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The locally integrable function $E(z)=1/(\pi z)$ on the plane is a fundamental solution for the Wirtinger derivative ${\partial }_{\bar{z}}=({\partial }_x+i{\partial }_y)/2$. Let $L=(4\pi {)}^{-1}\log (x^2+y^2)$ be the fundamental solution of the Laplace equation. Its first distributional derivatives agree with its ordinary ones: deleting a radius-$\epsilon$ disk in integration by parts gives an error $O(\epsilon |\log \epsilon |)$, tending to zero. Hence $E=4{\partial }_{z}L$ andThe analogous fundamental solution for ${\partial }_z$ is $1/(\pi \bar{z})$. The singular support of each kernel is exactly $\{0\}$: they are smooth elsewhere, and smoothness near zero would make their Wirtinger derivatives smooth there, contradicting the Dirac delta distribution identity.

Target: Distributional Cauchy-Riemann lemma. Parent: Cauchy-Riemann equations. Intended file: analysis.bigb.

If $H,G$ are holomorphic functions, both real and imaginary parts of $\bar{z}H(z)+G(z)$ solve the biharmonic equation. In Wirtinger derivatives, $\Delta =4{\partial }_z{\partial }_{\bar{z}}$, so $\Delta (\bar{z}H+G)=4H^{\prime }$ is harmonic and its next Laplacian is zero. This construction gives local representations of planar solutions of the biharmonic equation; it is useful for the biharmonic stream function for planar Stokes flow.

A distribution $u$ on a plane open set with ${\partial }_{\bar{z}}u=0$ is represented by a holomorphic function. Indeed $\Delta u=4{\partial }_z{\partial }_{\bar{z}}u=0$, and elliptic regularity makes $u$ smooth. For a direct proof, mollify locally with a radial mollifier ${\eta }_{\epsilon }$. The resulting smooth functions are harmonic. Their mean value property gives $(u\ast {\eta }_{\epsilon })\ast {\eta }_r=u\ast {\eta }_{\epsilon }$ on a smaller ball for fixed small $r$. Passing to the weak convergence of distributions limit gives $u\ast {\eta }_r=u$ there. This proves smoothness without assuming local integrability beforehand. The ordinary Cauchy-Riemann equations now prove holomorphy.

The equation ${\partial }_{z}u=0$ similarly characterizes antiholomorphic functions. On the whole plane, the homogeneous solutions are respectively arbitrary entire functions $h(z)$ and the functions $h(\bar{z})$ with $h$ entire. Consequently any two fundamental solutions for one of these Wirtinger derivatives differ by the corresponding homogeneous solution.

Let $g$ be a holomorphic function on an open set $\Omega \subseteq \mathbb{C}$. Among continuously differentiable functions, all solutions of ${\partial }_{\bar{z}}f=g$ arewhere $h$ is an arbitrary holomorphic function on $\Omega$. The Wirtinger derivatives satisfy ${\partial }_{\bar{z}}\bar{z}=1$ and ${\partial }_{\bar{z}}g=0$, so the product rule verifies the particular solution $\bar{z}g(z)$. Subtracting it from any other solution leaves a continuously differentiable function with vanishing Wirtinger derivative, hence a holomorphic function by the Cauchy-Riemann equations. Conversely adding such a holomorphic function changes no prescribed derivative. For $g(z)=z$ on the complex plane, the solutions are $|z{|}^2+h(z)$ with $h$ any entire function.

Write $z=x+iy$, $r=|z|$, and $L=(4\pi {)}^{-1}\log (x^2+y^2)$. The Wirtinger derivatives satisfy $4{\partial }_{\bar{z}}{\partial }_z=\Delta$. Away from zero, ordinary differentiation givesThis equality also holds for distributional derivatives. To check that no Dirac delta distribution was lost at zero, integrate by parts outside the disk $r\le \epsilon$ against a test function. The new boundary term has absolute value at most $C\epsilon |\log \epsilon |\Vert \varphi {\Vert }_{\infty }\to 0$. Both $L$ and its displayed first derivatives, of size $O(r^{-1})$, are locally integrable functions in two dimensions, so the remaining integrals converge to their full-plane values.

Using the supplied fundamental solution of the Laplace equation identity $\Delta L={\delta }_0$, and commutation of distributional derivatives, now givesThus the Cauchy kernel fundamental solution has exactly the required normalization.

