Proving a Partial Differential Equation: Is the Sum of Second Derivatives Zero?

[ibysaiyan]

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)

Homework Equations

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 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)

Homework Equations

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.

 

[ibysaiyan]

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]

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?

 

[ibysaiyan]

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]

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}

 

[ibysaiyan]

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]

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.

 

[ibysaiyan]

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!