# Partial differential equation

## Homework Statement

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)

## 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

Dick
Homework Helper

## Homework Statement

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)

## 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

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

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

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...

Dick
Homework Helper
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...

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?

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?

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 ?

Dick
Homework Helper
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:
$$\frac{1}{(x^2+y^2)}-\frac{2 x^2}{(x^2+y^2)^2}$$

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:
$$\frac{1}{(x^2+y^2)}-\frac{2 x^2}{(x^2+y^2)^2}$$

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 ?

Dick
Homework Helper
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 ?

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

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

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