MHB How to Solve a Harmonic Function Problem Involving Partial Derivatives?

[jaychay]

Can you please help me how to do it ?
I am really struggle with this question.

Thank you in advance

 

[I like Serena]

Hint: A harmonic conjugate function $v$ must have that $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$, so that the function given by $f(x+iy)=u(x,y)+iv(x,y)$ is holomorphic (complex differentiable).

 

[jaychay]

Hint: A harmonic conjugate function $v$ must have that $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$, so that the function given by $f(x+iy)=u(x,y)+iv(x,y)$ is holomorphic (complex differentiable).

Can you tell what is the next step that I should do please ?

 

[I like Serena]

Can you tell what is the next step that I should do please ?

It's a partial derivative (note the round d's). When we partially differentiate with respect to $x$, then that means that we treat $y$ as a constant.
More concisely:
$$\frac{\partial}{\partial x}(y)=0$$
So you can simplify your expression for $\frac{\partial u}{\partial x}$ a bit.

Next step is to integrate $\frac{\partial u}{\partial x}$ with respect to $y$ to find $v(x,y)$.
When we do so, we treat $x$ as a constant.

Let me give an example.
Suppose we have $u(x,y)=xy$. Then we have $\frac{\partial u}{\partial x}=y$.
And $v(x,y)=\int \frac{\partial u}{\partial x} \,dy = \int y\,dy = \frac 12y^2 + C(x)$, where $C(x)$ is some function of $x$.
We can verify by evaluating $\frac{\partial v}{\partial y} = \frac{\partial}{\partial y}\Big(\frac 12y^2 + C(x)\Big) = y$.
As we can see, we get indeed that $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$.