Bateman transform

Remarks

The function $u$ is harmonic on $U$, that is, it solves Laplace's equation $\sum_{k=1}^4 \partial_k^2 u = 0$. Indeed, by the chain rule, $\sum_k \partial_k^2$ applied to $f(A, B, \zeta)$ is a combination of second derivatives of $f$ with coefficients $\sum_k (\partial_k A)^2$, $\sum_k \partial_k A\, \partial_k B$ and $\sum_k (\partial_k B)^2$. The gradients are $\nabla A = (1, i, i\zeta, \zeta)$ and $\nabla B = (-\zeta, i\zeta, -i, 1)$, and all three sums vanish: $1 - 1 - \zeta^2 + \zeta^2 = 0$, $-\zeta - \zeta + \zeta + \zeta = 0$ and $\zeta^2 - \zeta^2 - 1 + 1 = 0$. Differentiation under the integral sign then gives $\Delta u = 0$. When $\Omega$ also contains the points $(A(x, \zeta), B(x, \zeta), \zeta)$ with $x_4$ replaced by $i t$, the same integral defines $u(x_1, x_2, x_3, i t)$. Replacing $x_4$ by $i t$ turns $\partial_{x_4}^2$ into $-\partial_t^2$, so $u(x_1, x_2, x_3, it)$ solves the wave equation $\partial_t^2 u = \partial_1^2 u + \partial_2^2 u + \partial_3^2 u$.

Example: $f(w_1, w_2, \zeta) = \frac{w_1}{\zeta^2}$ and $C$ the unit circle. Then $u(x) = \oint_C \Big( \frac{x_1 + i x_2}{\zeta^2} + \frac{x_4 + i x_3}{\zeta} \Big) d\zeta = 2\pi i\, (x_4 + i x_3)$, a linear, hence harmonic, function.

Non-example: if $f(w_1, w_2, \zeta) = \overline{w_1}$, which is not holomorphic, the function $\oint_C \overline{A(x, \zeta)}\, d\zeta$ is not covered by the definition, and the chain rule computation above fails.