Mumbai University > Information Technology, Computer Engineering > Sem 3 > Applied Mathematics 3

Marks: 8M

Year: Dec 2015

Part-1

Laplace equation in polar form in polar form is

$\frac{∂^2 u}{∂^2 r} + \frac{1}{r} \frac{∂u}{∂r} + \frac{1}{r^2} \frac{∂^2 u}{∂θ^2} =0$

We have

$u = [r+ \frac{a^2}{r}cosθ = (r + a^2 r^{-1} )cosθ$

Differentiating with respect to r

Therefore,

$\frac{∂u}{∂r} = (1 + (-1) a^2 r^{-2} )cosθ\\ \frac{∂u}{∂r} = (1 - a^2 r^{-2} )cosθ$ ---------------------------(1)

Again, differentiating with respect to r

$\frac{∂^2 u}{∂r^2} = (o - a^2 r^{-3} (-2))cosθ$

$\frac{∂^2 u}{∂r^2} = -2a^2 r^{-3} cosθ$ -----------------------------(2)

Now, differentiating u with respect to θ

$ \frac{∂u}{∂θ} = (r+a^2 r^{-1})(-sinθ)$

Again, differentiating with respect to θ

$\frac{∂^2 u}{∂θ^2} = (r + a^2 r^{-1})(-cosθ)$ -----------------------(3)

Consider Laplace equation in polar for

$\frac{∂^2 u}{∂r^2} + \frac{1}{r} \frac{∂u}{∂r} + \frac{1}{r^2} \frac{∂^2 u}{∂θ^2} = \frac{2a^2}{r^3} cosθ + \frac{1}{r} (1 - \frac{a^2}{r^2} cosθ + \frac{1}{r^2} (- (r + \frac{a^2}{r})cosθ)$

From (1)&(2)&(3)

$\frac{∂^2 u}{∂r^2} + \frac{1}{r} \frac{∂u}{∂r} + \frac{1}{r^2} \frac{∂^2 u}{∂θ^2} = cosθ \bigg[\frac{2a^2}{r^3} + \frac{1}{r} - \frac{a^2}{r^3} = \frac{1}{r} - {a^2}{r^3} \bigg] = 0$

Therefore, u satisfies Laplace equation. Hence, u is harmonic

Part-2

Using Cauchy Riemann’s equation in polar form

Therefore,

$\frac{∂u}{∂r} = \frac{1}{r} \frac{∂u}{∂θ}$

$(1 - \frac{a^2}{r^2}) cosθ = \frac{1}{r} \frac{∂u}{∂θ}$ from (1)

$∂u = r(1 - \frac{a^2}{r^2}cosθ ∂θ$

Integrating on both sides

Therefore, $u = (r - \frac{a^2}{r})sinθ$

Hence,

$f(z) = u + i v $

$= (r + \frac{a^2}{r}) cosθ + i(r = \frac{a^2}{r}) sinθ$

$= rcosθ + \frac{a^2}{r} cosθ + i(rsinθ - \frac{a^2}{r} sinθ)$

$= rcosθ + \frac{a^2}{r} cosθ + irsinθ - i \frac{a^2}{r} sinθ$

$= r(cosθ+isinθ) + \frac{a^2}{r} (cosθ-isinθ)$

$= re^{iθ} + \frac{a^2}{r} e^{-iθ}4 $f(z) = z + \frac{a^2}{z} \hspace{0.2cm} and \hspace{0.2cm} v = (r - \frac{a^2}{r})sinθ$