Finding the harmonic conjugate of a function

Question

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

Answer 1

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.