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A Harmonic Function is Zero on an open portion of the boundary, help! 
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#1
Nov111, 08:38 PM

P: 20

A harmonic function in a region is zero on an open portion of the boundary, and its normal derivative is also zero on the same part, and it is continuously differentiable on the boundary. I have to show that the function is zero everywhere, but I have no idea how. I have tried this for hours and hours, and haven't come up with anything useful. The best answer I've had so far involves the Cauchy Kovelevskaya theorem, but that was shown to be flawed. Can anyone help? This is very difficult..
I'm pretty sure this implies the gradient is zero on the boundary, is there anything I can possible do with that? 


#2
Nov211, 07:03 PM

P: 2

Hi lackrange,
This is actually a pretty easy corollary to Gronwall's inequality. Just play around with parameters β as shown on the wikipedia page 


#3
Nov411, 02:13 PM

P: 2

Actually I'm just trolling, I think Gronwall's inequality is completely irrelevant to this problem but it has a really cool name. This post came up after a google search on the day before my takehome midterm was due, and I assumed you were writing the same takehome as me since that question was on the test. I didn't think it was fair that you were asking people on the internet to solve your takehome midterm for you so I posted that reply as as joke.
... This didn't turn out as funny as I expected since I thought you'd reply asking me what the hell Gronwall's inequality has anything to do with this and then I'd tell you that it was all a joke. So I'm sorry if this caused any confusion to anyone, mods feel free to delete these two posts. 


#4
Jul1112, 10:53 AM

P: 1

A Harmonic Function is Zero on an open portion of the boundary, help!
Do you guys know the solution finally?



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