Q:

Harmonic Functions and Partial Derivatives with Chain Rule (Complex Variables)

Suppose that an analytic function $w= f(z) = u(x,y) + iv(x,y)$ maps a domain $D_z$ in the $z$ plane onto a domain $D_w$ in the $w$ plane. Let a function $h(u,v)$ with continuous first and second partial derivatives be defined on $D_w$. Show using chain rule that if
  $H(x,y) = h\left( u(x,y) , v(x,y) \right)$ then
  $$H_{xx} (x,y) + H_{yy} (x,y) = \left( h_{uu}(u,v) + h_{vv}(u,v) \right) |f’(z)|^2.$$
  How does it follow from this that $H(x,y)$ is harmonic in $D_z$ when $h(u,v)$ is harmonic in $D_w?$

I feel that this is just a huge messy chain rule using partial derivatives problem, is there another approach to this or at least a way to do this elegantly without creating a big mess of terms? We know that since $f$ is analytic, the Cauchy-Riemann equations hold: $u_x = v_y \text{ and } u_y = -v_x$. Also, the functions $u$ and $v$ satisfy Laplace’s Equation. The continuity conditions on the partial derivatives yield
$h_{vu} = h_{uv}$. (I am having trouble putting these together as well.)

A:

For example:
$\begin{align}
H_x&=h_uu_x+h_vv_x\\
\Rightarrow
H_{xx}&=(h_uu_x)_x+(h_vv_x)_x\\
&=(h_{uu}u_x+h_{uv}v_x)u_x+h_uu_{xx}+(h_{vv}v_x+h_{uv}u_x)v_x+h_vv_{xx}\\
&=h_{uu}u_x^2+h_{uv}v_xu_x+h_uu_{xx}+h_{vv}v_x^2+h_{uv}u_xv_x+h_vv_{xx}
\end{align}$
By symmetry:
$$H_{yy}=h_{uu}u_y^2+h_{uv}v_yu_y+h_uu_{yy}+h_{vv}v_y^2+h_{uv}u_yv_y+h_vv_{yy}$$
change some of the $y$'s to $x$'s with CR:
$$H_{yy}=h_{uu}u_y^2-h_{uv}v_xu_x+h_uu_{yy}+h_{vv}v_y^2-h_{uv}u_xv_x+h_vv_{yy}$$
Combine:
$$H_{xx}+H_{yy}= h_{uu}(u_x^2+u_y^2)+h_u(u_{xx}+u_{yy})+h_v(v_{xx}+v_{yy})+h_{vv}(v_x^2+v_y^2)$$
But $u$ and $v$ are harmonic, so this simplifies further:$$H_{xx}+H_{yy}= h_{uu}(u_x^2+u_y^2)+h_{vv}(v_x^2+v_y^2)$$
Use CR again:
$$H_{xx}+H_{yy}= (h_{uu}+h_{vv})(u_x^2+u_y^2)$$
To see that $(u_x^2+u_y^2)=|f'(z)|^2$, (Complex chain rule for complex valued functions)
$$f'(z)=u_x-iu_y\Rightarrow |f'(z)|^2=u_x^2+u_y^2$$
as desired.

