Prove that the only solution to $x frac partial f partial y - y frac partial f partial x = f$ is $f = 0$

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Suppose $f: mathbbR^2 setminus (0,0) rightarrow mathbbR$ is a smooth function, such that
$$ x dfrac partial f partial y - y dfrac partial f partial x = f$$
Show that $f=0$.

I haven't made any progress on the question, mostly just taking partial derivatives to see if anything noticeable happens; I haven't noticed anything.
It was given as a question which could be solved by someone who hasn't done a course in partial differential equations, so there's likely a fairly simple trick that will work.
If it helps, $e^xy $ satisfies the left hand side set to zero, but I doubt that's useful.

real-analysis multivariable-calculus pde

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edited Jul 26 at 23:04

Michael McGovern

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asked Jul 26 at 23:02

Cataline

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up vote
0
down vote

favorite

Suppose $f: mathbbR^2 setminus (0,0) rightarrow mathbbR$ is a smooth function, such that
$$ x dfrac partial f partial y - y dfrac partial f partial x = f$$
Show that $f=0$.

I haven't made any progress on the question, mostly just taking partial derivatives to see if anything noticeable happens; I haven't noticed anything.
It was given as a question which could be solved by someone who hasn't done a course in partial differential equations, so there's likely a fairly simple trick that will work.
If it helps, $e^xy $ satisfies the left hand side set to zero, but I doubt that's useful.

real-analysis multivariable-calculus pde

share|cite|improve this question

edited Jul 26 at 23:04

Michael McGovern

5701314

asked Jul 26 at 23:02

Cataline

1,071715

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Suppose $f: mathbbR^2 setminus (0,0) rightarrow mathbbR$ is a smooth function, such that
$$ x dfrac partial f partial y - y dfrac partial f partial x = f$$
Show that $f=0$.

I haven't made any progress on the question, mostly just taking partial derivatives to see if anything noticeable happens; I haven't noticed anything.
It was given as a question which could be solved by someone who hasn't done a course in partial differential equations, so there's likely a fairly simple trick that will work.
If it helps, $e^xy $ satisfies the left hand side set to zero, but I doubt that's useful.

real-analysis multivariable-calculus pde

share|cite|improve this question

edited Jul 26 at 23:04

Michael McGovern

5701314

asked Jul 26 at 23:02

Cataline

1,071715

Suppose $f: mathbbR^2 setminus (0,0) rightarrow mathbbR$ is a smooth function, such that
$$ x dfrac partial f partial y - y dfrac partial f partial x = f$$
Show that $f=0$.

I haven't made any progress on the question, mostly just taking partial derivatives to see if anything noticeable happens; I haven't noticed anything.
It was given as a question which could be solved by someone who hasn't done a course in partial differential equations, so there's likely a fairly simple trick that will work.
If it helps, $e^xy $ satisfies the left hand side set to zero, but I doubt that's useful.

real-analysis multivariable-calculus pde

share|cite|improve this question

edited Jul 26 at 23:04

Michael McGovern

5701314

asked Jul 26 at 23:02

Cataline

1,071715

share|cite|improve this question

share|cite|improve this question

edited Jul 26 at 23:04

Michael McGovern

5701314

edited Jul 26 at 23:04

Michael McGovern

5701314

edited Jul 26 at 23:04

Michael McGovern

5701314

5701314

asked Jul 26 at 23:02

Cataline

1,071715

asked Jul 26 at 23:02

Cataline

1,071715

asked Jul 26 at 23:02

Cataline

1,071715

1,071715

add a comment | 

add a comment | 

2 Answers
2

active

oldest

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up vote
5
down vote



accepted

HINT: Note that the integral curves of the vector field $(-y,x)$ are circles centered at the origin. The left-hand side is the directional derivative of $f$ along such circles. Consider $g(t) = f(acos t,asin t)$ for any $a>0$. What is $g'(t)$?

share|cite|improve this answer

answered Jul 26 at 23:13

Ted Shifrin

59.4k44386

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the hint. So I've gotten that $g'(t) = g(t)$ which means that $g(t) = Ce^t$ for some real $C$. But since $g(t+2pi) = g(t)$, we must have that $C=0$, and so we must have $f = 0$ since $a$ was arbitrary. Is that correct?
  – Cataline
  Jul 26 at 23:29
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Perfect !!! There you go. Often with elementary linear PDE it's useful to think about directional derivatives :)
  – Ted Shifrin
  Jul 26 at 23:30
  

  
  
  
  

add a comment | 

up vote
0
down vote

This is not an answer to the question, but a comment but too long to be posted in the comments section.

$$ x dfrac partial f partial y - y dfrac partial f partial x = f tag 1$$
Thanks to the method of characteristics, one can express the general solution on the form :
$$f(x,y)=expleft(tan^-1left(fracxyright)right)Fleft(x^2+y^2right)tag 2$$
where $F$ is an arbitrary fonction.

