Turn boundary problems into coercive variational equations

Try

A rough source may not produce a solution with classical second derivatives everywhere. Nevertheless, multiplying Poisson’s equation by a test function and integrating by parts requires only first derivatives of the unknown. This weaker formulation is also the foundation of finite-element approximations.

For −u″=f on [0,1] with zero endpoints, the integrated equation is ∫u′v′=∫fv for every suitable test function v vanishing at the endpoints. The boundary term disappears because of the test condition, not because u′ must vanish.

Predict

Approximate u by a single shape a x(1−x). For constant f=2, predict whether solving one integrated equation can recover the exact solution. Then explain why that success would not prove that one shape suffices for arbitrary sources.

Experiment

Let v=x(1−x) and use it as the test shape. The scalar equation is a∫(v′)²=∫2v.

Quantity	Integral	Value
Stiffness	∫₀¹(1−2x)²dx	1/3
Load	∫₀¹2x(1−x)dx	1/3
Coefficient	Load divided by stiffness	1

The resulting u=x(1−x) happens to solve the differential equation exactly. Change the source to 2x. The new load is 1/6, so the one-shape coefficient is 1/2. Its residual is not zero pointwise, but it is orthogonal to the chosen test shape. This distinction is the essence of a Galerkin approximation.

Understand

The space H¹ consists of square-integrable functions with square-integrable weak first derivatives. H₀¹ is the H¹ closure of smooth compactly supported functions in the domain; on suitable domains this corresponds to zero boundary trace. Define a(u,v)=∫∇u·∇v and ℓ(v)=∫fv. A weak Dirichlet solution satisfies a(u,v)=ℓ(v) for every v in H₀¹.

Lax–Milgram states that on a real Hilbert space, a bounded bilinear form with a(v,v)≥α∥v∥² for some α positive and a bounded linear functional determine a unique solution. For bounded-domain Dirichlet Poisson, Poincaré’s inequality controls the L² norm by the gradient norm, yielding coercivity; Cauchy–Schwarz bounds the load when f is L². The proof below establishes its symmetric case directly. The later contraction argument proves the nonsymmetric theorem as well.

For this symmetric problem, the mechanism can also be seen through energy E(v)=a(v,v)/2−ℓ(v). Its first variation at u in direction h is a(u,h)−ℓ(h). At a weak solution, expanding the square gives E(u+h)−E(u)=a(h,h)/2. Coercivity therefore makes u the unique minimizer. The next argument constructs the minimizer using completeness and a Cauchy estimate; it does not assume the existence of a weakly convergent subsequence or invoke Riesz representation.

Weak existence does not automatically imply C² regularity. Source smoothness, boundary geometry and elliptic regularity estimates control that upgrade. Corners or rough data can obstruct classical derivatives even when the weak solution is unique.

Complete existence proof for a symmetric coercive form

Let V be a real Hilbert space with norm ∥·∥. Assume a is symmetric and bilinear, |a(v,w)|≤M∥v∥∥w∥, and a(v,v)≥α∥v∥² with α>0. Let ℓ be linear with |ℓ(v)|≤L∥v∥. The constants M, α and L are uniform over V. We prove that exactly one u in V satisfies a(u,h)=ℓ(h) for every h in V.

1. The energy has a finite infimum. For E(v)=a(v,v)/2−ℓ(v), completing a scalar square gives

E(v)≥(α/2)∥v∥²−L∥v∥=(α/2)(∥v∥−L/α)²−L²/(2α).

Since E(0)=0, the infimum m is a real number between −L²/(2α) and 0. By the definition of infimum, choose vⱼ with m≤E(vⱼ)≤m+1/j. This is an existence argument for a sequence, not an algorithm that knows m.

2. Near-minimizers are close to each other. Symmetry and bilinearity give the exact midpoint identity

E((v+w)/2)=(E(v)+E(w))/2−a(v−w,v−w)/8.

Because every midpoint belongs to V and has energy at least m,

α∥vⱼ−vₖ∥²≤a(vⱼ−vₖ,vⱼ−vₖ)≤4(E(vⱼ)+E(vₖ)−2m)≤4(1/j+1/k).

