Can you have solution for the equation $|nabla f|^2=fDelta f$, for a homogenous polynomial $f$ with...
$begingroup$
Consider the following equation
begin{equation}
|nabla f|^2=fDelta f,
end{equation}
where $nabla f$ is the gradient of $f$ and $Delta f$ is the Laplacian of $f$.
Does this equation have a solution $fcolon mathbb{R}^nto mathbb{R}$ such that $f$ is a homogenous polynomial with $deg(f)>2$ and $n>1$?
ordinary-differential-equations pde
edited Jan 7 at 21:47
jgon
13.6k22041
asked Jan 7 at 16:32
user73454user73454
487314
$endgroup$
add a comment |
$begingroup$
Consider the following equation
begin{equation}
|nabla f|^2=fDelta f,
end{equation}
where $nabla f$ is the gradient of $f$ and $Delta f$ is the Laplacian of $f$.
Does this equation have a solution $fcolon mathbb{R}^nto mathbb{R}$ such that $f$ is a homogenous polynomial with $deg(f)>2$ and $n>1$?
ordinary-differential-equations pde
edited Jan 7 at 21:47
jgon
13.6k22041
asked Jan 7 at 16:32
user73454user73454
487314
$endgroup$
5
$begingroup$
Essentially you are asking for $Deltaln|f|=0$?
$endgroup$
– LutzL
Jan 7 at 21:55
$begingroup$
Maybe you are right also, but I think the equation above is equivalent to $Deltaln f^2=0$.
$endgroup$
– user73454
Jan 8 at 13:48
$begingroup$
As $ln f^2=2ln|f|$, this constant factor does not change the equation, as the Laplace operator is linear.
$endgroup$
– LutzL
Jan 8 at 14:02
$begingroup$
@LutzL, yes, you are right, thanks! Do you have any idea if my question has a positive or negative answer?
$endgroup$
– user73454
Jan 8 at 14:34
$begingroup$
@user73454 In case the solutions suggested previously look pathological for you for some reason, I updated my answer to include solutions given by symmetric polynomials.
$endgroup$
– Start wearing purple
Jan 9 at 16:15
add a comment |
$begingroup$
Consider the following equation
begin{equation}
|nabla f|^2=fDelta f,
end{equation}
where $nabla f$ is the gradient of $f$ and $Delta f$ is the Laplacian of $f$.
Does this equation have a solution $fcolon mathbb{R}^nto mathbb{R}$ such that $f$ is a homogenous polynomial with $deg(f)>2$ and $n>1$?
ordinary-differential-equations pde
edited Jan 7 at 21:47
jgon
13.6k22041
asked Jan 7 at 16:32
user73454user73454
487314
$endgroup$
Consider the following equation
begin{equation}
|nabla f|^2=fDelta f,
end{equation}
where $nabla f$ is the gradient of $f$ and $Delta f$ is the Laplacian of $f$.
Does this equation have a solution $fcolon mathbb{R}^nto mathbb{R}$ such that $f$ is a homogenous polynomial with $deg(f)>2$ and $n>1$?
ordinary-differential-equations pde
ordinary-differential-equations pde
edited Jan 7 at 21:47
jgon
13.6k22041
asked Jan 7 at 16:32
user73454user73454
487314
edited Jan 7 at 21:47
jgon
13.6k22041
asked Jan 7 at 16:32
user73454user73454
487314
edited Jan 7 at 21:47
jgon
13.6k22041
edited Jan 7 at 21:47
jgon
13.6k22041
edited Jan 7 at 21:47
jgon
13.6k22041
13.6k22041
asked Jan 7 at 16:32
user73454user73454
487314
asked Jan 7 at 16:32
user73454user73454
487314
asked Jan 7 at 16:32
user73454user73454
487314
487314
5
$begingroup$
Essentially you are asking for $Deltaln|f|=0$?
$endgroup$
– LutzL
Jan 7 at 21:55
$begingroup$
Maybe you are right also, but I think the equation above is equivalent to $Deltaln f^2=0$.