The function $F$ has to be determined according to some conditions generally specified in the wording of the problem.

We can directly prove that the function $(2)$ is solution of equation $(1)$ :

$$fracpartial fpartial x=-fracyx^2+y^2expleft(tan^-1left(fracxyright)right)F+2xF' tag 3$$

$$fracpartial fpartial y=fracxx^2+y^2expleft(tan^-1left(fracxyright)right)F+2yF'tag 4$$

Thus
$$ x dfrac partial f partial y - y dfrac partial f partial x =\
=xleft(fracxx^2+y^2expleft(tan^-1left(fracxyright)right)F+2yF' right)-yleft(-fracyx^2+y^2expleft(tan^-1left(fracxyright)right)F+2xF' right)$$
After simplification :
$$x dfrac partial f partial y - y dfrac partial f partial x =expleft(tan^-1left(fracxyright)right)F=f$$
The equation $(1)$ is satisfied. So the general solution of the PDE is :
$$f(x,y)=expleft(tan^-1left(fracxyright)right)Fleft(x^2+y^2right)qquadtextwith arbitrary function F.$$
I let you think about the consequence of the wording of the question : "Suppose $f: mathbbR^2 setminus (0,0) rightarrow mathbbR$ is a smooth function."

share|cite|improve this answer

answered Jul 27 at 6:53

JJacquelin

39.8k21649

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

HINT: Note that the integral curves of the vector field $(-y,x)$ are circles centered at the origin. The left-hand side is the directional derivative of $f$ along such circles. Consider $g(t) = f(acos t,asin t)$ for any $a>0$. What is $g'(t)$?

share|cite|improve this answer

answered Jul 26 at 23:13

Ted Shifrin

59.4k44386

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the hint. So I've gotten that $g'(t) = g(t)$ which means that $g(t) = Ce^t$ for some real $C$. But since $g(t+2pi) = g(t)$, we must have that $C=0$, and so we must have $f = 0$ since $a$ was arbitrary. Is that correct?
  – Cataline
  Jul 26 at 23:29
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Perfect !!! There you go. Often with elementary linear PDE it's useful to think about directional derivatives :)
  – Ted Shifrin
  Jul 26 at 23:30
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

HINT: Note that the integral curves of the vector field $(-y,x)$ are circles centered at the origin. The left-hand side is the directional derivative of $f$ along such circles. Consider $g(t) = f(acos t,asin t)$ for any $a>0$. What is $g'(t)$?

share|cite|improve this answer

answered Jul 26 at 23:13

Ted Shifrin

59.4k44386

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the hint. So I've gotten that $g'(t) = g(t)$ which means that $g(t) = Ce^t$ for some real $C$. But since $g(t+2pi) = g(t)$, we must have that $C=0$, and so we must have $f = 0$ since $a$ was arbitrary. Is that correct?
  – Cataline
  Jul 26 at 23:29
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Perfect !!! There you go. Often with elementary linear PDE it's useful to think about directional derivatives :)
  – Ted Shifrin
  Jul 26 at 23:30
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

HINT: Note that the integral curves of the vector field $(-y,x)$ are circles centered at the origin. The left-hand side is the directional derivative of $f$ along such circles. Consider $g(t) = f(acos t,asin t)$ for any $a>0$. What is $g'(t)$?

share|cite|improve this answer

answered Jul 26 at 23:13

Ted Shifrin

59.4k44386

HINT: Note that the integral curves of the vector field $(-y,x)$ are circles centered at the origin. The left-hand side is the directional derivative of $f$ along such circles. Consider $g(t) = f(acos t,asin t)$ for any $a>0$. What is $g'(t)$?