Thus vⱼ is Cauchy. Boundedness alone would not have established convergence; this estimate is the missing mechanism. Completeness gives vⱼ→u in V. Boundedness of a and ℓ makes E continuous: for example |a(vⱼ,vⱼ)−a(u,u)|≤M∥vⱼ−u∥(∥vⱼ∥+∥u∥), and a convergent sequence is bounded. Therefore E(u)=m.

3. The minimizer solves every test equation. For any h in V and real t,

E(u+th)−E(u)=t(a(u,h)−ℓ(h))+(t²/2)a(h,h)≥0.

A nonzero linear coefficient would make this negative for sufficiently small t of the opposite sign. Hence a(u,h)=ℓ(h) for every h. Conversely, such a solution has E(u+h)−E(u)=a(h,h)/2≥α∥h∥²/2, so it is the unique minimizer. If two weak solutions existed, their difference w would satisfy a(w,w)=0; coercivity forces w=0. Testing with u also gives α∥u∥²≤ℓ(u)≤L∥u∥ and therefore ∥u∥≤L/α, with u=0 handled separately.

Symmetry is essential to this energy proof. For a nonsymmetric form, differentiating a(v,v)/2 produces its symmetric part. Nonsymmetric Lax–Milgram is true under its own hypotheses, but cannot be claimed from this argument.

The symmetric variational formulation and Dirichlet application can be compared with Guermond, Chapter 25, Main results on well-posedness. The Cauchy argument above supplies the existence mechanism directly.

Why the hypotheses hold for Dirichlet Poisson

For any bounded open Ω contained in a box (−R,R)ᵈ, take V=H₀¹(Ω), defined as the closure of smooth compactly supported functions under ∥v∥²_H¹=∥v∥²_L²+∥∇v∥²_L². H¹ is complete: if the functions and their weak derivatives are Cauchy in L², their L² limits satisfy the weak derivative identity by passing to the limit against a smooth compactly supported test function. Its closed subspace H₀¹ is therefore complete. This uses completeness of L², not boundary smoothness.

For a smooth compactly supported v, extend it by zero to the box. Along the first coordinate, v(x₁,x′)=∫ from −R to x₁ of ∂₁v(s,x′)ds. Cauchy–Schwarz and integration over the box give ∥v∥_L²≤2R∥∂₁v∥_L²≤2R∥∇v∥_L². Approximation extends this inequality to H₀¹. Thus the gradient norm is equivalent to the H¹ norm on V and is complete. With that norm, a(v,w)=∫Ω∇v·∇w has M=α=1, while ℓ(v)=∫Ωfv has L≤2R∥f∥_L². The existence proof now gives a unique weak solution for every f in L²(Ω), and ∥∇u∥_L²≤2R∥f∥_L².

No smooth-boundary trace theorem is needed for this closure definition. Identifying H₀¹ with all zero-trace functions on a chosen boundary, or upgrading u to a classical C² solution, requires additional results. For example the solution for the step source in the experiment below has a continuous first derivative but a jump in the second derivative; the weak equation still holds.

Zero tested residual need not mean zero error

Use v₁=x(1−x), v₂=x(1−x)(2x−1), and candidate w=a v₁+b v₂ on [0,1]. Predict whether making both tested residuals zero makes w exact, then compare all three sources.

Source f 22x1 for x<1/2, otherwise 0 Coefficient a -0.500000-0.1562500.0000000.1666670.2500000.5000001.000000 Coefficient b -0.500000-0.1562500.0000000.1666670.2500000.5000001.000000

Stiffness diagonal (off-diagonal zero)

1/3, 1/5

Load vector

(0.16666667, 0.03333333)

Tested residual (a/3−ℓ(v₁), b/5−ℓ(v₂))

(0.00000000, -0.03333333)

Candidate energy E(w); exact minimum E(u)

-0.04166667; -0.04444444

Exact squared derivative error ∫|w′−u′|²=2(E(w)−E(u))

0.00555556

Use the two-shape Galerkin coefficients

The best possible squared error in this two-shape space is 0.00000000. For this source, the exact solution lies in the trial space, so the minimum error is zero.

