advanced31 min studyLesson 402 in Mathematics

Weak solutions and Galerkin approximation

Construct an energy-class solution without assuming the differentiability one hopes to prove.

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What you’ll understand

  • State a weak parabolic problem in space and time function spaces.
  • Derive a finite-dimensional Galerkin system and its uniform energy bound.
  • Distinguish weak convergence, strong initial traces, and later regularity conclusions.

An energy estimate can control a solution even when the initial data or coefficients do not permit classical second derivatives. To use that control for existence, start with finite-dimensional problems where the calculations are legitimate, obtain bounds independent of dimension, and pass to a suitable limit. Galerkin approximation implements this strategy. The argument succeeds only when the function spaces retain enough information to identify both the equation and the initial condition.

The equation makes sense through a dual pairing

Take a bounded Lipschitz domain Ω, a finite time interval, and homogeneous Dirichlet boundary data. Consider uₜ−div(A(x,t)∇u)=f, where A is a bounded measurable real symmetric matrix field satisfying λ|ξ|²≤ξᵀAξ≤Λ|ξ|² for fixed positive λ and Λ. For u₀ in L² and f in L² in time with values in H⁻¹, the natural solution belongs to L² in time with values in H₀¹, and its time derivative belongs to L² with values in H⁻¹. No pointwise second spatial derivative is required by this formulation.

Testing against v in H₀¹ gives a dual pairing for uₜ and an ordinary integral of A∇u dotted with ∇v. The boundary condition is encoded by the zero Sobolev trace, not by assigning pointwise boundary values to an arbitrary equivalence class. The solution has a representative continuous in time with values in L², allowing u(0)=u₀ to be imposed strongly in that norm. Spatial almost-everywhere representatives alone would not make an isolated time slice meaningful; the time regularity is doing essential work.

⟨ut,v⟩H−1,H01+∫ΩA∇u⋅∇v dx=⟨f,v⟩,u∈L2(0,T;H01),ut∈L2(0,T;H−1)\langle u_t,v\rangle_{H^{-1},H_0^1}+\int_\Omega A\nabla u\cdot\nabla v\,dx=\langle f,v\rangle,\qquad u\in L^2(0,T;H_0^1),\quad u_t\in L^2(0,T;H^{-1})
Assume Ω bounded Lipschitz, bounded measurable symmetric uniformly elliptic A, f∈L²(0,T;H⁻¹), and initial datum u₀∈L². The identity holds for almost every time and every spatial test v.↔ Scroll wide equations horizontally.

Finite-dimensional testing produces a uniform bound

Choose finite-dimensional spaces spanned by Dirichlet Laplacian eigenfunctions, orthonormal in L² and orthogonal in the gradient inner product. Set uN to a finite combination with time-dependent coefficients and require the weak equation against every basis function. This gives a linear finite-dimensional system of ordinary differential equations with measurable bounded matrix coefficients and integrable forcing. Its initial coefficients are the L² projection of u₀. Using uN itself as test produces the energy estimate because all coefficients are sufficiently regular in time for that calculation.

Uniform ellipticity controls the gradient term from below. The forcing is bounded by its H⁻¹ norm times the gradient norm, using the gradient norm on H₀¹ and its dual norm. Young's inequality absorbs half of the gradient term into the left side. Integrating in time yields bounds on the L∞-in-time L² norm and L²-in-time gradient norm independent of N. The equation also controls the time derivative in the corresponding dual space when the projection is chosen with the needed H₀¹ boundedness, as this eigenfunction construction provides.

12∥uN(t)∥22+λ2∫0t∥∇uN∥22 ds≤12∥uN(0)∥22+12λ∫0t∥f∥H−12 ds\frac12\lVert u_N(t)\rVert_2^2+\frac{\lambda}{2}\int_0^t\lVert\nabla u_N\rVert_2^2\,ds\le\frac12\lVert u_N(0)\rVert_2^2+\frac{1}{2\lambda}\int_0^t\lVert f\rVert_{H^{-1}}^2\,ds
The H⁻¹ norm here is dual to the gradient norm on H₀¹. The inequality follows from coercivity and Young's inequality, and its constant does not grow with the Galerkin dimension.↔ Scroll wide equations horizontally.

