The starting point of the proof is the fact that if \(\bfG = \nabla f\) for some \(f\) of class \(C^1\), then for any oriented curve \(C\) that starts at a point \(\bfp\) and ends at a point \(\bfq\), \[\begin{equation}\label{FFF} \int_C \bfG\cdot d\bfx = \int_C \nabla f\cdot d\bfx = f(\bfq) - f(\bfp). \end{equation}\] We saw this in Section 5.1, where we called it the Fundamental Theorem of Line Integrals.
1 implies 2
2 implies 3
(sketch) Let \(C_1, C_2\) be two piecewise smooth curves in \(U\) that both start at the same point \(\bfp\in U\) and end at the same point \(\bfq\in U\).
Given these, we may define a piecewise smooth closed curve \(C\) to be the curve formed by
- following \(C_1\) from \(\bfp\) to \(\bfq\), then
- following \(C_2\) backwards, from \(\bfq\) to \(\bfp\),
3 implies 1
This is the hardest and most interesting part of the theorem. Suppose that \(\bfG\) is a vector field that satisfies condition (3). If this function \(f\) exists, it must satisfy . Thus we can try to use formula to reconstruct \(f\) (if it exists) from \(\bfG\). That is, we can fix some \(\bfp\in U\), and we can look for \(f\) such that \(\bfG=\nabla f\) and \(f(\bfp)=0\). According to \(\eqref{FFF}\), such a function \(f\) must satisfy \[\begin{equation} f(\bfq) = \int_{C_{\bfp,\bfq}} \bfG\cdot d\bfx\qquad\text{ for any curve }C_{\bfp,\bfq}\text{ starting at }\bfp\text{ and ending at }\bfq. \label{FFFbis}\end{equation}\] Thus, given \(\bfG\) satisfying condition (3), we use equation \(\eqref{FFFbis}\) to define \(f:U\to \R\), and then verify that \(\nabla f = \bfG\). Condition (3) implies that the definition of \(f\) makes sense (that is, is independent of the choice of path \(C_{\bfp, \bfq}\) connecting \(\bfp\) to \(\bfq\)). Thus the proof of the theorem is completed by…
\(f\) satisfies \(\nabla f = \bfG\). (sketch)
Let \(f\) be the function defined in \(\eqref{FFFbis}\). We first claim that for any \(\bfq \in U\), any vector \(\bfv\), and any \(h\in \R\) such that the line segment from \(\bfq\) to \(\bfq+h\bfv\) is contained in \(U\), \[\begin{equation}\label{ffbb} f(\bfq + h\bfv) = f(\bfq) + \int_0^h \bfG(\bfq + t\bfv)\cdot \bfv \, dt. \end{equation}\] To see that this is true, fix some curve \(C_{\bfp, \bfq}\) that starts at \(\bfp\) and ends at \(\bfq\). Let \(\ell_{\bfq, \bfq+h\bfv}\) be the line segment that starts at \(\bfq\) and ends at \(\bfq+h\bfv\), and let \(C_{\bfp, \bfq+h\bfv}\) be the curve obtained by
- first following \(C_{\bfp, \bfq}\) from \(\bfp\) to \(\bfq\);
- then following \(\ell_{\bfq, \bfq+h\bfv}\) from \(\bfq\) to \(\bfq+h\bfv\).
Thus \(C_{\bfp, \bfq+h\bfv}\) is a piecewise smooth curve that starts at \(\bfp\) and ends at \(\bfq+h\bfv\). It follows from that \[ \begin{aligned} f(\bfq+h\bfv) = \int_{C_{\bfp, \bfq+h\bfv}} \bfG\cdot d\bfx &= \int_{C_{\bfp,\bfq}} \bfG\cdot d\bfx + \int_{\ell_{\bfq, \bfq+h\bfv}} \bfG\cdot d\bfx\\ &= f(\bfq) + \int_{\ell_{\bfq, \bfq+h\bfv}} \bfG\cdot d\bfx\\ \end{aligned} \] If we express the line integral over \(\ell_{\bfq, \bfq+h\bfv}\) in terms of the parametrization \(\bfg(t) = \bfq+t\bfv\), \(0\le t \le h\), this reduces to .
