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Just need to check I have this right, very simple.

Attachment 27253

I know the first two terms cancel and i'm guessing the other 3 do too... however, wasn't sure if the 3rd and 4th terms =0, so would just be left with the 5th?

Very stupid, but doesn't seem to work with the rest of the problem I'm solving...

Thank you for any help.

Re: Just need to check I have this right, very simple.

Quote:

Originally Posted by

**susie9393** Attachment 27253

I know the first two terms cancel and i'm guessing the other 3 do too... however, wasn't sure if the 3rd and 4th terms =0, so would just be left with the 5th?

Very stupid, but doesn't seem to work with the rest of the problem I'm solving...

Thank you for any help.

Hi susie9393! :)

The last 3 terms appear to have the same derivatives... combined with their coefficient they sum to zero. :confused:

So only the first 2 terms remain.

Those 2 make up the so called laplacian of f.

That will only be zero if you have more information than you have shown.

Re: Just need to check I have this right, very simple.

Thank you! I think I'm going a bit crazy...

So with having only that information, I would presume the first 2 do not sum to 0? What would make it 0?

Re: Just need to check I have this right, very simple.

Quote:

Originally Posted by

**susie9393** Thank you! I think I'm going a bit crazy...

So with having only that information, I would presume the first 2 do not sum to 0? What would make it 0?

This may sound a bit lame, but it would be zero if $\displaystyle \frac{\partial^2}{\partial u^2}f(u,v)=-\frac{\partial^2}{\partial v^2}f(u,v)$.

Re: Just need to check I have this right, very simple.

Ha, yes ok. I think I've got a bit confused with some simple math!

Thank you very much for your help.

1 Attachment(s)

Re: Just need to check I have this right, very simple.

Can you see an obvious way to simplify this?

Attachment 27265

Re: Just need to check I have this right, very simple.

Quote:

Originally Posted by

**susie9393**

I do not see an obvious way to simplify it.

However, if you treat it as a differential equation that is equal to zero, the solution is:

$\displaystyle f(u,v) = g(u+iv) + h(u-iv)$

where g and h are arbitrary functions.

Here is the solution of Wolfram|Alpha.

Re: Just need to check I have this right, very simple.

That's such help, thank you!