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Use Minkowski spacetime signature $(-,+,+,+)$ and a closed spatial parameter of period $2\pi$. Choose the future-directed branch with positive lapse $e$.

Canonical dynamics and gauge freedom. Variation of momentum and embedding in the Nambu-Goto phase-space action givesThe Lagrange multipliers impose the Nambu–Goto phase-space constraints $P^2+T^2{X}^{\prime 2}=0$ and $P\cdot X^{\prime }=0$. These two first-class constraints reflect freedom to relabel time and space on the same string worldsheet. The canonical Hamiltonian is a linear combination of constraints, with arbitrary multiplier functions. These functions specify a coordinate description rather than additional propagating fields. No explicit gauge transformation or Poisson-bracket calculation is required.

Eliminating auxiliary variables. Let $V=\dot{X}-u{X}^{\prime }$. Eliminating momentum using $P=V/e$ leavesOn a patch with spacelike spatial tangent, variation of $u$ gives $V\cdot X^{\prime }=0$, hence $u=(\dot{X}\cdot X^{\prime })/X^{\prime 2}$. For the induced worldsheet metric $g_{ab}={\partial }_{a}X\cdot {\partial }_{b}X$, this gives $V^2=\det g/X^{\prime 2}$. Variation of $e$ then gives $e=\sqrt{-\det g}/(T{X}^{\prime 2})$ on the positive branch. Substitution yieldsThis is minus string tension times Lorentzian worldsheet area. Vary the area using $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{ab}\delta g_{ab}$ and integrate by parts. The Nambu–Goto equations of motion areThe auxiliary-variable elimination and this metric form apply on nondegenerate timelike patches.

Circular motion, length and energy. For the circular embedding, direct differentiation givesAway from collapse, $\sqrt{-g}{g}^{ab}=\mathrm{diag}(-1,1)$, so the equations become ${\ddot{X}}^m-X^{\prime \prime m}=0$. The time and out-of-plane coordinates satisfy them immediately. For $Z=X^1+i{X}^2$, both $\ddot{Z}$ and $Z^{\prime \prime }$ are $-Z$. The relations $\dot{X}\cdot X^{\prime }=0$ and ${\dot{X}}^2+X^{\prime 2}=0$ verify the Virasoro constraints. Equivalently, $e=1/T$, $u=0$, $P=T\dot{X}$ solve the phase-space equations and constraints at all times.

The ring stays in a fixed plane, with radius $R|\cos t|$. It contracts to a point and re-expands. Each labeled point moves radially with speed $|\sin t|$ in target time $X^0=Rt$. The geometric ring repeats after target-time interval $\pi R$, although the labels have then shifted by half a circumference. It is not rigidly rotating. Velocity is perpendicular to the tangent, so simultaneous spatial arclength is also the local proper length along the string. Its length and conserved energy areAway from collapse, the same energy follows from $T\int dl/\sqrt{1-v^2}$. Increasing kinetic energy compensates the shrinking length. At collapse the induced metric degenerates; the area-form equation alone is undefined there. The regular phase-space solution supplies the continuation and the limiting constant energy. This is a pulsating circular string.

Endpoint variation. For an open string, integration by parts produces the boundary termIts coefficient is the open-string endpoint momentum flux. Evaluate this ungauge-fixed variation in a boundary-adapted temporal gauge for a string $X^0=t$. Then $X^{\prime 0}=0$ and the time equation gives $P^0=1/e\ne 0$. The free variation of time forces $u{P}_0=0$, hence $u=0$ at each end. Since $e$ is nonzero, the remaining spatial term requiresFree variation in every spatial direction gives ${\vec{X}}^{\prime }=0$, and together with $X^{\prime 0}=0$ gives the free-end string boundary condition $X^{\prime m}=0$. The general starting condition is the flux condition; the shift term should not be silently discarded before choosing the gauge.

Target-space charges. The Noether charges for translations and Lorentz transformations are the target-space Noether charges of a stringWriting $F_m=e{T}^2{X}_m^{\prime }+u{P}_m$, the canonical equations implyFor the second identity, antisymmetry cancels the $e{P}_m{P}_n$ terms. Integration by parts then cancels the $u{X}^{\prime }P$ terms and the symmetric $e{T}^2{X}_m^{\prime }{X}_n^{\prime }$ terms. For free ends $F_m=0$ individually, so both charges are constant. In conformal gauge, the same result follows from $u=0$ and $X^{\prime m}=0$. Target-space energy is $-{\mathcal{P}}_0$.

