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.