
What I try:
begin{align}
Lu_0^2
&=-frac{1}{2}frac{d^2u_0^2}{dx^2}+lambda u_0^2 -3u_0^4 \
&=-(frac{du_0}{dx})^2 -(u_0 frac{d^2 u_0}{dx^2}) + lambda u_0^2 -3 u_0^4 \
&=-(frac{du_0}{dx})^2-frac{1}{2}(u_0 frac{d^2 u_0}{dx^2})-2u_0^4
+ [-frac{1}{2}(u_0 frac{d^2 u_0}{dx^2})+ lambda u_0^2 - u_0^4] \
&=-(frac{du_0}{dx})^2-frac{1}{2}(u_0 frac{d^2 u_0}{dx^2})-2u_0^4
end{align}
then I don't how to deal the $-(frac{du_0}{dx})^2$, what should I do ?
differential-equations eigenvalues-eigenvectors eigenfunctions elliptic-equations
asked Nov 15 at 13:57
lanse7pty
1,7651823
If $u_0$ is a positive radial symmetric nontrival solution of
$$
-frac{1}{2}frac{d^2u}{dx^2}+lambda u -u^3=0
$$
Then how to show $-3lambda$ is a eigenvalue of
$$
Lu=-frac{1}{2}frac{d^2u}{dx^2}+lambda u -3u_0^2 u
$$
and the corresponding eigenfunction is $u_0^2$ ?
What I try:
begin{align}
Lu_0^2
&=-frac{1}{2}frac{d^2u_0^2}{dx^2}+lambda u_0^2 -3u_0^4 \
&=-(frac{du_0}{dx})^2 -(u_0 frac{d^2 u_0}{dx^2}) + lambda u_0^2 -3 u_0^4 \
&=-(frac{du_0}{dx})^2-frac{1}{2}(u_0 frac{d^2 u_0}{dx^2})-2u_0^4
+ [-frac{1}{2}(u_0 frac{d^2 u_0}{dx^2})+ lambda u_0^2 - u_0^4] \
&=-(frac{du_0}{dx})^2-frac{1}{2}(u_0 frac{d^2 u_0}{dx^2})-2u_0^4
end{align}
then I don't how to deal the $-(frac{du_0}{dx})^2$, what should I do ?
differential-equations eigenvalues-eigenvectors eigenfunctions elliptic-equations
differential-equations eigenvalues-eigenvectors eigenfunctions elliptic-equations
asked Nov 15 at 13:57
lanse7pty
1,7651823
asked Nov 15 at 13:57
lanse7pty
1,7651823
edited 2 days ago
asked Nov 15 at 13:57
lanse7pty
1,7651823
asked Nov 15 at 13:57
lanse7pty
1,7651823
asked Nov 15 at 13:57
lanse7pty
1,7651823
1,7651823
You could replace the second derivative term with the help of the original equation and then use and "integral of motion" of the same equation.
– minimax
Nov 15 at 14:33
1
Snookie is right. Something is missing. If $u_0=sqrt{lambda}$ (provided that $lambda>0$), then $u_0$ is positive and satisfies the first DE. But $Lu_0^2=-2lambda u_0^2$. So, $u_0^2$ is an eigenfunction of $L$ with eigenvalue $-2lambda$.
– Zvi
Nov 15 at 15:10
add a comment |
You could replace the second derivative term with the help of the original equation and then use and "integral of motion" of the same equation.
– minimax
Nov 15 at 14:33
1
Snookie is right. Something is missing. If $u_0=sqrt{lambda}$ (provided that $lambda>0$), then $u_0$ is positive and satisfies the first DE. But $Lu_0^2=-2lambda u_0^2$. So, $u_0^2$ is an eigenfunction of $L$ with eigenvalue $-2lambda$.
– Zvi
Nov 15 at 15:10
You could replace the second derivative term with the help of the original equation and then use and "integral of motion" of the same equation.
– minimax
Nov 15 at 14:33
You could replace the second derivative term with the help of the original equation and then use and "integral of motion" of the same equation.
– minimax
Nov 15 at 14:33
1
1
Snookie is right. Something is missing. If $u_0=sqrt{lambda}$ (provided that $lambda>0$), then $u_0$ is positive and satisfies the first DE. But $Lu_0^2=-2lambda u_0^2$. So, $u_0^2$ is an eigenfunction of $L$ with eigenvalue $-2lambda$.
– Zvi
Nov 15 at 15:10
Snookie is right. Something is missing. If $u_0=sqrt{lambda}$ (provided that $lambda>0$), then $u_0$ is positive and satisfies the first DE. But $Lu_0^2=-2lambda u_0^2$. So, $u_0^2$ is an eigenfunction of $L$ with eigenvalue $-2lambda$.
– Zvi
Nov 15 at 15:10
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Here is my try. Let $v_0=frac{du_0}{dx}$. From
$$-frac12frac{d^2u_0}{dx^2}+lambda u_0-u_0^3=0implies frac{d^2u_0}{dx^2}=2lambda u_0-2u_0^4,$$
we have as you computed
$$Lu_0^2=-v_0^2-u_0frac{d^2u_0}{dx^2}+lambda u_0^2-3u_0^4=-v_0^2-u_0(2lambda u_0-2u_0^4)+lambda u_0^2-3u_0^4.$$
So $Lu_0^2=-v_0^2-lambda u_0^2+u_0^4$. Also from
$$-frac12frac{d^2u_0}{dx^2}+lambda u_0-u_0^3=0implies-frac{1}{2}v_0frac{dv_0}{dx}+lambda u_0frac{du_0}{dx}-u_0^3frac{du_0}{dx}=0,$$
we get by integrating wrt $x$ that
$$-frac{v_0^2}{4}+frac{lambda u_0^2}{2}-frac{u_0^4}{4}=c$$
for some constant $c$. So $-v_0^2+2lambda u_0^2-u_0^4=4c$. That is,
$$Lu_0^2=-v_0^2-lambda u_0^2+u_0^4=(-v_0^2+2lambda u_0^2-u_0^4)-3lambda u_0^2=4c-3lambda u_0^2.$$
Presumably, you have some conditions to make $c=0$ (is there anything missing?), because if $c=0$, $Lu_0^2=-3lambda u_0^2$ as required.
answered Nov 15 at 14:36
Snookie
3159
New contributor
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I think the idea is that $u_0$ is not just an arbitrary positive solution to the first DE. It has to be one that satisfies $$left(frac{du_0}{dx}right)^2=2lambda u_0^2-u_0^4.$$ That is, $c=0$.