Interchange the two Wirtinger derivatives in part (i). The same real fundamental solution of the Laplace equation $L$ givesThe calculation is valid for distributions by the same vanishing boundary term estimate. Any other fundamental solution differs by a distribution $H$ with ${\partial }_{z}H=0$. Then $\Delta H=0$, so the regularity argument of part (ii) makes $H$ smooth, and its Cauchy-Riemann equations make it an antiholomorphic function. On the whole plane it can be written $H(z)=h(\bar{z})$ with $h$ an entire function. Conversely every such $H$ has ${\partial }_{z}H=0$. Therefore

Write $\chi =e^{-ikz-\lambda \bar{z}/(ik)}$. In terms of Wirtinger derivatives, the elliptic equation is $q_{z\bar{z}}=\lambda q$. If $W=Adz+Bd\bar{z}$, its coefficients are $A=\chi q_z$ and $B=-\lambda \chi q/(ik)$. Direct differentiation givesTherefore $dW=(B_z-A_{\bar{z}})dz\wedge d\bar{z}=0$, proving that $W$ is a closed differential one-form. The parameter restriction $k\ne 0$ makes its coefficients well defined.

With the Wirtinger derivatives, $q_{z\bar{z}}=(q_{xx}+q_{yy})/4$. Put $W=e^{-ikz-\lambda \bar{z}/(ik)}$, $k\ne 0$. The PDF has the factor $\lambda$ in this exponential; omitting it, as the TeX does, would give the wrong local relation. ChooseIndeed, $W_{\bar{z}}=-\lambda W/(ik)$ and $W_z=-ikW$, so direct differentiation givesThe unwanted $q_z$ terms cancel exactly. Thus the divergence identity is equivalent to the modified Helmholtz equation, including its Laplace equation case $\lambda =0$. Equivalently,is a closed differential one-form. Its oriented boundary integral supplies the polygonal modified Helmholtz global relation.

Use the Wirtinger derivatives ${\partial }_z=({\partial }_x-i{\partial }_y)/2$ and ${\partial }_{\bar{z}}=({\partial }_x+i{\partial }_y)/2$. In particular $q_{z\bar{z}}=\Delta q/4$. Let $E=e^{-ikz-\lambda \bar{z}/(ik)}$, $A=E(q_z+ikq)$ and $B=E(q_{\bar{z}}+\lambda q/(ik))$, so the printed differential form is $W=Adz-Bd\bar{z}$. Since $E_z=-ikE$ and $E_{\bar{z}}=-\lambda E/(ik)$,Thus the exterior derivative isBecause $E$ never vanishes and $k\ne 0$, this proves both directions: $W$ is a closed differential form if and only if $q_{z\bar{z}}=\lambda q$. The latter is the modified Helmholtz equation $\Delta q-4\lambda q=0$ for positive $\lambda$. The associated global relations on a bounded region $\Omega$ are, for every nonzero spectral parameter,The boundary orientation is positive, by Stokes theorem. Each admissible spectral parameter gives a separate identity between the traces. Interchanging $z$ and $\bar{z}$ gives the dual family, equivalently the same family at $k\mapsto -\lambda /k$ with an overall minus sign. No assumption that $q$ is real is needed.

For later use with $\lambda >0$, as in the remaining parts, make the global relation explicit in the first quadrant. PutLet $h_2(x)=q(x,0)$, $n_2(x)=q_y(x,0)$ and $h_1(y)=q(0,y)$, $n_1(y)=q_x(0,y)$. These normals are inward-coordinate derivatives. The bottom is traversed from zero to infinity and the left side from infinity to zero. Their oriented spectral integrals areHere ${\rho }_x$ is analytic for $\mathrm{Im}k<0$ and ${\rho }_y$ for $\mathrm{Re}k<0$, with suitable boundary limits. Indeed $\mathrm{Im}a=(r+\lambda /r)\sin \theta$ and $\mathrm{Re}b=(r+\lambda /r)\cos \theta$ for $k=r{e}^{i\theta }$. On their common third quadrant, both exponential weights decay. Apply the finite-domain relation to expanding rectangles; sufficient decay of the solution and traces eliminates the distant edges. This derivesThe limiting relations on the negative real and negative imaginary axes will remove the unknown derivative traces.

Use the Wirtinger derivatives $q_z=(q_x-i{q}_y)/2$ and $q_{\bar{z}}=(q_x+i{q}_y)/2$. The Laplace equation gives $q_{z\bar{z}}=0$, so writewhere $\Phi$ is holomorphic in the first quadrant and $\Psi$ is holomorphic in the fourth quadrant. They are independent functions: the data may be complex, so a complex-conjugation relation between them cannot be assumed.

Let $ϵ=(-1{)}^n$, where ${\beta }_1+{\beta }_2=n\pi /2$. The oblique derivative boundary conditions becomeThe hinted complementary traces would giveHere $u_2$ is a function of $x>0$ on the horizontal side; the printed $y$ range in that hint is a variable typo. We shall eliminate the same unknown traces by reflection of the holomorphic derivatives.

