Determine whether the function f(x,y)=ln(x^2 +y^2) is harmonic.
well, does ∇²f = 0?
Fxx = 2(y^2-x^2)/(x^2+y^2)^2
Fyy = 2(x^2-y^2)/(x^2+y^2)^2
so, what do you think?
No, but how do you know they equal zero or not?
To determine whether the function f(x, y) = ln(x^2 + y^2) is harmonic, we need to check if it satisfies Laplace's equation.
Laplace's equation states that a function is harmonic if its second-order partial derivatives with respect to x and y satisfy the equation:
∂^2f/∂x^2 + ∂^2f/∂y^2 = 0
Let's calculate the second-order partial derivatives of f(x, y):
∂^2f/∂x^2 = (∂/∂x) [ (∂f/∂x) ]
= (∂/∂x) [ (∂/∂x) (ln(x^2 + y^2)) ]
= (∂/∂x) [ (2x) / (x^2 + y^2) ]
= 2(x^2 + y^2 - x^2) / (x^2 + y^2)^2
= 2y^2 / (x^2 + y^2)^2
∂^2f/∂y^2 = (∂/∂y) [ (∂f/∂y) ]
= (∂/∂y) [ (∂/∂y) (ln(x^2 + y^2)) ]
= (∂/∂y) [ (2y) / (x^2 + y^2) ]
= 2(x^2 + y^2 - y^2) / (x^2 + y^2)^2
= 2x^2 / (x^2 + y^2)^2
Now, let's calculate the sum of the second-order partial derivatives:
∂^2f/∂x^2 + ∂^2f/∂y^2 = 2y^2 / (x^2 + y^2)^2 + 2x^2 / (x^2 + y^2)^2
= (2y^2 + 2x^2) / (x^2 + y^2)^2
Since the sum of the second-order partial derivatives is not equal to zero, i.e., (2y^2 + 2x^2) / (x^2 + y^2)^2 ≠ 0, the function f(x, y) = ln(x^2 + y^2) is not harmonic.