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For a vector field $\vec{F}(x,y,z) = \langle F_1(x,y,z), F_2(x,y,z), F_3(x,y,z) \rangle$ in $\mathbb{R}^3$, how can I use mixed second-order partial derivatives of each of the components to determine whether it is conservative? Which partial derivatives should I compare?

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    $\begingroup$ Since the domain is $\mathbb R^3$, you should check that $\mathrm{curl}(\textbf{F}) = \textbf{0}$. $\endgroup$ – dannum Apr 25 '17 at 17:46
  • $\begingroup$ The key word in your question is “mixed.” That should give you a strong clue as to which derivatives to examine. $\endgroup$ – amd Apr 25 '17 at 18:19
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You should check that

$$ \frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}, \\ \frac{\partial F_1}{\partial z} = \frac{\partial F_3}{\partial x}, \\ \frac{\partial F_2}{\partial z} = \frac{\partial F_3}{\partial y}. $$

An easy way to remember this is to call your variables $x^1,x^2,x^3$ (instead of $x,y,z$). Then you need to check that

$$ \frac{\partial F_i}{\partial x^j} = \frac{\partial F_j}{\partial x^i} $$

for all $i \neq j$. This works in all dimensions.

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  • $\begingroup$ I'd like to know how you came to these comparisons. $\endgroup$ – Ksquared Apr 25 '17 at 21:05
  • $\begingroup$ @Ksquared: This is the basic theorem about conservative vector spaces in $\mathbb{R}^n$... If $F(x^1,\dots,x^n) = (F_1(x^1,\dots,x^n), \dots, F_n(x^1,\dots,x^n))$ is a smooth enough vector field and $\frac{\partial F_i}{\partial x^j} = \frac{\partial F_j}{\partial x^i}$ for all $i \neq j$ then $F$ is locally conservative (and globally conservative if it is defined on a simply connected domain). $\endgroup$ – levap Apr 25 '17 at 21:12
  • $\begingroup$ It is easy to see that this is a necessary condition because if $F = \nabla f$ then $\frac{\partial f}{\partial x^i x^j} = \frac{\partial F_j}{\partial x^i} = \frac{\partial f}{\partial x^j x^i} = \frac{\partial F_i}{\partial x^j}$. $\endgroup$ – levap Apr 25 '17 at 21:14
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Assume the derivatives of the components of $\vec F$ are continuous functions. If $\vec F = \left<P(x,y,z),Q(x,y,z),R(x,y,z)\right>$ were the gradient of a function $f(x,y,z)$, then $$ \frac{\partial f}{\partial x} = P \text{,}\qquad \frac{\partial f}{\partial y} = Q \text{,}\qquad\text{and}\qquad \frac{\partial f}{\partial z} = R $$ Taking the derivative of both sides of the first equation with respect to $y$, and the second equation with respect to $x$, we get $$ \frac{\partial^2 f}{\partial y\,\partial x} = \frac{\partial P}{\partial y} \qquad\text{and}\qquad \frac{\partial^2 f}{\partial x\,\partial y} = \frac{\partial Q}{\partial x} $$ By Clairaut's theorem the left-hand sides of the two equations are equal. Therefore $$ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} $$ Similarly, it can be shown that $$ \frac{\partial P}{\partial z} = \frac{\partial R}{\partial x} \qquad\text{and}\qquad \frac{\partial Q}{\partial z} = \frac{\partial R}{\partial y} $$ These are all necessary conditions for $\vec F$ to be conservative. They can be written in a form that doesn't assume only three variables, or, in the case of three variables, in a vector form. If these equations do not hold, $\vec F$ cannot be conservative.

The big question is when these conditions are sufficient as well. The answer is that as long as the domain of $\vec F$ is connected and simply connected, they are sufficient. This follows from Stokes's Theorem.

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Hint: show that $$\nabla\times\vec F(x,y,z) = \vec 0$$

Then, it easily follows that $\vec F $ is conservative.

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If $\nabla G = <G_x,G_y,G_z> \ = F = \ <F_1,F_2,F_3>$ for some $G(x,y,z)$, then $F$ is conservative.

$(F_1)_y = (G_x)_y = G_{xy} = G_{yx} = (G_y)_x = (F_2)_x$.

So, $\displaystyle \frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}$ must hold.

The other two equations are similar, and given in other answers.

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The criterion is $$ \DeclareMathOperator{curl}{curl} \curl F = \epsilon_{ijk} e_i \partial_j F_k = 0 $$ where we sum over same indices, and $\epsilon_{ijk}$ is the sign of the permutation $(ijk)$ or zero if index values repeat.

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