The problem:
Find the harmonic conjugate of $G(x,y)= 2v^2(x,y)-2u^2(x,y)$
My attempt to solving it
I know that
"If two given functions u and v are harmonic in a domain D and their
first-order partial derivatives satisfy the Cauchy–Riemann equations
throughout D, then v is said to be a harmonic conjugate of u."
So having $G(x,y)$ I'm searching for a function let's call it $H(x,y)$ that is the harmonic conjugate of $G(x,y)$. Applying the Cauchy-Rieman equations I'd need to fulfill these conditions:
$G_x=H_y$ and $G_y=-H_x$
I'm stuck at this point. I know how to do this if I am given a defined function, but not something as $v(x,y)$ or $u(x,y)$.
If I continue with the "normal" method of solving these problems (when functions are defined) I'd derivate G in terms of either x and y and then integrate in terms of the other variable and so on. But for example:
If I derivate G in terms of x I'd get:
$G_x= 4vv_x-4uu_x$
My idea would be to integrate $G_x$ in terms of x… But that's where I get stuck.
I don't know if I'm missing some property that could help me in this case.
Any help will be much appreciated
Best Answer
This problem can in principle be solved by integrating the Cauchy-Riemann equations, but I think it is a lot less work to simply find a holomorphic function with convenient and appropriate real and imaginary parts, to wit:
Recall that two harmonic functions $u$ and $v$ are conjugate if and only if they form the real and imaginary parts of a holomorphic function
$f = u + iv; \tag 1$
since $f$ is holomorphic, so is
$-2f^2 = -2(u + iv)^2 = -2(u^2 - v^2 + 2iuv) = 2v^2 - 2u^2 - 4iuv; \tag 2$
since
$2v^2 - 2u^2 = \Re(-2f^2), \tag 3$
it follows that $2v^2 - 2u^2$ is harmonic, as is its conjugate
$-4uv = \Im(-2f^2). \tag 4$
Note Added in Edit, Saturday 9 February 2019 2:33 PM PST: It seems only fair, and a good idea to boot, to follow through with the notion our OP Jonathan Perales introduced in the text of the question; he found that
$G_x = 4vv_x - 4uu_x; \tag 5$
if we deploy the Cauchy-Riemann equations for $u$ and $v$,
$u_x = v_y, \; u_y = -v_x, \tag 6$
then we may convert (5) to
$G_x = -4vu_y - 4uv_y = -4(vu_y + uv_y) = -4(uv)_y = H_y; \tag 7$
then, integrating with respect to $y$,
$H = -4uv + \phi(x), \tag 8$
whence,
$H_x = -4u_xv - 4uv_x + \phi'(x); \tag 9$
again using Cauchy-Riemann,
$H_x = -G_y = 4uu_y - 4vv_y = -4vu_x - 4uv_x; \tag{10}$
combining (9) and (10) we find
$-4u_xv - 4uv_x + \phi'(x) = -4vu_x - 4uv_x, \tag{11}$
yielding
$\phi'(x) = 0; \tag{12}$
thus
$\phi(x) = \phi_0, \; \text{a constant}; \tag{13}$
returning to (8),
$H = -4uv + \phi_0; \tag{14}$
since harmonic conjuates are only defined up to an addiive constant, we see that (14) describes a conjugate of $2v^2 - 2u^2$. End of Note.