Other possibilities fix selected spatial directions by Dirichlet boundary conditions, while leaving Neumann boundary conditions in the permitted tangent directions. These mixed conditions describe endpoints on a D-brane; the two ends can lie on different branes. Time stays free under the given assumption. Fixed supports may absorb momentum in their Dirichlet directions and break corresponding translations or Lorentz symmetries. More general force-law conditions require an additional boundary interaction.

Use the Dirichlet Laplacian eigenvalues and eigenfunctions on $(-1,1)$:An expansion $u={\sum }_j{q}_j(t)e_j$ reduces the wave equation to independent ordinary differential equationsWhen $\alpha <{\lambda }_1$, put ${\omega }_j=\sqrt{{\lambda }_j-\alpha }$; each coefficient isFor finite-energy data $\varphi \in H_0^1(-1,1)$ and $\psi \in L^2(-1,1)$, a uniform spatial bound follows from the energy method, rather than from unjustified absolute summation of the Fourier series. The conserved quantity isThe Poincaré inequality gives $\Vert u{\Vert }_2^2\le {\lambda }_1^{-1}\Vert u_x{\Vert }_2^2$. Hence $2E\ge c_{\alpha }\Vert u_x{\Vert }_2^2$ with $c_{\alpha }=1-\alpha /{\lambda }_1>0$ for $0\le \alpha <{\lambda }_1$, and with $c_{\alpha }=1$ for $\alpha <0$. Since the Dirichlet boundary condition gives $u(x,t)={\int }_{-1}^x{u}_x(s,t)ds$, the Cauchy-Schwarz inequality yieldsAt $\alpha ={\lambda }_1$, the first coefficient is $q_1(t)={\varphi }_1+t{\psi }_1$, which is unbounded for some admissible data. For $\alpha >{\lambda }_1$, that coefficient has an exponentially growing component for some admissible data. Thus all-time boundedness of a wave equation with a reaction term requiresAt equality, boundedness for a particular data set requires ${\psi }_1=0$; the remaining modes are bounded by their spectral gap. Above the threshold, every unstable mode must have its growing component canceled, and any zero-frequency mode must have zero initial velocity. There is no condition on $\alpha$ alone for arbitrary specially chosen data.

The printed reference to a limit needs qualification. Bounded oscillations generally have no limit as $t\to \infty$. If actual existence of that limit for every admissible initial datum is required, no real $\alpha$ works: for every $\alpha$, choose $j$ with ${\lambda }_j>\alpha$ and nonzero pure oscillatory data in that eigenfunction. The boxed inequality answers the intended long-time boundedness question.

There is a genuine conflict in the printed coefficient assumptions: no uniformly positive coefficient in a zero-boundary Sobolev space exists. A function $a\in H_0^1(\Omega )$ cannot also obey $a\ge a_{-}>0$ almost everywhere. Indeed, the Lipschitz truncation $g(s)=\min (\max (s,0),a_{-})$ has $g(0)=0$, so Lipschitz truncation preserves zero-boundary Sobolev spaces, giving $g(a)\in H_0^1(\Omega )$. But $g(a)$ would be the nonzero constant $a_{-}$. Its zero gradient contradicts the Poincaré inequality in the zero-boundary Sobolev space. Thus the literal coefficient class is empty.

For the meaningful uniformly elliptic problem, take $a\in L^{\infty }(\Omega )$ with $0<a_{-}\le a\le a_{+}$, and impose the Dirichlet boundary condition on the unknown and test functions: $V=H_0^1((0,1{)}^2)$. Additional $H^1$ regularity of $a$ is harmless, but a zero trace for $a$ must be removed. The divergence-form elliptic operator is $L=-\mathrm{div}(a\nabla )$. Integration by parts defines the symmetric bounded bilinear formIn particular,Here $2{\pi }^2$ is the first Dirichlet Laplacian eigenvalue on the unit square; the Poincaré inequality follows, for example, by applying the one-dimensional inequality in each coordinate and adding. Thus $B$ is coercive in the gradient norm on $V$, and the Dirichlet realization of $L$ is a positive definite symmetric operator. On its operator domain, $⟨Lv,v⟩=B(v,v)$.

