advanced32 min studyLesson 392 in Mathematics

Lax–Milgram and weak Dirichlet problems

Construct a weak solution from a bounded coercive bilinear form without assuming classical second derivatives.

On this page

What you’ll understand

  • Translate a divergence-form Dirichlet equation into a Hilbert-space identity.
  • Verify the exact boundedness and coercivity assumptions for Lax–Milgram.
  • Separate weak existence from classical regularity and nonhomogeneous boundary lifting.

A source need not be an ordinary continuous function, and a material coefficient may jump at an interface. Demanding a twice continuously differentiable solution before asking whether a boundary problem makes sense would discard useful models. A weak formulation moves one derivative onto test functions and seeks a finite-energy solution. Lax–Milgram turns that formulation into an existence and uniqueness theorem when its bilinear form is bounded and coercive. The theorem supplies a solution in the chosen Hilbert space; it does not secretly promise additional derivatives.

Choose the space before integrating by parts

Let Omega be a bounded Lipschitz domain, let A be bounded measurable, real symmetric, and uniformly elliptic, and let c be a bounded nonnegative function. For homogeneous Dirichlet data choose H-zero-one, the closure of compactly supported smooth functions in H-one. Poincaré's inequality makes the gradient norm a complete equivalent norm on this space. Take the source F to be a continuous linear functional on that gradient-normed Hilbert space, commonly denoted an H-minus-one datum.

For smooth functions with zero boundary test data, integration by parts changes the equation minus divergence of A gradient u plus c u equals F into a bilinear identity. That identity remains meaningful for bounded measurable A and H-one functions because products of square-integrable gradients are integrable. Its extension to every H-zero-one test function is the definition of the weak solution. No normal flux trace or pointwise second derivative is required to write this particular homogeneous Dirichlet formulation.

a(u,v)=∫ΩA∇u⋅∇v+cuv dx=F(v)for every v∈H01(Ω)a(u,v)=\int_\Omega A\nabla u\cdot\nabla v+cuv\,dx=F(v)\qquad\text{for every }v\in H_0^1(\Omega)
The source acts as a dual functional. If F is represented by an L² function f, then F(v)=∫fv; the weak definition also admits more general continuous functionals.↔ Scroll wide equations horizontally.

Verify the two inequalities separately

Cauchy–Schwarz, the coefficient bound, and Poincaré show that the form is bounded by a constant times the product of the two gradient norms. Uniform ellipticity and c nonnegative show coercivity with constant lambda in that same norm. Lax–Milgram states that any bounded coercive bilinear form on a real Hilbert space represents every bounded linear functional by one unique solution. Symmetry is not required by the abstract theorem, although it holds in the simple PDE family used here.

Testing the solution identity with u gives an immediate stability estimate: its gradient norm is at most the dual source norm divided by lambda. This argument also proves uniqueness by applying it to the difference of two solutions with the same source. Keep the norm convention explicit, since using the full H-one norm changes the coercivity constant through Poincaré. Existence is not a consequence of integration by parts alone; completeness, boundedness, and coercivity are doing the functional-analytic work.

∣a(u,v)∣≤M∥u∥H∥v∥H,a(v,v)≥α∥v∥H2⟹∥u∥H≤α−1∥F∥H∗|a(u,v)|\leq M\|u\|_H\|v\|_H,\quad a(v,v)\geq\alpha\|v\|_H^2\quad\Longrightarrow\quad\|u\|_H\leq\alpha^{-1}\|F\|_{H^*}
Lax–Milgram gives existence and uniqueness under both inequalities, for α>0. Here H=H₀¹ with gradient norm, and α can be the ellipticity lower bound λ.↔ Scroll wide equations horizontally.

Boundary lifting and regularity are further tasks

For nonzero Dirichlet data, first require an H-one extension G with the prescribed boundary trace. Write u=w+G, with w in H-zero-one. The new functional is F(v)−a(G,v), which is continuous, so the homogeneous theorem applies to w. A boundary function stated only pointwise without the required trace-space membership need not admit such a lift. The Lipschitz-domain assumption supports the usual trace framework and makes this distinction precise.

The resulting weak solution might have no classical second derivatives, particularly when coefficients jump or the boundary has corners. Interior regularity, boundary regularity, and classical solvability require stronger assumptions and later arguments. In the symmetric case the same weak solution uniquely minimizes the strictly convex energy one half a(v,v) minus F(v) over the homogeneous space. That variational interpretation is useful, but should not be imposed on a general nonsymmetric bilinear form merely because Lax–Milgram still applies.

Verify a weak solution and its energy

Solve −2u″=6 on (0,1) with u(0)=u(1)=0. Verify the weak identity and compute a(u,u).

  1. Integrating twice with the endpoint conditions gives u(x)=32x(1−x)u(x)=\tfrac32x(1-x). Its derivative is 3/2−3x, so it lies in H₀¹ and its second derivative is −3. Therefore it also happens to be a classical solution in this smooth example.

  2. For any smooth compactly supported test v, integration by parts gives ∫012u′v′ dx=∫016v dx\int_0^1 2u'v'\,dx=\int_0^1 6v\,dx. Density and continuity extend the identity to every H₀¹ test function, which is the required weak statement.

  3. The squared gradient integral is ∫01(3/2−3x)2dx=3/4\int_0^1(3/2-3x)^2dx=3/4, hence a(u,u)=3/2. On the source side, 6∫01u dx=6(1/4)=3/26\int_0^1u\,dx=6(1/4)=3/2, agreeing with the energy identity. The midpoint value is 3/8.

The result

The weak and classical answers agree here, while the existence theorem only needed bounded coefficients, a dual source, and the coercive function-space structure.

Explore the model

Interactive lab

Interactive lab · function

See the quadratic energy's strict positivity

Pause, Reset, then set a=1 and x=1. Repeat from Reset with both a=2 and x=1. Predict energies one and two and slopes two and four. The graph represents a single scalar coefficient in a positive quadratic energy, not the spatial solution profile of the boundary problem or a proof of infinite-dimensional coercivity.

See the quadratic energy's strict positivityThis 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. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

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 forA positive quadratic form controls amplitudes; the PDE theorem needs that control uniformly over every function in its Hilbert space.

Check your understanding

Which conclusion follows directly from Lax–Milgram for a bounded coercive form on H₀¹ and an H⁻¹ source?

Try it yourself

Solve −3u″=12 on (0,1) with zero endpoint values, then verify the energy identity with the test v=u.

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

Give me a hint

Integrate the constant source twice and use both endpoints before evaluating the two integrals.

Compare with the solution

The solution is u=2x(1−x), with derivative 2−4x and second derivative −4. Its energy is ∫013(2−4x)2dx=4\int_0^1 3(2-4x)^2dx=4. The source pairing is 12∫012x(1−x)dx=12/3=412\int_0^1 2x(1-x)dx=12/3=4. These equal values verify the identity at v=u; the full weak identity follows by integration by parts and density, and coercivity gives uniqueness.

Take it with you

  • Weak Dirichlet problems are defined by an identity against all H₀¹ tests.
  • Lax–Milgram requires a complete Hilbert space, boundedness, and positive coercivity.
  • Trace lifting and classical regularity require hypotheses beyond weak existence.

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