M4.3 Harmonioc Property & Orthogonal Property of a Complex Function

Harmonic Function

A function $\phi$ is said to be harmonic if it satisfies the Laplace equation $\nabla^2 \phi=0.$ i.e., $$\frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}=0 \longrightarrow Cartesian Form$$ $$\frac{\partial^2 \phi}{\partial r^2}+\frac{1}{r}\frac{\partial \phi}{\partial r}+\frac{1}{r^2}\frac{\partial \phi^2}{\partial \theta^2}=0 \longrightarrow Polar Form$$

Harmonic Property

The real and imaginary parts of an analytic function are harmonic
Proof: Cartesian Form
Let $f(z)=u(x,y)+iv(x,y)$ be analytic. We have to show that $u$ and $v$ satisfies the Laplace equations. i.e., $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$ & $\frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2}=0$
Since $f(z)$ is analytic, $u$ and $v$ should satisfies CR-equations $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial x}\,\,\,\,\,\,\,(1)$$ $$\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}\,\,\,\,\,\,\,(2)$$ Differentiate (1) with respect to $x$ and differentiate (2) with respect to $y$ partially, we get $$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 v}{\partial x \partial y};\,\,\,\, \frac{\partial^2 v}{\partial y \partial x}=-\frac{\partial^2 u}{\partial y^2}$$ But we know that $\frac{\partial^2 v}{\partial x \partial y}=\frac{\partial^2 v}{\partial y \partial x}$ Hence $$\frac{\partial^2 u}{\partial x^2}=-\frac{\partial^2 u}{\partial y^2}\Longrightarrow \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$$ Similarly, by differentiating (1) with respect to $y$ and differentiate (2) with respect to $x$ partially, we get $$\frac{\partial^2 u}{\partial y \partial x}=\frac{\partial^2 v}{\partial y^2};\,\,\,\, \frac{\partial^2 v}{\partial x^2}=-\frac{\partial^2 u}{\partial x \partial y}$$ Since $\frac{\partial^2 v}{\partial x \partial y}=\frac{\partial^2 v}{\partial y \partial x}$, $$\frac{\partial^2 v}{\partial y^2}=-\frac{\partial^2 v}{\partial x^2}\Longrightarrow \frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2}=0$$ Hence the real and imaginary parts of an analytic function are harmonic.
Polar Form
Let $f(z)=u(r,\theta)+iv(r,\theta)$ be analytic. We have to show that $u$ and $v$ satisfies the Laplace equations. i.e., $\frac{\partial^2 u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial u^2}{\partial \theta^2}=0$ and $\frac{\partial^2 v}{\partial r^2}+\frac{1}{r}\frac{\partial v}{\partial r}+\frac{1}{r^2}\frac{\partial v^2}{\partial \theta^2}=0$. Since $f(z)$ is analytic $u$ and $v$ should satisfies CR-equations $$\frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial v}{\partial \theta}$$ $$\frac{\partial v}{\partial r}=-\frac{1}{r}\frac{\partial u}{\partial \theta}$$ Differentiate (1) with respect to $r$ and differentiate (2) with respect to $\theta$ partially, we get $$r\frac{\partial^2 u}{\partial r^2}+\frac{\partial u}{\partial r}=\frac{\partial^2 v}{\partial r \partial\theta};\,\,\,\, r\frac{\partial^2 v}{\partial \theta \partial r}= -\frac{\partial^2 u}{\partial \theta^2}$$ It is true that $\frac{\partial^2 v}{\partial r \partial\theta}=\frac{\partial^2 v}{\partial \theta \partial r}$, hence by substituting one in another we get $$r\frac{\partial^2 u}{\partial r^2}+\frac{\partial u}{\partial r}=-\frac{1}{r}\frac{\partial^2 u}{\partial \theta^2}$$ Divide throughout by $r$ we get $$\frac{\partial^2 u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial u^2}{\partial \theta^2}=0.$$ Hence $u$ is Harmonic. Similarly we can obtain $\frac{\partial^2 v}{\partial r^2}+\frac{1}{r}\frac{\partial v}{\partial r}+\frac{1}{r^2}\frac{\partial v^2}{\partial \theta^2}=0$ by differentiating (1) with respect to $\theta$ and differentiate (2) with respect to $r$ partially.

Orthogonal Property

If $f(z)=u+iv$ is analytic then the family of curves $u(x,y)=c_1, v(x,y)=c_2, c_1$ and $c_2$ being constants, intersect each other orthogonally.
Proof: Cartesian Form
We say that two curves cuts orthogonally at some point if the product of the slope of the tangents ($\frac{dy}{dx}$) drawn to those curve at that point is equal to $-1$.
Let $m_1$ be slope of the tangent drawn to the curve $u(x,y)=c_1$. Differentiate $u(x,y)=c_1$ with respect to $x$ by treating $y$ as function of $x$, we get $$\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y} \cdot \frac{dy}{dx}=0 \Longrightarrow m_1=\frac{dy}{dx}=-\frac{\partial u}{\partial x}/\frac{\partial u}{\partial y}$$ Similarly let $m_2$ be slope of the tangent drawn to the curve $v(x,y)=c_2$. Differentiate $v(x,y)=c_2$ with respect to $x$ by treating $y$ as function of $x$, we get $$m_2=\frac{dy}{dx}=-\frac{\partial v}{\partial x}/\frac{\partial v}{\partial y}$$ $$\Longrightarrow m_1\cdot m_2=\frac{\partial u}{\partial x}/\frac{\partial u}{\partial y} \cdot \frac{\partial v}{\partial x}/\frac{\partial v}{\partial y}$$ Using CR-equation in the above equation we get $$\Longrightarrow m_1\cdot m_2=\frac{\partial v}{\partial y}/\frac{\partial u}{\partial y} \cdot -\frac{\partial u}{\partial y}/\frac{\partial v}{\partial y}=-1$$ Hence the curves intersect orthogonally.

• Converse is not true, for example the curves $u=\frac{x^2}{y}$ and $v=x^2+2y^2$ intersects orthogonally but it does not satisfies CR-equations, hence the function $f(z)=u+iv$ is not analytic.