Passing to a limit is an existence argument, not instant classical smoothness

Weak compactness supplies a subsequence converging in the energy spaces. Because the PDE is linear and A is fixed bounded data, the weak limits pass through the bilinear form when tested appropriately. Time integration by parts identifies the initial trace. Energy estimates then give uniqueness, so different convergent subsequences cannot describe different weak solutions. Starting with smoother initial approximations and using the same stability estimate extends the construction to L² initial data. The resulting existence statement is at the weak-solution level.

Nonlinear terms require additional care: weak convergence of uN does not generally imply convergence of a nonlinear function of uN. Compactness in a stronger topology or a suitable monotonicity argument may be needed. Likewise, weak existence alone does not justify evaluating a maximum at a point or differentiating the PDE repeatedly. Additional data, coefficient, and boundary regularity support later upgrades. Keeping these proof stages separate prevents a formal integration by parts from silently assuming the very solution regularity the construction is meant to establish.

Two independent diffusion coefficients in a Galerkin basis

On (0,π) with zero Dirichlet boundary, solve uₜ=(1/2)uₓₓ from u₀=sin x+2sin3x. Find the squared L² error if only sin x is retained.

  1. The sine eigenfunctions have Laplacian eigenvalues −n². Their coefficients satisfy cₙ′=−(n²/2)cₙ, so c₁=exp(−t/2) and c₃=2exp(−9t/2).

  2. The exact solution is exp(−t/2)sin x+2exp(−9t/2)sin3x. Orthogonality gives squared norm (π/2)(e⁻ᵗ+4e⁻⁹ᵗ).

  3. The one-mode approximation omits only the third mode. Its squared error is (π/2)·4e⁻⁹ᵗ=2πe⁻⁹ᵗ, decaying faster than the retained first-mode energy.

The result

The initial projection error is real but rapidly damped here. A Galerkin approximation need not equal the full solution at finite dimension to converge as its space grows.

Explore the model

Interactive lab

Interactive lab · heat

Inspect an omitted Galerkin mode

Reset, set n=3 and κ=0.5, then Step once to t=0.5. Predict amplitude exp(−2.25) and unnormalized periodic L² norm √π exp(−2.25). Compare the readouts. The worked example's omitted mode has coefficient 2 and interval (0,π), so its norm must be rescaled accordingly; this instrument uses a unit-amplitude sine on a periodic interval of length 2π.

Inspect an omitted Galerkin modeThe exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Current readings: Current amplitude: 1; Spatial L² norm: 1.772; Decay rate · κn²: 3.6 /s. All parameters are adjustable with the labeled controls below.Periodic heat equation mode 3 at time 0: the amplitude is 1.0π/2π3π/22π−1−0.500.51x · periodic intervalTemperature perturbation uExact mode · n = 3
Current temperature modeInitial sineCurrent amplitude
Current amplitude
1
Spatial L² norm
1.772
Decay rate · κn²
3.6 /s
t = 0 s

The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. 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 forMode decay is exact, while coefficient normalization and domain length still matter to an error norm.

Check your understanding

Why can an energy-class weak solution have a meaningful initial condition in L²?

Try it yourself

On (0,π), solve uₜ=uₓₓ+sin2x with zero Dirichlet boundary and zero initial data using a sine Galerkin ansatz. What does an approximation retaining only sin x predict?

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 forcing is orthogonal to sin x but directly excites the second sine mode.

Compare with the solution

The only nonzero coefficient solves c₂′+4c₂=1 with c₂(0)=0, giving u=(1−e⁻⁴ᵗ)sin2x/4. A space containing only sin x projects the forcing to zero and predicts uN=0. That is a limitation of the selected subspace, not nonuniqueness of the PDE. Adding the second mode recovers this exact solution, and the boundary and initial conditions hold in their stated senses.

Take it with you

  • Weak formulations trade pointwise second derivatives for energy and dual-space control.
  • Dimension-independent estimates enable a Galerkin existence argument.
  • Passing nonlinear limits and proving classical regularity require additional arguments.

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