Proving $\nabla^2(|f|^{2})=4 |\partial f / \partial z|^2$ via a "Laplacian Approach"

1.9k Views Asked by At

In the text "Function Theory of a Complex Variable", I'm having trouble proving the following relation in $(1.)$

$(1.)$

Prove that if $f$ is holomorphic on $\text{U} \subset \mathbb{C}$, then

$$\nabla^2(|f|^{2})=4 |\partial f / \partial z|^2$$

$$\text{Lemma}$$

The following observations can be made on $(1.)$

Recall:

$$\nabla^2(|f|^{2})=4 |\partial f / \partial z|^2$$

One can observe in $(2.)$

$(2.)$

$$\nabla^2(|u(x,y) + iv(x,y)||u(x,y) + iv(x,y)|) = 4 |\partial f / \partial z|\partial f / \partial z|$$

Furthermore in $(3.)$

$(3.)$

$$\nabla^2(|u(x,y) + iv(x,y)||u(x,y) + iv(x,y)|) = 4|( \nabla(f(z)||\nabla (f(z)|$$ $$\text{Remark}$$ The formal manipulation of Partial Derivatives , used within the results of this proof can be formally notioned as follows:

$$\nabla f(a) = ( \partial f / \partial x_{i}(a),..., \partial f / \partial x_{n})$$ Finally in $(4.)$

$(4)$

$$\partial f / \partial x_{}(|u(x,y) + iv(x,y)|\partial f / \partial y_{i}|u(x,y) + iv(x,y)|) = 4|( \nabla(f(z)|( \nabla (f(z)|$$

$$\partial f / \partial x_{i}|u(x,y) + iv(x,y)|\partial f / \partial y_{i}|u(x,y) + iv(x,y)|) = 4 |\partial f / \partial x_{}(|u(x,y) + iv(x,y)|\partial f / \partial y_{}|u(x,y) + iv(x,y)|)$$

$$\text{Remark}$$ The recent developments are only true when our function $f(z)$, satisfies Laplace's Equation: $$\nabla^2 \phi = 0$$

I'm initially have trouble, at this phase of the proof, it looks like $(1.)$ holds true is their an alternative approach that doesn't draw so heavily on the Laplacian Operator ?

1

There are 1 best solutions below

1
On BEST ANSWER

Compute $|\partial f/\partial z|^2$ in a couple ways:

$$ \begin{array}{lll} \displaystyle \frac{\partial f}{\partial z} & = & \displaystyle \frac{\partial f}{\partial x} & = & \displaystyle \frac{\partial f}{\partial (iy)} \\ \displaystyle \left|\frac{\partial f}{\partial z}\right|^2 & = & u_x^2+v_x^2 & = & u_y^2+v_y^2 \end{array} $$

and then compute

$$ \begin{array}{ll} \Delta|f|^2 & \displaystyle =\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)(u^2+v^2) \\[5pt] & \displaystyle = \frac{\partial^2}{\partial x^2}(u^2+v^2)+\frac{\partial^2}{\partial y^2}(u^2+v^2) \\[5pt] & \displaystyle =\frac{\partial}{\partial x}(2uu_x+2vv_x)+\frac{\partial}{\partial y}(2uu_y+2vv_y) \\[5pt] & = 2(u_x^2+v_x^2+uu_{xx}+vv_{xx})+2(u_y^2+v_y^2+uu_{yy}+vv_{yy}) \\[5pt] & =2(u_x^2+v_x^2+u_y^2+v_y^2)+2u(u_{xx}+u_{yy})+2v(v_{xx}+v_{yy}) \\[5pt] & \displaystyle = 4\left|\frac{\partial f}{\partial z}\right|^2. \end{array} $$