$endgroup$
– user73454
Jan 8 at 13:48
$begingroup$
As $ln f^2=2ln|f|$, this constant factor does not change the equation, as the Laplace operator is linear.
$endgroup$
– LutzL
Jan 8 at 14:02
$begingroup$
@LutzL, yes, you are right, thanks! Do you have any idea if my question has a positive or negative answer?
$endgroup$
– user73454
Jan 8 at 14:34
$begingroup$
@user73454 In case the solutions suggested previously look pathological for you for some reason, I updated my answer to include solutions given by symmetric polynomials.
$endgroup$
– Start wearing purple
Jan 9 at 16:15
add a comment |
5
$begingroup$
Essentially you are asking for $Deltaln|f|=0$?
$endgroup$
– LutzL
Jan 7 at 21:55
$begingroup$
Maybe you are right also, but I think the equation above is equivalent to $Deltaln f^2=0$.
$endgroup$
– user73454
Jan 8 at 13:48
$begingroup$
As $ln f^2=2ln|f|$, this constant factor does not change the equation, as the Laplace operator is linear.
$endgroup$
– LutzL
Jan 8 at 14:02
$begingroup$
@LutzL, yes, you are right, thanks! Do you have any idea if my question has a positive or negative answer?
$endgroup$
– user73454
Jan 8 at 14:34
$begingroup$
@user73454 In case the solutions suggested previously look pathological for you for some reason, I updated my answer to include solutions given by symmetric polynomials.
$endgroup$
– Start wearing purple
Jan 9 at 16:15
5
5
$begingroup$
Essentially you are asking for $Deltaln|f|=0$?
$endgroup$
– LutzL
Jan 7 at 21:55
$begingroup$
Essentially you are asking for $Deltaln|f|=0$?
$endgroup$
– LutzL
Jan 7 at 21:55
$begingroup$
Maybe you are right also, but I think the equation above is equivalent to $Deltaln f^2=0$.
$endgroup$
– user73454
Jan 8 at 13:48
$begingroup$
Maybe you are right also, but I think the equation above is equivalent to $Deltaln f^2=0$.
$endgroup$
– user73454
Jan 8 at 13:48
$begingroup$
As $ln f^2=2ln|f|$, this constant factor does not change the equation, as the Laplace operator is linear.
$endgroup$
– LutzL
Jan 8 at 14:02
$begingroup$
As $ln f^2=2ln|f|$, this constant factor does not change the equation, as the Laplace operator is linear.
$endgroup$
– LutzL
Jan 8 at 14:02
$begingroup$
@LutzL, yes, you are right, thanks! Do you have any idea if my question has a positive or negative answer?
$endgroup$
– user73454
Jan 8 at 14:34
$begingroup$
@LutzL, yes, you are right, thanks! Do you have any idea if my question has a positive or negative answer?
$endgroup$
– user73454
Jan 8 at 14:34
$begingroup$
@user73454 In case the solutions suggested previously look pathological for you for some reason, I updated my answer to include solutions given by symmetric polynomials.
$endgroup$
– Start wearing purple
Jan 9 at 16:15
$begingroup$
@user73454 In case the solutions suggested previously look pathological for you for some reason, I updated my answer to include solutions given by symmetric polynomials.
$endgroup$
– Start wearing purple
Jan 9 at 16:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Answer: Yes. Moreover the equation has homogeneous polynomial solutions of any even degree.
Proof. It is easy to check that the function $u(x_1,ldots,x_n)=left(x_1^2+x_2^2right)^N$ satisfies the equation $$u_{x_1}^2+u_{x_2}^2=uleft(u_{x_1x_1}+u_{x_2x_2}right)$$
for any $Ninmathbb N$ (and, in fact, for any $Ninmathbb R$). This may also be understood without calculation using the first comment and that $ln(x^2+y^2)$ is essentially the Green's function of the 2D Laplacian. But then $u$ also satisfies $|nabla u|^2=uDelta u$.
