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An additive Riemann-Hilbert problem seeks a holomorphic function on each side of an oriented contour $\Gamma$, with prescribed difference of boundary valuesThe plus side is on the left of the orientation. Conditions at infinity and at contour endpoints are part of the problem; the jump alone does not usually give uniqueness.

On a positively oriented smooth simple closed contour, for Hölder-continuous data $g$, the Cauchy transformhas the required jump by the Sokhotski–Plemelj theorem and vanishes at infinity. For the unit circle and $g=1$, the Cauchy integral formula gives $F=1$ inside and $F=0$ outside. This concrete solution checks both the sign of the jump and the normalization. Subtracting two solutions removes the jump, so analytic continuation and growth conditions can establish uniqueness.^[ref]

The value assigned to or approached by a function at a point of the boundary of its domain. Prescribing a boundary value is a Dirichlet boundary condition.

If $f$ is holomorphic on a neighborhood of a closed disc and $C$ is its positively oriented boundary circle, then every interior point $z$ satisfiesThus boundary values determine interior values. Differentiation under the integral gives $f^{(n)}(z)=n!{\oint }_{C}f(\zeta )(\zeta -z{)}^{-n-1}d\zeta /(2\pi i)$, showing that one complex derivative forces derivatives of every order. For $f\equiv 1$, parametrizing a small circle about $z$ gives the normalization ${\oint }_{C}d\zeta /(\zeta -z)=2\pi i$; the general formula follows by separating this singular term from the removable remainder.

Suppose $w$ is holomorphic off $[0,1]$, tends to zero at infinity, has sufficiently regular boundary values in the interval interior, and has endpoint growth $O(|z-a{|}^{-\alpha })$ for some $\alpha <1$ at each endpoint $a$. If its jump is integrable, the displayed reconstruction follows by deforming a Cauchy integral formula contour onto the two sides of the interval. An endpoint circle of radius $\epsilon$ contributes $O({\epsilon }^{1-\alpha })$, which vanishes. The two oppositely oriented banks leave precisely the jump. Without decay at infinity, an entire remainder can survive; a nonzero constant is the simplest example.

The principal square root maps the upper $Z$ half-plane bijectively onto $\Re z>0$ with the ray $[1,\infty )$ removed. The prevertex zero is the slit tip, and $\pm 1$ represent the two ends at infinity. The inverse is the unique root $Z^2=1-z^{-2}$ with positive imaginary part. Pulling back ${\pi }^{-1}\arg ((Z-1)/(Z+1))$ gives the bounded harmonic function with value one on the slit banks and zero on the imaginary axis. The harmonic function $\Im [Z/(1-Z^2)]$ has zero finite boundary values but grows at infinity, explaining the role of boundedness.

A Cauchy integral solves a prescribed boundary difference, not a prescribed sum. If $W_{+}+W_{-}=g$ on $(0,c)$, choose $\rho (z)=\sqrt{z(z-c)}\sim z$ at infinity. Its boundary values satisfy ${\rho }_{-}=-{\rho }_{+}$, so $V=W/\rho$ has difference $V_{+}-V_{-}=g/{\rho }_{+}$. Thus $W=\rho [C(g/{\rho }_{+})+H]$, where $C$ is the Cauchy integral with coefficient $1/(2\pi i)$ and $H$ has no jump. Endpoint integrability and normalization at infinity determine permissible regularizations and homogeneous terms. Simply using $C(g)$ for a sum jump is incorrect.

For $0<a<b$ and an integer $n\ge 1$, the harmonic function on $a<r<b$ with $u(a,\theta )=0$ and $u(b,\theta )=\cos (n\theta )$ isSeparation of variables gives the radial combination $A{r}^n+B{r}^{-n}$; the two boundary values determine $A,B$. Substitution at both radii checks the normalization. The zero mode must be handled separately: for outer value one it is $u(r)=\log (r/a)/\log (b/a)$. The maximum principle supplies uniqueness among harmonic functions continuous on the closed annulus.

