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

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

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.

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

Originally Posted by susie9393

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.
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.

3. ## 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?

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

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 $\frac{\partial^2}{\partial u^2}f(u,v)=-\frac{\partial^2}{\partial v^2}f(u,v)$.

5. ## 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.

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

Can you see an obvious way to simplify this?

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

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

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

That's such help, thank you!