Explain the difference

The Galerkin equations test only v₁ and v₂. The weak equation tests every H₀¹ function. For the step source, u=3x/8−x²/2 on [0,1/2] and u=(1−x)/8 on [1/2,1]. Its value and first derivative match at the join. Integrating these explicit formulas gives ∫|u′|²=5/192 and the displayed exact error; numerical sampling is not used as proof.

Reset energy experiment

Exact integral formulas evaluated in floating-point arithmetic; tiny displayed roundoff is not mathematical error. Settings are temporary and do not award mastery.

A changed-case certificate rather than a residual slogan

On (0,1), replace −u″=f by −(c(x)u′)′=f, with measurable coefficient 1≤c(x)≤4 almost everywhere and f in L². Prove existence, uniqueness and an energy-error identity. Then remove the lower bound by setting c=0 on (1/3,2/3), and decide whether the same proof works. Write your argument before opening the feedback.

Feedback, failed hypothesis and revision criteria

Use a(v,w)=∫c v′w′ and the gradient norm on H₀¹. Cauchy–Schwarz gives M=4, and the lower bound gives α=1. The load is bounded by the interval Poincaré estimate. The proof above yields a unique weak u. For any candidate w, E(w)−E(u)=∫c|w′−u′|²/2. Zero tested residual on a finite trial space does not establish this error is zero.

If c vanishes on the middle interval, any nonzero smooth bump h supported there has a(h,h)=0, so coercivity fails. With f=0, both zero and every such bump are weak solutions; uniqueness actually fails. With f=h, testing a proposed solution against h would require 0=∫h², so no weak solution exists. A complete answer supplies the bump counterexample, distinguishes lack of a theorem from actual failure, and states the norm and all test functions. An initial/revised argument and feedback can be retained in your own document; this public practice does not record or grade proof quality.

Apply

Choose two trial shapes with zero endpoints, write the two-by-two stiffness matrix and load vector, and explain why linear independence makes the stiffness positive definite. Compare a small integrated residual with a pointwise residual. State which norm and test space your evidence actually controls.

Remove symmetry without minimizing the wrong energy

Let V be a real Hilbert space, a:V×V→R bilinear with |a(u,v)|≤M∥u∥∥v∥ and a(v,v)≥α∥v∥² for fixed M≥α>0. Let ℓ be bounded linear. Symmetry is not assumed. We prove that exactly one u satisfies a(u,v)=ℓ(v) for every v, and ∥u∥≤∥ℓ∥/α.

The symmetric existence theorem already proved above, applied to the Hilbert inner product, represents every bounded functional by an inner product. Thus there are a bounded linear operator A and vector f such that (Au,v)=a(u,v) and (f,v)=ℓ(v). In particular ∥Au∥≤M∥u∥ and (Au,u)≥α∥u∥². This is the Riesz representation step, here obtained from the earlier symmetric theorem rather than assumed from matrix intuition.

For τ=α/M², define F(u)=u−τ(Au−f). For w=u−v,

$$\|F(u)-F(v)\|^2=\|w\|^2-2\tau(Aw,w)+\tau^2\|Aw\|^2\le\left(1-\frac{\alpha^2}{M^2}\right)\|w\|^2.$$

The square-root factor q is below 1 (possibly zero). Iterates uₙ₊₁=F(uₙ) satisfy ∥uₙ₊₁−uₙ∥≤qⁿ∥u₁−u₀∥. The geometric tail makes them Cauchy, completeness supplies a limit u, and continuity gives F(u)=u, hence Au=f. Two fixed points have distance at most q times that distance, so are equal. Testing the weak equation with u gives the stability bound. This proves nonsymmetric Lax–Milgram by a fully specified contraction, not by minimizing a(v,v)/2−ℓ(v), whose derivative would discard the skew part.

For a closed finite-dimensional trial space V_h, the same proof yields u_h. Galerkin orthogonality a(u−u_h,v_h)=0 gives, for every w_h,

$$\alpha\|u-u_h\|^2\le a(u-u_h,u-u_h)=a(u-u_h,u-w_h)\le M\|u-u_h\|\|u-w_h\|.$$

Cancel only when the error is nonzero; otherwise the conclusion is immediate. Taking the infimum yields the quasi-optimal bound ∥u−u_h∥≤(M/α)inf∥u−w_h∥. The best-approximation constant need not be 1 without symmetry and the appropriate energy norm.

