Find a vector field $G$ with curl ($G$) = $F$
$begingroup$
Let $F(x, y, z) = (y, z, x^2)$ on $mathbb{R}^3$.
We know that
$$y = frac{partial G_3}{ partial y} - frac{partial G_2 }{partial z}, \ z = frac{partial G_1}{ partial z} - frac{partial G_3 }{partial x}, \
x^2 = frac{partial G_2}{ partial x} - frac{partial G_1 }{partial y}.$$
How do I go further?
multivariable-calculus vector-fields
asked Nov 30 '15 at 23:56
user286826user286826
443211
$endgroup$
add a comment |
$begingroup$
Let $F(x, y, z) = (y, z, x^2)$ on $mathbb{R}^3$.
We know that
$$y = frac{partial G_3}{ partial y} - frac{partial G_2 }{partial z}, \ z = frac{partial G_1}{ partial z} - frac{partial G_3 }{partial x}, \
x^2 = frac{partial G_2}{ partial x} - frac{partial G_1 }{partial y}.$$
How do I go further?
multivariable-calculus vector-fields
asked Nov 30 '15 at 23:56
user286826user286826
443211
$endgroup$
add a comment |
$begingroup$
Let $F(x, y, z) = (y, z, x^2)$ on $mathbb{R}^3$.
We know that
$$y = frac{partial G_3}{ partial y} - frac{partial G_2 }{partial z}, \ z = frac{partial G_1}{ partial z} - frac{partial G_3 }{partial x}, \
x^2 = frac{partial G_2}{ partial x} - frac{partial G_1 }{partial y}.$$
How do I go further?
multivariable-calculus vector-fields
asked Nov 30 '15 at 23:56
user286826user286826
443211
$endgroup$
Let $F(x, y, z) = (y, z, x^2)$ on $mathbb{R}^3$.
We know that
$$y = frac{partial G_3}{ partial y} - frac{partial G_2 }{partial z}, \ z = frac{partial G_1}{ partial z} - frac{partial G_3 }{partial x}, \
x^2 = frac{partial G_2}{ partial x} - frac{partial G_1 }{partial y}.$$
How do I go further?
multivariable-calculus vector-fields
multivariable-calculus vector-fields
asked Nov 30 '15 at 23:56
user286826user286826
443211
asked Nov 30 '15 at 23:56
user286826user286826
443211
edited Dec 1 '15 at 0:21
user286826
asked Nov 30 '15 at 23:56
user286826user286826
443211
asked Nov 30 '15 at 23:56
user286826user286826
443211
asked Nov 30 '15 at 23:56
user286826user286826
443211
443211
add a comment |
add a comment |
1 Answer
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$begingroup$
Knowing that $vec G(x,y,z)=(G_1(x,y,z),G_2(x,y,z),G_3(x,y,z))$ can only be determined up to a gradient of a scalar function, we can without loss of generality take $ G_1(x,y,z)=0$
$$
x^2 = frac{partial G_2}{ partial x} - frac{partial G_1 }{partial y}. implies G_2(x,y,z) = frac 13 x^3+f(y,z)$$
$$ z = frac{partial G_1}{ partial z} - frac{partial G_3 }{partial x}implies G_3(x,y,z) =-xz+h(y)$$
for functions $f$ and $h$ to be determined by the remaining equation ...
$$ y = frac{partial G_3}{ partial y} - frac{partial G_2 }{partial z}implies frac{partial h(y)}{partial y}-frac{partial f(y,z)}{partial z}=y $$
one solution that clearly works is $h(y)=frac 12 y^2$ and $f(x,y)=0$
So $vec G(x,y,z)=(0, frac 13 x^3 + frac 12 y^2, -xz)$ is a vector field satisfying $vecnabla times vec G(x,y,z)=(y, z, x^2)$
answered Dec 1 '15 at 0:50
WW1WW1
7,3151712
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1 Answer
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1 Answer
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$begingroup$
Knowing that $vec G(x,y,z)=(G_1(x,y,z),G_2(x,y,z),G_3(x,y,z))$ can only be determined up to a gradient of a scalar function, we can without loss of generality take $ G_1(x,y,z)=0$
$$
x^2 = frac{partial G_2}{ partial x} - frac{partial G_1 }{partial y}. implies G_2(x,y,z) = frac 13 x^3+f(y,z)$$
$$ z = frac{partial G_1}{ partial z} - frac{partial G_3 }{partial x}implies G_3(x,y,z) =-xz+h(y)$$
for functions $f$ and $h$ to be determined by the remaining equation ...