A jump matrix is a prescribed invertible matrix-valued function $J$ on an oriented contour in a matrix Riemann-Hilbert problem. With plus boundary on the left of the contour, the unknown holomorphic matrix function satisfiesMatrix multiplication order matters. Where $M_{-}$ is invertible, $J=M_{-}^{-1}{M}_{+}$, rather than $M_{+}{M}_{-}^{-1}$ in general. Reversing contour orientation swaps the sides and replaces $J$ by $J^{-1}$.

For the positively oriented unit circle, a constant invertible matrix $J=A$ is realized by $M=A$ inside and $M=I$ outside. The exterior solution is normalized to $I$ at infinity. More complicated jumps encode boundary values of spectral or scattering data; prescribed poles, endpoint behavior and normalization must also be specified to define the full Riemann-Hilbert problem.^[ref]

The two approaches to the slit base are distinct boundary points, each with interior angle $\pi /2$, whereas the slit tip has interior angle $2\pi$. Their exponents in the Schwarz-Christoffel formula are therefore $-1/2$, $1$ and $-1/2$. Write $a=x_2\in (-1,1)$ and takeon the upper half-plane, with the branch normalized by $S(z)\sim z$ at infinity. The derivative has the formOn $-1<x<1$, the upper boundary value is $S_{+}(x)=i\sqrt{1-x^2}$. Both endpoints of this interval map to the slit base, soThe odd term integrates to zero, and $c\ne 0$; hence $\boxed{x_2=0}$. Now $S^{\prime }(z)=z/S(z)$, so $f=cS+b$. The endpoint values force $b=0$, and the tip value, using $S(0)=i$ as a boundary limit, forces $c=1$. ThusThe branch is important: on $x>1$ its boundary value is positive real, on $x<-1$ negative real, and on $(-1,1)$ positive imaginary. The intervals $(-1,0)$ and $(0,1)$ traverse the two sides of the slit, meeting at its tip.

The map is injective in the upper half-plane: equal images imply $z_1^2=z_2^2$, and the alternative $z_1=-z_2$ cannot put both points in that half-plane. Its derivative has no interior zero. The image has positive imaginary part, by continuity of the branch and the fact that an interior real image would require $z^2-1$ to be nonnegative real, which is impossible for $\mathrm{Im}z>0$. Conversely, a point $\zeta$ in the upper half-plane outside the slit gives a unique upper-half-plane root of $z^2={\zeta }^2+1$. A point $\zeta =iy$ with $0<y\le 1$ gives only real roots or zero and is excluded. These observations verify that the image is exactly the required slit domain, rather than just matching its boundary.

Put ${\alpha }_j=1-{\beta }_j$, so the interior angle at the $j$th polygon vertex is $\pi {\alpha }_j$. A conformal map straightens that corner by a power: locally,This follows by composing with the power $1/{\alpha }_j$ that maps the corner to a half-plane and using Schwarz reflection along its straight edges. The reflected map has a nonzero first derivative in the straightened coordinates.

For an upper-half-plane branch of the complex logarithm, defineAlong each open interval of the real axis, $P$ has constant argument. Crossing $x_j$ from left to right decreases $\arg (z-x_j)$ from $\pi$ to zero, so the argument of $P$ increases by ${\beta }_j\pi$. This is exactly the exterior turning angle of the polygon boundary. Thus $f^{\prime }/P$ has one constant boundary argument on all the intervals. The corner asymptotics remove its apparent singularity at each prevertex.

For a bounded polygon, the sum of exterior angles gives ${\sum }_j{\beta }_j=2$. Since every vertex has a finite prevertex, infinity maps to an ordinary point of an edge. In the local coordinate $1/z$, Schwarz reflection gives $f(z)=f(\infty )+d/z+O(z^{-2})$ with $d\ne 0$. Hence $f^{\prime }/P$ has a finite nonzero limit at infinity. After multiplication by one unit complex number, its boundary values are real. Reflecting across the real axis extends it to an entire function, including each removed prevertex, and it is bounded at infinity. Liouville's theorem makes it constant. Therefore the Schwarz-Christoffel formula isIntegration introduces one further additive constant.