Keep the skew part of the equation

Skew parameter β: 1.0

For A = [[1, −β], [β, 1]] and f = (1, 0), u = (0.500000, -0.500000). The contraction step is 0.500000, with factor 0.707107.

Multiply A by this vector. Compare with (1,0), the minimizer obtained by incorrectly dropping the skew term. A finite matrix check illustrates but does not prove the Hilbert-space theorem.

Temporary practice only. No saved response, automatic proof grade or mastery award.

Apply it to a drift term

On(0,1), consider −u″+b u′=f with zero endpoint data and constant real b. In V=H₀¹ with gradient norm, set a(u,v)=∫u′v′+b∫u′v. The Poincaré inequality gives continuity with M≤1+|b|C_P. For a smooth compactly supported v, ∫v′v=[v²/2]₀¹=0; density and continuity extend this identity to V. Thus a(v,v)=∥v′∥² and α=1, whatever the sign of constant b. The load is bounded for f in L². A unique weak solution follows. If b depends on x, the identity becomes −(1/2)∫b′v² when b is sufficiently regular; do not carry over coercivity without checking this new term.

For a finite-dimensional countercheck take A=[[1,−β],[β,1]], with a(u,v)=(Au)·v and f=(1,0). Coercivity is 1 and operator norm is√(1+β²). The actual solution is u=(1,−β)/(1+β²), whereas minimizing ∥u∥²/2−u₁ gives(1,0). At β=2 the correct answer is(1/5,−2/5); direct multiplication verifies it. The skew term changes the solution even though it vanishes in a(u,u).

Primary theorem comparator: Oxford C6.4, Hilary 2021, §4.2, checked 30 October 2026. This argument is for a coercive real Hilbert-space form. Noncoercive Fredholm or inf-sup problems require other hypotheses.

When a weak solution is actually classical

A complete one-dimensional upgrade can be proved directly. Let c be continuously differentiable and strictly positive on[0,1], and f continuous. For −(c u′)′=f, u(0)=u(1)=0, define F(x)=∫₀ˣf(s)ds and

$$C=\frac{\int_0^1F(s)/c(s)\,ds}{\int_0^1 1/c(s)\,ds},\qquad u(x)=\int_0^x\frac{C-F(s)}{c(s)}\,ds.$$

The positive denominator is finite. Direct differentiation gives u′=(C−F)/c continuously differentiable, so u is C² and −(cu′)′=f pointwise. The choice of C makes both boundary values zero. Integration by parts shows this is also the unique weak solution from coercivity. This establishes classical existence under these explicit assumptions, not merely weak existence plus an unsupported smoothness claim.

For c=1,f=1, C=1/2 and u=x(1−x)/2. For a step source f=1 on(0,1/2), f=0 on(1/2,1), the same integral construction gives a C¹ state whose second derivative is−1 on the first half and 0 on the second; it is not C² across 1/2. The weak identity still holds. For a discontinuous positive c, the flux c u′ can be continuous while u′ jumps: with c=1 on the first half,2 on the second and f=0 but boundary data u(0)=0,u(1)=1, the constant flux is 4/3, giving slopes 4/3 and 2/3. Nonzero boundary data here are handled by subtracting a fixed lifting; this example is a regularity comparison, not the earlier homogeneous problem.

The multidimensional interior upgrade is proved below. It does not assert smoothness up to arbitrary boundaries.

Changed-case assessment: prove or refuse

A colleague uses the nonsymmetric theorem with a(v,v)≥0, replacing strict coercivity, and claims uniqueness. Refute the claim with V=R and a=0: ℓ=0 gives every u as a solution, and ℓ(v)=v gives none. Next solve the matrix example with β=−1 and f=(1,0): u=(1/2,1/2). Finally explain why a step source blocks the C² claim without blocking H₀¹ existence. Review criteria: identify α>0, substitute the matrix solution, and distinguish existence, uniqueness, regularity and numerical approximation as separate conclusions. Keep a written first proof and correction; these public examples do not certify independent mastery.

Prove the multidimensional interior upgrade

Prediction. If the coefficients and source are smooth inside a planar domain with a reentrant corner, must the weak solution have square-integrable second derivatives all the way to that corner? Keep this prediction: the interior proof below deliberately loses control as its support approaches the boundary.

