Conforming elements and boundary conditions
Build globally admissible functions and impose essential data through an affine trial set and correct load modification.
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The question for today
What you’ll understand
- Explain why matching face traces makes scalar piecewise polynomials H¹-conforming.
- Use a continuous and discrete lifting for nonzero Dirichlet data.
- Eliminate known boundary degrees of freedom while preserving the correct interior equations.
Local polynomial formulas become a conforming finite element space only after their traces match across neighboring cells. Boundary conditions then determine which globally conforming functions are admissible as trials and tests. For nonzero Dirichlet data the trial set is affine, not a zero-trace vector space. Treating those prescribed values as if they were ordinary unknowns, or zeroing matrix entries without adjusting the load, changes the mathematical problem even if the resulting linear solve succeeds.
Conformity is a trace-matching property
For a fitted conforming simplicial mesh, continuous piecewise affine functions lie in H¹ because their elementwise weak derivatives have no distributional jump contributions across interior faces. With scalar Lagrange elements of higher degree, matching the full polynomial face traces follows from matching the appropriate shared degrees of freedom. For piecewise smooth functions the essential condition is matching traces almost everywhere on each interior face, not merely checking agreement at one arbitrary point or relying on a visually continuous plot.
A zero-Dirichlet conforming subspace is obtained by imposing zero boundary trace. Other function spaces impose different continuity: H(div) elements match normal components and H(curl) elements match tangential components under their respective transformations. They need not be globally continuous as vector fields. Conformity always refers to a chosen continuous space. A collection of discontinuous cell polynomials is not automatically H¹-conforming just because it is finite-dimensional or has many local degrees of freedom.
Lift essential boundary data before applying zero-trace theory
For a bounded Lipschitz domain and admissible boundary data g in H to the one-half trace space, choose a lifting G in H¹ with trace g. Write u=G+w, where w has zero trace. The correction satisfies the homogeneous-space variational equation with load F(v)−a(G,v). This is a change of unknown, not a removal of the boundary data. The resulting right-hand side includes the influence of the lifting even when it does not look like an ordinary volume source.
A practical finite element construction uses a discrete lifting G_h whose trace equals the represented boundary data g_h. If that trace approximates rather than equals g, the discrete solution does not live in the exact affine trial set, and a lifting or boundary-data error must enter the analysis. Nodal boundary values require enough regularity to define them; rough trace data may need a suitable projection and stable lifting. One cannot apply homogeneous Céa directly to an error with a nonzero boundary trace and pretend no extra term exists.
Known coefficients change the interior right-hand side
Partition an assembled stiffness matrix into interior and prescribed boundary coefficients. The interior system is K_II u_I equal to f_I minus K_IB g_B. Eliminating the known variables in this way preserves the symmetric interior matrix when the variational form is symmetric. Replacing only boundary rows by identity while leaving their columns untouched can destroy symmetry, and zeroing those columns without the matching right-hand-side correction changes the interior equations. The boundary implementation must match the chosen algebraic convention.
Natural Neumann data enter the weak load through the boundary integral after integration by parts, rather than by prescribing every associated value degree of freedom. With only Neumann conditions, constants remain an energy kernel and a source-flux compatibility condition and normalization are needed. Mixed boundary problems require a sufficient essential boundary portion or another mechanism for coercivity. The name of the boundary condition therefore determines both the space and the equations; it is not a cosmetic flag added after assembly.
Work it through
A nonzero endpoint lifting on two cells
Solve −u″=2 on (0,1) with u(0)=2,u(1)=4 on nodes 0,1/2,1. Compute the interior value by elimination and by lifting.
The interior stiffness entry is four and its two boundary couplings are both negative two. The interior load is one, so the reduced equation is 4u_mid=1−(−2·2−2·4)=13.
Thus u_mid=13/4=3.25. The linear lifting G=2+2x has midpoint three. Its gradient is constant, and its variational coupling with the interior zero-trace hat is zero.
The homogeneous correction has midpoint one quarter, as in the earlier source-two problem. Adding it to the lifting gives 3.25. The exact solution 2+3x−x² has the same midpoint but differs from the piecewise linear function between nodes.
Both procedures represent the same boundary contribution. Omitting the K_IB g_B term would produce the wrong interior solution.
The living notebook
Explore the model
Interactive lab · function
Reconstruct a lifted solution from its pieces
Reset and set a=−1,x=0.5; predict the quadratic contribution −0.25. On paper add x=0.5 to obtain the homogeneous correction x−x²=0.25, then add the lifting 2+2x=3 to obtain 3.25. The instrument plots only a x², so the affine pieces are deliberate manual additions, not hidden model outputs or a PDE solve. Control changes retain time; Reset restores both settings and zero time.
- 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 forBoundary lifting is an explicit decomposition whose contributions must be retained in both the variational load and reconstructed solution.
Pause & reason
Check your understanding
Your turn
Try it yourself
For −u″=0 with u(0)=1,u(1)=0 on the same three nodes, use the reduced equation to find the midpoint coefficient. Is the finite element solution exact?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The interior volume load is zero, but the boundary coupling is not. The exact solution is affine.
Compare with the solution
The reduced equation is 4u_mid=0−(−2·1−2·0)=2, so u_mid=1/2. The piecewise linear function through values one, one half and zero is the global line 1−x, which is the exact solution. A zero source does not justify a zero interior solution when nonzero essential data drive the problem. Exactness here follows because the affine solution belongs to the conforming trial set.
Keep the idea
Take it with you
- Scalar H¹ conformity requires matching traces across shared faces.
- Nonzero Dirichlet data define an affine trial set and require a lifting or equivalent elimination.
- Boundary-data approximation, natural loads, and pure-Neumann kernels must be tracked explicitly.
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