### Author Topic: TT1 Problem 3 (morning)  (Read 2215 times)

#### Victor Ivrii

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##### TT1 Problem 3 (morning)
« on: October 19, 2018, 03:55:03 AM »
(a) Show that $u(x,y)= 8xy^3 -8x^3 y+ 5x$ is a harmonic function

(b) Find the harmonic conjugate function $v(x,y)$.

(c) Consider $u(x,y)+iv(x,y)$ and write it as a function $f(z)$ of $z=x+iy$.

#### Vedant Shah

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##### Re: TT1 Problem 3 (morning)
« Reply #1 on: October 19, 2018, 09:23:34 AM »
(a)
$U_{xx} = \frac{\partial}{\partial x} \frac{\partial}{\partial x} U \\ = \frac{\partial}{\partial x} 8y^3 -24x^2 y +5 \\ = -48xy \\ \\ U_{yy} = \frac{\partial}{\partial y} \frac{\partial}{\partial y} U \\ = \frac{\partial}{\partial y} 24x y^2 - 8x^3 \\ = 48xy \\ U_{xx} + U_{yy} = -48xy + 48xy = 0$
Thus, U is harmonic.

(b)
By Cauchy Reimann:
$V_y = U_x = 8y^3 - 24x^2 y + 5 \Rightarrow V = 2y^4 - 12 x^2 y^2 +5y +h(x)\\ V_x = -U_y = -24x y^2 + 8x^3 \Rightarrow V = -12 x^2 y^2 + 2x^4 + g(y) \\ \Rightarrow V(x,y) = 2x^4 - 12 x^2 y^2 + 2y^4 + 5y$

(c)
$$f(x,y) = U(x,y) + iV(x,y) = 8xy^3 - 8x^3y+5x + i(2x^4 - 12 x^2 y^2 + 2y^4 + 5y) \\ f(x,y) = 2i(x^4 + 4ix^3y - 6x^2y^2 -4ixy^3 +y^4) + 5(x+iy) \\ f(x,y) = 2i(x+iy)^4 + 5(x+iy) \\ f(z) = 2i ({z}) ^4 + 5z\color{red}{+Ci}.$$

« Last Edit: October 20, 2018, 03:08:41 PM by Victor Ivrii »