# Partial differential equation

1. Jan 14, 2012

### ibysaiyan

1. The problem statement, all variables and given/known data

Hi
I need some help in proving that $\partial^2 V / \partial x^2$ + $\partial^2 V / \partial y^2$ = 0 , when V = 1/2ln(x^2+y^2)
2. Relevant equations

3. The attempt at a solution
For $\partial V / \partial x$ here's what I get: (1/x^2+y^2) *2x*1/2 = x * (x^2+y^2)^-1
and for $\partial V / \partial y$
(x^2+y^2)^-1 *y ....

As you can see if i differentiate them once , I don't end up with any terms which may cancel each other to get zero..
what am I over looking ? PDE is new to me. I know we keep one variable constant in respect to other,etc..
Thanks

2. Jan 14, 2012

### Dick

The equation has second derivatives of V. Why are you stopping with the first derivative? You were doing fine.

3. Jan 14, 2012

### ibysaiyan

I should have posted my complete answer which I know is wrong...
Here is the second derivative for x , keeping y fixed.

For $\partial V / \partial x$ here's what I get: (1/x^2+y^2) *2x*1/2 = x * (x^2+y^2)^-1

=>
-(x^2+y^2)^-1 * (x^2+y^2)^-2 *2x^2...

Which seems wrong...

4. Jan 14, 2012

### Dick

It is wrong. But I can see parts of the right answer in it. Did you use the product rule on the first derivative? Can you show your steps?

5. Jan 14, 2012

### ibysaiyan

Yes, that I have. Sure :
Let u = x , v = (x^2+y^2)^-1
Then using product rule :

f'' (x,v) = u' (v) + (u)v' ( Differentiating w.r.t x , keeping y constant)
=>
1* (x^2+y^2)^-1 + x [ -1 ( x^2+y^2)^-2 * 2x) ]

There's something about this expression which makes me think that it's wrong.. ( (x^2+y^2)^-2 , shouldn't the right version be : -1(x^2)^-2 * 2x ?

6. Jan 14, 2012

### Dick

No, it's right. That doesn't look a lot like what you posted before. Were you just being sloppy? Here I'll simplify it a little and translate into TeX:
$$\frac{1}{(x^2+y^2)}-\frac{2 x^2}{(x^2+y^2)^2}$$

7. Jan 14, 2012

### ibysaiyan

Oh silly me.. ok then in that case my second partial derivative of y is :

$$\frac{1}{(x^2+y^2)}-\frac{2 y^2}{(x^2+y^2)^2}$$
It still doesn't answer the question.. they want me to prove that the sum of these two derivatives = 0 , am I right ?

8. Jan 14, 2012

### Dick

Yes, prove the sum is zero. Add them and do some algebra. Put stuff over a common denominator.

9. Jan 14, 2012

### ibysaiyan

Oh.. that makes sense.. Thanks for your help!