If partial derivatives of a harmonic function are constant, is the function linear?

Question

Let $u: \mathbb{R}^2 \to \mathbb{R}$ be a harmonic function.

If $\frac{\partial u(x,y)}{\partial x} = k_1$ and $\frac{\partial u(x,y)}{\partial y} = k_2,\forall (x,y) \in \mathbb{R}^2, k_1, k_2 \in \mathbb{R}$, may I say that $u$ is a linear function?

Answer 1

No.

If $u(x,y) = k_{1}x+k_{2}y+C$, then:

$$\dfrac{\partial u(x,y)}{\partial x} = k_{1}$$ $$\dfrac{\partial u(x,y)}{\partial y} = k_{2}$$ But $u(x,y)$ is not linear - this kind of function is an affine function. $u$ is linear iff $C = 0$.

Answer 2

A simple counterexample is $u\equiv 1.$

The general result here, which has nothing to do with harmonicity, is this: If $a,b\in \mathbb R,$ and $\partial u/\partial x \equiv a,$ $\partial u/\partial y \equiv b,$ then $u(x,y) = ax + by + u(0,0)$ for all $(x,y).$

Proof: $u(x,y)- u(0,0) = u(x,y)- u(x,0) +u(x,0)-u(0,0).$ Apply the mean value theorem to each difference.