– Zvi
Nov 15 at 15:13
Thanks very much. The $u_0$ is positive radial symmetry solution, since I feel the radial symmetry is not useful, I miss it. But , in fact, I still don't know how the radial symmetry make $c=0$.
– lanse7pty
Nov 16 at 1:34
@Zvi I miss the $u_0$ is radial symmetry. But I still don't know how to use it get $c=0$.
– lanse7pty
Nov 16 at 1:35
1
@lanse7pty It is probably a good idea if you could include the complete information of your question. The constant function $u_0equivsqrt{lambda}$ given by Zvi above is just about as symmetric as you can get, and it does not give $c=0$ for $lambda>0$ (i.e., $c=dfrac{lambda^2}{4}$). There has to be something else missing.
– Batominovski
Nov 16 at 23:07
1
@lanse7pty Using WolframAlpha, I cannot see why $c=0$ is the only possible answer. There is a nontrivial solution with $lambda=2$, $u_0(0)=1$, and $v_0(0)=u'_0(0)=0$ (for a radially symmetric solution, $v_0(0)=0$ must hold). For this solution, $c=dfrac34$. See here.
– Batominovski
2 days ago
|
show 1 more comment
up vote
1
down vote
This answer is a supplement to Snookie's answer above. All notations are borrowed from there. Some information still seems to be missing. Therefore, I assume that you want a non-periodic solution $u_0$. (Alternatively, you can require that $u_0$ is a nontrivial solution such that $limlimits_{xtopminfty},u_0(x)=0$ and the proof is essentially unchanged.)
Assuming that $x$ is the radius, then radial symmetry of $u_0$ implies that $v_0(0)=u_0'(x)=0$. Therefore, if $s:=u_0(0)$, then from Snookie's answer, we have $$2lambda s^2-s^4=4c,.$$
That is,
$$-v_0^2+2lambda u_0^2-u_0^4=2lambda s^2-s^4,.$$
Hence,
$$v_0=pmsqrt{(u_0^2-s^2),left(2lambda-u_0^2-s^2right)},.$$
That is,
$$frac{1}{sqrt{(u_0^2-s^2),left(2lambda-s^2-u_0^2right)}},left(frac{text{d}u_0}{text{d}x}right)=pm1tag{*},.$$
Note from $-v_0^2+2lambda,u_0-u_0^4=4c$, we have
$$(u_0^2-lambda)^2+v_0^2=lambda^2-4c,.$$
Thus, if we look at the $(u_0,v_0)$-diagram, then we see that any solution is either trivial or lies in a closed loop. A nonperiodic solution can only arise if the period of the trajectory in the corresponding closed loop in the $(u_0,v_0)$-diagram is infinite.
Without loss of generality, suppose that $sgeq 0$ (otherwise, note that swapping the sign of $u_0$ also yields a solution). If $sneq 0$ or $s^2neq 2lambda$, then there are three cases: $s^2<2lambda -s^2$, $s^2=2lambda-s^2$, and $s^2>2lambda-s^2$. If $s^2<2lambda-s^2$, then note from (*) that
$$pm x=int_{s}^{u_0(x)},frac{1}{sqrt{(t^2-s^2),left(2lambda-s^2-t^2right)}},text{d}t,.$$
Therefore, $u_0(x)$ lies between $s$ and $sqrt{2lambda-s^2}$, and so
$$|x|leq int_{s}^{sqrt{2lambda-s^2}},frac{1}{sqrt{(t^2-s^2),left(2lambda-s^2-t^2right)}},text{d}t<infty,,tag{#}$$
whence $u_0$ is periodic. (I omit the proof of (#) here, but it is due to the fact that $displaystyleint_0^y,frac{1}{sqrt{t}},text{d}t$ is finite for every $y>0$.)
Similarly, if $s^2>2lambda-s^2$, then according to (*), we have
$$pm x=int_s^{u_0(x)},frac{1}{sqrt{(t^2-2lambda+s^2),(s^2-t^2)}},text{d}t,.$$
This means $u_0(x)$ lies between $sqrt{max{0,2lambda-s^2}}$ and $s$, making
$$|x|leq int_{sqrt{max{0,2lambda-s^2}}}^{s},frac{1}{sqrt{(t^2-2lambda+s^2),(s^2-t^2)}},text{d}t<infty,.$$ Ergo, $u_0$ is periodic. Consequently, $s^2=lambda$ is the only possibility.
However, there exists a unique solution $u:=u_0$ to
$$-frac{1}{2},frac{text{d}^2u}{text{d}x^2}+lambda,u-u^3=0$$
provided that $u_0(0)$ and $u_0'(0)$ is known (this is due to the Picard-Lindelöf Theorem). In the case that $s^2=lambda$, we have $s=sqrt{lambda}$, and we already have one solution $u_0(x)=sqrt{lambda}$ for all $x$. Therefore, this is the only solution, which is a constant (whence periodic) solution. Therefore, the assumption that $sneq 0$ and $s^2neq 2lambda$ is false. However, if $s=0$, then we have another constant (whence periodic) solution $u_0equiv 0$. Hence, $s^2=2lambda$, which gives $c=0$. This is the only case that yields a non-periodic solution $u_0$.
For $lambda>0$ and $s=sqrt{2lambda}$, we have
$$u_0(x)=sqrt{2lambda},text{sech}(sqrt{2lambda},x),.$$
This solution is indeed non-periodic (and $limlimits_{xtopminfty},u_0(x)=0$). For $lambda leq 0$, there does not exist a nontrivial, non-periodic, radially symmetric solution $u_0$.
answered 2 days ago
Batominovski
31.3k23187
Thanks， what is $(u_0,v_0)$-diagram ? And what is closed loop ?
