Weak solutions and Sobolev spaces
Move derivatives onto test functions and distinguish a distributional derivative from an integrable weak derivative.
On this page
The question for today
What you’ll understand
- Define a weak derivative through integration against test functions.
- Check whether a nonsmooth function belongs to a Sobolev space.
- Formulate a zero-Dirichlet Poisson problem in H1_0.
A PDE may have useful solutions whose second derivatives do not exist as ordinary functions. Requiring classical differentiability can exclude corners, finite-energy initial data, and limits of approximations that still obey the intended balance law. Weak formulations retain the equation through integration against smooth probes. Sobolev spaces then specify which of the resulting derivatives are represented by integrable functions. This extra requirement distinguishes a usable energy class from the much larger universe of distributions.
An integration identity replaces a pointwise derivative
For a locally integrable function u, a locally integrable function v is its weak derivative in direction i when ∫u ∂_iφ=−∫vφ for every smooth compactly supported test function φ. Compact support removes boundary terms, so this definition initially describes interior differentiation. If a classical derivative is continuous, integration by parts shows it is also the weak derivative. The converse fails: absolute value has a weak derivative even though its classical derivative is undefined at zero. Weak derivatives represented by functions are unique up to equality almost everywhere.
Every locally integrable function has distributional derivatives, but those derivatives need not be locally integrable functions. A step function has a delta distribution as its derivative, which cannot be represented by an Lp function on an interval containing the jump. It therefore fails the first-order Sobolev condition there. By contrast, the first derivative of |x| is sign(x), an essentially bounded function. Changing the assigned value of sign at zero changes no integrals and no Sobolev class. The second distributional derivative of |x| is 2δ₀, so membership at one derivative order does not imply membership at two.
A norm records the regularity actually controlled
For p=2, W1,2 is written H1 and has norm squared equal to the L2 norm squared of the function plus that of its gradient. It is a Hilbert space, allowing orthogonality and bounded functional arguments. Convergence in H1 controls both quantities, while L2 convergence alone can hide increasingly rapid oscillation in derivatives. Sobolev embedding theorems provide stronger conclusions only with the dimension, derivative order, exponent, and domain hypotheses stated. In particular, an arbitrary H1 function in several dimensions need not have a continuous representative.
Boundary conditions need a comparable level of care. Define H1_0 as the closure of smooth compactly supported functions in the H1 norm. On a bounded Lipschitz domain, this agrees with H1 functions having zero trace. A trace is a continuous boundary restriction operator in the appropriate spaces, rather than a license to choose arbitrary point values for an equivalence class. Nonzero Dirichlet data can be handled by a suitable extension and subtraction. These definitions keep a boundary condition meaningful even when the solution is not continuous up to every boundary point.
The weak PDE asks for a balance against every test
For −Δu=f and homogeneous Dirichlet data, multiply by a zero-boundary test function and integrate by parts. The weak problem asks for u in H1_0 such that ∫∇u·∇v=∫fv for every v in H1_0, assuming for example f in L2 and a bounded Lipschitz domain. Only one derivative of u appears. This enlarges the admissible solution class without changing the sign of the original operator. A smooth weak solution recovers the classical equation in the interior by testing against compactly supported probes. Whether a particular weak solution is smooth requires a separate regularity theorem.
In the spectral lab, fix a modest decay p and increase mode count N. The added amplitudes may look small while their short wavelengths noticeably sharpen the profile. Each spatial derivative multiplies the nth sine coefficient by n, so visual amplitude alone misses the quantities entering an H1 norm. The lab displays a finite sum and every such sum is smooth; it neither creates a genuine jump nor proves an infinite series is in a Sobolev space. Use it to motivate the derivative-weighted estimate, then apply an actual convergence calculation when passing to infinitely many modes.
Work it through
A corner with finite first-derivative energy
On (−1,1), determine whether u(x)=|x| belongs to H1 and compute its H1 norm squared. Does it belong to H2?
Split the test integral at zero. Integration by parts on the two intervals yields . The terms at zero vanish because |x| is zero there.
Both u and sign(x) lie in L2(−1,1). Their squared norms are ∫x² dx=2/3 and ∫1 dx=2, so .
The derivative sign(x) jumps by two at zero. Its distributional derivative is 2δ₀, not an L2 function. Thus u is in H1 but not H2 on this interval.
The endpoint traces equal one rather than zero, so this u is not in H1_0(−1,1). Sobolev regularity and the homogeneous boundary subspace answer different questions.
The corner has finite H1 norm squared 8/3, but lacks an L2 second weak derivative and does not satisfy the zero-trace condition.
The living notebook
Explore the model
Interactive lab · spectral
Small amplitudes can hide large derivatives
Increase N in Σ sin(nx−t)/nᵖ and compare several values of decay p. Watch the short spatial scales supplied by higher modes. Every displayed finite sum is smooth; the lab does not generate a distribution or certify convergence of an infinite series in H1.
- Sum at x = π/2
- 0.867
- Smallest wavelength
- 2π / 5
- Highest-mode coefficient
- 0.2
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 forSobolev control measures derivatives as well as the visible size of a function.
Pause & reason
Check your understanding
Your turn
Try it yourself
For u(x)=x(1−x) on (0,1), verify the weak form of −u''=2 with zero Dirichlet data and calculate ∫₀¹|u′|² dx.
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 (1−2x)v′ by parts for v with zero endpoint trace.
Compare with the solution
The polynomial has zero endpoint trace and lies in H1_0. Since u'=1−2x, integration by parts gives ∫₀¹u′v′ dx=2∫₀¹v dx. The boundary term vanishes for zero-trace tests. Its gradient energy is . The identity extends from smooth tests to H1_0 tests by density and continuity.
Keep the idea
Take it with you
- Distributional differentiation is broader than having an Lp weak derivative.
- Sobolev spaces specify exactly which derivative norms remain finite.
- A weak equation and a regularity theorem make separate claims.
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