The Möbius transformations preserving the upper half-plane have three real parameters:with common nonzero scaling irrelevant. They can send any cyclically ordered triple of boundary points to another such triple. Consequently three prevertices may be fixed independently; the other $n-3$ are constrained by the polygon's shape, through its side-length ratios. The constants of integration and multiplication account for translation, rotation and scale of the image, rather than further prevertex freedom.

If the last prevertex is infinity, use $1/z$ as its local coordinate. Its factor is absorbed into the overall normalization, leavingFor a finite image vertex at infinity's preimage, this has asymptotic order $z^{{\beta }_n-2}$, because $f(z)-f(\infty )$ is of order $z^{-(1-{\beta }_n)}$. Thus the omitted factor is not evaluated by substituting infinity into the finite product; its angle is encoded by the behavior at infinity.

Let $\zeta =\zeta (z,t)$ be the inverse conformal map. With $Q$ real and nonzero, the function $Q\log |\zeta |$ is harmonic away from the source, vanishes on the moving boundary, and has the required logarithmic singularity. Indeed $F(\zeta ,t)=a_1(t)\zeta +O({\zeta }^2)$ with $a_1\ne 0$, so $\log |\zeta |=\log |z|-\log |a_1|+o(1)$ near the origin. Subtracting this candidate from the physical velocity potential leaves a bounded harmonic function near the source. Its singularity is removable, and the difference has zero boundary values. The maximum principle for harmonic functions makes it identically zero. ThereforeOnly the real part is needed, so the multivalued imaginary part of the logarithm causes no ambiguity in $\varphi$.

Differentiate $F(\zeta (z,t),t)=z$ while holding the physical point $z$ fixed:Also the local analytic complex potential of a two-dimensional potential flow has derivativeThe distinction between differentiating at fixed $z$ and fixed $\zeta$ is essential because the boundary map depends on time. Substituting into the boundary evolution law and multiplying by $|\zeta F_{\zeta }{|}^2/Q$ givesTaking the conjugate inside a real part leaves it unchanged, so the desired conformal map equation for planar Darcy growth isAs a normalization check, parameterize the boundary by $\zeta =e^{i\theta }$. Its outward normal vector times arclength is $\zeta F_{\zeta }d\theta$, so its area growth rate is ${\int }_0^{2\pi }\mathrm{Re}(F_t\overline{\zeta F_{\zeta }})d\theta =2\pi Q$, exactly the specified source flux. For an expanding circular map $F=R(t)\zeta$, the equation reduces to $R\dot{R}=Q$.

Orient the contour and call its left-hand side the plus side. At a smooth interior point $t$, take a local arclength parameter $\gamma (s)$ with $\gamma (0)=t$. The Cauchy principal value means the symmetric deletion limitThe two local pieces are deleted with the same arclength cutoff. For a sufficiently smooth regular contour, an equal-radius deletion about $t$ gives the same limit. The point is assumed to be away from endpoints and self-intersections.

To derive the Sokhotski–Plemelj formula, split the contour into a short smooth arc through $t$ and its complement. The complementary integral has the same limit from both sides. On the short arc subtract $f(z)$ from the numerator. Because $f$ is holomorphic nearby, $(f(\zeta )-f(z))/(\zeta -z)$ extends continuously as $\zeta$ approaches $z$; this part also has the same limit from either side. The remaining local term is $f(z)\int d\zeta /(\zeta -z)$. Straightening the oriented tangent shows its extra contribution is $+i\pi f(t)$ from the left and $-i\pi f(t)$ from the right: on a real segment this is the direct limit of ${\int }_{-\delta }^{\delta }du/(u\mp i\epsilon )$. Dividing by $2\pi i$ yieldsIn particular the jump is $w_{+}-w_{-}=f$, while the sum is the principal-value integral divided by $\pi i$. Smooth densities on a real interval satisfy the same formulas at every interior point, by the identical subtraction step with a Hölder-continuous remainder.