One may also construct examples of symmetric homogeneous polynomial solutions in the form of products of the above elementary solutions. Specifically,
$$ v(x_1,ldots,x_n)=prod_{1leq i<jleq n}left(x_i^2+x_j^2right)^N$$
gives a homogeneous polynomial solution of degree $n(n-1)N$ for any $Ninmathbb N$. For $n>2$, one may set $N=1$ and still have a solution with required properties.
answered Jan 8 at 16:57
Start wearing purpleStart wearing purple
47.1k12135192
$endgroup$
$begingroup$
(+1) This is all so clear once we realize that $fDelta f=|nabla f|^2$ is equivalent to $Deltalog(f)=0$. Then the product of two solutions is also a solution.
$endgroup$
– robjohn♦
Jan 9 at 16:42
add a comment |
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1 Answer
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$begingroup$
Answer: Yes. Moreover the equation has homogeneous polynomial solutions of any even degree.
Proof. It is easy to check that the function $u(x_1,ldots,x_n)=left(x_1^2+x_2^2right)^N$ satisfies the equation $$u_{x_1}^2+u_{x_2}^2=uleft(u_{x_1x_1}+u_{x_2x_2}right)$$
for any $Ninmathbb N$ (and, in fact, for any $Ninmathbb R$). This may also be understood without calculation using the first comment and that $ln(x^2+y^2)$ is essentially the Green's function of the 2D Laplacian. But then $u$ also satisfies $|nabla u|^2=uDelta u$.
One may also construct examples of symmetric homogeneous polynomial solutions in the form of products of the above elementary solutions. Specifically,
$$ v(x_1,ldots,x_n)=prod_{1leq i<jleq n}left(x_i^2+x_j^2right)^N$$
gives a homogeneous polynomial solution of degree $n(n-1)N$ for any $Ninmathbb N$. For $n>2$, one may set $N=1$ and still have a solution with required properties.
answered Jan 8 at 16:57
Start wearing purpleStart wearing purple
47.1k12135192
$endgroup$
$begingroup$
(+1) This is all so clear once we realize that $fDelta f=|nabla f|^2$ is equivalent to $Deltalog(f)=0$. Then the product of two solutions is also a solution.
$endgroup$
– robjohn♦
Jan 9 at 16:42
add a comment |
$begingroup$
Answer: Yes. Moreover the equation has homogeneous polynomial solutions of any even degree.
Proof. It is easy to check that the function $u(x_1,ldots,x_n)=left(x_1^2+x_2^2right)^N$ satisfies the equation $$u_{x_1}^2+u_{x_2}^2=uleft(u_{x_1x_1}+u_{x_2x_2}right)$$
for any $Ninmathbb N$ (and, in fact, for any $Ninmathbb R$). This may also be understood without calculation using the first comment and that $ln(x^2+y^2)$ is essentially the Green's function of the 2D Laplacian. But then $u$ also satisfies $|nabla u|^2=uDelta u$.
One may also construct examples of symmetric homogeneous polynomial solutions in the form of products of the above elementary solutions. Specifically,
$$ v(x_1,ldots,x_n)=prod_{1leq i<jleq n}left(x_i^2+x_j^2right)^N$$
gives a homogeneous polynomial solution of degree $n(n-1)N$ for any $Ninmathbb N$. For $n>2$, one may set $N=1$ and still have a solution with required properties.
answered Jan 8 at 16:57
Start wearing purpleStart wearing purple
47.1k12135192
$endgroup$
$begingroup$
(+1) This is all so clear once we realize that $fDelta f=|nabla f|^2$ is equivalent to $Deltalog(f)=0$. Then the product of two solutions is also a solution.
$endgroup$
– robjohn♦
Jan 9 at 16:42
add a comment |
$begingroup$
Answer: Yes. Moreover the equation has homogeneous polynomial solutions of any even degree.
Proof. It is easy to check that the function $u(x_1,ldots,x_n)=left(x_1^2+x_2^2right)^N$ satisfies the equation $$u_{x_1}^2+u_{x_2}^2=uleft(u_{x_1x_1}+u_{x_2x_2}right)$$
for any $Ninmathbb N$ (and, in fact, for any $Ninmathbb R$). This may also be understood without calculation using the first comment and that $ln(x^2+y^2)$ is essentially the Green's function of the 2D Laplacian. But then $u$ also satisfies $|nabla u|^2=uDelta u$.