– lanse7pty
yesterday
I posted the image. The said diagram is the contour plot of $(u_0,v_0)$ with different $c$. In the plot above, I used $lambda:=1$. Note that each contour line forms a loop. There is a special contour line that seems to cross itself over in the figure-$8$ shape. That one corresponds to $c=0$.
– Batominovski
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Here is my try. Let $v_0=frac{du_0}{dx}$. From
$$-frac12frac{d^2u_0}{dx^2}+lambda u_0-u_0^3=0implies frac{d^2u_0}{dx^2}=2lambda u_0-2u_0^4,$$
we have as you computed
$$Lu_0^2=-v_0^2-u_0frac{d^2u_0}{dx^2}+lambda u_0^2-3u_0^4=-v_0^2-u_0(2lambda u_0-2u_0^4)+lambda u_0^2-3u_0^4.$$
So $Lu_0^2=-v_0^2-lambda u_0^2+u_0^4$. Also from
$$-frac12frac{d^2u_0}{dx^2}+lambda u_0-u_0^3=0implies-frac{1}{2}v_0frac{dv_0}{dx}+lambda u_0frac{du_0}{dx}-u_0^3frac{du_0}{dx}=0,$$
we get by integrating wrt $x$ that
$$-frac{v_0^2}{4}+frac{lambda u_0^2}{2}-frac{u_0^4}{4}=c$$
for some constant $c$. So $-v_0^2+2lambda u_0^2-u_0^4=4c$. That is,
$$Lu_0^2=-v_0^2-lambda u_0^2+u_0^4=(-v_0^2+2lambda u_0^2-u_0^4)-3lambda u_0^2=4c-3lambda u_0^2.$$
Presumably, you have some conditions to make $c=0$ (is there anything missing?), because if $c=0$, $Lu_0^2=-3lambda u_0^2$ as required.
answered Nov 15 at 14:36
Snookie
3159
New contributor
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I think the idea is that $u_0$ is not just an arbitrary positive solution to the first DE. It has to be one that satisfies $$left(frac{du_0}{dx}right)^2=2lambda u_0^2-u_0^4.$$ That is, $c=0$.
– Zvi
Nov 15 at 15:13
Thanks very much. The $u_0$ is positive radial symmetry solution, since I feel the radial symmetry is not useful, I miss it. But , in fact, I still don't know how the radial symmetry make $c=0$.
– lanse7pty
Nov 16 at 1:34
@Zvi I miss the $u_0$ is radial symmetry. But I still don't know how to use it get $c=0$.
– lanse7pty
Nov 16 at 1:35
1
@lanse7pty It is probably a good idea if you could include the complete information of your question. The constant function $u_0equivsqrt{lambda}$ given by Zvi above is just about as symmetric as you can get, and it does not give $c=0$ for $lambda>0$ (i.e., $c=dfrac{lambda^2}{4}$). There has to be something else missing.
– Batominovski
Nov 16 at 23:07
1
@lanse7pty Using WolframAlpha, I cannot see why $c=0$ is the only possible answer. There is a nontrivial solution with $lambda=2$, $u_0(0)=1$, and $v_0(0)=u'_0(0)=0$ (for a radially symmetric solution, $v_0(0)=0$ must hold). For this solution, $c=dfrac34$. See here.
– Batominovski
2 days ago
|
show 1 more comment
up vote
2
down vote
accepted
Here is my try. Let $v_0=frac{du_0}{dx}$. From
$$-frac12frac{d^2u_0}{dx^2}+lambda u_0-u_0^3=0implies frac{d^2u_0}{dx^2}=2lambda u_0-2u_0^4,$$
we have as you computed
$$Lu_0^2=-v_0^2-u_0frac{d^2u_0}{dx^2}+lambda u_0^2-3u_0^4=-v_0^2-u_0(2lambda u_0-2u_0^4)+lambda u_0^2-3u_0^4.$$
So $Lu_0^2=-v_0^2-lambda u_0^2+u_0^4$. Also from
$$-frac12frac{d^2u_0}{dx^2}+lambda u_0-u_0^3=0implies-frac{1}{2}v_0frac{dv_0}{dx}+lambda u_0frac{du_0}{dx}-u_0^3frac{du_0}{dx}=0,$$
we get by integrating wrt $x$ that
$$-frac{v_0^2}{4}+frac{lambda u_0^2}{2}-frac{u_0^4}{4}=c$$
for some constant $c$. So $-v_0^2+2lambda u_0^2-u_0^4=4c$. That is,
$$Lu_0^2=-v_0^2-lambda u_0^2+u_0^4=(-v_0^2+2lambda u_0^2-u_0^4)-3lambda u_0^2=4c-3lambda u_0^2.$$
Presumably, you have some conditions to make $c=0$ (is there anything missing?), because if $c=0$, $Lu_0^2=-3lambda u_0^2$ as required.
answered Nov 15 at 14:36
Snookie
3159
New contributor
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I think the idea is that $u_0$ is not just an arbitrary positive solution to the first DE. It has to be one that satisfies $$left(frac{du_0}{dx}right)^2=2lambda u_0^2-u_0^4.$$ That is, $c=0$.
– Zvi
Nov 15 at 15:13
Thanks very much. The $u_0$ is positive radial symmetry solution, since I feel the radial symmetry is not useful, I miss it. But , in fact, I still don't know how the radial symmetry make $c=0$.
– lanse7pty
Nov 16 at 1:34
@Zvi I miss the $u_0$ is radial symmetry. But I still don't know how to use it get $c=0$.
– lanse7pty
Nov 16 at 1:35
1
@lanse7pty It is probably a good idea if you could include the complete information of your question. The constant function $u_0equivsqrt{lambda}$ given by Zvi above is just about as symmetric as you can get, and it does not give $c=0$ for $lambda>0$ (i.e., $c=dfrac{lambda^2}{4}$). There has to be something else missing.
