# Linear harmonic functions

Let $$u(x,y)$$ be a harmonic function on $$\mathbb{R^2}$$ and $$v(x,y)$$ be a harmonic conjugate of $$u(x,y)$$ on $$\mathbb{R^2}$$. Suppose that the partial derivative $$v_x(x,y) for a real constant $$C$$. Show that $$u(x,y)$$ is a linear function.

I have derived the following:

Since $$v$$ is the harmonic conjugate of $$u$$, we have $$u_x=v_y$$ & $$u_y=-v_x$$

Since both $$u$$ and $$v$$ are harmonic, we also have $$u_{xx}+u_{yy}=0$$ & $$v_{xx}+v_{yy}=0$$

However, I cant establish a prove given the following identities.

$$v_x$$ is harmonic as well: $$(v_x)_{xx} + (v_x)_{yy} = (v_{xx})_x + (v_{yy})_x = (v_{xx} + v_{yy})_x = 0$$ behause a harmonic function is infinitely often differentiable and the order of partial derivatives can be changed arbitrarily.
Then $$C - v_x$$ is a non-negative harmonic function in the plane, and therefore constant. It follows that $$v$$ is a linear function. Now use the Cauchy Riemann equations to conclude that $$u$$ is a linear function.
• May I know why is $v_x$ harmonic? – Jayne Leblanc Mar 17 '19 at 10:00
Let $$f=u+iv$$, then $$f$$ is entire and $$f'=u_x+iv_x$$. Since $$v_x , $$f'$$ is constant, by Picard.