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How to prove that there is no differentiable function with given partial derivatives
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Let $U=(x,y)inmathbbR^2:(x,y)neq(0,0)$. Show that there is no differentiable function $f:UrightarrowmathbbR$ satisfying
$$fracpartial fpartial x=fracyx^2+y^2, fracpartial fpartial y=-fracxx^2+y^2.$$
My initial thought was to show that these partials are not continuous at some point $(x,y)neq(0,0)$, but since it is possible to have a differentiable function that is not $C^1$, this is not enough. Any suggestions are greatly appreciated.
real-analysis multivariable-calculus derivatives
asked Aug 25 at 17:01
Atsina
674115
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up vote
8
down vote
favorite
Let $U=(x,y)inmathbbR^2:(x,y)neq(0,0)$. Show that there is no differentiable function $f:UrightarrowmathbbR$ satisfying
$$fracpartial fpartial x=fracyx^2+y^2, fracpartial fpartial y=-fracxx^2+y^2.$$
My initial thought was to show that these partials are not continuous at some point $(x,y)neq(0,0)$, but since it is possible to have a differentiable function that is not $C^1$, this is not enough. Any suggestions are greatly appreciated.
real-analysis multivariable-calculus derivatives
asked Aug 25 at 17:01
Atsina
674115
Was this a Calculus II exercise? It looks like it's a hard one: it is easy to get the right idea if you have seen something more advanced, but to someone who is just learning partial derivatives and has never heard about exact/closed forms it is difficult to come up with it.
– Federico Poloni
Aug 26 at 6:47
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Let $U=(x,y)inmathbbR^2:(x,y)neq(0,0)$. Show that there is no differentiable function $f:UrightarrowmathbbR$ satisfying
$$fracpartial fpartial x=fracyx^2+y^2, fracpartial fpartial y=-fracxx^2+y^2.$$
My initial thought was to show that these partials are not continuous at some point $(x,y)neq(0,0)$, but since it is possible to have a differentiable function that is not $C^1$, this is not enough. Any suggestions are greatly appreciated.
real-analysis multivariable-calculus derivatives
asked Aug 25 at 17:01
Atsina
674115
Let $U=(x,y)inmathbbR^2:(x,y)neq(0,0)$. Show that there is no differentiable function $f:UrightarrowmathbbR$ satisfying
$$fracpartial fpartial x=fracyx^2+y^2, fracpartial fpartial y=-fracxx^2+y^2.$$
My initial thought was to show that these partials are not continuous at some point $(x,y)neq(0,0)$, but since it is possible to have a differentiable function that is not $C^1$, this is not enough. Any suggestions are greatly appreciated.
real-analysis multivariable-calculus derivatives
asked Aug 25 at 17:01
Atsina
674115
asked Aug 25 at 17:01
Atsina
674115
asked Aug 25 at 17:01
Atsina
674115
asked Aug 25 at 17:01
Atsina
674115
674115
Was this a Calculus II exercise? It looks like it's a hard one: it is easy to get the right idea if you have seen something more advanced, but to someone who is just learning partial derivatives and has never heard about exact/closed forms it is difficult to come up with it.
– Federico Poloni
Aug 26 at 6:47
add a comment |Â
Was this a Calculus II exercise? It looks like it's a hard one: it is easy to get the right idea if you have seen something more advanced, but to someone who is just learning partial derivatives and has never heard about exact/closed forms it is difficult to come up with it.
– Federico Poloni
Aug 26 at 6:47
Was this a Calculus II exercise? It looks like it's a hard one: it is easy to get the right idea if you have seen something more advanced, but to someone who is just learning partial derivatives and has never heard about exact/closed forms it is difficult to come up with it.
– Federico Poloni
Aug 26 at 6:47
Was this a Calculus II exercise? It looks like it's a hard one: it is easy to get the right idea if you have seen something more advanced, but to someone who is just learning partial derivatives and has never heard about exact/closed forms it is difficult to come up with it.
– Federico Poloni
Aug 26 at 6:47
add a comment |Â
2 Answers
2
active
oldest
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up vote
12
down vote
accepted
In my opinion, I think this exercise is the most important exercise in calculus since it is a starting point of the de Rham cohomology, which shows an interaction between topology and functions on it. Anyway, this question is equivalent to the following: can we find $f:U to mathbbR$ s.t.
$$
nabla f = left(fracyx^2+y^2, -fracxx^2+y^2right)?
$$
If such $f$ exists, the vector field on the right-hand side will be called as a conservative field, and by the fundamental theorem of a line integral, if we integrate the vector field over any given (simple) closed curve, the result should be zero.
