It's quite easy to show that two functions which satisfy the Cauchy-Riemann equations must be harmonic, but is the reverse true?
Specifically, if I have two functions u and v, which satisfy u_xx + u_yy=0 and v_xx+ v_yy =0, is it possible to show that they must also satisfy
u_x=v_y and u_y=-v_x??
Thanks for your help!!
Yes, a good point.
There's a bit more to the problem, I don't know if it will make a difference:
I'm doing a model problem in fluid dynamics, solving for pressure and vorticity in a square at large times.
Both satisfy laplace's equation within the square, and both are zero along the vertical sides. Along the top I'm prescribing a condition (have chosen pressure and vorticity to equal sin^2(pi x)) and along the bottom (y=0) pressure and vorticity satisfy the Cauchy Riemann equations.
I'm pretty convinced from dimensional analysis of the Navier stokes that the system should satisfy the Cauchy Riemann equations everywhere but this isn't matching up with my numerical code.
(I've got
u_t = -p_x -zeta_y
v_t = -p_y +zeta_x
And dimensional analysis points to velocities becoming steady at large time)
My analysis is rusty to say the least.. But trying to decide which is correct C-R everywhere or not..
I would really appreciate any pointers-Thanks for your help!
As Opalg said, no, two arbitrary harmonic functions do not necessarily satify the Cauchy-Riemann equations. However, what is true is this: if f(x,y) is harmonic in a region, then there exist a (unique up to a constant) function g (sometimes called the "conjugate" of f) such that and .
To use Opalg's example, if f(x,y)= x, then so that g must depend upon y only. Then so g(x,y)= y+ C. f(x,y)= x and g(x,y)= y are harmonic, form a "Caucy-Riemann" pair and so f(z)= f(x+ iy)= x+ iy= z is an analytic function.