The symbols Hᵐ and Hᵐ_loc mean, respectively, that all weak derivatives through integer order m are square integrable, and that this holds on every compactly contained subdomain. The notation V⋐U means the closure of V is compact and contained in U. No boundary trace is involved in an interior assertion.

The local theorem, including the term needed for bootstrapping

Let Ω be open in Rᵈ. On each relatively compact subdomain assume the matrix A(x) is bounded and Lipschitz, and there is λ>0 such that ξ·A(x)ξ≥λ|ξ|² for every real vector ξ. Symmetry is unnecessary. Let u belong to H¹_loc, f to L²_loc and the vector field G to H¹_loc. Suppose, for every smooth compactly supported v,

$$\int A\nabla u\cdot\nabla v=\int fv-\int G\cdot\nabla v,\qquad -\operatorname{div}(A\nabla u)=f+\operatorname{div}G.$$

Then u belongs to H²_loc. More precisely, if V⋐U⋐Ω, there is a constant depending only on these nested sets, d, λ, the bound on A and its Lipschitz constant on a slightly larger neighborhood, such that

$$\|u\|_{H^2(V)}\le C\bigl(\|u\|_{H^1(U)}+\|f\|_{L^2(U)}+\|G\|_{H^1(U)}\bigr).$$

One can choose all cutoffs and translated supports strictly inside U, so the coefficient bounds there suffice. The H¹ norm on the right is intentional; no sharper L²-only estimate is being asserted. The ordinary Poisson case is A=I and G=0. An L² source gives H² locally, not automatically C².

Step 1: the discrete calculus, with no assumed second derivative. In coordinate direction eₖ set δₕw(x)=(w(x+h eₖ)−w(x))/h, for nonzero h of either sign. All translations below stay inside U. Substitution in the integral gives

$$\int p\,\delta_{-h}q=-\int(\delta_hp)q,\qquad \delta_h(pq)=p(x+he_k)\delta_hq+(\delta_hp)q.$$

For smooth w, the fundamental theorem of calculus gives δₕw(x)=∫₀¹∂ₖw(x+t h eₖ)dt. Jensen’s inequality and translation of the integral show that its L² norm on a set is at most the L² norm of ∂ₖw on a neighborhood containing every intervening segment. The same statement holds for H¹ functions: multiply by a cutoff equal to one near the segments and convolve with a smooth unit-mass kernel, apply the smooth inequality, and pass in H¹. This approximation follows because convolution commutes with weak derivatives and translations are continuous in L². To see the last fact without assuming pointwise continuity of w, first approximate an L² function by finite linear combinations of indicators of boxes, for which translated symmetric differences have measure tending to zero, and then use the translation isometry and a three-term estimate. Density here is the ordinary simple-function/box approximation from Lebesgue integration. For fixed h, δₕ is bounded on L² by 2/|h|, so passage to the limit is valid.

Step 2: use an admissible test instead of differentiating u twice. Choose a smooth cutoff η supported inside U, equal to one on a neighborhood of V, with 0≤η≤1. Restrict |h| so its translated support and all intervening segments stay inside U. Set z=δₕu and w=η²z. For each fixed h, z is H¹, w is H¹ with compact support, and v=−δ₋ₕw is an admissible H₀¹ test. The original identity extends to these tests by the same local smoothing and continuity of every integral. It does not require a uniform bound on v before it is used.

Discrete integration by parts gives the exact identity

$$\int\bigl(A(x+he_k)\nabla z+(\delta_hA)\nabla u\bigr)\cdot\nabla(\eta^2z)=\int f[-\delta_{-h}(\eta^2z)]-\int(\delta_hG)\cdot\nabla(\eta^2z).$$

The sign of the G term follows from the displayed weak equation; it is not a formal second derivative of G. Expand ∇(η²z)=η²∇z+2ηz∇η. Write X=∥η∇z∥₂ and B=∥u∥_H¹(U). Step 1 gives ∥z∥₂ on the cutoff neighborhood at most B, independently of h. Also ∥δₕA∥∞ is at most the Lipschitz constant L, and ∥δₕG∥₂ is at most ∥∂ₖG∥_L²(U). The first quadratic term on the left is at least λX². Every remaining term has an explicit bound:

