Finding the conjugate harmonic of a function u(x,y)

[Mmmm]

Homework Statement

Let u(x,y) be harmonic in a simply connected domain $\Omega$. Use the Cauchy-Riemann equations to obtain the formula for the conjugate harmonic

$$v(x,y)=\int^{(x,y)}_{x_0,y_0} (u_xdy-u_ydx)$$

where $(x_0,y_0)$ is any fixed point of $\Omega$ and the integration is along any path in $\Omega$ joining $(x_0,y_0)$ and (x,y).

Homework Equations

Cauchy Riemann eqns

$$u_x=v_y, u_y=v_x$$

The Attempt at a Solution

At first this just looks like a simple bit of integration but for some reason I just cannot get the result. How do I get rid of the dependence on x and y of the constants of integration?

$$u_x=v_y$$
$$\Rightarrow v(x,y) = \int^y_{y_0} u_xdy$$
$$v(x,y) = \int^x_{x_0} u_ydx$$

Differentiating each wrt the other variable in an attempt to link the two eqns doesn't seem to get me anywhere...

 

[mighty2000]

Hello,

Thank you for posting your question on this forum.

To obtain the formula for the conjugate harmonic v(x,y), we can use the Cauchy-Riemann equations as follows:

Since u(x,y) is harmonic, we know that it satisfies the Laplace equation:

u_{xx} + u_{yy} = 0

We can also write u in terms of its conjugate harmonic v as follows:

u(x,y) = u(x,y) + iv(x,y)

Now, using the Cauchy-Riemann equations, we can write:

u_x = v_y

and

u_y = -v_x

Substituting these into the Laplace equation, we get:

v_{xx} + v_{yy} = 0

This shows that v(x,y) is also harmonic.

Next, we can write v(x,y) in terms of its partial derivatives as follows:

v(x,y) = \int^y_{y_0} v_xdx + \int^x_{x_0} v_ydy

Now, using the Cauchy-Riemann equations, we can substitute for v_x and v_y to get:

v(x,y) = \int^y_{y_0} (-u_y)dx + \int^x_{x_0} u_xdy

= -\int^y_{y_0} u_ydx + \int^x_{x_0} u_xdy

= \int^x_{x_0} (u_xdy - u_ydx)

which is the desired formula for the conjugate harmonic.

I hope this helps. Please let me know if you have any further questions or if anything is unclear.