Proof: There is no function $f in C^2(mathbb{R^3})$ with gradient $nabla f(x,y,z) = (yz, xz, xy^2)$
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How can one show that there is no function, which is a continuously partially derivable function $f in C^2(mathbb{R^3})$ with this gradient
$$nabla f(x,y,z) = (yz, xz, xy^2)$$
I thought about using the Hessian matrix since one has to calculate all second partial derivatives of $f$ there.
Since only the gradient is given, can I calculate the antiderivatives first:
$yz = xyz$
$xz = xyz$
$xy^2 = xy^2z$
Now I want to calculate the antiderivatives of the antiderivatives:
$xyz = frac{yzx^2}{2}$
$xyz = frac{yzx^2}{2}$
$xy^2z = dfrac{y^2zx^2}{2}$
I didn't calculate the antiderivatives of the partial derivatives and I don't even know if that way is correct...
analysis functions derivatives continuity partial-derivative
asked Nov 22 at 22:13
Bad At Math
203
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up vote
2
down vote
favorite
How can one show that there is no function, which is a continuously partially derivable function $f in C^2(mathbb{R^3})$ with this gradient
$$nabla f(x,y,z) = (yz, xz, xy^2)$$
I thought about using the Hessian matrix since one has to calculate all second partial derivatives of $f$ there.
Since only the gradient is given, can I calculate the antiderivatives first:
$yz = xyz$
$xz = xyz$
$xy^2 = xy^2z$
Now I want to calculate the antiderivatives of the antiderivatives:
$xyz = frac{yzx^2}{2}$
$xyz = frac{yzx^2}{2}$
$xy^2z = dfrac{y^2zx^2}{2}$
I didn't calculate the antiderivatives of the partial derivatives and I don't even know if that way is correct...
analysis functions derivatives continuity partial-derivative
asked Nov 22 at 22:13
Bad At Math
203
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
How can one show that there is no function, which is a continuously partially derivable function $f in C^2(mathbb{R^3})$ with this gradient
$$nabla f(x,y,z) = (yz, xz, xy^2)$$
I thought about using the Hessian matrix since one has to calculate all second partial derivatives of $f$ there.
Since only the gradient is given, can I calculate the antiderivatives first:
$yz = xyz$
$xz = xyz$
$xy^2 = xy^2z$
Now I want to calculate the antiderivatives of the antiderivatives:
$xyz = frac{yzx^2}{2}$
$xyz = frac{yzx^2}{2}$
$xy^2z = dfrac{y^2zx^2}{2}$
I didn't calculate the antiderivatives of the partial derivatives and I don't even know if that way is correct...
analysis functions derivatives continuity partial-derivative
asked Nov 22 at 22:13
Bad At Math
203
How can one show that there is no function, which is a continuously partially derivable function $f in C^2(mathbb{R^3})$ with this gradient
$$nabla f(x,y,z) = (yz, xz, xy^2)$$
I thought about using the Hessian matrix since one has to calculate all second partial derivatives of $f$ there.
Since only the gradient is given, can I calculate the antiderivatives first:
$yz = xyz$
$xz = xyz$
$xy^2 = xy^2z$
Now I want to calculate the antiderivatives of the antiderivatives:
$xyz = frac{yzx^2}{2}$
$xyz = frac{yzx^2}{2}$
$xy^2z = dfrac{y^2zx^2}{2}$
I didn't calculate the antiderivatives of the partial derivatives and I don't even know if that way is correct...
analysis functions derivatives continuity partial-derivative
analysis functions derivatives continuity partial-derivative
asked Nov 22 at 22:13
Bad At Math
203
asked Nov 22 at 22:13
Bad At Math
203
asked Nov 22 at 22:13
Bad At Math
203
asked Nov 22 at 22:13
Bad At Math
203
asked Nov 22 at 22:13
Bad At Math
203
203
add a comment |
add a comment |
2 Answers
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2
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accepted
If $$nabla f=(f_1(x,y,z),f_2(x,y,z),f_3(x,y,z))$$and $f_i$s are twice differentiable, we must have $${partial ^2 f_1over partial ypartial z}={partial ^2 f_2over partial xpartial z}={partial ^2 f_3over partial xpartial y}$$for all $x,y,z$ but this doesn't hold here since $$1=1ne 2yqquad forall x,y,zin Bbb R$$therefore such a function doesn't exist.