The required functional and weak equation areBecause $f\in L^2$, the right side is a bounded linear functional by the Cauchy-Schwarz inequality and Poincaré inequality. The Lax-Milgram theorem gives a unique weak solution, and part (b) proves that it uniquely minimizes $I$. These formulas prove the intended conclusion under the repaired coefficient hypothesis; under the literal hypothesis there is no coefficient to which the conclusion can be applied.

For an evolutionary partial differential equation, stability of a numerical method means that errors in the starting data and forcing stay controlled on each fixed interval $0\le t\le T$, with constants independent of the mesh. Fourier stability analysis is especially effective for a constant-coefficient finite difference method on a uniform infinite or periodic grid: translation invariance makes different Fourier modes evolve independently.

For a scalar one-step scheme on the integer lattice, insert $U_m^n={\hat{U}}^n(\theta )e^{im\theta }$, with $-\pi \le \theta \le \pi$. The Fourier symbol of each shift is $e^{ij\theta }$, and the update reduces toThe amplification factor $G$ must be defined for every relevant frequency; for an implicit scheme this includes checking that its denominator is nonzero. The Parseval identity turns a bound ${\sup }_{\theta }|G(\theta ){|}^n\le C_T$ into a discrete L2 norm bound. Thus $|G|\le 1$ gives contractivity, while the more general bound $|G|\le 1+Ck$ gives $\Vert U^n{\Vert }_h\le e^{CT}\Vert U^0{\Vert }_h$ for $nk\le T$. Conversely, frequencies with amplification uniformly greater than one produce unstable wave packets or periodic Fourier modes under refinement.

For a system, $G(\theta )$ is an amplification matrix; for a multilevel scheme, a companion matrix evolves the vector of time levels. The actual requirement is a uniform bound on matrix powers, not just their spectral radius. A nontrivial Jordan block at a unit-modulus eigenvalue creates polynomial growth in the time index. Even simple eigenvalues inside the unit disk can fail to give a uniform bound if the eigenvector matrices become ill-conditioned as the mesh changes. Uniformly controlled diagonalization of a matrix, or a suitable quadratic energy estimate, resolves this issue. The power boundedness of a two-level Fourier scheme illustrates why root multiplicities and conditioning matter.

For the heat equation, put $\mu =k/h_x^2$. Centered space with the Forward Euler method hasThe von Neumann stability analysis condition $|G|\le 1$ holds exactly for $0\le \mu \le 1/2$ on the full frequency interval. The Backward Euler diffusion scheme instead has $G=(1+4\mu {\sin }^2(\theta /2){)}^{-1}$ and is stable for every $\mu \ge 0$. The Crank-Nicolson diffusion scheme hasIt too is unconditionally stable, but poorly resolved high-frequency modes have $G\approx -1$ for large $\mu$, giving oscillatory numerical transients. A-stability therefore does not guarantee strong damping; L-stability distinguishes the damping of the Backward Euler method.

For the advection equation $u_t+c{u}_x=0$, let $\nu =ck/h_x$. Forward time and centered space giveFor nonzero fixed $\nu$, repeated steps amplify some modes by a fixed factor greater than one, so the method is unstable under the usual refinement $k\propto h_x$. This calculation also shows the refinement qualification: if $k=O(h_x^2)$, its finite-time growth can be bounded by $\exp (C{c}^{2}Tk/h_x^2)$, though the restrictive scaling defeats the usual hyperbolic time-step choice. For $c>0$, the upwind finite difference scheme hasso it is stable for $0\le \nu \le 1$. The Lax-Wendroff advection scheme instead hasgiving $|\nu |\le 1$. These examples separate the effects of the spatial stencil, temporal approximation and Courant number; a higher order of a numerical method by itself does not establish stability.

A periodic boundary condition is ideal for this analysis. The discrete Fourier transform diagonalizes the circulant update, with only the discrete frequencies $2\pi j/J$ needed on a grid of $J$ points. A bound on all $\theta$ is a convenient guarantee over every such grid. On the whole line, the Fourier transform uses a continuous frequency interval and the Parseval identity proves the corresponding square-summable-data estimate.