The polygonal ray formula can first be viewed as the Cauchy integral formula. For an oriented edge point $\zeta$, choose its spectral ray so that $e^{ik(z-\zeta )}$ decays for interior $z$. ThenInterchanging the edge and ray integrals in the supplied formula gives $\Phi (z)=(2\pi i{)}^{-1}{\int }_{\partial \Omega }\Phi (\zeta )/(\zeta -z)d\zeta$, with counterclockwise edge orientation. Thus a quadrant limit of that formula can be evaluated by the corresponding holomorphic boundary jumps, provided the additional arcs at infinity and at the corner give no contribution.

Define an holomorphic function separately in the four quadrants byThe angle condition gives $e^{2i({\beta }_1+{\beta }_2)}=e^{-2i({\beta }_1+{\beta }_2)}=ϵ$. It makes the reflected definitions consistent, with parity $H(-z)=-ϵH(z)$. On each axis oriented outward from zero, let $J=H_{\mathrm{left}}-H_{\mathrm{right}}$. The two known boundary equations give all four jumps:For example the vertical equation gives $\Phi (iy)-e^{-2i{\beta }_1}\Psi (-iy)=-i{e}^{-i{\beta }_1}{g}_1(y)$; reversing the jump order on the upward axis accounts for its displayed sign. The horizontal equation similarly supplies the positive real-axis jump, and reflection supplies the other two.

Apply the Cauchy integral formula in the four quadrants and combine their common edges. The edge contributions are exactly the known jumps of this scalar Riemann-Hilbert problem. If the corner and large-arc integrals vanish, the representation in the first quadrant isThis is the rational-angle oblique derivative problem on a quadrant representation, with all unknown traces removed. The Cauchy integrals have precisely the displayed jumps by the Sokhotski–Plemelj formula. Their reflection parity then reconstructs $\Psi$ from the fourth-quadrant value of $H$, so the two original boundary equations hold. Integrating $\Phi (z)$ and $\Psi (\bar{z})$ gives a harmonic function $q$ up to an additive constant.

To put the answer back on the spectral rays of the question, defineUse $1/(s-z)=i{\int }_0^{\infty }{e}^{ik(z-s)}dk$, $1/(s+z)=-i{\int }_0^{\infty }{e}^{ik(z+s)}dk$, and the same identities with $s$ replaced by $is$ and the ray $0\to i\infty$. The spectral representation isAll transforms used on these rays are ordinary boundary values of convergent half-line transforms. For $z=x+iy$ with $x,y>0$, the real-ray integrand decays through $e^{-ky}$ and the imaginary-ray integrand through $e^{-\mathrm{Im}kx}$.

There is an admissibility qualification in passing from a bounded polygon to this unbounded quadrant. Sufficient conditions are $H(z)\to 0$ at infinity uniformly away from the axes, with the corresponding large-arc integral tending to zero, and $|z|H(z)\to 0$ uniformly on small corner arcs. The latter allows logarithmic gradient singularities but excludes a pole with nonzero corner contribution. Smooth decaying data alone, as printed, do not impose these conditions on $q$.

For a concrete homogeneous corner ambiguity in a quadrant Laplace problem, with ${\beta }_1={\beta }_2=0$ and zero data, $q=xy$ satisfies both prescribed tangential derivative conditions but has $q_z=-iz/2$, while the displayed particular representation is zero. With ${\beta }_1={\beta }_2=\pi /2$, the zero-data solution $q=\log |z|$ has $q_z=1/(2z)$: it decays at infinity but has an excluded corner pole. Thus no unique representation of every unrestricted solution follows from the printed data alone. Without the arc conditions, one must add a homogeneous holomorphic term $P(z)$ to the displayed $q_z$, with reflected parity $P(-z)=-ϵP(z)$; permitted corner singularities are included if the admissible class allows them. With the stated decay and corner conditions that term vanishes, giving the boxed formulas. This identifies the missing solution-class conditions without assuming real-valued data.

Use the Wirtinger derivatives ${\partial }_z=({\partial }_x-i{\partial }_y)/2$ and ${\partial }_{\bar{z}}=({\partial }_x+i{\partial }_y)/2$, and area measure $dA=dxdy=(i/2)dz\wedge d\bar{z}$. The supplied boundary-integral identity is the planar Generalized Stokes theorem; the usual Poincaré lemma is a different local exactness result.

Let $f\in C^1(\overline{D})$ and $z\in D$. Remove a disk of radius $ϵ$ around $z$, and apply Generalized Stokes theorem to the one-form $f(\zeta )d\zeta /(\zeta -z)$ on the punctured domain. Away from the puncture,The outer boundary is counterclockwise and the small inner circle clockwise. Its counterclockwise integral tends to $2\pi if(z)$. Passing to the limit gives the Cauchy-Pompeiu formulaThe weak $1/|\zeta -z|$ singularity is locally integrable. If the boundary term vanishes on expanding $D$ to the whole plane, the formula becomes $f(z)={\pi }^{-1}{\int }_{\mathbb{C}}{f}_{\bar{\zeta }}(\zeta )/(z-\zeta )dA(\zeta )$. In particular it yields the distributional normalization ${\partial }_{\bar{z}}[1/(\pi z)]={\delta }_0$. For holomorphic functions the area term vanishes and one recovers the Cauchy integral formula.