Next, for any \(j\in \left\{1,\ldots, n\right\}\), if (as usual) \({\bf e}_j\) denotes the unit vector in the \(j\)th coordinate direction, then it follows from that \[ \begin{aligned} \frac{\partial f}{\partial x_j}(\bfq) = \lim_{h\to 0}\frac{f(\bfq+h {\bf e}_j) - f(\bfq)}h &= \lim_{h\to 0} \frac 1 h \int_0^h \bfG(\bfq+t {\bf e}_j)\cdot {\bf e}_j dt \\ &= \lim_{h\to 0} \frac 1 h \int_0^h G_j(\bfq+t {\bf e}_j)\, dt . \end{aligned} \] It is then an exercise to prove (using our assumption that \(\bfG\) is continuous) then the limit exists and equals \(G_j(\bfq)\). Since \(j\) is arbitrary, this proves that \(\nabla f(\bfq) = \bfG(\bfq)\).One drawback of Theorem 1 is that, given a vector field \(\bfG\), it might be hard to check whether it satisfies condition (2) or (3). In order to do this, one would need to evaluate line integrals of \(\bfG\) over every possible closed curve (for (2)) or pair of curves that start and end at the same point (for (3)). However, in \(3\) dimensions, and if \(\bfG\) is \(C^1\), there is sometimes a much easier way to check whether it is conservative:
If \(U\) is convex, then the converse is true:
However, on some non-convex sets, there exist non-conservative vector fields \(\bfG\) that satisfy \(\curl \bfG = \bf 0\). This is a special case of a much more general theorem that we will neither state nor discuss.
Sketch of proof.
We already know that if \(\bfG = \grad f\), then \(\curl \bfG = \curl \grad f = \bf 0\).
For the converse, suppose that \(\bfG\) is a \(C^1\) vector field on a convex set \(U\subset \R^3\) such that \(\curl \bfG = {\bf 0}\).
Fix \(\bfp\in U\), and for \(\bfq\in U\), let \(\ell_{\bfp, \bfq}\) denote the line segment starting at \(\bfp\) and ending at \(\bfq\). Since \(U\) is convex, this line segment is entirely contained in \(U\).
We now define \[ f(\bfq) = \int_{\ell_{\bfp, \bfq}} \bfG\cdot d\bfx. \] This is clearly well-defined, since we have specified exactly which path we follow to get from \(\bfp\) to \(\bfq\), for every \(\bfq\).
In order to mimic the proof of Theorem 1, all we need is to be able to check that \[\begin{equation}\label{ffbbb} f(\bfq + h\bfv) = f(\bfq) + \int_0^h \bfG(\bfq + t\bfv)\cdot \bfv \, dt. \end{equation}\] holds. If we know this, then every other argument in the earlier proof can be repeated with no change.
To prove it, let’s suppose that \(\bfp, \bfq\) and \(\bfq+h\bfv\) are not colinear (in that case one can give a different, easier argument). Then they form the vertices of a triangle. Let’s call the triangle \(S\). For any choice of orientation (that is, of the direction \(\bfn\) of the unit normal), \[ \iint_{S }(\curl \bfG)\cdot \bfn \, dA = 0 \] since \(\curl \bfG = \bf 0\) by assumption. Thus Stokes’ Theorem implies that \[ \int_{\partial S} \bfG\cdot d\bfx = 0 \] for any choice of the orientation of \(\partial S\). Note however that \(\partial S\) consists of the three line segments connecting \(\bfp\) to \(\bfq\) to \(\bfq+h\bfv\), and back to \(\bfp\). If it is oriented in that order (\(\bfp\) to \(\bfq\) to \(\bfq+h\bfv\) to \(\bfp\)), then \[ \begin{aligned} 0 = \int_{\partial S} \bfG\cdot d\bfx \ &= \ \int_{\ell_{\bfp,\bfq}}\bfG\cdot d\bfx + \int_{\ell_{\bfq,\bfq+h\bfv}}\bfG\cdot d\bfx + \int_{\ell_{\bfq+h\bfv,\bfp}}\bfG\cdot d\bfx \\ &= f(\bfq) + \int_0^h \bfG(\bfq + t\bfv)\cdot \bfv \, dt -f(\bfq+h\bfv). \end{aligned} \] We obtain by rearranging this. \(\quad \Box\)
An example of a non-conservative vector field \(\bfG\) on a convex set that satisfies \(\curl \bfG = \bf 0\) can be found in the exercises.
For conservative \(\bfG\), how to find \(f\) such that \(\nabla f = \bfG\)?
Now suppose \(\bfG\) is a continuous vector field on an open set \(U\subset \R^n\), and that we somehow know that \(\bfG\) is conservative. How can we find \(f\) such that \(\nabla f = \bfG\)?