share|cite|improve this answer

answered Jul 26 at 23:13

Ted Shifrin

59.4k44386

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 23:13

Ted Shifrin

59.4k44386

answered Jul 26 at 23:13

Ted Shifrin

59.4k44386

answered Jul 26 at 23:13

Ted Shifrin

59.4k44386

59.4k44386

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the hint. So I've gotten that $g'(t) = g(t)$ which means that $g(t) = Ce^t$ for some real $C$. But since $g(t+2pi) = g(t)$, we must have that $C=0$, and so we must have $f = 0$ since $a$ was arbitrary. Is that correct?
  – Cataline
  Jul 26 at 23:29
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Perfect !!! There you go. Often with elementary linear PDE it's useful to think about directional derivatives :)
  – Ted Shifrin
  Jul 26 at 23:30
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the hint. So I've gotten that $g'(t) = g(t)$ which means that $g(t) = Ce^t$ for some real $C$. But since $g(t+2pi) = g(t)$, we must have that $C=0$, and so we must have $f = 0$ since $a$ was arbitrary. Is that correct?
  – Cataline
  Jul 26 at 23:29
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Perfect !!! There you go. Often with elementary linear PDE it's useful to think about directional derivatives :)
  – Ted Shifrin
  Jul 26 at 23:30
  

  
  
  
  

Thanks for the hint. So I've gotten that $g'(t) = g(t)$ which means that $g(t) = Ce^t$ for some real $C$. But since $g(t+2pi) = g(t)$, we must have that $C=0$, and so we must have $f = 0$ since $a$ was arbitrary. Is that correct?
– Cataline
Jul 26 at 23:29

Thanks for the hint. So I've gotten that $g'(t) = g(t)$ which means that $g(t) = Ce^t$ for some real $C$. But since $g(t+2pi) = g(t)$, we must have that $C=0$, and so we must have $f = 0$ since $a$ was arbitrary. Is that correct?
– Cataline
Jul 26 at 23:29

3

3

Perfect !!! There you go. Often with elementary linear PDE it's useful to think about directional derivatives :)
– Ted Shifrin
Jul 26 at 23:30

Perfect !!! There you go. Often with elementary linear PDE it's useful to think about directional derivatives :)
– Ted Shifrin
Jul 26 at 23:30

add a comment | 

up vote
0
down vote

This is not an answer to the question, but a comment but too long to be posted in the comments section.

$$ x dfrac partial f partial y - y dfrac partial f partial x = f tag 1$$
Thanks to the method of characteristics, one can express the general solution on the form :
$$f(x,y)=expleft(tan^-1left(fracxyright)right)Fleft(x^2+y^2right)tag 2$$
where $F$ is an arbitrary fonction.

The function $F$ has to be determined according to some conditions generally specified in the wording of the problem.

We can directly prove that the function $(2)$ is solution of equation $(1)$ :

$$fracpartial fpartial x=-fracyx^2+y^2expleft(tan^-1left(fracxyright)right)F+2xF' tag 3$$

$$fracpartial fpartial y=fracxx^2+y^2expleft(tan^-1left(fracxyright)right)F+2yF'tag 4$$

Thus
$$ x dfrac partial f partial y - y dfrac partial f partial x =\
=xleft(fracxx^2+y^2expleft(tan^-1left(fracxyright)right)F+2yF' right)-yleft(-fracyx^2+y^2expleft(tan^-1left(fracxyright)right)F+2xF' right)$$
After simplification :
$$x dfrac partial f partial y - y dfrac partial f partial x =expleft(tan^-1left(fracxyright)right)F=f$$
The equation $(1)$ is satisfied. So the general solution of the PDE is :
$$f(x,y)=expleft(tan^-1left(fracxyright)right)Fleft(x^2+y^2right)qquadtextwith arbitrary function F.$$
I let you think about the consequence of the wording of the question : "Suppose $f: mathbbR^2 setminus (0,0) rightarrow mathbbR$ is a smooth function."

share|cite|improve this answer

answered Jul 27 at 6:53

JJacquelin

39.8k21649

add a comment | 

up vote
0
down vote

This is not an answer to the question, but a comment but too long to be posted in the comments section.

$$ x dfrac partial f partial y - y dfrac partial f partial x = f tag 1$$
Thanks to the method of characteristics, one can express the general solution on the form :
$$f(x,y)=expleft(tan^-1left(fracxyright)right)Fleft(x^2+y^2right)tag 2$$
where $F$ is an arbitrary fonction.

The function $F$ has to be determined according to some conditions generally specified in the wording of the problem.

