Determining whether the Function is harmonic ?

Question

Determining whether the Function is harmonic ?

 

f(x,y) = 1/(x²+y²)

 

 

- What I learned from class was that if fxx + fyy = 0 then the function is harmonic. 

 

My attempt was that....

 

fx = -2x/(x²+y²)²

 

then

 

fxx = (6x²-2y²)/(x²+y²)³  I think that how you do it

 

But I am stuck trying to determine the fyy....

 

Answer 1

Hi Mikey,

 

That looks right to me for fxx.

 

Since f(x,y) is symmetric in x and y, you can find fyy by simply replacing the x's with y's.

 

fyy=(6y^2-2x^2)/(....

 

So fxx+fyy=4/(x^2+y^2)^2 which is not equal to zero.  The function is not harmonic.

 

Do you know the polar form of this condition (Laplace equation)?

 

(1/r)∂(r.fr)/∂r +(1/r^2)fθθ=0

 

Now f=1/r^2, fr=-2/r^3,(1/r)∂(r.fr)/∂r +(1/r^2)fθθ=(4/r^3) which is not identically 0 and matches the earlier result.

 

 

 

 

Answer 2

I agree with your fxx.

But why wouldn't fyy simply be (6y² - 2x²)/(x² + y²)³ , since it's just the partial derivative (twice) for y, instead of x?

 

However, adding these according to the Laplace equation doesn't give 0, but rather 4(x² + y²)/(x² + y²)³

 

So, if we did it correctly, then we would conclude that f(x,y) is not in fact a harmonic function.