– Batominovski
Nov 16 at 23:07
1
@lanse7pty Using WolframAlpha, I cannot see why $c=0$ is the only possible answer. There is a nontrivial solution with $lambda=2$, $u_0(0)=1$, and $v_0(0)=u'_0(0)=0$ (for a radially symmetric solution, $v_0(0)=0$ must hold). For this solution, $c=dfrac34$. See here.
– Batominovski
2 days ago
|
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Here is my try. Let $v_0=frac{du_0}{dx}$. From
$$-frac12frac{d^2u_0}{dx^2}+lambda u_0-u_0^3=0implies frac{d^2u_0}{dx^2}=2lambda u_0-2u_0^4,$$
we have as you computed
$$Lu_0^2=-v_0^2-u_0frac{d^2u_0}{dx^2}+lambda u_0^2-3u_0^4=-v_0^2-u_0(2lambda u_0-2u_0^4)+lambda u_0^2-3u_0^4.$$
So $Lu_0^2=-v_0^2-lambda u_0^2+u_0^4$. Also from
$$-frac12frac{d^2u_0}{dx^2}+lambda u_0-u_0^3=0implies-frac{1}{2}v_0frac{dv_0}{dx}+lambda u_0frac{du_0}{dx}-u_0^3frac{du_0}{dx}=0,$$
we get by integrating wrt $x$ that
$$-frac{v_0^2}{4}+frac{lambda u_0^2}{2}-frac{u_0^4}{4}=c$$
for some constant $c$. So $-v_0^2+2lambda u_0^2-u_0^4=4c$. That is,
$$Lu_0^2=-v_0^2-lambda u_0^2+u_0^4=(-v_0^2+2lambda u_0^2-u_0^4)-3lambda u_0^2=4c-3lambda u_0^2.$$
Presumably, you have some conditions to make $c=0$ (is there anything missing?), because if $c=0$, $Lu_0^2=-3lambda u_0^2$ as required.
answered Nov 15 at 14:36
Snookie
3159
New contributor
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Here is my try. Let $v_0=frac{du_0}{dx}$. From
$$-frac12frac{d^2u_0}{dx^2}+lambda u_0-u_0^3=0implies frac{d^2u_0}{dx^2}=2lambda u_0-2u_0^4,$$
we have as you computed
$$Lu_0^2=-v_0^2-u_0frac{d^2u_0}{dx^2}+lambda u_0^2-3u_0^4=-v_0^2-u_0(2lambda u_0-2u_0^4)+lambda u_0^2-3u_0^4.$$
So $Lu_0^2=-v_0^2-lambda u_0^2+u_0^4$. Also from
$$-frac12frac{d^2u_0}{dx^2}+lambda u_0-u_0^3=0implies-frac{1}{2}v_0frac{dv_0}{dx}+lambda u_0frac{du_0}{dx}-u_0^3frac{du_0}{dx}=0,$$
we get by integrating wrt $x$ that
$$-frac{v_0^2}{4}+frac{lambda u_0^2}{2}-frac{u_0^4}{4}=c$$
for some constant $c$. So $-v_0^2+2lambda u_0^2-u_0^4=4c$. That is,
$$Lu_0^2=-v_0^2-lambda u_0^2+u_0^4=(-v_0^2+2lambda u_0^2-u_0^4)-3lambda u_0^2=4c-3lambda u_0^2.$$
Presumably, you have some conditions to make $c=0$ (is there anything missing?), because if $c=0$, $Lu_0^2=-3lambda u_0^2$ as required.
answered Nov 15 at 14:36
Snookie
3159
New contributor
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 15 at 14:36
Snookie
3159
New contributor
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 15 at 14:36
Snookie
3159
answered Nov 15 at 14:36
Snookie
3159
3159
New contributor
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I think the idea is that $u_0$ is not just an arbitrary positive solution to the first DE. It has to be one that satisfies $$left(frac{du_0}{dx}right)^2=2lambda u_0^2-u_0^4.$$ That is, $c=0$.
– Zvi
Nov 15 at 15:13
Thanks very much. The $u_0$ is positive radial symmetry solution, since I feel the radial symmetry is not useful, I miss it. But , in fact, I still don't know how the radial symmetry make $c=0$.
– lanse7pty
Nov 16 at 1:34
@Zvi I miss the $u_0$ is radial symmetry. But I still don't know how to use it get $c=0$.
– lanse7pty
Nov 16 at 1:35
1
@lanse7pty It is probably a good idea if you could include the complete information of your question. The constant function $u_0equivsqrt{lambda}$ given by Zvi above is just about as symmetric as you can get, and it does not give $c=0$ for $lambda>0$ (i.e., $c=dfrac{lambda^2}{4}$). There has to be something else missing.
– Batominovski
Nov 16 at 23:07
1
@lanse7pty Using WolframAlpha, I cannot see why $c=0$ is the only possible answer. There is a nontrivial solution with $lambda=2$, $u_0(0)=1$, and $v_0(0)=u'_0(0)=0$ (for a radially symmetric solution, $v_0(0)=0$ must hold). For this solution, $c=dfrac34$. See here.
– Batominovski
2 days ago
|
show 1 more comment
1
I think the idea is that $u_0$ is not just an arbitrary positive solution to the first DE. It has to be one that satisfies $$left(frac{du_0}{dx}right)^2=2lambda u_0^2-u_0^4.$$ That is, $c=0$.
– Zvi
Nov 15 at 15:13
Thanks very much. The $u_0$ is positive radial symmetry solution, since I feel the radial symmetry is not useful, I miss it. But , in fact, I still don't know how the radial symmetry make $c=0$.
– lanse7pty
Nov 16 at 1:34
@Zvi I miss the $u_0$ is radial symmetry. But I still don't know how to use it get $c=0$.
– lanse7pty
Nov 16 at 1:35
1
@lanse7pty It is probably a good idea if you could include the complete information of your question. The constant function $u_0equivsqrt{lambda}$ given by Zvi above is just about as symmetric as you can get, and it does not give $c=0$ for $lambda>0$ (i.e., $c=dfrac{lambda^2}{4}$). There has to be something else missing.
