### Author Topic: TT1 Problem 3 (night)  (Read 3498 times)

#### Victor Ivrii

• Elder Member
• Posts: 2599
• Karma: 0
##### TT1 Problem 3 (night)
« on: October 19, 2018, 04:14:27 AM »
(a) Show that $u(x,y)= x^3 - 3xy^2 +2y +3x$ 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 $z=x+iy$.

#### Meng Wu

• Elder Member
• Posts: 91
• Karma: 36
• MAT3342018F
##### Re: TT1 Problem 3 (night)
« Reply #1 on: October 19, 2018, 06:35:55 AM »
$(a).$
\begin{align}\frac{\partial u(x,y)}{\partial x}&=3x^2-3y^2+3\\ \frac{\partial^2u(x,y)}{\partial x^2}&=6x\ \\ \frac{\partial u(x,y)}{\partial y}&=-6xy+2\\ \frac{\partial^2u(x,y)}{\partial y^2}&=-6x\end{align}
Where first and second partial derivatives are continuous with respect to both $x$ and $y$.
$$\Delta u= \frac{\partial^2u(x,y)}{\partial x^2}+ \frac{\partial^2u(x,y)}{\partial y^2}= 6x+(-6x)=0$$
Therefore, $u(x,y)=x^3-3xy^2+2y+3x$ is a harmonic function.
$\\$
$\\$
$(b).$
$\\$
Use CR-equation to find the harmonic conjugate.
$$\frac{\partial u}{\partial x }=3x^2-3y^2+3= \frac{\partial v}{\partial y }$$
\begin{align}\Rightarrow v(x,y)&= \int(3x^2-3y^2+3)dy \\&=3x^2y-y^3+3y+h(x) \\ \Rightarrow \frac{\partial v}{\partial x }=6x+h'(x)\end{align}
Hence,$$\frac{\partial u}{\partial y } =-6xy+2=-\frac{\partial v}{\partial x }=-(6x+h'(x)) \\ \Rightarrow h'(x)=-2 \\ \Rightarrow h(x)= -2x$$
Therefore the harmonic conjugate $v(x,y)=3x^2y-y^3+3y-2x.$
$\\$
$\\$
$(c).$
$\\$
\begin{align}u(x,y)+iv(x,y)&=x^3-3xy^2+2y+3x+i(3x^2y-y^3+3y-2x)\\ &=x^3-3xy^2+i3x^2y-iy^3+3x+i3y+2y-i2x\end{align}
Consider $$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$
Thus $$(x+iy)^3=x^3+i3x^2y-3xy^2-iy^3=z^3$$
$$3x+i3y=3(x+iy)=3z$$
$$2y-i2x=-i2(x+iy)=-2iz$$
Therefore, $f(z)=z^3+3z-2iz$.
« Last Edit: October 19, 2018, 11:46:32 AM by Meng Wu »

#### Meng Wu

• Elder Member
• Posts: 91
• Karma: 36
• MAT3342018F
##### Re: TT1 Problem 3 (night)
« Reply #2 on: October 19, 2018, 10:07:07 AM »
I think $h'(x)=-2 \Rightarrow h(x)=-2x+C$, where $C$ is an arbitrary real constant.
Thus, $v(x,y)=3x^2y-y^3+3y-2x+C$.
Therefore, $f(z)=f(z)=z^3+3z-2iz+iC$
« Last Edit: October 19, 2018, 11:52:25 PM by Meng Wu »

#### Zihan Wan

• Jr. Member
• Posts: 7
• Karma: 1
##### Re: TT1 Problem 3 (night)
« Reply #3 on: October 19, 2018, 10:55:04 AM »
that should be i2z

#### Zihan Wan

• Jr. Member
• Posts: 7
• Karma: 1
##### Re: TT1 Problem 3 (night)
« Reply #4 on: October 19, 2018, 11:28:22 AM »
For(c), C is an arbitrary real number that is derived when looking for the formula for v, a real-valued function. So in the end, there will a term of the form Ci, a pure imaginary number, not an arbitrary constant.
Just my thought.

#### Victor Ivrii

• Elder Member
• Posts: 2599
• Karma: 0
##### Re: TT1 Problem 3 (night)
« Reply #5 on: October 20, 2018, 03:01:57 PM »
Zihan, it will be $-2iz = -2ix +2y$, with the real part $2y$. So, Meng is right
Also "$Ci$" with an arbitrary real constant $C$ is exactly an arbitrary imaginary constant $Ci$. So Meng and you are saynig the same