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

• Feb 25th 2013, 02:00 PM
susie9393
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.
• Feb 25th 2013, 03:05 PM
ILikeSerena
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.
• Feb 26th 2013, 03:16 AM
susie9393
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?
• Feb 26th 2013, 04:05 AM
ILikeSerena
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 $\frac{\partial^2}{\partial u^2}f(u,v)=-\frac{\partial^2}{\partial v^2}f(u,v)$.
• Feb 26th 2013, 04:39 AM
susie9393
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.
• Feb 26th 2013, 09:31 AM
susie9393
Re: Just need to check I have this right, very simple.
Can you see an obvious way to simplify this?
Attachment 27265
• Feb 26th 2013, 10:30 AM
ILikeSerena
Re: Just need to check I have this right, very simple.
Quote:

Originally Posted by susie9393
Can you see an obvious way to simplify this?
Attachment 27265

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:

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

where g and h are arbitrary functions.
Here is the solution of Wolfram|Alpha.
• Feb 26th 2013, 11:15 AM
susie9393
Re: Just need to check I have this right, very simple.
That's such help, thank you!