• The cutoff part of A(x+h eₖ)∇z is at most 2∥A∥∞∥∇η∥∞XB.
• The coefficient-difference term is at most LB(X+2∥∇η∥∞B).
• The source term is at most ∥f∥₂∥∂ₖ(η²z)∥₂, hence at most ∥f∥₂(X+2∥∇η∥∞B). Here Step 1 was applied to w extended by zero, not to f; f need not have a derivative.
• The G term is at most ∥∂ₖG∥₂(X+2∥∇η∥∞B).

Consequently, with D=B+∥f∥₂+∥G∥_H¹(U), λX²≤C₁DX+C₂D². Young’s inequality C₁DX≤λX²/2+C₁²D²/(2λ) yields X≤C₃D. The constants are independent of h. This is the decisive estimate; simply writing a formally differentiated equation would not have proved that its derivatives exist.

Step 3: turn the estimate into actual weak derivatives. For each j, the difference quotients δₕ∂ⱼu are bounded in L²(V). For any smooth φ supported in V, discrete integration by parts and uniform convergence of δ₋ₕφ to ∂ₖφ give

$$\left|\int (\partial_j u)\partial_k\phi\right|=\lim_{h\to0}\left|\int(\delta_h\partial_j u)\phi\right|\le C_3D\|\phi\|_2.$$

Thus φ↦−∫(∂ⱼu)∂ₖφ extends by density to a bounded functional on L²(V). The Hilbert representation already proved in the nonsymmetric-existence section supplies an L² function representing it. That function is exactly the weak derivative ∂ₖ∂ⱼu. Doing this for every j,k proves H²(V) and the claimed estimate, without assuming weak subsequential compactness or a classical solution in advance.

Smooth coefficients really do give smooth interior solutions

Take G=0 initially. For an integer m≥0, if A is Cᵐ⁺¹ locally and f is Hᵐ_loc, the conclusion is u in Hᵐ⁺²_loc. Here is the induction rather than an unexplained appeal to elliptic regularity.

The case m=0 is proved above. Suppose the corresponding conclusion is established through m−1, so u is Hᵐ⁺¹_loc. For a multi-index α of length m, w=Dᵅu is H¹_loc. Differentiate the equation distributionally using the product rule:

$$-\operatorname{div}(A\nabla w)=D^\alpha f+\operatorname{div}G_\alpha,\qquad G_\alpha=\sum_{0<\beta\le\alpha}\binom{\alpha}{\beta}(D^\beta A)\nabla D^{\alpha-\beta}u.$$

The binomial coefficient counts the coordinatewise product-rule choices. Every term of Gα is H¹ on a compactly contained subdomain: differentiating it once uses derivatives of A through m+1 and derivatives of u through at most m+1, all available. The product rule follows first for local mollifications and then by convergence of these weak derivatives, so it does not presume the desired new derivative. The local theorem with source Dᵅf in L² and vector Gα in H¹ gives w in H². All derivatives of u through order m+2 therefore exist locally. At each induction step choose a slightly larger compactly contained subdomain; a finite induction needs only finitely many nested neighborhoods.

If A and f are C∞, this proves u in Hˢ_loc for every integer s. We still must connect weak derivatives to continuous ones. The following elementary, non-sharp embedding is sufficient and avoids hiding that step. For a smooth function v supported in a fixed box Q, repeated one-dimensional integration from −∞ in each coordinate gives

$$v(x)=\int_{-\infty}^{x_1}\!\cdots\!\int_{-\infty}^{x_d}(\partial_1\cdots\partial_d v)(t)\,dt_d\cdots dt_1,\qquad \|v\|_\infty\le |Q|^{1/2}\|\partial_1\cdots\partial_d v\|_2.$$

Apply the inequality also to each Dᵅv for |α|≤k. Multiply u by a smooth compactly supported cutoff χ that is one on the region of interest and mollify χu. The mollifications have support in one fixed box and converge in Hᵏ⁺ᵈ. The inequality makes them and all derivatives through k uniformly Cauchy. Their uniform limits are continuous; integrating along coordinate segments shows the limit of each first derivative is the derivative of the preceding limit, and iteration gives a Cᵏ representative. Its L² limit is χu, so it represents the original weak solution almost everywhere. Since k was arbitrary, u has a C∞ representative inside Ω. The weak equation then becomes a pointwise equation because all its terms are continuous and a continuous function that is zero as a distribution is zero everywhere.

