Partial differential equation

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Homework Statement



Hi
I need some help in proving that [itex]\partial^2 V / \partial x^2[/itex] + [itex]\partial^2 V / \partial y^2[/itex] = 0 , when V = 1/2ln(x^2+y^2)

Homework Equations





The Attempt at a Solution


For [itex]\partial V / \partial x[/itex] here's what I get: (1/x^2+y^2) *2x*1/2 = x * (x^2+y^2)^-1
and for [itex]\partial V / \partial y[/itex]
(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
 

Answers and Replies

  • #2
Dick
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Homework Statement



Hi
I need some help in proving that [itex]\partial^2 V / \partial x^2[/itex] + [itex]\partial^2 V / \partial y^2[/itex] = 0 , when V = 1/2ln(x^2+y^2)

Homework Equations





The Attempt at a Solution


For [itex]\partial V / \partial x[/itex] here's what I get: (1/x^2+y^2) *2x*1/2 = x * (x^2+y^2)^-1
and for [itex]\partial V / \partial y[/itex]
(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.
 
  • #3
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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 [itex]\partial V / \partial x[/itex] 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
Dick
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I should have posted my complete answer which I know is wrong...
Here is the second derivative for x , keeping y fixed.

For [itex]\partial V / \partial x[/itex] 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?
 
  • #5
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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 ?
 
  • #6
Dick
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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:
[tex]\frac{1}{(x^2+y^2)}-\frac{2 x^2}{(x^2+y^2)^2}[/tex]
 
  • #7
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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:
[tex]\frac{1}{(x^2+y^2)}-\frac{2 x^2}{(x^2+y^2)^2}[/tex]

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

[tex]\frac{1}{(x^2+y^2)}-\frac{2 y^2}{(x^2+y^2)^2}[/tex]
It still doesn't answer the question.. they want me to prove that the sum of these two derivatives = 0 , am I right ?
 
  • #8
Dick
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Oh silly me.. ok then in that case my second partial derivative of y is :

[tex]\frac{1}{(x^2+y^2)}-\frac{2 y^2}{(x^2+y^2)^2}[/tex]
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.
 
  • #9
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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!
 

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