There is a nice explicit solution: $u(x,y)=\operatorname{Im}\sqrt{x+iy}$. This uses the principal branch of square root, for which $\sqrt{x}$ is real when $x>0$ (hence $u$ vanishes there). Being the imaginary part of a holomorphic function, $u$ is harmonic. To see that the normal derivative vanishes on the negative half-axis, note that the harmonic conjugate, $v(x,y)=-\operatorname{Re}\sqrt{x+iy}$, is identically zero on that axis. Thus $u_y=v_x = 0$.
An explicit real-algebraic formula for $u$ can be found here.
As originally written, with both conditions for $x>0$, the only solution is $u(x,y)\equiv 0$. Indeed, one extend $u$ to lower half-plane in two ways:
- by letting $u_1(x,-y)=u(x,y)$ when $y<0$;
- by letting $u_2(x,y) = -u(x,y)$ when $y<0$;
Either way, we get a harmonic function on $\mathbb{R}^2 \setminus \{(x,0):x\le 0\}$, since both extensions are $C^1$ across the ray $\{(x,0):x > 0\}$. Since $u_1=u_2$ in the upper half-plane, it follows that $u_1\equiv u_2$ everywhere (if a harmonic function vanishes on an open set, it vanishes everywhere). Thus, $u\equiv 0$.
The same idea can be used to show that requiring both the value and the normal derivative of a harmonic function to be zero on the same, however small, part of the boundary, makes the function identically zero.
