Toronto Math Forum
MAT3342018F => MAT334Tests => Term Test 1 => Topic started by: Victor Ivrii 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 $z=x+iy$.

$(a).$
$$\begin{align}\frac{\partial u(x,y)}{\partial x}&=3x^23y^2+3\\ \frac{\partial^2u(x,y)}{\partial x^2}&=6x\ \\ \frac{\partial u(x,y)}{\partial y}&=6xy+2\\ \frac{\partial^2u(x,y)}{\partial y^2}&=6x\end{align}$$
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, $u(x,y)=x^33xy^2+2y+3x$ is a harmonic function.
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$(b).$
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Use CRequation to find the harmonic conjugate.
$$\frac{\partial u}{\partial x }=3x^23y^2+3= \frac{\partial v}{\partial y }$$
$$\begin{align}\Rightarrow v(x,y)&= \int(3x^23y^2+3)dy \\&=3x^2yy^3+3y+h(x) \\ \Rightarrow \frac{\partial v}{\partial x }=6x+h'(x)\end{align}$$
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 $v(x,y)=3x^2yy^3+3y2x.$
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$(c).$
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$$\begin{align}u(x,y)+iv(x,y)&=x^33xy^2+2y+3x+i(3x^2yy^3+3y2x)\\ &=x^33xy^2+i3x^2yiy^3+3x+i3y+2yi2x\end{align}$$
Consider $$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$
Thus $$(x+iy)^3=x^3+i3x^2y3xy^2iy^3=z^3$$
$$3x+i3y=3(x+iy)=3z$$
$$2yi2x=i2(x+iy)=2iz$$
Therefore, $f(z)=z^3+3z2iz$.

I think $h'(x)=2 \Rightarrow h(x)=2x+C$, where $C$ is an arbitrary real constant.
Thus, $v(x,y)=3x^2yy^3+3y2x+C$.
Therefore, $f(z)=f(z)=z^3+3z2iz+iC$

that should be i2z

For(c), C is an arbitrary real number that is derived when looking for the formula for v, a realvalued function. So in the end, there will a term of the form Ci, a pure imaginary number, not an arbitrary constant.
Just my thought.

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