Recipe

Find a harmonic conjugate of a harmonic function

Use when: you have one harmonic function, u(x,y) or v(x,y), and need the other half of an analytic f=u+iv

Goal: the harmonic conjugate, correct up to an added real constant, built directly from the Cauchy-Riemann equations

Taught in Harmonic conjugates

1. 1

   Choose one of the two Cauchy-Riemann equations. Integrate its known side with respect to the variable it does not fix, and hold the other variable constant. Either way, the leftover "constant" of integration is really a function. It depends on the variable you did not integrate over.

   Example

   For u(x,y)=x^2-y^2, start from v_y=u_x. Here u_x=2x. Integrating with respect to y gives v=\int 2x\,dy+g(x)=2xy+g(x).

   • starting from v_y=u_x — integrate u_x with respect to y. The leftover piece is a function g(x) of x alone (Steps 2 through 4 below follow this path).
   • starting from v_x=-u_y — integrate -u_y with respect to x instead. The leftover piece is a function h(y) of y alone, and x and y trade places for the rest of the recipe.

   Why: each Cauchy-Riemann equation fixes how v changes in only one direction. Integrating recovers v along that direction. What happens in the other direction stays unknown.

2. 2

   Differentiate Step 1's expression for v with respect to the variable you did not just integrate over. Then the derivative of the unknown function shows up in plain view.

   Example

   Differentiate v=2xy+g(x) with respect to x: v_x=2y+g'(x).

   Why: this exposes g'(x) (or h'(y)) as an unknown. The other Cauchy-Riemann equation can now solve for it.

3. 3

   Set Step 2's expression equal to what the other Cauchy-Riemann equation demands. Then solve for the derivative of the unknown function. It must depend only on its own variable.

   Example

   Here u_y=-2y, so -u_y=2y. Set Step 2's v_x=2y+g'(x) equal to 2y: 2y+g'(x)=2y \quad\Longrightarrow\quad g'(x)=0.

   • the result depends only on the correct variable (for instance g'(x) comes out free of y) — continue to Step 4.
   • the result still has the other variable in it (for instance g'(x) depends on y) — an earlier step has an arithmetic slip. Recheck Step 1's integration and Step 2's differentiation before you go on.

   Why: this is where the second Cauchy-Riemann equation gets used. Without it, Step 1 alone never pins down the leftover function.

4. 4

   Integrate the derivative of the unknown function to get the function itself. Then put it back into Step 1's expression for v.

   Example

   Integrating g'(x)=0 gives g(x)=C for a real constant C, so v(x,y)=2xy+C.

   Why: the result is unique only up to an added real constant. A constant has derivative 0 in either variable. So no amount of Cauchy-Riemann checking can pin the constant down further.