Homogeneous Dirichlet boundary conditions do not admit arbitrary complex exponential Fourier modes. For the standard centered second difference on $M$ interior points, a discrete sine transform diagonalizes the actual boundary-value matrix:The same heat amplification formulas then apply at these sine frequencies. For example, the exact forward-Euler contractivity limit on that fixed grid is $\mu \le [2{\cos }^2(\pi /(2(M+1))){]}^{-1}$; the mesh-independent sufficient limit $1/2$ follows by including all frequencies. Homogeneous Neumann boundary conditions can similarly permit a discrete cosine transform, provided the endpoint discretization and its weighted inner product are chosen consistently. The constant mode is then present, reflecting conservation of the heat equation's spatial mean. For inhomogeneous boundary conditions, subtract a suitable lifting of the prescribed boundary data and estimate the induced source using the homogeneous evolution's bound and a Duhamel principle estimate. Bounds for the lifting and source must themselves be uniform in the mesh.

General boundary conditions require separate boundary stability of a finite-difference method. For an advection equation, incoming data are prescribed at the inflow boundary, while an outflow closure must respect the outgoing characteristics. An interior periodic Fourier symbol cannot detect a growing mode confined near a boundary. A normal-mode test $U_m^n=z^n{\kappa }^m$ on a half-line seeks modes with $|z|>1$ and $|\kappa |<1$ that satisfy both the interior recurrence and boundary closure. Their existence proves instability, and uniform control requires more than merely excluding isolated growing roots; the Uniform Kreiss--Lopatinskii condition addresses boundary resolvent bounds. Alternatively, a direct energy method using summation by parts can include the boundary terms and establish the required estimate.

Variable coefficients and nonuniform meshes usually destroy exact Fourier transform diagonalization. Frozen-coefficient Fourier stability analysis is then a useful diagnostic, but it is not automatically a proof for the full variable-coefficient boundary problem. The mesh-uniform energy method in question 5 is an example of a direct proof. Also, an L2 norm proof is not automatically a mesh-uniform maximum-norm proof: finite-dimensional norm-equivalence constants may grow with the number of grid points.

Finally, stability measures error propagation; consistency of a numerical method measures the defect of inserting the exact solution. For a well-posed linear initial-value problem in the chosen norm, the Lax equivalence theorem says that a consistent approximation is convergent exactly when it is stable. Boundary discretization and starting data must be included in that assertion. A Fourier calculation proves convergence only after consistency, well-posedness and the actual boundary treatment have also been checked.

Splitting the free Schrodinger equation into real and imaginary parts givesDifferentiating the first relation in time and using the second yields the Euler-Bernoulli beam equation $u_{tt}+u_{xxxx}=0$. This is the Schrodinger factorization of the elastic beam equation.

Assume the initial velocity has an integrable first spatial moment, as allowed by sufficient decay, and defineThen $v_0^{\prime \prime }=-u_1$ and $v_0$ decays at infinity. To encode the second boundary datum, defineThe Dirichlet boundary condition for the resulting free Schrodinger equation is compatible at the corner, since $q_0(0)=g_0(0)$. Insert these explicit $q_0,g_0$ into the data-only complex integral in part (b), with ${\hat{q}}_0$ and $G_0$ defined as in part (a). The required displacement is the real part of that integral. Equivalently, the uniformly convergent lifted integral in part (b) may be used with the same complex data.

The Schrodinger factorization of the elastic beam equation verifies every condition: $u(x,0)=u_0(x)$, $u_t(x,0)=-v_0^{\prime \prime }(x)=u_1(x)$, $u(0,t)={\tilde{u}}_0(t)$, andThe corner requirements on $u_0^{\prime \prime }$ and ${\tilde{u}}_0^{\prime }$ ensure consistency of these derivative traces; the natural interpretation of the last printed compatibility is ${\tilde{u}}_0^{\prime }(0)=u_1(0)$. If its prime were instead imposed for every $t$, that would simply be an extra restriction on the data, and the same construction would still solve them.

Fix the convention $\Delta q-m^{2}q=0$, $m>0$, and let $c_m=m^2/4$. With $z=x+iy$, two families of modified Helmholtz adjoint plane waves areBoth satisfy $\Delta E_{\pm }=m^2{E}_{\pm }$, since the products of their $z$ and $\bar{z}$ exponents are $m^2/4$. They obey $E_{-}(z,k)=\overline{E_{+}(z,\bar{k})}$. Green second identity gives the two conjugate global relations for the modified Helmholtz equationFor real boundary traces, ${\mathcal{G}}_{-}(k)=\overline{{\mathcal{G}}_{+}(\bar{k})}$. Thus the second identity is the conjugate spectral companion, rather than an unrelated extra boundary condition. In complex differential form the first relation is equivalentlyIts integrand is a closed one-form: differentiating its coefficients gives $2E_{+}(q_{z\bar{z}}-c_{m}q)=0$. This supplies a direct derivation of the global relation as well as its outward-normal sign convention.

