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Supose we have a vector field $E:R^3\rightarrow R^3$ with the property $\nabla\times E=0 \Longleftrightarrow E=-\nabla \phi$ where $\phi:R^3\rightarrow R$

How does the boundedness of $E$ imply the continuity of $\phi$

I can solve this physically for a certain case by assuming a rectangular curve through a surface across which $E$ is discontinuous as $\nabla\times E=0 \Longleftrightarrow \oint E.l dl=0$. So even though $E$ is discontinuous its associated scalar ($\phi$) is still continuous. But the argument above has been hinted to apply in general and I am having trouble getting it mathematically.

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Can't you use the property that $\phi$ is simply given by $\phi(x) = \int_0^x E \cdot dx + C$? –  Rahul Mar 19 '11 at 18:54
    
Yes, for fields and potentials acrossa surface with some charge. What you write is what I applied when I wrote the line integral is zero. Decomposing the line integrals over the edges of the rectangle, and using the fundamental theorem of calculus gives me that the potential is unchanged "on either side". But even for this specific case, how does boundedness of E imply this? –  Please Delete Account Mar 19 '11 at 19:03
    
The integral of a bounded function is continuous, isn't it? –  Rahul Mar 19 '11 at 19:27
    
Also, what I wrote is not some specific case, as you seem to imply, but in general. If you have any irrotational vector field on $\mathbb R^3$, you can define its potential by fixing an origin, choosing the value of the potential there, and determining the value on any other point using the line integral of $E$ on any curve joining it to the origin. (Since $E$ is irrotational, the line integral is independent of the choice of curve.) –  Rahul Mar 19 '11 at 19:27
    
@Rahul I see. so how do I prove that integral fo a bounded function is continuous. –  Please Delete Account Mar 19 '11 at 20:39

2 Answers 2

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Actually, you don't need all that I said in my comment. From $E = -\nabla\phi$, it follows that $\int_x^y E \cdot dx = \phi(x) - \phi(y)$ over any path joining points $x$ and $y$. Use the fact that $\lVert E \rVert \le B$ for some bound $B$ to show that you can make the size of $\{\phi(y) - \phi(x) : \lVert y - x \rVert < \delta\}$ arbitrarily small.

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While I'm fine with the part of the argument you're giving, I wonder: How does $\nabla \times E = 0$ and $E$ bounded imply that $E$ is integrable (in other words: how do the conditions imply that $E$ is measurable in the first place)? –  t.b. Mar 19 '11 at 22:16
    
@Theo: Do you mean, why should there exist a $\phi$ such that $E = -\nabla\phi$? –  Rahul Mar 20 '11 at 0:11
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Yes, exactly. Maybe it's really stupid, but I don't see how the hypotheses imply that you can integrate. Assuming measurability would definitely be fine with me and then your argument is perfect, but I don't see how it follows from the hypotheses. –  t.b. Mar 20 '11 at 0:17
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I understand that you are saying but why is the integral of $E$ defined? I'm very willing to believe that this is the case but I just don't see how to go about it at the moment. Is it true that the points of discontinuity of $E$ must be isolated and if so why? I'm tired and currently having a huge blockade concerning this. –  t.b. Mar 20 '11 at 2:41
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@Theo: Oh, I'm sorry, I see I completely misunderstood your question! I confess I haven't thought seriously about integrability at all; I learned vector calculus in a physics context where such issues are usually swept under the rug. –  Rahul Mar 20 '11 at 6:26

Well, the result is well known: If a function has bounded partial derivatives in a neighbourhood of a point then it's continuous at that point.

The proof I know involves using the mean value theorem on line segments parallel to the coordinate axis (joining said point and others in the neighbourhood).

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