However, if you try to integrate it over a unit circle centered at origin, the result became $2pi$.
answered Aug 25 at 17:17
Seewoo Lee
4,945824
I'm getting $-2pi$ when I integrate over the unit circle centered at the origin. It's relatively straightforward to convert to polar and simplify, but I get $-int_0^2pidt$. Is there some reason this must be positive?
– Atsina
Aug 25 at 17:58
3
It's a question of orientation. Simply go around the circle in the other direction, by changing the parameterization. What is important is only that the integral is nonzero.
– Stephan Kolassa
Aug 25 at 18:00
1
@Atsina Yes you are right, it gives $-2pi$. The important point is that 1-form $$ fracyx^2+y^2dx - fracxx^2+y^2dy $$ is a closed 1-form which is not exact. Also, this is a basis of the de Rham cohomology group $H^1(U;mathbbR)simeq mathbbR$.
– Seewoo Lee
Aug 25 at 18:17
add a comment |Â
up vote
6
down vote
Assume that such a $f$ exists and let $gammacolon[0,1]to U$ defined by $gamma(t)=(cos(2pi t),sin(2pi t))$, then:
$$int_gammamathrmdf=f(gamma(1))-f(gamma(0))=0,$$
however using the given formula for $mathrmdf$, one finds that:
$$int_gammamathrmdf=int_0^1 mathrmdf_gamma(t)cdotdotgamma(t),mathrmdt=2piint_0^1frac-sin(2pi t)^2-cos^2(2pi t)cos^2(2pi t)+sin^2(2pi t)=-2pi,$$
whence a contradiction.
The intuition behind this reasoning is that the angular form $fracymathrmdx-xmathrmdyx^2+y^2$ is not exact on $U$.
answered Aug 25 at 17:12
C. Falcon
14.7k41748
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
In my opinion, I think this exercise is the most important exercise in calculus since it is a starting point of the de Rham cohomology, which shows an interaction between topology and functions on it. Anyway, this question is equivalent to the following: can we find $f:U to mathbbR$ s.t.
$$
nabla f = left(fracyx^2+y^2, -fracxx^2+y^2right)?
$$
If such $f$ exists, the vector field on the right-hand side will be called as a conservative field, and by the fundamental theorem of a line integral, if we integrate the vector field over any given (simple) closed curve, the result should be zero.
However, if you try to integrate it over a unit circle centered at origin, the result became $2pi$.
answered Aug 25 at 17:17
Seewoo Lee
4,945824
I'm getting $-2pi$ when I integrate over the unit circle centered at the origin. It's relatively straightforward to convert to polar and simplify, but I get $-int_0^2pidt$. Is there some reason this must be positive?
– Atsina
Aug 25 at 17:58
3
It's a question of orientation. Simply go around the circle in the other direction, by changing the parameterization. What is important is only that the integral is nonzero.
– Stephan Kolassa
Aug 25 at 18:00
1
@Atsina Yes you are right, it gives $-2pi$. The important point is that 1-form $$ fracyx^2+y^2dx - fracxx^2+y^2dy $$ is a closed 1-form which is not exact. Also, this is a basis of the de Rham cohomology group $H^1(U;mathbbR)simeq mathbbR$.
– Seewoo Lee
Aug 25 at 18:17
add a comment |Â
up vote
12
down vote
accepted
In my opinion, I think this exercise is the most important exercise in calculus since it is a starting point of the de Rham cohomology, which shows an interaction between topology and functions on it. Anyway, this question is equivalent to the following: can we find $f:U to mathbbR$ s.t.
$$
nabla f = left(fracyx^2+y^2, -fracxx^2+y^2right)?
$$
If such $f$ exists, the vector field on the right-hand side will be called as a conservative field, and by the fundamental theorem of a line integral, if we integrate the vector field over any given (simple) closed curve, the result should be zero.
However, if you try to integrate it over a unit circle centered at origin, the result became $2pi$.
answered Aug 25 at 17:17
Seewoo Lee
4,945824
I'm getting $-2pi$ when I integrate over the unit circle centered at the origin. It's relatively straightforward to convert to polar and simplify, but I get $-int_0^2pidt$. Is there some reason this must be positive?
– Atsina
Aug 25 at 17:58
3
It's a question of orientation. Simply go around the circle in the other direction, by changing the parameterization. What is important is only that the integral is nonzero.