Write the known Dirichlet boundary condition on side $j$ as $f_j(s)$ and the unknown outward normal derivative as $h_j(s)$, with $-1<s<1$. For any linear combination $v$ of the adjoint waves, the global relation isExpanding the $h_j$ in a finite basis and enforcing these identities at selected collocation points for a global relation gives a linear system with a known right side. The square permits especially simple paired tests.

Let $p_n=n\pi /2$, ${\eta }_n=\sqrt{m^2+p_n^2}$, and ${\phi }_n(s)=\sin [p_n(s+1)]$ for $n\ge 1$. Define four real adjoint solutionsThey satisfy the modified Helmholtz equation because ${\eta }_n^2-p_n^2=m^2$. Each is a linear combination of two adjoint exponentials, hence is obtained from the two spectral global relations. More explicitly, put ${\rho }_n^{\pm }=({\eta }_n\pm p_n)/2$, so ${\rho }_n^{+}{\rho }_n^{-}=c_m$. For the $E_{+}$ convention above, the required spectral pairs areAppropriate phase-weighted differences produce $\sin [p_n(s+1)]$. For example the top test is $e^{-{\eta }_n}[e^{i{p}_n}{E}_{+}({\rho }_n^{-})-e^{-i{p}_n}{E}_{+}({\rho }_n^{+})]/(2i)$. The conjugate relation ensures a real system for real data. This is sine collocation of square modified Helmholtz global relations.

The functions ${\phi }_n$ are an orthonormal Fourier sine basis on $[-1,1]$. Set $h_{j,n}={\int }_{-1}^1{h}_j{\phi }_{n}ds$ and compute the known quantity $R_{j,n}={\int }_{\partial \Omega }f{\partial }_n{v}_{j,n}ds$ by numerical integration. Since ${\phi }_n(\pm 1)=0$, each adjoint test vanishes on both adjacent sides. On its own side it equals ${\phi }_n$, and on the opposite side it equals ${\delta }_n{\phi }_n$, where ${\delta }_n=e^{-2{\eta }_n}$. Thus the square modified Helmholtz Dirichlet-to-Neumann coefficients obeyEach block is inverted explicitly: for opposite sides $j,j^{\prime }$,Compute these coefficients for $1\le n\le N$ and reconstruct $h_j^{(N)}={\sum }_{n=1}^N{h}_{j,n}{\phi }_n$. This is a semi-analytical scheme: the spectral basis integrals and two-by-two inverses are explicit, while the known boundary integrals are evaluated numerically. Increase $N$ and the quadrature resolution until the desired convergence is observed; compatible smooth side data have convergent normal-trace expansions, while corner singularities require the usual weaker trace interpretation and more careful quadrature.

The requested weak interaction between sides is particularly clear: the adjacent-side unknown traces contribute exactly zero, and the opposite-side coefficient is $e^{-2\sqrt{m^2+(n\pi /2{)}^2}}\to 0$. The own-side coefficient stays equal to one. Thus diagonal dominance of paired square global-relation collocation is genuine after pairing; individual uncombined exponential samples need not have the same conditioning. The eigenvalues of each block are $1\pm {\delta }_n$, so its condition number is $(1+{\delta }_n)/(1-{\delta }_n)$.

There is also a direct localization check for an individual adjoint wave. On side $j$ with outward unit normal $n_j$ and tangent $t_j$, the normalized test $e^{\eta (n_j\cdot r-1)+ip{t}_j\cdot r}$ has $\eta =\sqrt{m^2+p^2}$. Its opposite-side magnitude is $e^{-2\eta }$, and its magnitude integrates to at most $1/\eta$ along either adjacent side. This side localization of modified Helmholtz plane waves explains why suitable large spectral parameters suppress remote-side effects even before the exact sine cancellation.

