advanced34 min studyLesson 396 in Mathematics

Interior regularity by difference quotients

Earn a second weak derivative through uniform translation estimates, and see how a coefficient jump can prevent it.

On this page

What you’ll understand

  • Use a localized difference-quotient test without assuming second derivatives.
  • State coefficient and source hypotheses for an interior H² estimate.
  • Check a uniformly elliptic rough-coefficient example whose solution is not H² locally.

An energy solution initially has one weak derivative. Differentiating its equation as though two derivatives already existed would assume the conclusion we want to prove. Difference quotients avoid that circular step: translate the function by a small distance, subtract, divide by the distance, and obtain estimates independent of the translation. A compactness argument then identifies an actual weak derivative. The method also reveals exactly where coefficient regularity is used, which makes it possible to diagnose why uniform ellipticity alone does not guarantee second derivatives.

Translate inside the domain

Choose a coordinate direction and a small nonzero translation h. Define the difference quotient only where both the original and translated points remain inside the domain. If an H-one function has a weak derivative in that direction, its difference quotients are bounded locally in L-two by an appropriate gradient norm. Conversely, a locally uniform L-two bound on the difference quotients of an L-two function yields a weak derivative by testing, weak compactness, and passage to the limit. The bound must be independent of h.

For a weak solution of minus divergence of A gradient u equals f, insert a cutoff supported strictly inside the domain. A useful test is minus the backward difference quotient of the cutoff squared times the forward difference quotient of u. Discrete integration by parts transfers a translation between factors. The leading resulting term controls the squared gradient of the difference quotient, while cutoff terms are handled as in a localized energy estimate. Every translated support must remain admissible for the weak identity.

δhku(x)=u(x+hek)−u(x)h,v=−δ−hk(η2δhku)\delta_h^ku(x)=\frac{u(x+he_k)-u(x)}h,\qquad v=-\delta_{-h}^k\bigl(\eta^2\delta_h^ku\bigr)
The compactly supported cutoff η and sufficiently small h keep the test inside the domain. The backward quotient uses its own signed denominator, so the displayed minus sign is the discrete integration-by-parts sign.↔ Scroll wide equations horizontally.

Coefficient differences are the price of the argument

Assume A is real symmetric, uniformly elliptic, and C-one on the larger interior region, with bounded first derivatives there; assume f is L-two and u is an H-one weak solution. Differences of A are then bounded by its first-derivative bound times the translation length. This controls the commutator terms after dividing by h. Cauchy–Schwarz and Young's inequality absorb part of the unknown gradient-quotient term into ellipticity, leaving a uniform estimate in terms of known norms.

Consequently the solution belongs to H-two on every smaller compactly contained region, with an estimate depending on the ellipticity bounds, coefficient derivative bound, dimension, and separation from the outer region. The result is an interior theorem and imposes no boundary condition. Reaching H-two all the way to the boundary requires extra boundary geometry and compatible boundary data. Repeating the argument yields higher regularity only when coefficients and source supply the additional derivatives it consumes.

∥u∥H2(B1/2)≤C(∥f∥L2(B1)+∥u∥H1(B1)),−div⁡(A∇u)=f\|u\|_{H^2(B_{1/2})}\leq C\bigl(\|f\|_{L^2(B_1)}+\|u\|_{H^1(B_1)}\bigr),\qquad -\operatorname{div}(A\nabla u)=f
This interior estimate assumes bounded uniformly elliptic C¹ coefficients with a bounded first-derivative norm on B₁, an L² source, and an H¹ weak solution. The constant depends on those bounds, not on the translation h.↔ Scroll wide equations horizontally.

A flux can be smooth while the gradient jumps

Across a material interface, the weak equation controls the flux A gradient u. If A jumps, a continuous flux can coexist with a jumping gradient. The weak second derivative of u then contains an interface delta, which is not an L-two function. Such a solution may be perfectly valid in H-one while failing H-two near the interface. This behavior is not a contradiction of elliptic regularity: the coefficient derivative hypothesis of this particular H-two theorem has failed.

A finite-difference computation can expose the same issue. Differences of a step-like gradient have height proportional to one over h on an interval of width h, so their squared L-two norm grows like one over h rather than remaining bounded. Refining a grid will not turn that interface delta into an ordinary square-integrable derivative. One must choose a regularity framework and numerical formulation consistent with the weak transmission condition, rather than interpret every large discrete derivative as an implementation bug.

A uniform ellipticity bound does not prevent an interface singularity

On (−1,1), let a=1 for x<0 and a=2 for x>0. Define u=x for x≤0 and u=x/2 for x≥0. Check the weak equation and the second-derivative obstruction.

  1. The function is continuous and piecewise affine with bounded first derivative, so it lies in H¹. Its flux a u′ equals one on both sides. For every compactly supported smooth test v, ∫au′v′=∫v′=0\int au'v'=\int v'=0, proving the homogeneous weak equation.

  2. The derivative jumps from one to one half. Its distributional derivative is u′′=−12δ0u''=-\tfrac12\delta_0, which is not an L² function. Thus u does not belong to H² on any interval containing zero, despite ellipticity bounds λ=1 and Λ=2.

  3. For 0<h<1/2, the forward difference of u′ is −1/(2h) on (−h,0) and zero away from the crossing. Its squared integral near zero is h/(4h2)=1/(4h)h/(4h^2)=1/(4h), which diverges as h tends to zero. The missing uniform quotient bound precisely detects the obstruction.

The result

The flux equation is satisfied exactly; coefficient smoothness, rather than ellipticity, is the missing hypothesis for this H² conclusion.

Explore the model

Interactive lab

Interactive lab · spectral

See how differentiation amplifies short spatial scales

Pause, Reset, and set mode count N=3 and coefficient decay p=1. Repeat from Reset with both N=12 and p=1. Predict more small-scale oscillations even though the new coefficient amplitudes are smaller. Differentiating the nth sine mode multiplies its amplitude by n, but this instrument plots the finite sum itself and does not compute derivative norms or simulate the interface example.

See how differentiation amplifies short spatial scalesA finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forSmall amplitudes at short wavelengths can still matter strongly to derivative estimates; a finite smooth plot cannot establish a uniform weak-derivative bound.

Check your understanding

What does the interface example show about a bounded measurable uniformly elliptic coefficient?

Try it yourself

For a(x)=1+x on (0,1), verify that u(x)=log(1+x)/log 2 solves −(a u′)′=0 with boundary values zero and one. Determine whether u belongs to H².

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compute the flux before differentiating it, and check whether the denominator of u″ can vanish on the closed interval.

Compare with the solution

One has u′=1/((1+x)log⁡2)u'=1/((1+x)\log2), hence a u′=1/log 2 is constant and its derivative is zero. The endpoint values are zero and one. Also u′′=−1/((1+x)2log⁡2)u''=-1/((1+x)^2\log2) is bounded on (0,1), so it is L² and u is H². Here a is smooth and bounded below by one; there is no interface jump to create a singular second derivative.

Take it with you

  • Difference quotients justify regularity without assuming the desired derivatives exist.
  • Uniform estimates need coefficient difference bounds and an interior support margin.
  • Weak flux continuity does not force gradient continuity across rough coefficients.

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