– Stephan Kolassa
Aug 25 at 18:00
1
@Atsina Yes you are right, it gives $-2pi$. The important point is that 1-form $$ fracyx^2+y^2dx - fracxx^2+y^2dy $$ is a closed 1-form which is not exact. Also, this is a basis of the de Rham cohomology group $H^1(U;mathbbR)simeq mathbbR$.
– Seewoo Lee
Aug 25 at 18:17
add a comment |Â
up vote
12
down vote
accepted
up vote
12
down vote
accepted
In my opinion, I think this exercise is the most important exercise in calculus since it is a starting point of the de Rham cohomology, which shows an interaction between topology and functions on it. Anyway, this question is equivalent to the following: can we find $f:U to mathbbR$ s.t.
$$
nabla f = left(fracyx^2+y^2, -fracxx^2+y^2right)?
$$
If such $f$ exists, the vector field on the right-hand side will be called as a conservative field, and by the fundamental theorem of a line integral, if we integrate the vector field over any given (simple) closed curve, the result should be zero.
However, if you try to integrate it over a unit circle centered at origin, the result became $2pi$.
answered Aug 25 at 17:17
Seewoo Lee
4,945824
In my opinion, I think this exercise is the most important exercise in calculus since it is a starting point of the de Rham cohomology, which shows an interaction between topology and functions on it. Anyway, this question is equivalent to the following: can we find $f:U to mathbbR$ s.t.
$$
nabla f = left(fracyx^2+y^2, -fracxx^2+y^2right)?
$$
If such $f$ exists, the vector field on the right-hand side will be called as a conservative field, and by the fundamental theorem of a line integral, if we integrate the vector field over any given (simple) closed curve, the result should be zero.
However, if you try to integrate it over a unit circle centered at origin, the result became $2pi$.
answered Aug 25 at 17:17
Seewoo Lee
4,945824
answered Aug 25 at 17:17
Seewoo Lee
4,945824
answered Aug 25 at 17:17
Seewoo Lee
4,945824
answered Aug 25 at 17:17
Seewoo Lee
4,945824
4,945824
I'm getting $-2pi$ when I integrate over the unit circle centered at the origin. It's relatively straightforward to convert to polar and simplify, but I get $-int_0^2pidt$. Is there some reason this must be positive?
– Atsina
Aug 25 at 17:58
3
It's a question of orientation. Simply go around the circle in the other direction, by changing the parameterization. What is important is only that the integral is nonzero.
– Stephan Kolassa
Aug 25 at 18:00
1
@Atsina Yes you are right, it gives $-2pi$. The important point is that 1-form $$ fracyx^2+y^2dx - fracxx^2+y^2dy $$ is a closed 1-form which is not exact. Also, this is a basis of the de Rham cohomology group $H^1(U;mathbbR)simeq mathbbR$.
– Seewoo Lee
Aug 25 at 18:17
add a comment |Â
I'm getting $-2pi$ when I integrate over the unit circle centered at the origin. It's relatively straightforward to convert to polar and simplify, but I get $-int_0^2pidt$. Is there some reason this must be positive?
– Atsina
Aug 25 at 17:58
3
It's a question of orientation. Simply go around the circle in the other direction, by changing the parameterization. What is important is only that the integral is nonzero.
– Stephan Kolassa
Aug 25 at 18:00
1
@Atsina Yes you are right, it gives $-2pi$. The important point is that 1-form $$ fracyx^2+y^2dx - fracxx^2+y^2dy $$ is a closed 1-form which is not exact. Also, this is a basis of the de Rham cohomology group $H^1(U;mathbbR)simeq mathbbR$.
– Seewoo Lee
Aug 25 at 18:17
I'm getting $-2pi$ when I integrate over the unit circle centered at the origin. It's relatively straightforward to convert to polar and simplify, but I get $-int_0^2pidt$. Is there some reason this must be positive?
– Atsina
Aug 25 at 17:58
I'm getting $-2pi$ when I integrate over the unit circle centered at the origin. It's relatively straightforward to convert to polar and simplify, but I get $-int_0^2pidt$. Is there some reason this must be positive?
– Atsina
Aug 25 at 17:58
3
3
It's a question of orientation. Simply go around the circle in the other direction, by changing the parameterization. What is important is only that the integral is nonzero.
– Stephan Kolassa
Aug 25 at 18:00
It's a question of orientation. Simply go around the circle in the other direction, by changing the parameterization. What is important is only that the integral is nonzero.