One method is simply to carry out a concrete version of the abstract proof, sketched above, that uses formula to demonstrate of the existence of \(f\). This works particularly well if \(U\) is a rectangle. Then given \(\bfp = (a,b,c)\) and \(\bfq = (x,y,z)\), we can always join \(\bfp\) to \(\bfq\) as follows:
- Change the first component from \(a\) to \(x\), holding the other two components fixed. This straight-line path is parametrized by \[ (t,b,c), \qquad t\text{ moves from }a\text{ to }x. \]
- Next change the second component from \(b\) to \(y\), holding the other two components fixed. Since the first component has already been changed to \(x\), this straight-line path is parametrized by \[ (x,t,c), \qquad t\text{ moves from }b\text{ to }y. \]
- Finally, change the third component from \(c\) to \(z\), holding the other two components fixed. This straight-line path is parametrized by \[ (x,y,t), \qquad t\text{ moves from }c\text{ to }z. \]
Let \(C_{\bfp, \bfq}\) be the piecewise linear curve obtained in this way. Then \[ \int_{C_{\bfp, \bfq} }\bfG\cdot d\bfx = \int_a^x G_1(t,b,c)\, dt + \int_b^y G_2(x,t,c)\, dt + \int_c^z G_3(x,y,t)\, dt. \] So one way to implement formula is by: fix \((a,b,c)\), and define \[\begin{equation}\label{inp} \boxed{ f(x,y,z) = \int_a^x G_1(t,b,c)\, dt + \int_b^y G_2(x,t,c)\, dt + \int_c^z G_3(x,y,t)\, dt.} \end{equation}\] (Note, we could also change the components in a different order, if we prefer…)
The above theoretical considerations guarantee that if \(\bfG\) is conservative, then \(f\) defined in this way must satisfy \(\nabla f = \bfG\).
Example 1.
Let \(\bfG:\R^3\to \R^3\) be defined by \[ \bfG(x,y,z) = (-y \sin xy , -x \sin xy + z\cos yz, y\cos yz). \] One can check that \(\curl \bfG = \bf0\). Thus \(\bfG\) is conservative. Let us try to find \(f\) such that \(\nabla f = \bfG\). We will simply use the above formula, with \((a,b,c) = (0,0,0)\). (We could choose any point, but \((0,0,0)\) is convenient.)
Then \[ \int_a^x G_1(t,b,c)\, dt = \int_0^x 0\, dt = 0, \] \[ \left. \int_b^y G_2(x,t,c)\, dt = \int_0^y(-x\sin xt\, + 0)dt = \cos xt \right|_{t=0}^{t=y} = \cos xy - 1, \] and \[ \left. \int_c^zG_3(x,y,t)\, dt = \int_0^z y\cos yt \, dt = \sin yt\right|_{t=0}^z = \sin yz. \] By adding the contributions from these three terms, we conclude that \[ f(x,y,z) = \cos xy + \sin yz -1 \text{ satisfies }\nabla f = \bfG. \] It is easy to check that this is indeed the case. (The constant \(-1\) appears because, in choosing \(\bfp = (0,0,0)\), we implicitly arranged that \(f(0,0,0) = 0\), and \(-1\) is the constant that makes this be the case. If we had chosen a different point \(\bfp = (a,b,c)\), then chances are that we would have gotten a different constant.
Remark. Note that in writing down a concrete implementation of \(\eqref{FFFbis}\), we could have chosen any curve from \((a,b,c)\) to \((x,y,z)\), such as (for example) a straight line. If the domain of \(\bfG\) is a convex set containing the origin, then we can take \((a,b,c) = (0,0,0)\), and the straight line to \((x,y,z)\)is parametrized by \(\bfg(t) = (tx, ty, tz)\) for \(0\le t\le 1\). This leads to a the general formula \[ f(x,y,z) = \int_0^1 \bfG(tx, ty, tz)\cdot (x,y,z)\, dt. \] For example, for \(\bfG(x,y,z) = (-y \sin xy , -x \sin xy + z\cos yz, y\cos yz)\) as above, this becomes (after some computation) \[ f(x,y,z) = \int_0^1 - 2txy\sin(t^2 xy) + 2tyz\cos(t^2yz)\, dt \] This is easily evaluated by considering the two halves separately, and making th substitutions \(u = t^2xy\) in the first half and \(u =t^2yz\) in the second half.