$$ y = frac{partial G_3}{ partial y} - frac{partial G_2 }{partial z}implies frac{partial h(y)}{partial y}-frac{partial f(y,z)}{partial z}=y $$
one solution that clearly works is $h(y)=frac 12 y^2$ and $f(x,y)=0$
So $vec G(x,y,z)=(0, frac 13 x^3 + frac 12 y^2, -xz)$ is a vector field satisfying $vecnabla times vec G(x,y,z)=(y, z, x^2)$
answered Dec 1 '15 at 0:50
WW1WW1
7,3151712
$endgroup$
add a comment |
$begingroup$
Knowing that $vec G(x,y,z)=(G_1(x,y,z),G_2(x,y,z),G_3(x,y,z))$ can only be determined up to a gradient of a scalar function, we can without loss of generality take $ G_1(x,y,z)=0$
$$
x^2 = frac{partial G_2}{ partial x} - frac{partial G_1 }{partial y}. implies G_2(x,y,z) = frac 13 x^3+f(y,z)$$
$$ z = frac{partial G_1}{ partial z} - frac{partial G_3 }{partial x}implies G_3(x,y,z) =-xz+h(y)$$
for functions $f$ and $h$ to be determined by the remaining equation ...
$$ y = frac{partial G_3}{ partial y} - frac{partial G_2 }{partial z}implies frac{partial h(y)}{partial y}-frac{partial f(y,z)}{partial z}=y $$
one solution that clearly works is $h(y)=frac 12 y^2$ and $f(x,y)=0$
So $vec G(x,y,z)=(0, frac 13 x^3 + frac 12 y^2, -xz)$ is a vector field satisfying $vecnabla times vec G(x,y,z)=(y, z, x^2)$
answered Dec 1 '15 at 0:50
WW1WW1
7,3151712
$endgroup$
add a comment |
$begingroup$
Knowing that $vec G(x,y,z)=(G_1(x,y,z),G_2(x,y,z),G_3(x,y,z))$ can only be determined up to a gradient of a scalar function, we can without loss of generality take $ G_1(x,y,z)=0$
$$
x^2 = frac{partial G_2}{ partial x} - frac{partial G_1 }{partial y}. implies G_2(x,y,z) = frac 13 x^3+f(y,z)$$
$$ z = frac{partial G_1}{ partial z} - frac{partial G_3 }{partial x}implies G_3(x,y,z) =-xz+h(y)$$
for functions $f$ and $h$ to be determined by the remaining equation ...
$$ y = frac{partial G_3}{ partial y} - frac{partial G_2 }{partial z}implies frac{partial h(y)}{partial y}-frac{partial f(y,z)}{partial z}=y $$
one solution that clearly works is $h(y)=frac 12 y^2$ and $f(x,y)=0$
So $vec G(x,y,z)=(0, frac 13 x^3 + frac 12 y^2, -xz)$ is a vector field satisfying $vecnabla times vec G(x,y,z)=(y, z, x^2)$
answered Dec 1 '15 at 0:50
WW1WW1
7,3151712
$endgroup$
Knowing that $vec G(x,y,z)=(G_1(x,y,z),G_2(x,y,z),G_3(x,y,z))$ can only be determined up to a gradient of a scalar function, we can without loss of generality take $ G_1(x,y,z)=0$
$$
x^2 = frac{partial G_2}{ partial x} - frac{partial G_1 }{partial y}. implies G_2(x,y,z) = frac 13 x^3+f(y,z)$$
$$ z = frac{partial G_1}{ partial z} - frac{partial G_3 }{partial x}implies G_3(x,y,z) =-xz+h(y)$$
for functions $f$ and $h$ to be determined by the remaining equation ...
$$ y = frac{partial G_3}{ partial y} - frac{partial G_2 }{partial z}implies frac{partial h(y)}{partial y}-frac{partial f(y,z)}{partial z}=y $$
one solution that clearly works is $h(y)=frac 12 y^2$ and $f(x,y)=0$
So $vec G(x,y,z)=(0, frac 13 x^3 + frac 12 y^2, -xz)$ is a vector field satisfying $vecnabla times vec G(x,y,z)=(y, z, x^2)$
answered Dec 1 '15 at 0:50
WW1WW1
7,3151712
edited Dec 28 '18 at 3:54
answered Dec 1 '15 at 0:50
WW1WW1
7,3151712
answered Dec 1 '15 at 0:50
WW1WW1
7,3151712
answered Dec 1 '15 at 0:50
WW1WW1
7,3151712
7,3151712
add a comment |
add a comment |
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