– Batominovski
Nov 16 at 23:07
1
@lanse7pty Using WolframAlpha, I cannot see why $c=0$ is the only possible answer. There is a nontrivial solution with $lambda=2$, $u_0(0)=1$, and $v_0(0)=u'_0(0)=0$ (for a radially symmetric solution, $v_0(0)=0$ must hold). For this solution, $c=dfrac34$. See here.
– Batominovski
2 days ago
1
1
I think the idea is that $u_0$ is not just an arbitrary positive solution to the first DE. It has to be one that satisfies $$left(frac{du_0}{dx}right)^2=2lambda u_0^2-u_0^4.$$ That is, $c=0$.
– Zvi
Nov 15 at 15:13
I think the idea is that $u_0$ is not just an arbitrary positive solution to the first DE. It has to be one that satisfies $$left(frac{du_0}{dx}right)^2=2lambda u_0^2-u_0^4.$$ That is, $c=0$.
– Zvi
Nov 15 at 15:13
Thanks very much. The $u_0$ is positive radial symmetry solution, since I feel the radial symmetry is not useful, I miss it. But , in fact, I still don't know how the radial symmetry make $c=0$.
– lanse7pty
Nov 16 at 1:34
Thanks very much. The $u_0$ is positive radial symmetry solution, since I feel the radial symmetry is not useful, I miss it. But , in fact, I still don't know how the radial symmetry make $c=0$.
– lanse7pty
Nov 16 at 1:34
@Zvi I miss the $u_0$ is radial symmetry. But I still don't know how to use it get $c=0$.
– lanse7pty
Nov 16 at 1:35
@Zvi I miss the $u_0$ is radial symmetry. But I still don't know how to use it get $c=0$.
– lanse7pty
Nov 16 at 1:35
1
1
@lanse7pty It is probably a good idea if you could include the complete information of your question. The constant function $u_0equivsqrt{lambda}$ given by Zvi above is just about as symmetric as you can get, and it does not give $c=0$ for $lambda>0$ (i.e., $c=dfrac{lambda^2}{4}$). There has to be something else missing.
– Batominovski
Nov 16 at 23:07
@lanse7pty It is probably a good idea if you could include the complete information of your question. The constant function $u_0equivsqrt{lambda}$ given by Zvi above is just about as symmetric as you can get, and it does not give $c=0$ for $lambda>0$ (i.e., $c=dfrac{lambda^2}{4}$). There has to be something else missing.
– Batominovski
Nov 16 at 23:07
1
1
@lanse7pty Using WolframAlpha, I cannot see why $c=0$ is the only possible answer. There is a nontrivial solution with $lambda=2$, $u_0(0)=1$, and $v_0(0)=u'_0(0)=0$ (for a radially symmetric solution, $v_0(0)=0$ must hold). For this solution, $c=dfrac34$. See here.
– Batominovski
2 days ago
@lanse7pty Using WolframAlpha, I cannot see why $c=0$ is the only possible answer. There is a nontrivial solution with $lambda=2$, $u_0(0)=1$, and $v_0(0)=u'_0(0)=0$ (for a radially symmetric solution, $v_0(0)=0$ must hold). For this solution, $c=dfrac34$. See here.
– Batominovski
2 days ago
|
show 1 more comment
up vote
1
down vote
This answer is a supplement to Snookie's answer above. All notations are borrowed from there. Some information still seems to be missing. Therefore, I assume that you want a non-periodic solution $u_0$. (Alternatively, you can require that $u_0$ is a nontrivial solution such that $limlimits_{xtopminfty},u_0(x)=0$ and the proof is essentially unchanged.)
Assuming that $x$ is the radius, then radial symmetry of $u_0$ implies that $v_0(0)=u_0'(x)=0$. Therefore, if $s:=u_0(0)$, then from Snookie's answer, we have $$2lambda s^2-s^4=4c,.$$
That is,
$$-v_0^2+2lambda u_0^2-u_0^4=2lambda s^2-s^4,.$$
Hence,
$$v_0=pmsqrt{(u_0^2-s^2),left(2lambda-u_0^2-s^2right)},.$$
That is,
$$frac{1}{sqrt{(u_0^2-s^2),left(2lambda-s^2-u_0^2right)}},left(frac{text{d}u_0}{text{d}x}right)=pm1tag{*},.$$
Note from $-v_0^2+2lambda,u_0-u_0^4=4c$, we have
$$(u_0^2-lambda)^2+v_0^2=lambda^2-4c,.$$
Thus, if we look at the $(u_0,v_0)$-diagram, then we see that any solution is either trivial or lies in a closed loop. A nonperiodic solution can only arise if the period of the trajectory in the corresponding closed loop in the $(u_0,v_0)$-diagram is infinite.
Without loss of generality, suppose that $sgeq 0$ (otherwise, note that swapping the sign of $u_0$ also yields a solution). If $sneq 0$ or $s^2neq 2lambda$, then there are three cases: $s^2<2lambda -s^2$, $s^2=2lambda-s^2$, and $s^2>2lambda-s^2$. If $s^2<2lambda-s^2$, then note from (*) that
$$pm x=int_{s}^{u_0(x)},frac{1}{sqrt{(t^2-s^2),left(2lambda-s^2-t^2right)}},text{d}t,.$$
Therefore, $u_0(x)$ lies between $s$ and $sqrt{2lambda-s^2}$, and so
$$|x|leq int_{s}^{sqrt{2lambda-s^2}},frac{1}{sqrt{(t^2-s^2),left(2lambda-s^2-t^2right)}},text{d}t<infty,,tag{#}$$
whence $u_0$ is periodic. (I omit the proof of (#) here, but it is due to the fact that $displaystyleint_0^y,frac{1}{sqrt{t}},text{d}t$ is finite for every $y>0$.)
