Can you have solution for the equation $|nabla f|^2=fDelta f$, for a homogenous polynomial $f$ with...












3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














3












3








3


3



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 21:47









jgon

13.6k22041




13.6k22041










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














  • 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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+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













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+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


















2












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+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
















2












2








2





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 9 at 16:09

























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




















  • $begingroup$
    (+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




















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