# Toronto Math Forum

## MAT334--2020F => MAT334--Tests and Quizzes => Test 2 => Topic started by: RunboZhang on November 04, 2020, 06:36:41 PM

Title: 2020F Test2-ALT-F Q1
Post by: RunboZhang on November 04, 2020, 06:36:41 PM
$\textbf{Problem 1:} \\\\ \text{(a) Show that } \ 3x^{4} - 18x^{2}y^{2} + 3y^{4} - 3x \text{ is a harmonic function;} \\\\ \text{(b) Find the harmonic conjugate function } v(x,y) \\\\ \text{(c) Consider } u(x,y) + iv(x,y) \ \text{ and write it as a function } f(z) \text{ of } z=x+iy$

$\textbf{Solution for part (a): } \\\\ \text{Since we have:} \\\\$
\begin{gather} \begin{aligned} U(x,y) = 3x^{4} - 18x^{2}y^{2} + 3y^{4} - 3x \\\\ \end{aligned} \end{gather}

$\text{then we have the following:} \\\\$
\begin{gather} \begin{aligned} u_x &= 12x^{3} - 36xy^{2} - 3 \\\\ u_{xx} &= 36x^{2} - 36y^{2}\\\\ u_y &= -36x^{2}y + 12y^{3} \\\\ u_{yy} &= -36x^{2} + 36y{2} \end{aligned} \end{gather}

$\text{Thus we have: } \\\\$

\begin{gather} \begin{aligned} \Delta u &= u_{xx} + u_{yy} \\\\ &= 36x^{2} - 36y^{2} - 36x^{2} + 36y^{2} \\\\ &= 0 \end{aligned} \end{gather}
$\text{Therefore the function is harmonic. } \\\\$

\\\\ \\\\ \textbf{Solution for part (b): } \\\\ \text{We have:} \\\\ \begin{gather} \begin{aligned} v_x &= -u_y = 36x^{2}y - 12y^{3} \\\\ v_y &= u_x = 12x^{3} -36xy^{2} -3 \end{aligned} \end{gather}

\text{Then} \\\\ \begin{gather} \begin{aligned} v_{x,y} &= \int{v_x} \,dx + \phi(y) \\\\ &= 12x^{3}y-12xy^{3}+\phi(y) \end{aligned} \end{gather}

\text{Now we calculate } \phi(y) \text{ by following:} \\\\ \begin{gather} \begin{aligned} v_y &= (v_{x,y})' \\\\ &= 12x^{3} -36xy^{2}-3 \end{aligned} \end{gather}

\text{thus, } \\\\ \begin{gather} \begin{aligned} \phi'(y) &= -3 \\\\ \phi(y) &= \int{-3}\,dy \\\\ &= -3y + c \end{aligned} \end{gather}

\text{Therefore} \\\\ \begin{gather} \begin{aligned} v_{x,y} = 12x^{3}y-12xy^{3}-3y+c \end{aligned} \end{gather}

\\\\ \\\\ \textbf{Solution for part (c): } \\\\ \text{Let } \ z=x+iy \\\\ \begin{gather} \begin{aligned} u_{x,y} + iv_{x,y} &= 3x^{4} -12x^{2}y^{2} + 3y^{4} -3x +i(12x^{3}y-12xy^{3}-3y)+iC \\\\ &= 3(3x^{4} -12x^{2}y^{2} + 3y^{4} + i12x^{3}y-i12xy^{3}) -3(x+iy) +iC\\\\ &= 3(x+iy)^{4}-3(x+iy)+iC\\\\ &= 3z^{4} -3z + iC \end{aligned} \end{gather}