Similarly, if $s^2>2lambda-s^2$, then according to (*), we have
$$pm x=int_s^{u_0(x)},frac{1}{sqrt{(t^2-2lambda+s^2),(s^2-t^2)}},text{d}t,.$$
This means $u_0(x)$ lies between $sqrt{max{0,2lambda-s^2}}$ and $s$, making
$$|x|leq int_{sqrt{max{0,2lambda-s^2}}}^{s},frac{1}{sqrt{(t^2-2lambda+s^2),(s^2-t^2)}},text{d}t<infty,.$$ Ergo, $u_0$ is periodic. Consequently, $s^2=lambda$ is the only possibility.
However, there exists a unique solution $u:=u_0$ to
$$-frac{1}{2},frac{text{d}^2u}{text{d}x^2}+lambda,u-u^3=0$$
provided that $u_0(0)$ and $u_0'(0)$ is known (this is due to the Picard-Lindelöf Theorem). In the case that $s^2=lambda$, we have $s=sqrt{lambda}$, and we already have one solution $u_0(x)=sqrt{lambda}$ for all $x$. Therefore, this is the only solution, which is a constant (whence periodic) solution. Therefore, the assumption that $sneq 0$ and $s^2neq 2lambda$ is false. However, if $s=0$, then we have another constant (whence periodic) solution $u_0equiv 0$. Hence, $s^2=2lambda$, which gives $c=0$. This is the only case that yields a non-periodic solution $u_0$.
For $lambda>0$ and $s=sqrt{2lambda}$, we have
$$u_0(x)=sqrt{2lambda},text{sech}(sqrt{2lambda},x),.$$
This solution is indeed non-periodic (and $limlimits_{xtopminfty},u_0(x)=0$). For $lambda leq 0$, there does not exist a nontrivial, non-periodic, radially symmetric solution $u_0$.
answered 2 days ago
Batominovski
31.3k23187
Thanks， what is $(u_0,v_0)$-diagram ? And what is closed loop ?
– lanse7pty
yesterday
I posted the image. The said diagram is the contour plot of $(u_0,v_0)$ with different $c$. In the plot above, I used $lambda:=1$. Note that each contour line forms a loop. There is a special contour line that seems to cross itself over in the figure-$8$ shape. That one corresponds to $c=0$.
– Batominovski
yesterday
add a comment |
up vote
1
down vote
This answer is a supplement to Snookie's answer above. All notations are borrowed from there. Some information still seems to be missing. Therefore, I assume that you want a non-periodic solution $u_0$. (Alternatively, you can require that $u_0$ is a nontrivial solution such that $limlimits_{xtopminfty},u_0(x)=0$ and the proof is essentially unchanged.)
Assuming that $x$ is the radius, then radial symmetry of $u_0$ implies that $v_0(0)=u_0'(x)=0$. Therefore, if $s:=u_0(0)$, then from Snookie's answer, we have $$2lambda s^2-s^4=4c,.$$
That is,
$$-v_0^2+2lambda u_0^2-u_0^4=2lambda s^2-s^4,.$$
Hence,
$$v_0=pmsqrt{(u_0^2-s^2),left(2lambda-u_0^2-s^2right)},.$$
That is,
$$frac{1}{sqrt{(u_0^2-s^2),left(2lambda-s^2-u_0^2right)}},left(frac{text{d}u_0}{text{d}x}right)=pm1tag{*},.$$
Note from $-v_0^2+2lambda,u_0-u_0^4=4c$, we have
$$(u_0^2-lambda)^2+v_0^2=lambda^2-4c,.$$
Thus, if we look at the $(u_0,v_0)$-diagram, then we see that any solution is either trivial or lies in a closed loop. A nonperiodic solution can only arise if the period of the trajectory in the corresponding closed loop in the $(u_0,v_0)$-diagram is infinite.
Without loss of generality, suppose that $sgeq 0$ (otherwise, note that swapping the sign of $u_0$ also yields a solution). If $sneq 0$ or $s^2neq 2lambda$, then there are three cases: $s^2<2lambda -s^2$, $s^2=2lambda-s^2$, and $s^2>2lambda-s^2$. If $s^2<2lambda-s^2$, then note from (*) that
$$pm x=int_{s}^{u_0(x)},frac{1}{sqrt{(t^2-s^2),left(2lambda-s^2-t^2right)}},text{d}t,.$$
Therefore, $u_0(x)$ lies between $s$ and $sqrt{2lambda-s^2}$, and so
$$|x|leq int_{s}^{sqrt{2lambda-s^2}},frac{1}{sqrt{(t^2-s^2),left(2lambda-s^2-t^2right)}},text{d}t<infty,,tag{#}$$
whence $u_0$ is periodic. (I omit the proof of (#) here, but it is due to the fact that $displaystyleint_0^y,frac{1}{sqrt{t}},text{d}t$ is finite for every $y>0$.)
Similarly, if $s^2>2lambda-s^2$, then according to (*), we have
$$pm x=int_s^{u_0(x)},frac{1}{sqrt{(t^2-2lambda+s^2),(s^2-t^2)}},text{d}t,.$$
This means $u_0(x)$ lies between $sqrt{max{0,2lambda-s^2}}$ and $s$, making
$$|x|leq int_{sqrt{max{0,2lambda-s^2}}}^{s},frac{1}{sqrt{(t^2-2lambda+s^2),(s^2-t^2)}},text{d}t<infty,.$$ Ergo, $u_0$ is periodic. Consequently, $s^2=lambda$ is the only possibility.
However, there exists a unique solution $u:=u_0$ to
$$-frac{1}{2},frac{text{d}^2u}{text{d}x^2}+lambda,u-u^3=0$$
provided that $u_0(0)$ and $u_0'(0)$ is known (this is due to the Picard-Lindelöf Theorem). In the case that $s^2=lambda$, we have $s=sqrt{lambda}$, and we already have one solution $u_0(x)=sqrt{lambda}$ for all $x$. Therefore, this is the only solution, which is a constant (whence periodic) solution. Therefore, the assumption that $sneq 0$ and $s^2neq 2lambda$ is false. However, if $s=0$, then we have another constant (whence periodic) solution $u_0equiv 0$. Hence, $s^2=2lambda$, which gives $c=0$. This is the only case that yields a non-periodic solution $u_0$.
