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The same theorem is true for vector fields in a plane.
Since a conservative vector field is the gradient of a scalar function, the previous theorem says that $\text{curl}\phantom{\rule{0.2em}{0ex}}\left(\text{\u2207}f\right)=0$ for any scalar function $f.$ In terms of our curl notation, $\nabla \phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\nabla \left(f\right)=0.$ This equation makes sense because the cross product of a vector with itself is always the zero vector. Sometimes equation $\nabla \phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\nabla \left(f\right)=0$ is simplified as $\nabla \phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\nabla =0.$
Let $\text{F}=\u27e8P,Q,R\u27e9$ be a vector field in space on a simply connected domain. If $\text{curl}\phantom{\rule{0.2em}{0ex}}\text{F}=0,$ then F is conservative.
Since $\text{curl}\phantom{\rule{0.2em}{0ex}}\text{F}=0,$ we have that ${R}_{y}={Q}_{z},{P}_{z}={R}_{x},$ and ${Q}_{x}={P}_{y}.$ Therefore, F satisfies the cross-partials property on a simply connected domain, and [link] implies that F is conservative.
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The same theorem is also true in a plane. Therefore, if F is a vector field in a plane or in space and the domain is simply connected, then F is conservative if and only if $\text{curl}\phantom{\rule{0.2em}{0ex}}\text{F}=0.$
Use the curl to determine whether $\text{F}\left(x,y,z\right)=\u27e8yz,xz,xy\u27e9$ is conservative.
Note that the domain of F is all of ${\mathbb{R}}^{3},$ which is simply connected ( [link] ). Therefore, we can test whether F is conservative by calculating its curl.
The curl of F is
Thus, F is conservative.
We have seen that the curl of a gradient is zero. What is the divergence of a gradient? If $f$ is a function of two variables, then $\text{div}(\text{\u2207}f)=\nabla \xb7(\text{\u2207}f)={f}_{xx}+{f}_{yy}.$ We abbreviate this “double dot product” as ${\nabla}^{2}.$ This operator is called the Laplace operator , and in this notation Laplace’s equation becomes ${\nabla}^{2}f=0.$ Therefore, a harmonic function is a function that becomes zero after taking the divergence of a gradient.
Similarly, if $f$ is a function of three variables then
Using this notation we get Laplace’s equation for harmonic functions of three variables:
Harmonic functions arise in many applications. For example, the potential function of an electrostatic field in a region of space that has no static charge is harmonic.
Is it possible for $f(x,y)={x}^{2}+x-y$ to be the potential function of an electrostatic field that is located in a region of ${\mathbb{R}}^{2}$ free of static charge?
If $f$ were such a potential function, then $f$ would be harmonic. Note that ${f}_{xx}=2$ and ${f}_{yy}=0,$ and so ${f}_{xx}+{f}_{yy}\ne 0.$ Therefore, $f$ is not harmonic and $f$ cannot represent an electrostatic potential.
Is it possible for function $f(x,y)={x}^{2}-{y}^{2}+x$ to be the potential function of an electrostatic field located in a region of ${\mathbb{R}}^{2}$ free of static charge?
Yes
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