The imaginary part $~v~$ is equal to $~\frac{1}{2} \log(x+y)~$ . Verify whether it is harmonic or not?

Question

The imaginary part $~v~$ is equal to $~\frac{1}{2} \log(x+y)~$ . Verify whether it is harmonic or not?

Relayed to complex functions ie harmonic functions

Answer 1

$$v=\frac{1}{2} \log(x+y)$$ Since $v(x,y)$ is a polynomial function, so all partial verivatives $v_x,~v_y,~v_{xx},~v_{yy},~v_{xy}$ are defined and continuous where $(x,y)\ne (0,0)$.

Now $$v_x=\frac{1}{2}\cdot \frac{1}{x+y},~~~~~~v_y=\frac{1}{2}\cdot \frac{1}{x+y}$$ $$v_{xx}=-~\frac{1}{2}\cdot \frac{1}{(x+y)^2},~~~~~~v_{yy}=-~\frac{1}{2}\cdot \frac{1}{(x+y)^2}$$

Here $$v_{xx}+v_{yy}=-\frac{1}{(x+y)^2}\ne 0\qquad \text{when}\quad (x,y)\ne (0,0)$$

Hence $~v~$ is not harmonic when $~(x,y)\ne (0,0)~$.

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Harmonic function:

If $~\phi_x(x,y),~\phi_y(x,y),~\phi_{xx}(x,y),~\phi_{yy}(x,y),~\phi_{xy}(x,y)~$ are all continuous, and if $~\phi(x,y)~$ satisfies Laplace's equation$$\phi_{xx}+\phi_{yy}= 0~,$$ then $~\phi(x,y)~$ is called a harmonic function.