Finally the interior solution can be evaluated from the computed traces using the two-dimensional modified Helmholtz fundamental solutionHere $K_0$ is the Modified Bessel function of the second kind, and $(-\Delta +m^2){\Gamma }_m=\delta$. Replace $h$ by its computed finite Fourier sine series and use numerical integration; at interior points the boundary kernels are smooth. The sign follows from Green second identity with this fundamental-solution convention. This completes the numerical integration of the boundary value problem with a Dirichlet boundary condition, rather than stopping at an equation for its unknown boundary derivatives.

For the Hamiltonian $p^2/(2m)+V(q)$, the real-time configuration-space path integral iswith its measure defined by time slicing. Wick rotation gives the Euclidean path integral and the thermal trace.

For the infinite square well on $(0,L)$, the Dirichlet boundary conditions select the normalized energy eigenstatesThere is no $n=0$ state: the corresponding sine is identically zero. Taking the trace in this orthonormal basis gives the canonical partition function

For the real coefficients of the stated self-adjoint Sturm-Liouville problem, define the weighted inner productSince $w>0$, a nonzero eigenfunction has positive squared norm. Let $Ly=-(p{y}^{\prime }{)}^{\prime }+qy$. Integration by parts and the Dirichlet boundary conditions giveTaking $u=v$ to be an eigenfunction shows that its eigenvalue is real, since $\lambda \int w|u{|}^2=\int (p|u^{\prime }{|}^2+q|u{|}^2)$ is real. For two eigenfunctions with eigenvalues $\lambda ,\mu$, the same identity givesHence distinct eigenvalues imply orthogonality in the weighted inner product. This is the orthogonality of Sturm-Liouville eigenfunctions. The calculation assumes the regularity needed for the stated self-adjoint boundary problem, so all boundary terms and integrals are defined.

Multiply the equation by $x^2$ and combine the derivative terms:Thus the self-adjoint form isThe Dirichlet boundary conditions remain at $1$ and $e$, and the weight $w=x$ is positive on this interval, as required for the Sturm-Liouville problem.

With the positive logarithmic fundamental solution used here, a Dirichlet Green function for an interior pole ${\mathbf{r}}_0$ satisfiesEquivalently, $G-G_0$ is harmonic near the pole, $G$ is harmonic away from it, and it has the specified logarithmic singularity. Regularity up to the remaining boundary is understood; on an unbounded domain impose the appropriate condition at infinity as well.

Apply Green second identity with $f=\psi$ and $g=G$. Since $\psi$ is a harmonic function, the volume integral is $\int \psi {\delta }_{{\mathbf{r}}_0}dA=\psi ({\mathbf{r}}_0)$. The term $G{\partial }_n\psi$ vanishes on the boundary by the Dirichlet boundary condition. ThereforeOne may equivalently excise a small disc around the pole and take its radius to zero. The sign is positive because this question uses ${\nabla }^{2}G=+\delta$, not the alternative negative-Laplacian convention.

Write the pole as $(x_0,y_0)$ with $x_0,y_0>0$. The method of images uses three reflected poles with signs $-,-,+$. ThusThe logarithms cancel in pairs on each axis. Only the original pole is in the interior, so this is the required Dirichlet Green function in a quadrant.

On the horizontal boundary $(t,0)$ the outward normal is $-{\mathbf{e}}_y$. Differentiating the four logarithms givesMultiply by the horizontal Dirichlet boundary condition $e^{-t^2}$ and integrate to obtainBy exchanging the axes, the vertical boundary contributes $F(y_0,x_0)$. The representation in part (a) consequently givesFor the unbounded-domain condition, the same expression is the Poisson integral after the conformal change of variables $w=(x+iy{)}^2$ to the upper half-plane, with real-axis data $e^{-|s|}$. This continuous boundary function tends to zero at both ends. Its Poisson integral tends to zero as $|w|\to \infty$: split the data into a compact part, whose kernel integral tends to zero, and a tail uniformly smaller than any prescribed positive number. Hence the constructed function has the required decay, and its two boundary traces agree at the corner. The maximum principle for harmonic functions on expanding quarter-discs proves uniqueness among decaying solutions. The corner value is obtained by continuity as $1$.

A bounded-domain hypothesis is needed for the requested global construction and uniqueness. The printed question does not include it. For example, the upper half-plane satisfies the stated exterior cone condition, but both $u=0$ and $u(x_1,x_2)=x_2$ are harmonic functions with zero Dirichlet boundary condition. Thus part (iv), in its printed unrestricted solution class, is false. In this question's solutions we add that $\Omega$ is bounded; the exterior cone condition and all other data remain as printed.

