Multiscale finite element methods
Build coarse trial functions that satisfy local microscopic equations and identify the errors that remain.
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The question for today
What you’ll understand
- Define a conforming local multiscale basis with compatible boundary traces.
- Derive the exact one-dimensional local stiffness matrix from resistance.
- Distinguish approximation, resonance, local-solve, and quadrature errors.
A conventional coarse element cannot resolve microscopic oscillations in its polynomial trial functions. A multiscale finite element instead spends local computation on basis functions that already respond to the coefficient. The global problem then has relatively few degrees of freedom while its trial space contains microscopic structure. This is a numerical construction, not a consequence of homogenization by itself. Its accuracy depends on how local boundary data, representative sampling, and numerical integration interact with the coarse scale H and microscopic scale ε.
Local equations define the coarse trial space
Take a symmetric uniformly elliptic coefficient Aε on a shape-regular simplicial coarse mesh. For each element K and each local vertex, solve −div(Aε∇φiK)=0 inside K with the corresponding affine nodal hat function prescribed on its boundary. Adjacent elements use matching affine traces on their common face. Assembling the local functions therefore gives continuous H¹ trial functions, with boundary nodes treated consistently with the global Dirichlet condition. The sum of the local basis functions is one because its boundary value is one and the constant function solves the local equation uniquely.
Insert this conforming space into the original weak bilinear form, integrating the oscillatory coefficient against the multiscale basis gradients. Coercivity gives a well-posed discrete system; Galerkin orthogonality and Céa's lemma compare its energy error to the best approximation in this particular space. Those statements do not estimate that best approximation automatically. The local PDEs and the bilinear integrals must also be solved accurately. An exact mathematical basis does not justify using an unresolved point sample for a rapidly oscillating stiffness integral.
In one dimension the basis is an integrated resistance
Let K be an interval with endpoints left and right. A local harmonic function has constant flux, so the right-endpoint basis is the accumulated resistance from the left endpoint divided by the total interval resistance RK. Its complement gives the left-endpoint basis. Differentiating shows the basis gradient is inversely proportional to the local conductivity. This redistributes strain into the low-conductivity phase exactly as the cell corrector does, while matching the prescribed coarse nodal values.
The resulting two-by-two stiffness matrix is one divided by RK times the usual matrix with diagonal entries one and off-diagonal entries minus one. Its energy for a nodal jump c is c²/(2RK). In contrast, a conventional affine basis uses a uniform gradient and gives a stiffness involving the arithmetic conductivity integral divided by the square of the element length. The two coincide for constant coefficients but can differ substantially in layered media. Exact local integration therefore changes the response even though both spaces have the same nodal degrees of freedom.
Local accuracy does not eliminate scale interaction
When periodic resistance has a bounded mean-zero primitive, integrating over an interval of length H leaves an endpoint remainder of order ε. Thus RK equals H times mean resistance plus an order-ε phase-dependent remainder. The relative discrepancy can be of order ε/H. This elementary estimate explains why a local window containing few cells can misrepresent the bulk response. Exact alignment with whole periods can cancel that remainder in one dimension, but such cancellation is a special geometry rather than a general numerical guarantee.
In several dimensions, artificial local boundary conditions can create resonance errors between the coarse element and the oscillations. Oversampling computes local functions on enlarged patches to reduce sensitivity to those artificial boundaries, but the resulting space and conformity treatment must be specified. No universal rate follows merely from using that name. A complete error argument balances coarse approximation, microscopic boundary effects, local discretization, quadrature, and any oversampling parameters. Homogenization suggests effective behavior; numerical analysis must still show that the chosen algorithm actually captures it.
Work it through
Compute the local stiffness of a layered element
Let K=(0,1) contain conductivity one on its first half and four on its second. Compare an exact local multiscale basis with an ordinary affine basis.
The resistance is RK=(1/2)/1+(1/2)/4=5/8. The exact local right basis has derivative 1/(aRK), hence slopes 8/5 and 2/5 on the two halves.
Its stiffness is 1/RK=8/5, so its matrix is (8/5)[[1,−1],[−1,1]]. A nodal jump c has energy (4/5)c².
For the ordinary affine right basis, the derivative is one everywhere. Its stiffness integral is (1/2)·1+(1/2)·4=5/2, giving energy (5/4)c². Matching the same endpoints does not make the two internal trial responses equivalent.
The multiscale basis reproduces the exact one-dimensional series response 8/5; an unresolved affine basis instead produces the arithmetic response 5/2.
The living notebook
Explore the model
Interactive lab · function
Inspect the effective local nodal energy
Reset, set coefficient a=0.8 and position x=1; predict energy y=0.8 and derivative 1.6. For a fresh comparison Reset, set a=0.8 and x=2; predict energy 3.2 and derivative 3.2. Here x denotes the nodal jump, so doubling the jump quadruples its quadratic energy. The instrument evaluates the already derived local energy; it does not solve the microscopic basis equation or check quadrature accuracy.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 forThe energy derivative is the nodal force; effective stiffness is the second derivative with respect to the jump.
Pause & reason
Check your understanding
Your turn
Try it yourself
Suppose an element of length H=ε/4 lies entirely in the conductivity-one half of a periodic cell. Compute its total resistance and represented local conductivity. Compare with the infinite periodic effective value.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use RK=H/a for a constant coefficient on that particular window, then distinguish H/RK from the stiffness 1/RK.
Compare with the solution
Every point of the element has a=1, so RK=H=ε/4. Its stiffness is 1/H=4/ε, while its represented conductivity is H/RK=1. The infinite equal-half periodic medium has conductivity 8/5. The local solve is exact for its small window, but that window contains no sample of the conductivity-four phase and is not representative of the bulk. This error is not repaired by solving the same tiny window more accurately; the sampling geometry or approximation strategy must change.
Keep the idea
Take it with you
- Multiscale bases incorporate coefficient-dependent gradients through local PDE solves.
- One-dimensional local stiffness is the reciprocal of total resistance.
- Coarse approximation and local scale, boundary, and integration errors need separate control.
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