For $lambda>0$ and $s=sqrt{2lambda}$, we have
$$u_0(x)=sqrt{2lambda},text{sech}(sqrt{2lambda},x),.$$
This solution is indeed non-periodic (and $limlimits_{xtopminfty},u_0(x)=0$). For $lambda leq 0$, there does not exist a nontrivial, non-periodic, radially symmetric solution $u_0$.
answered 2 days ago
Batominovski
31.3k23187
Thanks， what is $(u_0,v_0)$-diagram ? And what is closed loop ?
– lanse7pty
yesterday
I posted the image. The said diagram is the contour plot of $(u_0,v_0)$ with different $c$. In the plot above, I used $lambda:=1$. Note that each contour line forms a loop. There is a special contour line that seems to cross itself over in the figure-$8$ shape. That one corresponds to $c=0$.
– Batominovski
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
This answer is a supplement to Snookie's answer above. All notations are borrowed from there. Some information still seems to be missing. Therefore, I assume that you want a non-periodic solution $u_0$. (Alternatively, you can require that $u_0$ is a nontrivial solution such that $limlimits_{xtopminfty},u_0(x)=0$ and the proof is essentially unchanged.)
Assuming that $x$ is the radius, then radial symmetry of $u_0$ implies that $v_0(0)=u_0'(x)=0$. Therefore, if $s:=u_0(0)$, then from Snookie's answer, we have $$2lambda s^2-s^4=4c,.$$
That is,
$$-v_0^2+2lambda u_0^2-u_0^4=2lambda s^2-s^4,.$$
Hence,
$$v_0=pmsqrt{(u_0^2-s^2),left(2lambda-u_0^2-s^2right)},.$$
That is,
$$frac{1}{sqrt{(u_0^2-s^2),left(2lambda-s^2-u_0^2right)}},left(frac{text{d}u_0}{text{d}x}right)=pm1tag{*},.$$
Note from $-v_0^2+2lambda,u_0-u_0^4=4c$, we have
$$(u_0^2-lambda)^2+v_0^2=lambda^2-4c,.$$
Thus, if we look at the $(u_0,v_0)$-diagram, then we see that any solution is either trivial or lies in a closed loop. A nonperiodic solution can only arise if the period of the trajectory in the corresponding closed loop in the $(u_0,v_0)$-diagram is infinite.
Without loss of generality, suppose that $sgeq 0$ (otherwise, note that swapping the sign of $u_0$ also yields a solution). If $sneq 0$ or $s^2neq 2lambda$, then there are three cases: $s^2<2lambda -s^2$, $s^2=2lambda-s^2$, and $s^2>2lambda-s^2$. If $s^2<2lambda-s^2$, then note from (*) that
$$pm x=int_{s}^{u_0(x)},frac{1}{sqrt{(t^2-s^2),left(2lambda-s^2-t^2right)}},text{d}t,.$$
Therefore, $u_0(x)$ lies between $s$ and $sqrt{2lambda-s^2}$, and so
$$|x|leq int_{s}^{sqrt{2lambda-s^2}},frac{1}{sqrt{(t^2-s^2),left(2lambda-s^2-t^2right)}},text{d}t<infty,,tag{#}$$
whence $u_0$ is periodic. (I omit the proof of (#) here, but it is due to the fact that $displaystyleint_0^y,frac{1}{sqrt{t}},text{d}t$ is finite for every $y>0$.)
Similarly, if $s^2>2lambda-s^2$, then according to (*), we have
$$pm x=int_s^{u_0(x)},frac{1}{sqrt{(t^2-2lambda+s^2),(s^2-t^2)}},text{d}t,.$$
This means $u_0(x)$ lies between $sqrt{max{0,2lambda-s^2}}$ and $s$, making
$$|x|leq int_{sqrt{max{0,2lambda-s^2}}}^{s},frac{1}{sqrt{(t^2-2lambda+s^2),(s^2-t^2)}},text{d}t<infty,.$$ Ergo, $u_0$ is periodic. Consequently, $s^2=lambda$ is the only possibility.
However, there exists a unique solution $u:=u_0$ to
$$-frac{1}{2},frac{text{d}^2u}{text{d}x^2}+lambda,u-u^3=0$$
provided that $u_0(0)$ and $u_0'(0)$ is known (this is due to the Picard-Lindelöf Theorem). In the case that $s^2=lambda$, we have $s=sqrt{lambda}$, and we already have one solution $u_0(x)=sqrt{lambda}$ for all $x$. Therefore, this is the only solution, which is a constant (whence periodic) solution. Therefore, the assumption that $sneq 0$ and $s^2neq 2lambda$ is false. However, if $s=0$, then we have another constant (whence periodic) solution $u_0equiv 0$. Hence, $s^2=2lambda$, which gives $c=0$. This is the only case that yields a non-periodic solution $u_0$.
For $lambda>0$ and $s=sqrt{2lambda}$, we have
$$u_0(x)=sqrt{2lambda},text{sech}(sqrt{2lambda},x),.$$
This solution is indeed non-periodic (and $limlimits_{xtopminfty},u_0(x)=0$). For $lambda leq 0$, there does not exist a nontrivial, non-periodic, radially symmetric solution $u_0$.
answered 2 days ago
Batominovski
31.3k23187
This answer is a supplement to Snookie's answer above. All notations are borrowed from there. Some information still seems to be missing. Therefore, I assume that you want a non-periodic solution $u_0$. (Alternatively, you can require that $u_0$ is a nontrivial solution such that $limlimits_{xtopminfty},u_0(x)=0$ and the proof is essentially unchanged.)