Use the Wirtinger derivatives ${\partial }_z=({\partial }_x-i{\partial }_y)/2$ and ${\partial }_{\bar{z}}=({\partial }_x+i{\partial }_y)/2$, for which $\Delta =4{\partial }_z{\partial }_{\bar{z}}$. Since $H$ and $G$ are holomorphic,The Laplacian has real coefficients, so taking real and imaginary parts provesThis is the complex analytic representation of biharmonic functions; $F$ itself is generally not holomorphic because of its $\bar{z}$ factor.

In the displayed complex-coordinate convention, the sinh-Gordon equation is an elliptic partial differential equation. Write $z=x+iy$ and $\bar{z}=x-iy$. The Wirtinger derivatives obey ${\partial }_z{\partial }_{\bar{z}}=({\partial }_x^2+{\partial }_y^2)/4$, soThis Euclidean version must be distinguished from the Lorentzian nonlinear wave equation also called sinh-Gordon. The coordinate and sign conventions specify which version is meant.

For the one-dimensional reduction $u=u(x)$, put $\varphi =2u$. Then ${\varphi }^{\prime \prime }=4\sinh \varphi$, and multiplication by ${\varphi }^{\prime }$ gives the first integralIt reduces this stationary profile problem to a quadrature; its integration constant is fixed by the chosen initial data or boundary conditions.

For known attenuation $a(x)$ and a scalar source $f(x)$ in the plane, start with the directional transport equation $(\mathbf{\theta }\cdot \nabla +a)\psi =f$. Introduce a complex spectral parameter $k$ and Wirtinger derivatives ${\partial }_z=({\partial }_{x_1}-i{\partial }_{x_2})/2$, ${\partial }_{\bar{z}}=({\partial }_{x_1}+i{\partial }_{x_2})/2$:At $k=e^{i\theta }$ this is the original directional derivative $\cos \theta {\partial }_{x_1}+\sin \theta {\partial }_{x_2}$. Away from the unit circle, its fundamental solution supplies analytic spectral families. Choosing $h_k$ with $L_k{h}_k=a$ turns $(L_k+a){\psi }_k=f$ into $L_k(e^{h_k}{\psi }_k)=e^{h_k}f$, removing attenuation by an integrating factor.^[ref]

The boundary values from inside and outside the unit circle have a jump determined by directed attenuated Radon transform data and transverse Hilbert transforms. A normalized Riemann-Hilbert problem reconstructs the spectral family. In the exterior normalization ${\psi }_k={\psi }_{-1}/k+O(k^{-2})$, substituting into the transformed equation gives $f={\partial }_z{\psi }_{-1}$. Thus the desired source is recovered from a spectral coefficient, rather than from a direct angular average.

This construction assumes sufficient regularity and decay, known attenuation and the required directed angular measurements. General angle-dependent sources have a different identifiable content; partial angular data require additional hypotheses. The projection formula for inverse attenuated Radon transform makes one full-data normalization explicit.

The Weight-k hyperbolic Laplacian is a differential operator adapted to weight-$k$ transformation laws on the upper half-plane $z=x+iy$, $y>0$. Using the sign consistent with the nonnegative weight-zero hyperbolic Laplacian, and the holomorphic-weight convention, defineHere ${\partial }_z=\frac{1}{2}({\partial }_x-i{\partial }_y)$ and ${\partial }_{\bar{z}}=\frac{1}{2}({\partial }_x+i{\partial }_y)$ are Wirtinger derivatives. The cited paper uses the opposite overall sign, so its eigenvalues must also have their signs reversed.^[ref]

The correction terms let the operator preserve the transformation law $f((az+b)/(cz+d))=(cz+d{)}^{k}f(z)$ for determinant-one integer matrices, as used for modular forms and Maass forms. At $k=0$ it reduces to the ordinary hyperbolic Laplacian. Every holomorphic function is locally annihilated because ${\partial }_{\bar{z}}f=0$; being holomorphic alone does not supply modularity or growth conditions.

For a direct calculation, take $f(z)=y^s$. Its $x$ derivatives vanish, givingThus both $1$ and $y^{1-k}$ are annihilated; when $k=1$ these coincide and a second independent solution depending only on $y$ is $\log y$. These are local operator examples, not claims that all satisfy a modular transformation law. Conventions using a unit-modulus weight multiplier have a different first-derivative correction, so the displayed formula matters as much as the name.