Now put $h=w/\tilde{w}$. On the interval, the sign reversal of the auxiliary boundary values givesThus the required Cauchy transform density isWhenever this density is integrable at the endpoints, taking $h=C[F]$ constructs a function with the required jump. Multiplication back gives $w_{+}+w_{-}=G$. The jump of the constructed $w$ isTo conclude that this jump solves the final singular equation, one must additionally justify $w=C[f]$. This reconstruction requires endpoint and infinity conditions; the sign reversal of $\tilde{w}$ alone does not ensure it. Here is a complete valid normalization. LetThe branch is holomorphic off $[0,1]$, with $S_{+}(x)=i\sqrt{x(1-x)}$ and $S_{-}=-S_{+}$. If $G$ extends smoothly to $[0,1]$, the density $F=G/{\tilde{w}}_{+}=i\sqrt{x(1-x)}G$ vanishes like a square root at each endpoint. Its Cauchy transform $h$ is bounded at the endpoints and is $O(z^{-1})$ at infinity. Therefore $w=h/S$ is at most inverse-square-root singular at each endpoint and is $O(z^{-2})$ at infinity. Its jump $f$ is integrable.

For Cauchy reconstruction from a jump across a finite interval, apply the Cauchy integral formula to $w$ on a large slit contour, and shrink its banks to $[0,1]$. The large circle contributes zero in the limit; each small endpoint circle contributes $O(\sqrt{\epsilon })$ and also vanishes. The two banks give the jump $f$, proving $w=C[f]$. The sum part of the Sokhotski–Plemelj formula now givesThis proves the intended finite Hilbert transform inversion with admissible endpoint behavior. Adding a multiple of $1/\sqrt{x(1-x)}$ to $f$ gives a homogeneous solution; equivalently one can add a constant multiple of $1/S$ to $w$, whose boundary sum is zero.

The omitted normalization is a real limitation of the printed claim. To see it explicitly, choose instead $\tilde{w}=S$ and $G=1$. This auxiliary function satisfies all the stated holomorphicity, nonvanishing and sign-reversal requirements. ButThe displayed integral identity follows from $\xi =(1+\cos \theta )/2$ and its continuation from real $z>1$. Hence the proposed jump formula gives $f=0$, which does not solve the equation with $G=1$. The failure is the nonzero constant at infinity: $w-C[f]=1/2$. Thus the final implication is valid with suitable endpoint and decay conditions, but is false for an unrestricted auxiliary function as printed. The valid reciprocal-square-root normalization above supplies an actual solution for $G$ smooth up to the two endpoints.

True. The first of Green's identities and the Robin boundary condition ${\partial }_{n}u=-u$ giveBoth integrands are nonnegative, so $\nabla u=0$ in $V$ and $u=0$ on the boundary. By path connectedness, $u$ is a constant function, and its boundary value makes that constant zero. Therefore $\boxed{u=0}$.

True. Two applications of Green's identities, or two integrations by parts, giveThe biharmonic equation makes the left side zero, while the two clamped boundary conditions $u={\partial }_{n}u=0$ annihilate the boundary integral. Consequentlyso $\Delta u=0$. Apply the first of Green's identities once more with $\varphi =\psi =u$: because $u=0$ on the boundary, it gives ${\int }_V|\nabla u{|}^2=0$. By path connectedness, $u$ is constant, and the zero boundary value implies $\boxed{u=0}$. This argument assumes the classical differentiability needed for the displayed integrations by parts.