answered Nov 22 at 22:25
Mostafa Ayaz
13.5k3836
add a comment |
up vote
1
down vote
In other words, you are asking to show that your vector field is not conservative. It is enough to show that the curl of your vector field is different to zero.
answered Nov 22 at 22:27
Daniel Duque
267
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
2
down vote
accepted
If $$nabla f=(f_1(x,y,z),f_2(x,y,z),f_3(x,y,z))$$and $f_i$s are twice differentiable, we must have $${partial ^2 f_1over partial ypartial z}={partial ^2 f_2over partial xpartial z}={partial ^2 f_3over partial xpartial y}$$for all $x,y,z$ but this doesn't hold here since $$1=1ne 2yqquad forall x,y,zin Bbb R$$therefore such a function doesn't exist.
answered Nov 22 at 22:25
Mostafa Ayaz
13.5k3836
add a comment |
up vote
2
down vote
accepted
If $$nabla f=(f_1(x,y,z),f_2(x,y,z),f_3(x,y,z))$$and $f_i$s are twice differentiable, we must have $${partial ^2 f_1over partial ypartial z}={partial ^2 f_2over partial xpartial z}={partial ^2 f_3over partial xpartial y}$$for all $x,y,z$ but this doesn't hold here since $$1=1ne 2yqquad forall x,y,zin Bbb R$$therefore such a function doesn't exist.
answered Nov 22 at 22:25
Mostafa Ayaz
13.5k3836
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If $$nabla f=(f_1(x,y,z),f_2(x,y,z),f_3(x,y,z))$$and $f_i$s are twice differentiable, we must have $${partial ^2 f_1over partial ypartial z}={partial ^2 f_2over partial xpartial z}={partial ^2 f_3over partial xpartial y}$$for all $x,y,z$ but this doesn't hold here since $$1=1ne 2yqquad forall x,y,zin Bbb R$$therefore such a function doesn't exist.
answered Nov 22 at 22:25
Mostafa Ayaz
13.5k3836
If $$nabla f=(f_1(x,y,z),f_2(x,y,z),f_3(x,y,z))$$and $f_i$s are twice differentiable, we must have $${partial ^2 f_1over partial ypartial z}={partial ^2 f_2over partial xpartial z}={partial ^2 f_3over partial xpartial y}$$for all $x,y,z$ but this doesn't hold here since $$1=1ne 2yqquad forall x,y,zin Bbb R$$therefore such a function doesn't exist.
answered Nov 22 at 22:25
Mostafa Ayaz
13.5k3836
answered Nov 22 at 22:25
Mostafa Ayaz
13.5k3836
answered Nov 22 at 22:25
Mostafa Ayaz
13.5k3836
answered Nov 22 at 22:25
Mostafa Ayaz
13.5k3836
13.5k3836
add a comment |
add a comment |
up vote
1
down vote
In other words, you are asking to show that your vector field is not conservative. It is enough to show that the curl of your vector field is different to zero.
answered Nov 22 at 22:27
Daniel Duque
267
add a comment |
up vote
1
down vote
In other words, you are asking to show that your vector field is not conservative. It is enough to show that the curl of your vector field is different to zero.
answered Nov 22 at 22:27
Daniel Duque
267
add a comment |
up vote
1
down vote
up vote
1
down vote
In other words, you are asking to show that your vector field is not conservative. It is enough to show that the curl of your vector field is different to zero.
answered Nov 22 at 22:27
Daniel Duque
267
In other words, you are asking to show that your vector field is not conservative. It is enough to show that the curl of your vector field is different to zero.
answered Nov 22 at 22:27
Daniel Duque
267
answered Nov 22 at 22:27
Daniel Duque
267
answered Nov 22 at 22:27
Daniel Duque
267
answered Nov 22 at 22:27
Daniel Duque
267
267
add a comment |
add a comment |
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