We can directly prove that the function $(2)$ is solution of equation $(1)$ :

$$fracpartial fpartial x=-fracyx^2+y^2expleft(tan^-1left(fracxyright)right)F+2xF' tag 3$$

$$fracpartial fpartial y=fracxx^2+y^2expleft(tan^-1left(fracxyright)right)F+2yF'tag 4$$

Thus
$$ x dfrac partial f partial y - y dfrac partial f partial x =\
=xleft(fracxx^2+y^2expleft(tan^-1left(fracxyright)right)F+2yF' right)-yleft(-fracyx^2+y^2expleft(tan^-1left(fracxyright)right)F+2xF' right)$$
After simplification :
$$x dfrac partial f partial y - y dfrac partial f partial x =expleft(tan^-1left(fracxyright)right)F=f$$
The equation $(1)$ is satisfied. So the general solution of the PDE is :
$$f(x,y)=expleft(tan^-1left(fracxyright)right)Fleft(x^2+y^2right)qquadtextwith arbitrary function F.$$
I let you think about the consequence of the wording of the question : "Suppose $f: mathbbR^2 setminus (0,0) rightarrow mathbbR$ is a smooth function."

share|cite|improve this answer

answered Jul 27 at 6:53

JJacquelin

39.8k21649

add a comment | 

up vote
0
down vote

up vote
0
down vote

This is not an answer to the question, but a comment but too long to be posted in the comments section.

$$ x dfrac partial f partial y - y dfrac partial f partial x = f tag 1$$
Thanks to the method of characteristics, one can express the general solution on the form :
$$f(x,y)=expleft(tan^-1left(fracxyright)right)Fleft(x^2+y^2right)tag 2$$
where $F$ is an arbitrary fonction.

The function $F$ has to be determined according to some conditions generally specified in the wording of the problem.

We can directly prove that the function $(2)$ is solution of equation $(1)$ :

$$fracpartial fpartial x=-fracyx^2+y^2expleft(tan^-1left(fracxyright)right)F+2xF' tag 3$$

$$fracpartial fpartial y=fracxx^2+y^2expleft(tan^-1left(fracxyright)right)F+2yF'tag 4$$

Thus
$$ x dfrac partial f partial y - y dfrac partial f partial x =\
=xleft(fracxx^2+y^2expleft(tan^-1left(fracxyright)right)F+2yF' right)-yleft(-fracyx^2+y^2expleft(tan^-1left(fracxyright)right)F+2xF' right)$$
After simplification :
$$x dfrac partial f partial y - y dfrac partial f partial x =expleft(tan^-1left(fracxyright)right)F=f$$
The equation $(1)$ is satisfied. So the general solution of the PDE is :
$$f(x,y)=expleft(tan^-1left(fracxyright)right)Fleft(x^2+y^2right)qquadtextwith arbitrary function F.$$
I let you think about the consequence of the wording of the question : "Suppose $f: mathbbR^2 setminus (0,0) rightarrow mathbbR$ is a smooth function."

share|cite|improve this answer

answered Jul 27 at 6:53

JJacquelin

39.8k21649

This is not an answer to the question, but a comment but too long to be posted in the comments section.

$$ x dfrac partial f partial y - y dfrac partial f partial x = f tag 1$$
Thanks to the method of characteristics, one can express the general solution on the form :
$$f(x,y)=expleft(tan^-1left(fracxyright)right)Fleft(x^2+y^2right)tag 2$$
where $F$ is an arbitrary fonction.

The function $F$ has to be determined according to some conditions generally specified in the wording of the problem.

We can directly prove that the function $(2)$ is solution of equation $(1)$ :

$$fracpartial fpartial x=-fracyx^2+y^2expleft(tan^-1left(fracxyright)right)F+2xF' tag 3$$

$$fracpartial fpartial y=fracxx^2+y^2expleft(tan^-1left(fracxyright)right)F+2yF'tag 4$$

Thus
$$ x dfrac partial f partial y - y dfrac partial f partial x =\
=xleft(fracxx^2+y^2expleft(tan^-1left(fracxyright)right)F+2yF' right)-yleft(-fracyx^2+y^2expleft(tan^-1left(fracxyright)right)F+2xF' right)$$
After simplification :
$$x dfrac partial f partial y - y dfrac partial f partial x =expleft(tan^-1left(fracxyright)right)F=f$$
The equation $(1)$ is satisfied. So the general solution of the PDE is :
$$f(x,y)=expleft(tan^-1left(fracxyright)right)Fleft(x^2+y^2right)qquadtextwith arbitrary function F.$$
I let you think about the consequence of the wording of the question : "Suppose $f: mathbbR^2 setminus (0,0) rightarrow mathbbR$ is a smooth function."

share|cite|improve this answer

answered Jul 27 at 6:53

JJacquelin

39.8k21649

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 6:53

JJacquelin

39.8k21649

answered Jul 27 at 6:53

JJacquelin

39.8k21649

answered Jul 27 at 6:53

JJacquelin

39.8k21649

39.8k21649

add a comment | 

add a comment | 

 

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