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The compact support just established and $f(0)=0$ permit differentiation of the antiderivative and integration of the scalar conservation law from $-\infty$:This is a Hamilton-Jacobi equation. There is no arbitrary function of time: the normalization at $-\infty$ and the zero exterior flux determine it.

Choose a star centre $a$ of the open star-shaped set $D$ and define the candidate antiderivative by a straight-segment contour integral:For each fixed $z\in D$ and all sufficiently small $h$, the filled triangle with vertices $a,z,z+h$ lies in $D$. Indeed, the compact segment $[a,z]$ has a neighbourhood of some positive radius contained in the open set $D$, and every point of this triangle lies within $|h|$ of that segment. The zero triangle integral consequently givesBy continuity, the last integral tends to $f(z)$ as $h\to 0$. HenceIn particular, this proves the existence of the antiderivative without incorrectly assuming the star-shaped set is convex.

On a general domain, the conclusion can fail. Take $D=\mathbb{C}\setminus \{0\}$ and $f(z)=1/z$. Every filled triangle contained in $D$ has zero boundary integral by Cauchy integral theorem, but the unit circle has integral $2\pi i$. An antiderivative would make every closed contour integral zero. This is the period obstruction to a holomorphic antiderivative: the local conclusion of Morera's theorem does not remove the global obstruction.

The Wirtinger derivative $u_z=(u_x-i{u}_y)/2$ is holomorphic, since ${\partial }_{\bar{z}}{u}_z=\Delta u/4=0$ by the Laplace equation. The conformal map $w=z^2$ takes the quadrant to the complex upper half-plane. Integrating the prescribed tangential boundary derivatives, choose a common corner value $c$ and define the real boundary functionThe decaying data are bounded under the usual regularity at the corner, so $f(s)=O(1+\sqrt{|s|})$. Thus its Poisson integral converges. A holomorphic function whose real part is that Poisson integral is supplied by the regularized Schwarz integral formula:For $u_{\ast }(z)=\mathrm{Re}W(z^2)$, the chain rule gives $(u_{\ast }{)}_z=z{W}^{\prime }(z^2)$. Apply integration by parts to obtainSubstituting $s=r^2$ and $s=-r^2$ gives the requested particular integral representation:Both integrals converge absolutely for an interior point of the quadrant. The Sokhotski–Plemelj formula gives $\mathrm{Re}(u_{\ast }{)}_z(x)=g_2(x)/2$ on the horizontal edge and $\mathrm{Im}(u_{\ast }{)}_z(iy)=-g_1(y)/2$ on the vertical edge, verifying the prescribed derivatives.

The printed conditions alone do not determine $u_z$ uniquely. The general answer adds a holomorphic function $H$ satisfyingEquivalently, write $H(z)=ih(z^2)/z$, where $h$ is holomorphic in the complex upper half-plane with real boundary values on the nonzero real axis. An antiderivative of $2H$ has a harmonic real part with zero tangential boundary derivatives. For example, $u=xy$ contributes $H=-iz/2$ without changing either datum. If the two edge constants differ by $c_2-c_1$, the angular harmonic function contributes $i(c_2-c_1)/(\pi z)$. The boxed expression selects the Poisson integral representative with equal edge constants; regularity and suitable growth conditions can be used to select that representative.

A point is ramified with multiplicity $m_i$ exactly when it is a zero of $f^{\prime }$ of multiplicity $m_i-1$. Since $f^{\prime }$ has degree $d-1$, the fundamental theorem of algebra gives ${\sum }_i(m_i-1)=d-1$.

Conversely, choose distinct complex numbers $a_i$ and put $g(z)=c{\prod }_i(z-a_i{)}^{m_i-1}$ with $c\ne 0$. The assumption makes $g$ a polynomial of degree $d-1$. Any antiderivative $f$ of $g$ has degree $d$ and precisely the prescribed ramification multiplicities.

Take the Fourier transform in $x$. The wave equation becomes the ordinary differential equationwith initial data $\tilde{u}(k,0)=\tilde{f}(k)$ and ${\partial }_t\tilde{u}(k,0)=\tilde{g}(k)$. Therefore

Part (c) and the Translation property of the Fourier transform invert the first term toFor the second term choose an antiderivative $p$ of $g$. By part (b), $\tilde{g}=ik\tilde{p}$, and henceIts inverse transform isCombining the terms gives the D'Alembert formula

Every holomorphic function $f$ on a simply connected domain has a single-valued antiderivative. Fixing $z_0$ and settinggives a path-independent function with $F^{\prime }=f$, because the Cauchy integral theorem makes the integral around every closed path zero.