The real and imaginary parts of an analytic function are not arbitrary. Differentiating the Cauchy–Riemann equations once more (using mixed partials) shows that each of them solves Laplace's equation, the equation of steady heat flow and electrostatics. This link is one reason complex analysis matters in physics.
Idea clave
A real function $u(x, y)$ with continuous second partials is harmonic if $u_{xx} + u_{yy} = 0$. If $f = u + iv$ is analytic, then $u$ and $v$ are harmonic. Conversely, if $u$ is harmonic on a disk, there is a harmonic conjugate $v$ (unique up to adding a constant) with $u + iv$ analytic; find it from $v_y = u_x$ and $v_x = -u_y$.
Por qué funciona
From $u_x = v_y$, differentiate in $x$: $u_{xx} = v_{yx}$. From $u_y = -v_x$, differentiate in $y$: $u_{yy} = -v_{xy}$. Mixed partials are equal (an analytic function has continuous partials of all orders, proved later in the course), so $u_{xx} + u_{yy} = v_{yx} - v_{xy} = 0$. The same argument works for $v$.
Worked examples
Ejemplo resuelto: Testing for harmonic
Is $u = x^3 - 3xy^2 + y$ harmonic? Is $u = x^2 + y^2$?
- First: $u_{xx} = 6x$ and $u_{yy} = -6x$, so the sum is $0$: harmonic (it is $\operatorname{Re}(z^3 - iz)$).
- Second: $u_{xx} + u_{yy} = 2 + 2 = 4 \ne 0$: not harmonic, so $x^2 + y^2$ is not the real part of any analytic function.
Respuesta: yes; no
Ejemplo resuelto: Finding a harmonic conjugate
Find $v$ with $v(0, 0) = 0$ so that $u + iv$ is analytic, for $u = x^2 - y^2 + 3x$.
- $v_y = u_x = 2x + 3$. Integrate in $y$: $v = 2xy + 3y + g(x)$, where $g$ is any function of $x$ (it disappears when we differentiate in $y$).
- $v_x = 2y + g'(x)$ must equal $-u_y = 2y$, so $g'(x) = 0$ and $g$ is a constant.
- $v(0, 0) = 0$ makes the constant $0$: $v = 2xy + 3y$. Check: $u + iv = z^2 + 3z$.
Respuesta: $v = 2xy + 3y$
Ejemplo resuelto: An exponential example
Find the harmonic conjugate of $u = e^x\cos y$ with $v(0, 0) = 0$.
- $v_y = u_x = e^x\cos y$, so $v = e^x\sin y + g(x)$.
- $v_x = e^x\sin y + g'(x)$ must equal $-u_y = e^x\sin y$, so $g$ is constant, and $v(0, 0) = 0$ makes it $0$.
- $v = e^x\sin y$, and $u + iv = e^x(\cos y + i\sin y) = e^z$.
Respuesta: $v = e^x\sin y$
A geometric bonus: where $f'(z) \ne 0$, the level curves $u = c_1$ and $v = c_2$ cross at right angles, because $\nabla u \cdot \nabla v = u_xv_x + u_yv_y = u_xv_x - v_xu_x = 0$ by the Cauchy–Riemann equations. In a steady flow, $u$ can be the potential and $v$ the stream function: the flow lines cross the equipotential lines at right angles.
Error común
Adding a plain constant after integrating. When you integrate $v_y = u_x$ with respect to $y$, the "constant" of integration can be any function $g(x)$, because $\frac{\partial}{\partial y}g(x) = 0$. Example: $u = x^2 - y^2 + 5y$. Then $v_y = 2x$ gives $v = 2xy + g(x)$, and $v_x = 2y + g'(x)$ must equal $-u_y = 2y - 5$, so $g(x) = -5x$ and $v = 2xy - 5x$. Writing $v = 2xy + C$ would miss the $-5x$. Always finish with the check $v_x = -u_y$.