 Solution

An isothermal parametrization is an immersion with first fundamental form $Ed{u}^2+2Fdudv+Gd{v}^2={\lambda }^2(d{u}^2+d{v}^2)$, where $\lambda >0$. Thus $E=G={\lambda }^2$ and $F=0$.

Write $W={\varphi }_{uu}+{\varphi }_{vv}$. Differentiating the metric identities givesFor example ${\varphi }_{vv}\cdot {\varphi }_u=({\varphi }_v\cdot {\varphi }_u{)}_v-{\varphi }_v\cdot {\varphi }_{uv}=-G_u/2$; the other identities are analogous. Hence $W$ is normal to the surface. For a unit normal $N$, its normal component is $e+g$. The supplied mean curvature formula reduces to $H=(e+g)/(2{\lambda }^2)$. ThereforeA surface is a minimal surface when its mean curvature vector vanishes. Since ${\lambda }^2>0$, the preceding identity proves minimality is equivalent to $\Delta \varphi =0$ in isothermal coordinates.