Assuming that $x$ is the radius, then radial symmetry of $u_0$ implies that $v_0(0)=u_0'(x)=0$. Therefore, if $s:=u_0(0)$, then from Snookie's answer, we have $$2lambda s^2-s^4=4c,.$$
That is,
$$-v_0^2+2lambda u_0^2-u_0^4=2lambda s^2-s^4,.$$
Hence,
$$v_0=pmsqrt{(u_0^2-s^2),left(2lambda-u_0^2-s^2right)},.$$
That is,
$$frac{1}{sqrt{(u_0^2-s^2),left(2lambda-s^2-u_0^2right)}},left(frac{text{d}u_0}{text{d}x}right)=pm1tag{*},.$$
Note from $-v_0^2+2lambda,u_0-u_0^4=4c$, we have
$$(u_0^2-lambda)^2+v_0^2=lambda^2-4c,.$$
Thus, if we look at the $(u_0,v_0)$-diagram, then we see that any solution is either trivial or lies in a closed loop. A nonperiodic solution can only arise if the period of the trajectory in the corresponding closed loop in the $(u_0,v_0)$-diagram is infinite.
Without loss of generality, suppose that $sgeq 0$ (otherwise, note that swapping the sign of $u_0$ also yields a solution). If $sneq 0$ or $s^2neq 2lambda$, then there are three cases: $s^2<2lambda -s^2$, $s^2=2lambda-s^2$, and $s^2>2lambda-s^2$. If $s^2<2lambda-s^2$, then note from (*) that
$$pm x=int_{s}^{u_0(x)},frac{1}{sqrt{(t^2-s^2),left(2lambda-s^2-t^2right)}},text{d}t,.$$
Therefore, $u_0(x)$ lies between $s$ and $sqrt{2lambda-s^2}$, and so
$$|x|leq int_{s}^{sqrt{2lambda-s^2}},frac{1}{sqrt{(t^2-s^2),left(2lambda-s^2-t^2right)}},text{d}t<infty,,tag{#}$$
whence $u_0$ is periodic. (I omit the proof of (#) here, but it is due to the fact that $displaystyleint_0^y,frac{1}{sqrt{t}},text{d}t$ is finite for every $y>0$.)
Similarly, if $s^2>2lambda-s^2$, then according to (*), we have
$$pm x=int_s^{u_0(x)},frac{1}{sqrt{(t^2-2lambda+s^2),(s^2-t^2)}},text{d}t,.$$
This means $u_0(x)$ lies between $sqrt{max{0,2lambda-s^2}}$ and $s$, making
$$|x|leq int_{sqrt{max{0,2lambda-s^2}}}^{s},frac{1}{sqrt{(t^2-2lambda+s^2),(s^2-t^2)}},text{d}t<infty,.$$ Ergo, $u_0$ is periodic. Consequently, $s^2=lambda$ is the only possibility.
However, there exists a unique solution $u:=u_0$ to
$$-frac{1}{2},frac{text{d}^2u}{text{d}x^2}+lambda,u-u^3=0$$
provided that $u_0(0)$ and $u_0'(0)$ is known (this is due to the Picard-Lindelöf Theorem). In the case that $s^2=lambda$, we have $s=sqrt{lambda}$, and we already have one solution $u_0(x)=sqrt{lambda}$ for all $x$. Therefore, this is the only solution, which is a constant (whence periodic) solution. Therefore, the assumption that $sneq 0$ and $s^2neq 2lambda$ is false. However, if $s=0$, then we have another constant (whence periodic) solution $u_0equiv 0$. Hence, $s^2=2lambda$, which gives $c=0$. This is the only case that yields a non-periodic solution $u_0$.
For $lambda>0$ and $s=sqrt{2lambda}$, we have
$$u_0(x)=sqrt{2lambda},text{sech}(sqrt{2lambda},x),.$$
This solution is indeed non-periodic (and $limlimits_{xtopminfty},u_0(x)=0$). For $lambda leq 0$, there does not exist a nontrivial, non-periodic, radially symmetric solution $u_0$.
answered 2 days ago
Batominovski
31.3k23187
edited yesterday
answered 2 days ago
Batominovski
31.3k23187
answered 2 days ago
Batominovski
31.3k23187
answered 2 days ago
Batominovski
31.3k23187
31.3k23187
Thanks， what is $(u_0,v_0)$-diagram ? And what is closed loop ?
– lanse7pty
yesterday
I posted the image. The said diagram is the contour plot of $(u_0,v_0)$ with different $c$. In the plot above, I used $lambda:=1$. Note that each contour line forms a loop. There is a special contour line that seems to cross itself over in the figure-$8$ shape. That one corresponds to $c=0$.
– Batominovski
yesterday
add a comment |
Thanks， what is $(u_0,v_0)$-diagram ? And what is closed loop ?
– lanse7pty
yesterday
I posted the image. The said diagram is the contour plot of $(u_0,v_0)$ with different $c$. In the plot above, I used $lambda:=1$. Note that each contour line forms a loop. There is a special contour line that seems to cross itself over in the figure-$8$ shape. That one corresponds to $c=0$.
– Batominovski
yesterday
Thanks， what is $(u_0,v_0)$-diagram ? And what is closed loop ?
– lanse7pty
yesterday
Thanks， what is $(u_0,v_0)$-diagram ? And what is closed loop ?
– lanse7pty
yesterday
I posted the image. The said diagram is the contour plot of $(u_0,v_0)$ with different $c$. In the plot above, I used $lambda:=1$. Note that each contour line forms a loop. There is a special contour line that seems to cross itself over in the figure-$8$ shape. That one corresponds to $c=0$.
– Batominovski
yesterday
I posted the image. The said diagram is the contour plot of $(u_0,v_0)$ with different $c$. In the plot above, I used $lambda:=1$. Note that each contour line forms a loop. There is a special contour line that seems to cross itself over in the figure-$8$ shape. That one corresponds to $c=0$.
– Batominovski
yesterday
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999730%2feigenvalue-of-a-given-operator%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
You could replace the second derivative term with the help of the original equation and then use and "integral of motion" of the same equation.
– minimax
Nov 15 at 14:33
1
Snookie is right. Something is missing. If $u_0=sqrt{lambda}$ (provided that $lambda>0$), then $u_0$ is positive and satisfies the first DE. But $Lu_0^2=-2lambda u_0^2$. So, $u_0^2$ is an eigenfunction of $L$ with eigenvalue $-2lambda$.
– Zvi
Nov 15 at 15:10