Does $a\ln(x^2 +y^ 2 )+b$ satisfy Laplace’s equation?

Question

I can't verify that $F(x,y) = a\ln(x^2 +y^ 2 )+b$ satisfies Laplace’s equation ($F_{xx}+F_{yy}=0$). Here is what I did: \begin{align*} F_x &= \frac{2ax}{x^2 + y^2} &F_y &= \frac{2ay}{x^2+y^2} \\ F_{xx} &= \frac{2a}{x^2+y^2} - \frac{4ax^2}{(x^2+y^2)^2} & F_{yy} &= \frac{2a}{x^2+y^2} - \frac{4ay^2}{(x^2+y^2)^2} \end{align*} and $F_{yy}\neq-F_{xx}$. So what should I do?

Answer 1

Let $F(x,y) = a \ln(x^2 + y^2) + b $. Then, Using quotient rule:

$$ F_x = \frac{ 2ax}{x^2 + y^2} \iff F_{xx} = \frac{2a(x^2+y^2)-2x(2ax)}{(x^2+y^2)^2} = \frac{2ay^2 - 2ax^2}{(x^2+y^2)^2}$$

Similarly,

$$ F_y = \frac{ 2ay}{x^2+y^2} \implies F_{yy} = \frac{2a(x^2+y^2) -2ay(2y)}{(x^2+y^2)^2} = \frac{2ax^2 - 2ay^2}{(x^2+y^2)^2}$$

Hence,

$$ F_{xx} + F_{yy} = 0 $$