The half-plane also rules out the printed global separated barrier. In its angular interval $(0,\pi )$, varying $r$ on a fixed ray forces $\nu =2$ if $r^{\nu -2}(h^{\prime \prime }+{\nu }^{2}h)\ge \delta >0$ for every $r>0$. We would then have $h^{\prime \prime }+4h\ge \delta$ and $h<0$. Put $a=\pi /4$ and $b=3\pi /4$. Twice integration by parts givescontradicting the positive integrand. Thus a source qualification is necessary for part (i) as well as part (iv).

Fix $x_0$ and its exterior cone of half-angle $\alpha$. Choose $0<{\alpha }_0<\min (\alpha ,\pi /2)$ and setOn the complement of the cone, unwrap the angle in polar coordinates as $\vartheta \in (\alpha ,2\pi -\alpha )$, measured from the cone axis. Then $\psi =\vartheta -\pi$ is the angle from the opposite axis and $|\psi |\le \pi -\alpha$ on $\overline{\Omega }\setminus \{x_0\}$. Define the power barrier for an exterior cone byThis has the requested separated form $r^{\nu }h(\vartheta )$. In the printed signed-angle convention, the same function is $-r^{\nu }\cos (\beta (\pi -|\theta |))$ outside the cone. It is smooth across the negative axis: near that axis the angular expression is the even function $\cos (\beta \psi )$.

Since $\beta (\pi -\alpha )<\pi /2$, put $m=\cos (\beta (\pi -\alpha ))>0$. The cosine is at least $m$, so $g\le -m{r}^{\nu }<0$ away from $x_0$, and $g$ is continuous at $x_0$. The Laplacian in polar coordinates givesLet $D={\sup }_{x\in \Omega }|x-x_0|<\infty$. Since $\nu <2$, the barrier for the Dirichlet problem satisfiesBoth the uniform lower bound and the later global comparison use boundedness. In particular, $-g$ is positive at every other boundary point, and is bounded away from zero on boundary sets staying a positive distance from $x_0$.

For the corrected bounded-domain problem, the interior theorem for the Perron method for the Dirichlet problem gives a classical solution of the Poisson equation, and the power barrier for an exterior cone argument gives its continuous extension with the required Dirichlet boundary condition. Hence existence holds in $C^2(\Omega )\cap C^0(\overline{\Omega })$.

If $u_1,u_2$ are two such classical solutions, then $w=u_1-u_2$ is a harmonic function with zero boundary trace. The weak maximum principle for elliptic operators, applied to $w$ and $-w$ on the bounded domain, gives $w\le 0$ and $-w\le 0$. ConsequentlyThere is exactly one solution after adding boundedness of $\Omega$. For the printed unbounded-domain version, the half-plane counterexample in part (i) disproves uniqueness; an additional condition at infinity or a suitable restricted solution class would instead have to be specified.

Choose a smooth cutoff function $\eta$ supported in $B_2$, equal to one on $B_1$, with $0\le \eta \le 1$ and $|D\eta |\le C$. For $i<n$, put $v_i=D_{i}u$. The zero Dirichlet boundary condition implies that the tangential boundary derivative $v_i$ vanishes on the flat face. Differentiating the Poisson equation gives $\Delta v_i=D_{i}f$. Test this equation with ${\eta }^2{v}_i$. The boundary terms vanish because $v_i=0$ on the flat face and $\eta =0$ near the curved face. Integration by parts in the tangential direction moves $D_i$ off $f$, yieldingThis is the essential Caccioppoli inequality step: it involves $f$, rather than $Df$. Expanding $D_i({\eta }^2{v}_i)$ and using the Cauchy-Schwarz inequality and the Young inequality givesThe term $f\eta (D_i\eta )v_i$ is bounded by $C{f}^2+C|D\eta {|}^2{v}_i^2$. Absorbing the first term and summing over $i<n$ proves the hinted estimate:All mixed second partial derivatives are now controlled, since $D_{ni}u=D_{in}u$. The remaining normal derivative is recovered from the equation:The Cauchy-Schwarz inequality bounds its squared L2 norm by $n$ times the sum of the squared L2 norms on the right. Including the already controlled zeroth and first partial derivatives, and taking square roots, gives the second-derivative estimate at a flat Dirichlet boundary:The full second-order Sobolev norm is controlled without any norm of $Df$.