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I'm studying in preparation for a Mathematical Analysis II examination and I'm solving past exam exercises. If it's any indicator of difficulty, the exercise is Exercise 4 of 4, part $b$ and graded for 10%.

Let $f$ be an arithmetic function of class $C^2$ with $f_{xx}+f_{yy}=0$. Prove that $\oint\vec{F}•\vec{dr}=0$ where $\vec{F}=(f_y,-f_x)$ for every simple, smooth, closed curve $c$, positively oriented.

Out of all the exercises that I've posted here in the past two days, this is the one that baffled me the most. I don't even know how to start the proof.

I thought of starting by expanding the given integral, but so far nothing came of it as integration happens over $\vec{dr}$. Should I be replacing variables with something else? I see that I should be trying to get $f_x$ and $f_y$ in the integral somehow, and integrate to get a relation with $f_{xx}$ and $f_{yy}$ in the same line, which would yield $0=0$ based on the exercise 's given info, but how can I get those in there?

Dimitris S.
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1 Answers1

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Consider the vector field

$$\mathbf{G}=(f_y,-f_x,0)$$

Then $\nabla \times {\mathbf{G}} = \mathbf{0}$ because of the condition $f_{xx}+f_{yy}=0$. (In two dimensions, the curl is simply the $z$-component.) By Stokes Theorem,

$$\oint_C d\mathbf{r} \mathbf{\cdot} \mathbf{G} = \mathbf{0}$$

for every simple closed curve $C$, including those in the $xy$ plane. Thus, it follows, when we just consider the two-dimensional (i.e., $xy$) plane and $\mathbf{r}=(x,y)$,

$$\oint_C d\mathbf{r} \mathbf{\cdot} \mathbf{F} = \mathbf{0}$$

Ron Gordon
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  • Thank you for the solution. Can you explain the $\nabla \times G = 0$ part? Isn't $\nabla G= f_{xx} - f_{yy}$? How did you consider the thought to get the product? – Dimitris S. Jan 23 '16 at 15:05
  • @DimitrisS. : yes, the curl of $G$ is indeed the laplacian of $f$ pointed in the $z$ direction. You need to look at Stokes' Theorem and what the $z$ component of the curl looks like. – Ron Gordon Jan 23 '16 at 15:11