Vector Fields that are Curls
There is a whole theory about vector fields \(\bfG:U\to \R^3\) (for \(U\) an open subset of \(\R^3\)) with the property that \(\bfG = \curl \bfF\) for some other vector field \(\bfF\) of class \(C^1\). It is very much parallel to the theory of gradient (= conservative) vector fields. However, we considered it in less detail.
Some main facts are summarized in the following:
If \(\bfG:U\to \R^3\) is a vector field of class \(C^1\) and \(\bfG = \curl \bfF\) for some vector field \(\bfF:U\to \R^3\) of class \(C^2\), then \(\operatorname{div}\bfG = 0.\)
Suppose \(\bfG\) is a \(C^1\) vector field in an open set \(U\subset \R^3\) such that \(\operatorname{div}\bfG = 0\).
a. If \(U\) is convex, then there exists a vector field \(\bfF\) such that \(\curl \bfF = \bfG\).
b. However, if \(U\) is not convex, it may be the case that no such vector field exists.
About the proof
1. We already know that \(\div\curl \bfF = 0\) for all \(\bfF\).
2b. Consider the vector field \[ \bfG(x,y,z) = (\frac {x}{r^3},\frac {y}{r^3}, \frac {z}{r^3}) ,\qquad\text{ where }\ \quad r = \sqrt {x^2+y^2+z^2}. \] This is an example showing that on a nonconvex set \(U\) (note that the domain of \(\bfG\) is \(U = \R^3\setminus\left\{ {\bf 0}\right\}\)), there can exist vector fields with zero divergence that are not curls. Imdeed, you can check that \(\div \bfG = 0\) for this \(\bfG\). You can also check that if \(S = \left\{(x,y,z)\in \R^3 : x^2+y^2+z^2=1\right\}\), oriented with the unit normal pointing outward, then \[ \iint_S \bfG\cdot \bfn \, dA = 4\pi. \] whereas if \(\bfG=\curl \bfF\) in \(U\), then we would have \(\iint_S \bfG\cdot \bfn \, dA =\iint_S \curl \bfF\cdot \bfn \, dA = 0\). (See Section 5.6.)
2a. The way the proof works is illustrates below for a concrete example. After that we will discus the abstract proof.Example 2.
Let \[ \bfG(x,y,z) = (xe^{-x^2z^2} -6x, 5y+2z, z - ze^{-x^2z^2}) . \] One can check that \(\operatorname{div}\bfG=0\), and the domain of \(\bfG\) is all of \(\R^3\), which is convex. So it must be possible to write \(\bfG\) as the curl of some vector field \(\bfF\).
It turns out that in this situation, it is always possible to find \(\bfF\) such that one of its components is zero everywhere. In this example, it turns out to be easiest to lok for \(\bfF\) of the form \(\bfF = (F_1, 0, F_3)\). (This can be discovered by trial and error.) Then the equation that we are trying to solve, \(\curl \bfF = \bfG\), can be written as three equations: \[ \begin{aligned} \partial_2 F_3 &= G_1 =xe^{-x^2z^2} -6x \\ \partial_3 F_1-\partial_1F_3 &= G_2 = 5y+2z\\ -\partial_2 F_1 &=G_3 = z - ze^{-x^2z^2}. \end{aligned} \] First we consider the first and third equations. For every fixed \(x,z\), we may take the antiderivative with respect to the \(y\) variable to obtain \[ \begin{aligned} F_3(x,y,z) &= \int(xe^{-x^2z^2} -6x)dy = y(xe^{-x^2z^2} -6x) + \phi_3(x,z), \\ F_1(x,y,z) &= -\int(z - ze^{-x^2z^2})dy = y(ze^{-x^2z^2}-z) + \phi_1(x,z), \end{aligned} \] where \(\phi_1(x,z)\) and \(\phi_3(x,z)\) are constants of integration, which are written in that way because they may depend on \(x\) and \(z\). Substituting these into the second equation and simplifying, we obtain \[\begin{multline} 5y+2z = G_2 = \partial_3 F_1-\partial_1F_3 = \partial_3 \phi_1-\partial_1\phi_3 \\ + y(e^{-x^2z^2}-1) -2y x^2z^2 e^{-x^2z^2} - \left[ y(e^{-x^2z^2}- 6) - 2yx^2z^2 e^{-x^2z^2} \right]. \nonumber \end{multline}\] After a lot of cancellation, this reduces to \[ \partial_3 \phi_1-\partial_1\phi_3 = 2z \] and we can see by inspection that ths is solved for example by \(\phi_1(x,z) = 0, \phi_3(x,z) = -2xz\). We conclude that \(\curl \bfF = \bfG\) for \[ \bfF (x,y,z) = (yze^{-x^2z^2}-yz , \ 0 , \ yxe^{-x^2z^2} -6xy -2xz ). \]
Proof of part 2a of Theorem 2
We will just follow the same procedure as in Example 2. We will also cheat a little cheat by proving the theorem not for an arbitrary convex set, but instead under the assumption that \(U\) is a rectangle, e.g. \((a_1,b_1)\times (a_2,b_2)\times (a_3,b_3)\).