This also yields a concrete non-sharp finite criterion: A in Cᵈ⁺¹_loc and f in Hᵈ_loc imply u in Hᵈ⁺²_loc and hence C² locally by the box argument. Sharper Sobolev exponents exist, but are not needed for the proved smooth conclusion.

Why this does not prove regularity at a corner

Consider the sector with radius r strictly between zero and one and angle θ strictly between zero and 3π/2 in the plane, and

$$u(r,\theta)=r^{2/3}\sin(2\theta/3).$$

The polar formula Δu=uᵣᵣ+r⁻¹uᵣ+r⁻²u_θθ gives Δu=0 at every interior point. The solution is continuous on the closed sector, vanishes on its radial sides, and has boundary value sin(2θ/3) on the arc. Its gradient has magnitude (2/3)r⁻¹/³, so its squared integral near the vertex is a constant times ∫₀^εr¹/³dr, which is finite. Thus u belongs to H¹. But uᵣᵣ=−(2/9)r⁻⁴/³sin(2θ/3). On an angular subinterval where the sine stays away from zero, the squared Hessian norm is at least |uᵣᵣ|², and its integral dominates a positive multiple of ∫₀^εr⁻⁵/³dr, which diverges. Hence u is not H² on the entire domain, despite a smooth constant coefficient matrix and zero source. The vertex is a boundary point, not an interior singularity, so there is no conflict with the theorem.

Global estimates need boundary hypotheses and compatible boundary data, together with a different boundary argument. Arbitrary-domain C² closure regularity is false, not an unfinished consequence of Lax–Milgram. This lesson does not assert a general boundary Schauder estimate, a global H² theorem on smooth domains, or nonlinear regularity.

Changed-case proof exercise and feedback

1. Replace A by the nonsymmetric constant matrix [[2,1],[−1,2]]. Does the local proof survive? Give λ and explain why symmetry was not used.
2. Replace the source by an L² step function. Which local conclusion is justified, and which upgrade needs more hypotheses?
3. Locate the precise obstruction if A has a jump across an interior interface.
4. Change the sector angle to π/2 and examine r²sin(2θ). Compare the corner conclusion with the reentrant example; do not infer a theorem for every convex domain from one polynomial.

Feedback. The symmetric part is 2I, so λ=2; the operator bound is √5 and δₕA=0. All estimates remain valid. L² forcing gives H² locally, not C² automatically; the earlier one-dimensional step solution, extended independently of the other coordinates on a box, supplies a counterexample to continuous second derivatives. With a coefficient jump, the uniform Lipschitz bound for δₕA fails; the earlier flux example shows why an H¹ solution can have a derivative jump. For the right-angle sector r²sin(2θ)=2xy is smooth. The reentrant example instead has exponent 2/3 and a divergent Hessian integral. A satisfactory retained proof identifies the admissible test, the h-independent bound, the weak-derivative representation, and the interior/boundary distinction. Revise any argument that differentiates u twice before establishing that those derivatives exist.

Source-scope check (30 October 2026). MIT 18.152 Fall 2011, calendar lists canonical Poisson tools in lectures 6–9. Stanford Math 220 Autumn 2018, public syllabus specifies Laplace energy estimates and disk boundary problems, not a full general-coefficient boundary-regularity theory. The local estimates above close the weak-to-classical mechanism explicitly rather than expanding that historical baseline into all graduate PDE theory. For a theorem-level comparator, Hunter, PDE notes, §§4.11–4.12 and 4.C, revised 18 June 2014, separates interior H², higher regularity and boundary estimates. The derivation here includes its own difference-quotient criterion, nonsymmetric/vector-source extension and non-sharp smoothness argument; it does not merely import the comparator’s conclusion.

Continue

Numerical evolution introduces stability restrictions in addition to spatial approximation. The next lesson derives the explicit heat time-step bound from both convex weights and Fourier modes.