Complex Variables and Partial Differential Equations (3130005)

Question-1c

BE | Semester-3 Winter-2019 | 26-11-2019

Q1) (c)

Definition:

A function

u (x, y)

is called harmonic if it is twice continuously differentiable and satisfies

\frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = 0

Verification:

u (x, y) = \sinh x \sin y

\Rightarrow \frac{\partial u}{\partial x} = \sin y \cosh x - - - - - (1)

\Rightarrow \frac{\partial^{2} u}{\partial x^{2}} = \sin y \sinh x

& \frac{\partial u}{\partial y} = \sinh x \cos y - - - - - (2)

\Rightarrow \frac{\partial^{2} u}{\partial y^{2}} = - \sinh x \sin y

So, \frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = \sinh x \sin y - \sinh x \sin y = 0

∴ u (x, y)

is a harmonic function.

Harmonic conjugate:

Let,

f (z) = u + i v

\Rightarrow f^{'} (z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}

\Rightarrow f^{'} (z) = \frac{\partial u}{\partial x} - i \frac{\partial v}{\partial y} (∵ CR Equation)

\Rightarrow f^{'} (z) = \sin y \cosh x - i \sinh x \cos y (∵ 1 & 2)

\Rightarrow f^{'} (z) = \sin 0 \cosh z - i \sinh z \cos 0 (∵ x = z & y = 0)

\Rightarrow f^{'} (z) = - i \sinh z (∵ \sin 0 = 0 & \cos 0 = 1)

\Rightarrow f (z) = - i \int \sinh z dz (∵ Taking Integration)

\Rightarrow f (z) = - i \cosh z + c

\Rightarrow f (z) = - i \cosh (x + i y) + c

\Rightarrow f (z) = - i (\cosh x \cosh iy + \sinh x \sinh iy) + c

\Rightarrow f (z) = - i (\cosh x \cos y + i \sinh x \sin y) + c (∵ \cosh iy = \cos y & \sinh iy = i \sin y)

\Rightarrow f (z) = - i \cosh x \cos y - i^{2} \sinh x \sin y + c

\Rightarrow f (z) = \sinh x \sin y - i \cosh x \cos y + c

So, harmonic conjugate is,

v (x, y) = - \cosh x \cos y + c