We suppose that \(\bfG\) is \(C^1\) in \(U\) and that \(\div \bfG = 0\). We want to find \(\bfF\) solving the system of equations \[ \begin{aligned} \partial_2F_3 - \partial_3F_2&= G_1 \\ \partial_3F_1 - \partial_1F_3&= G_2 \\ \partial_1F_2 - \partial_2F_1&= G_3 . \end{aligned} \] We will look for \(\bfF\) such that \(F_3 = 0\). Then the first two equations reduce to \[ \begin{aligned} - \partial_3F_2&= G_1 \\ \partial_3F_1 &= G_2 . \end{aligned} \] We fix some \(a\in (a_3, b_3)\) and integrate both sides with respect to the third variable (calling it \(t\) instead of \(z\) for purposes of integration) getting constants of integration that depend on \(x\) and \(y\): \[ \begin{aligned} F_2(x,y,z)&= -\int_a^z G_1(x,y,t)\, dt +\phi(x,y)\\ F_1(x,y,z)&= \int_a^z G_2(x,y,t)\,dt +\psi(x,y). \end{aligned} \] We now substitute these into the third equation, differentiate under the integral sign, then simplify using the fact that \(\div \bfG = 0\): \[ \begin{aligned} G_3(x,y,z) &= \partial_1F_2 - \partial_2F_1 \\ &= \partial_1 \phi - \partial_2\psi - \int_a^z (\partial_1 G_1 +\partial_2 G_2)(x,y,t)\, dt\\ &= \partial_1 \phi - \partial_2\psi + \int_a^z \partial_3 G_3(x,y,t) \ dt\\ &= \partial_1 \phi - \partial_2\psi + G_3(x,y,z) - G_3(x,y,a). \end{aligned} \] Thus the equation reduces to \[ \partial_1 \phi(x,y) - \partial_2\psi(x,y) = G_3(x,y,a). \] and this can be solves for example by fixing \(a'\in (a_1,b_1)\) and setting \[ \psi(x,y)=0, \qquad \phi(x,y) = \int_{a'}^x G_3(t,y,a)\, dt\ . \]Example 3
This is optional but interesting.
Suppose that \(R\subset \R^2\) is a regular region with piecewise smooth boundary, contained in \([-a,a]\times [-a,a]\) for some positive number \(a<\pi\), and let \[ S = \left\{ (x,y, \phi(x,y) : (x,y)\in R\right\} \quad\text{ for }\phi(x,y) = \ln(\frac {\cos x}{\cos y}). \]
Below, an image of a portion of the graph of \(\phi\). For a suitable choice of the region \(R\), this is what \(S\) would look like.
\(\qquad\qquad\)
Also define \[ \begin{aligned} \bfG(x,y,z) = \frac{(-\partial_x \phi, -\partial_y \phi , 1)}{\sqrt{1+|\nabla \phi|^2}}& = \frac{ (\tan x ,- \tan y, 1)}{ \sqrt{1+\tan^2 x +\tan^2 y}} %\\ %& = (\frac{\sin x}{\sqrt{1+\tan^2 y \cos^2x}} , \frac{-\sin y}{ %\sqrt{1+\tan^2 x\cos^2y}}, %\frac 1{\sqrt{1+\tan^2 x+\tan^2 y}}) \end{aligned} \] It is a fact that you can check, if you are unusually fond of differentiation, that \(\nabla \cdot \bfG = 0\), and hence that there exists some \(\bfF\) such that \(\bfG = \nabla\times \bfF\). It follows by Stokes Theorem that \[ \iint_{S}\bfG\cdot\bfn \, dA = \iint_{S'}\bfG\cdot\bfn \, dA \] for any other surface \(S'\) with the same boundary as \(S\).