One may also construct examples of symmetric homogeneous polynomial solutions in the form of products of the above elementary solutions. Specifically,
$$ v(x_1,ldots,x_n)=prod_{1leq i<jleq n}left(x_i^2+x_j^2right)^N$$
gives a homogeneous polynomial solution of degree $n(n-1)N$ for any $Ninmathbb N$. For $n>2$, one may set $N=1$ and still have a solution with required properties.
answered Jan 8 at 16:57
Start wearing purpleStart wearing purple
47.1k12135192
$endgroup$
Answer: Yes. Moreover the equation has homogeneous polynomial solutions of any even degree.
Proof. It is easy to check that the function $u(x_1,ldots,x_n)=left(x_1^2+x_2^2right)^N$ satisfies the equation $$u_{x_1}^2+u_{x_2}^2=uleft(u_{x_1x_1}+u_{x_2x_2}right)$$
for any $Ninmathbb N$ (and, in fact, for any $Ninmathbb R$). This may also be understood without calculation using the first comment and that $ln(x^2+y^2)$ is essentially the Green's function of the 2D Laplacian. But then $u$ also satisfies $|nabla u|^2=uDelta u$.
One may also construct examples of symmetric homogeneous polynomial solutions in the form of products of the above elementary solutions. Specifically,
$$ v(x_1,ldots,x_n)=prod_{1leq i<jleq n}left(x_i^2+x_j^2right)^N$$
gives a homogeneous polynomial solution of degree $n(n-1)N$ for any $Ninmathbb N$. For $n>2$, one may set $N=1$ and still have a solution with required properties.
answered Jan 8 at 16:57
Start wearing purpleStart wearing purple
47.1k12135192
edited Jan 9 at 16:09
answered Jan 8 at 16:57
Start wearing purpleStart wearing purple
47.1k12135192
answered Jan 8 at 16:57
Start wearing purpleStart wearing purple
47.1k12135192
answered Jan 8 at 16:57
Start wearing purpleStart wearing purple
47.1k12135192
47.1k12135192
$begingroup$
(+1) This is all so clear once we realize that $fDelta f=|nabla f|^2$ is equivalent to $Deltalog(f)=0$. Then the product of two solutions is also a solution.
$endgroup$
– robjohn♦
Jan 9 at 16:42
add a comment |
$begingroup$
(+1) This is all so clear once we realize that $fDelta f=|nabla f|^2$ is equivalent to $Deltalog(f)=0$. Then the product of two solutions is also a solution.
$endgroup$
– robjohn♦
Jan 9 at 16:42
$begingroup$
(+1) This is all so clear once we realize that $fDelta f=|nabla f|^2$ is equivalent to $Deltalog(f)=0$. Then the product of two solutions is also a solution.
$endgroup$
– robjohn♦
Jan 9 at 16:42
$begingroup$
(+1) This is all so clear once we realize that $fDelta f=|nabla f|^2$ is equivalent to $Deltalog(f)=0$. Then the product of two solutions is also a solution.
$endgroup$
– robjohn♦
Jan 9 at 16:42
add a comment |
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5
$begingroup$
Essentially you are asking for $Deltaln|f|=0$?
$endgroup$
– LutzL
Jan 7 at 21:55
$begingroup$
Maybe you are right also, but I think the equation above is equivalent to $Deltaln f^2=0$.
$endgroup$
– user73454
Jan 8 at 13:48
$begingroup$
As $ln f^2=2ln|f|$, this constant factor does not change the equation, as the Laplace operator is linear.
$endgroup$
– LutzL
Jan 8 at 14:02
$begingroup$
@LutzL, yes, you are right, thanks! Do you have any idea if my question has a positive or negative answer?
$endgroup$
– user73454
Jan 8 at 14:34
$begingroup$
@user73454 In case the solutions suggested previously look pathological for you for some reason, I updated my answer to include solutions given by symmetric polynomials.
$endgroup$
– Start wearing purple
Jan 9 at 16:15