– Stephan Kolassa
Aug 25 at 18:00
1
1
@Atsina Yes you are right, it gives $-2pi$. The important point is that 1-form $$ fracyx^2+y^2dx - fracxx^2+y^2dy $$ is a closed 1-form which is not exact. Also, this is a basis of the de Rham cohomology group $H^1(U;mathbbR)simeq mathbbR$.
– Seewoo Lee
Aug 25 at 18:17
@Atsina Yes you are right, it gives $-2pi$. The important point is that 1-form $$ fracyx^2+y^2dx - fracxx^2+y^2dy $$ is a closed 1-form which is not exact. Also, this is a basis of the de Rham cohomology group $H^1(U;mathbbR)simeq mathbbR$.
– Seewoo Lee
Aug 25 at 18:17
add a comment |Â
up vote
6
down vote
Assume that such a $f$ exists and let $gammacolon[0,1]to U$ defined by $gamma(t)=(cos(2pi t),sin(2pi t))$, then:
$$int_gammamathrmdf=f(gamma(1))-f(gamma(0))=0,$$
however using the given formula for $mathrmdf$, one finds that:
$$int_gammamathrmdf=int_0^1 mathrmdf_gamma(t)cdotdotgamma(t),mathrmdt=2piint_0^1frac-sin(2pi t)^2-cos^2(2pi t)cos^2(2pi t)+sin^2(2pi t)=-2pi,$$
whence a contradiction.
The intuition behind this reasoning is that the angular form $fracymathrmdx-xmathrmdyx^2+y^2$ is not exact on $U$.
answered Aug 25 at 17:12
C. Falcon
14.7k41748
add a comment |Â
up vote
6
down vote
Assume that such a $f$ exists and let $gammacolon[0,1]to U$ defined by $gamma(t)=(cos(2pi t),sin(2pi t))$, then:
$$int_gammamathrmdf=f(gamma(1))-f(gamma(0))=0,$$
however using the given formula for $mathrmdf$, one finds that:
$$int_gammamathrmdf=int_0^1 mathrmdf_gamma(t)cdotdotgamma(t),mathrmdt=2piint_0^1frac-sin(2pi t)^2-cos^2(2pi t)cos^2(2pi t)+sin^2(2pi t)=-2pi,$$
whence a contradiction.
The intuition behind this reasoning is that the angular form $fracymathrmdx-xmathrmdyx^2+y^2$ is not exact on $U$.
answered Aug 25 at 17:12
C. Falcon
14.7k41748
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Assume that such a $f$ exists and let $gammacolon[0,1]to U$ defined by $gamma(t)=(cos(2pi t),sin(2pi t))$, then:
$$int_gammamathrmdf=f(gamma(1))-f(gamma(0))=0,$$
however using the given formula for $mathrmdf$, one finds that:
$$int_gammamathrmdf=int_0^1 mathrmdf_gamma(t)cdotdotgamma(t),mathrmdt=2piint_0^1frac-sin(2pi t)^2-cos^2(2pi t)cos^2(2pi t)+sin^2(2pi t)=-2pi,$$
whence a contradiction.
The intuition behind this reasoning is that the angular form $fracymathrmdx-xmathrmdyx^2+y^2$ is not exact on $U$.
answered Aug 25 at 17:12
C. Falcon
14.7k41748
Assume that such a $f$ exists and let $gammacolon[0,1]to U$ defined by $gamma(t)=(cos(2pi t),sin(2pi t))$, then:
$$int_gammamathrmdf=f(gamma(1))-f(gamma(0))=0,$$
however using the given formula for $mathrmdf$, one finds that:
$$int_gammamathrmdf=int_0^1 mathrmdf_gamma(t)cdotdotgamma(t),mathrmdt=2piint_0^1frac-sin(2pi t)^2-cos^2(2pi t)cos^2(2pi t)+sin^2(2pi t)=-2pi,$$
whence a contradiction.
The intuition behind this reasoning is that the angular form $fracymathrmdx-xmathrmdyx^2+y^2$ is not exact on $U$.
answered Aug 25 at 17:12
C. Falcon
14.7k41748
edited Aug 25 at 20:17
answered Aug 25 at 17:12
C. Falcon
14.7k41748
answered Aug 25 at 17:12
C. Falcon
14.7k41748
answered Aug 25 at 17:12
C. Falcon
14.7k41748
14.7k41748
add a comment |Â
add a comment |Â
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Was this a Calculus II exercise? It looks like it's a hard one: it is easy to get the right idea if you have seen something more advanced, but to someone who is just learning partial derivatives and has never heard about exact/closed forms it is difficult to come up with it.
– Federico Poloni
Aug 26 at 6:47