It is easy to see that \(\bfG \cdot \bfn = 1\) on \(S\) (in fact \(\bfG\) exactly equals \(\bfn\) on \(S\)). As a result for any surface \(S'\) as above, \[
\text{area}(S)
\ = \
\iint_S 1\, dA
\ = \
\iint_S \bfG\cdot \bfn\, dA
\ = \
\iint_{S'} \bfG\cdot \bfn\, dA .
\] Also, it is pretty clear that \(|\bfG|=1\) everywhere, and hence that \(|\bfG \cdot \bfn| \le |\bfG|\, |\bfn| = 1\) everywhere on \(S'\). Therefore
\[
\iint_{S'} \bfG\cdot \bfn\, dA \
\le \ \iint_{S'} |\bfG\cdot \bfn|\, dA
\ \le \ \iint_{S'}1 \, dA = \text{area}(S').
\] Putting these together, we conclude that \[
\text{area}(S) \le \text{area}(S')\text{ for any surface }S'
\text{ such that }\partial S = \partial S'.
\] This shows that the particular surface \(S\) pictured above is an example of what is called a minimal surface: a surface that has the smallest possible area, among all surfaces with a given boundary.
Problems
Basic
Given a vector field \(\bfG:\R^3\to \R^3\), determine whether whether there exists any functions \(f\) such that \(\nabla f = \bfG\), and if so, find one. For example:
\(\bfG(x,y,z) = ( \frac {e^z} {1+y^2}, - \frac {2xy e^z}{(1+y^2)^2}, \frac {xe^z}{1+y^2} +\sin z)\).
\(\bfG(x,y,z) = (yz, xz,xy)\).
\(\bfG(x,y,z) = (x\sin y \cos z , \frac 12 x^2 \cos y \cos z , -\frac 12 x^2 \sin y \sin z).\)
\(\bfG(x,y,z) = (\frac 12, \frac{y^2-xz}{y^2z} , \frac {-y}{z^2}).\)
\(\bfG(x,y,z) = (\frac{\cos z}x , \frac {\cos z}y , -\sin z \ln(xyz) + \frac {\cos z}{z} )\), for \(x,y,z\) all positive.
\(\bfG\) is a vector field of the form \(\bfG(x,y,z) = (f(x), g(y), h(z))\), where \(f,g,h\) are all continuous functions of a single variable.
Given a vector field \(\bfG:\R^3\to \R^3\), determine whether there exist any vector fields \(\bfF\) such that \(\nabla \times \bfF = \bfG\), and if so find one. For example:
\(\bfG(x,y,z) = ( x,y , z )\).
\(\bfG(x,y,z) = ( -y,x, 0)\).
\(\bfG(x,y,z) = ( y, \frac 1 z \cos yz, \frac 1 y \cos yx)\).
\(\bfG(x,y,z) = ( yz, xz, xy)\).
\(\bfG\) is a vector field of the form \(\bfG(x,y,z) = (f(y,z), g(x,z), h(x,y))\), where \(f,g,h\) are all continuous functions of two variables.
Advanced
For \(U = \left\{(x,y,z)\in \R^3 : x^2+y^2>0\right\}\), define \(\bfG:U\to \R^3\) by \[\bfG(x,y,z) = (\frac {-y}{x^2+y^2}, \frac x{x^2+y^2}, 0)\].
- Prove that \(\curl \bfG = \bf 0\).
- Prove that \(\bfG\) is not conservative.
Hint
Consider \(\int_C \bfG\cdot d\bfx\), where \(C\) is a circle around the \(z\)-axis. - If we define \(U' = \left\{(x,y,z)\in \R^3 : x>0\right\}\), then we know that \(\bfG\) is conservative on \(U'\). (why?) Find \(f\) such that \(\nabla f = \bfG\) on \(U'\).
Prove that if \(g\) is a continuous function defined on an open set containing a point \(\bfq\), then for unit vector \(\bfe\), \[ \lim_{h\to 0} \frac 1 h \int_0^h g(\bfq+t {\bf e})\, dt = g(\bfq) \] (This point comes up in the proofs of Theorems 1 and 2).
Suppose that \(\bfG\) is a \(C^2\) vector field that is both a gradient and a curl.
Show that if \(f\) is a function such that \(\nabla f = \bfG\), then \(f\) is harmonic, that is, \(\nabla^2 f = 0\), where we recall that the Laplacian is \(\nabla^2 f = \dfrac{\partial^2 f}{\partial x_1^2}+\dfrac{\partial^2 f}{\partial x_2^2\dfrac{\partial^2 f}{\partial x_3^2}.\)
Conclude that each component \(G_i\) of \(\bfG\) is also harmonic.
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