Under the stated piecewise smooth hypotheses, the Dirichlet convergence theorem for Fourier series gives the average of the two one-sided limits of the corresponding periodic extension. The even extension is continuous at both endpoints, so the Fourier cosine series converges toThe odd extension agrees with $f$ on $(0,L)$, but its left and right limits are $-f(0),f(0)$ at zero and $f(L),-f(L)$ at the join $L$. Therefore the Fourier sine series converges toIndeed every partial sum of the Fourier sine series is already zero at both endpoints. The endpoint values agree with $f$ only when the corresponding boundary values vanish.

Take $\boxed{p(x)=x(L-x)}$. It vanishes at both endpoints. Its odd periodic extension is continuously differentiable across the joins, whereas its even extension has corners because $p^{\prime }(0)=L$ and $p^{\prime }(L)=-L$ are nonzero.

Two integrations by parts, with $q=n\pi /L$, giveThe Fourier coefficients are thereforeThe Fourier sine series has a uniform absolute tail $O(N^{-2})$, and each of its partial sums satisfies both zero boundary values exactly. The Fourier cosine series has an absolute tail bound $O(N^{-1})$; at $x=0$ its omitted even-index Fourier coefficients have the same sign and yield an error of order $N^{-1}$. This verifies a strictly faster uniform error bound for the sine approximation in this example.

Let $w$ be the difference of two sufficiently regular solutions of the wave equation with the prescribed data. By linearity, $w$ has zero initial displacement, zero initial velocity, and homogeneous Dirichlet boundary conditions. Define its wave energy byFor $\alpha =0$, differentiation under the integral sign and integration by parts yieldThe boundary term vanishes because $w=0$ at both fixed endpoints implies $w_t=0$ there. The zero initial data give $E(0)=0$, so $w_t=w_x=0$ for all times. Thus $w$ is spatially constant, and its zero boundary values give $\boxed{w=0}$. This proves uniqueness for the undamped wave equation.

For $\omega \ne 0$, substitution of the separated solutions into the heat equations givesThe zero exterior boundary values are incorporated byTo obtain a eigenvalue equation for heat conduction in two joined slabs, write $u=\omega {\alpha }_{1}L$ and $v=\omega {\alpha }_{2}L$. Continuity of temperature and heat flux at zero gives the homogeneous linear systemA nonzero amplitude pair exists precisely when its determinant vanishes, equivalentlyIf $\cos u\cos v\ne 0$, division by $k_1{k}_2{\alpha }_1{\alpha }_2\cos u\cos v$ gives the printed transcendental equationThe printed necessity claim needs the qualification that both tangents are defined. If just one of $\cos u,\cos v$ vanishes, $D(\omega )\ne 0$. If both cosines vanish, however, $D(\omega )=0$ and legitimate normal modes remain. Their temperature at the thermal interface need not be zero; the temperature condition fixes the amplitude ratio, while each heat flux is zero there.

For a concrete counterexample, take identical slabs: ${\rho }_1={\rho }_2=\rho$, $c_1=c_2=c$, $k_1=k_2=k$, and $\alpha =\sqrt{\rho c/k}$. Thenwith each formula restricted to its own slab, satisfies both heat equations and all four boundary conditions. But $\omega \alpha L=\pi /2$, so neither tangent is defined. This directly disproves the unqualified printed assertion. The determinant condition retains these missing normal modes.

Finally, $\omega =0$ must be checked before dividing by it. Then $G_1=A(x+L)$, $G_2=B(L-x)$, and interface continuity gives $A=B$, while heat flux continuity gives $k_{1}A=-k_{2}B$. Positivity of the thermal conductivities forces $A=B=0$. Thus $\boxed{\omega =0\text{ gives only the trivial solution}.}$ In the determinant formula, its apparent zero at $\omega =0$ is spurious because the chosen sine basis degenerates there.

Away from the boundary, derivatives of a smoothly varying solution are much smaller than $k^{2}f$ unless the outer solution vanishes. Nonzero boundary data therefore require a rapidly varying layer. If its thickness is $\delta$, the radial second derivative is of order $f/{\delta }^2$; balancing it with $k^{2}f$ gives $\delta =k^{-1}$.

Put $R=k(1-r)$ and write $f(r,\theta )=F(R,\theta )$. The equation becomesFor fixed $R$ and boundary data with angular variation independent of $k$, the leading equation is $F_{RR}=F$. Matching to the zero outer solution removes the growing exponential, and the boundary value at $R=0$ fixes its coefficient:This approximation describes the layer; it is not a uniform expansion in polar coordinates at the origin.

Laplace's method localizes an exponential integral near maxima of its real phase. At an isolated smooth interior maximum $t_0$ with $G^{\prime \prime }(t_0)<0$ and $F(t_0)\ne 0$, Taylor expansion and Gaussian integral giveAn endpoint maximum with nonzero inward phase slope instead gives an inverse-$k$ factor. Degenerate maxima require their first nonzero local phase term; continuous differentiability alone does not guarantee the quadratic formula.

Here $\cos (ix\sin \varphi )=\cosh (x\sin \varphi )$. In its two exponentials, the positive exponential dominates at the interior maximum $\varphi =\pi /2$ of $\sin \varphi$. Since $\sin (\pi /2+s)=1-s^2/2+O(s^4)$,The negative exponential has no comparable positive maximum and is negligible. Consequently, where $kr\to \infty$,In the boundary layer $r=1-R/k$, $r^{-1/2}=1+O(k^{-1})$, so this reduces to $e^{-R}$, agreeing with the formal result when $g=1$.

Choose each factor with its argument between zero and $\pi$ in the upper half-plane. On an interval between consecutive real prevertices, each factor has a constant boundary argument, so $f^{\prime }$ has constant argument and the image is a straight side. Crossing $X_j$ from left to right decreases $\arg (Z-X_j)$ by $\pi$, hence increases $\arg f^{\prime }$ by ${\beta }_j\pi$, giving the prescribed exterior turn. Locally,so an upper half-disc maps to angle $(1-{\beta }_j)\pi$, the required interior angle. The overall constant rotates and scales the polygon; integration adds translation. For a general prescribed polygon, the prevertex positions must also satisfy the side-length conditions: the derivative formula fixes angles but arbitrary choices do not fix arbitrary side lengths.

For the slit domain, the slit tip has interior angle $2\pi$, so its exterior parameter is $-1$. The two vertices at infinity give the factors at $\pm 1$, leading to $f^{\prime }(Z)$ proportional to $Z(1-Z^2{)}^{-3/2}$. Integrating and imposing the tip value givesThe square root is the principal one: for $\Im Z>0$, $1-Z^2$ never lies on the nonpositive real axis. Thus $\Re f>0$. Boundary points $-1<X<1$ map to the two banks of the slit, with $X=0$ at the tip; the outside intervals map to the imaginary axis. The derivative is nonzero in the open half-plane. Conversely each point of the slit domain determines exactly one root $Z$ of $Z^2=1-z^{-2}$ with positive imaginary part, proving bijectivity.

The Möbius map $(Z-1)/(Z+1)$ takes the upper half-plane to itself, so its principal argument lies in $(0,\pi )$. The imaginary part of its analytic logarithm is harmonic, equals $\pi$ on $(-1,1)$ and zero on the other real intervals. ThusIt lies between zero and one and has the required boundary limits away from the two discontinuities. Pulling this harmonic function back by the inverse conformal map yieldsThis specifies all branches and remains bounded even as $z$ tends to either end at infinity.

To exhibit nonuniqueness without the growth restriction, putThis is harmonic in the domain. On its boundary the corresponding $Z$ is real and the rational expression is real, except at $\pm 1$, which represent infinity. At $Z=\infty$, representing $z=0$, the expression tends to zero. Thus $H$ has zero finite boundary data. It is nonzero, since for $Z=iY$ its value is $Y/(1+Y^2)>0$, and it is unbounded as $Z$ approaches either pole $\pm 1$ non-tangentially. Therefore $u+cH$ gives distinct solutions with the same finite boundary data when boundedness at infinity is removed.