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Sobolev spaces and embeddings

Functions with weak derivatives in L^p, when they are bounded or continuous, and the scaling that predicts the exponents.

The Sobolev space \( W^{k,p}(\Omega) \) contains the functions in \( L^p(\Omega) \) whose weak derivatives up to order \( k \) are also in \( L^p \), with norm \( \|u\|_{W^{k,p}} = \big(\sum_{|\alpha|\le k}\|\partial^\alpha u\|_{L^p}^p\big)^{1/p} \). The case \( p = 2 \) is written \( H^k \); it is a Hilbert space. These are the spaces in which energy methods naturally live: an energy estimate bounds \( \int|\nabla u|^2 \), which is an \( H^1 \) norm, and Sobolev spaces are complete, so limits of approximate solutions stay inside.

Sobolev functions can be unbounded. On the unit ball in \( \mathbb R^3 \), let \( u = |x|^{-a} \) with \( a > 0 \). Then \( |\nabla u| = a|x|^{-a-1} \), and in spherical coordinates \( \int_B|\nabla u|^2 = 4\pi a^2\int_0^1r^{-2a}\,dr \), finite exactly when \( a < 1/2 \). The example evaluates the case \( a = 1/4 \): the integral is \( \pi/2 \), so \( |x|^{-1/4} \in H^1(B) \) although it is infinite at the origin. In \( \mathbb R^3 \), \( H^1 \) functions need not be bounded.

Embeddings say how much integrability or continuity derivatives buy. For \( 1 \le p < n \), the Gagliardo-Nirenberg-Sobolev inequality \( \|u\|_{L^{p^*}(\mathbb R^n)} \le C\|\nabla u\|_{L^p(\mathbb R^n)} \) holds with \( \frac1{p^*} = \frac1p - \frac1n \); in three dimensions \( H^1 \subset L^6 \). For \( p > n \), Morrey's inequality makes \( W^{1,p} \) functions Hölder continuous. And \( H^s(\mathbb R^n) \subset C^0 \) when \( s > n/2 \).

Picture it: the exponent \( p^* \) comes from scaling. Set \( u_\lambda(x) = u(\lambda x) \). Changing variables, \( \|u_\lambda\|_{L^q} = \lambda^{-n/q}\|u\|_{L^q} \) and \( \|\nabla u_\lambda\|_{L^p} = \lambda^{1 - n/p}\|\nabla u\|_{L^p} \). If \( \|u\|_{L^q} \le C\|\nabla u\|_{L^p} \) held with \( -n/q \ne 1 - n/p \), letting \( \lambda \to 0 \) or \( \lambda \to \infty \) would make one side arbitrarily small compared with the other. The only possible exponent is \( -n/q = 1 - n/p \), which is \( q = p^* \).

Think it: every inequality between norms on \( \mathbb R^n \) must balance under scaling, so exponents can be predicted before any proof. For Navier-Stokes in three dimensions the energy inequality puts \( u \) in \( L^\infty_tL^2_x \) and \( \nabla u \) in \( L^2_tL^2_x \); Sobolev gives \( u \in L^2_tL^6_x \). Whether that is enough to control the solution is decided by the scaling count in the last lesson, and the answer is: not quite.

Pracovný príklad · integrate 4*pi*r^2*(1/16)*r^(-5/2) dr from 0 to 1

Integrate pi/(4·sqrt(r)) from 0 to 1

\int_{0}^{1} \frac{\pi}{4 \sqrt{r}}\, dr

Krok za krokom

  1. \int_{0}^{1} \frac{\pi}{4 \sqrt{r}}\, dr

    First find an antiderivative F, then evaluate F(b) − F(a).

  2. \int \frac{\pi}{4 \sqrt{r}}\, dr = \frac{\pi}{4} \int \frac{1}{\sqrt{r}}\, dr

    Pull the constant \frac{\pi}{4} out of the integral.

  3. \int \frac{1}{\sqrt{r}}\, dr = 2 \sqrt{r}

    Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) (n ≠ −1).

  4. F(1) - F(0) = \left(\frac{\pi}{2}\right) - \left(0\right)

    Fundamental theorem of calculus: plug in the limits.

  5. = \frac{\pi}{2} \approx 1.5708

    Simplify.

Odhaliť odpoveď
\frac{\pi}{2}

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Ako nakupovať: Sobolev spaces and embeddings

  1. Decide which derivatives and which L^p norm the question is about.
  2. For power-type singularities |x|^{-a}, compute the integral in polar coordinates: the measure contributes r^{n-1}.
  3. Find the range of a for which each integral converges.
  4. For embeddings, balance the scaling exponents: -n/q = k - n/p.
  5. Check the side condition (p < n for Sobolev, p > n for Morrey, s > n/2 for continuity).

Otázky, ktoré ľudia kladú

Why not just work with continuously differentiable functions?

Because C¹ with an integral norm is not complete: a sequence of smooth approximate solutions can converge to something outside it. Sobolev spaces are the completions, so limits exist in them.

Is H¹ in one dimension made of continuous functions?

Yes: s = 1 > n/2 = 1/2. In one dimension an H¹ function is Hölder continuous with exponent 1/2, by Cauchy-Schwarz applied to u(y) - u(x) = ∫ u'.

What should I know before starting?

Partial derivatives, the divergence theorem and multiple integrals from multivariable calculus; linear second-order ODEs; eigenvalues from linear algebra; and enough analysis to be comfortable with uniform convergence and integrals over infinite intervals.

Why are there so few formulas for solutions?

Explicit solutions exist for linear equations with constant coefficients on simple domains. Almost everything else (curved domains, variable coefficients, nonlinear terms) has none, so the modern subject proves that a solution exists and estimates its size and smoothness without ever writing it down.

How does this course lead to the Navier-Stokes problem?

The Navier-Stokes equations are a heat equation for the velocity with a transport term and a pressure that is found from a Poisson equation. Reading the Millennium problem needs energy estimates, weak solutions, Sobolev spaces, local existence with a blow-up criterion, and the scaling argument that explains why three dimensions is hard. The last six lessons build those.

What do elliptic, parabolic and hyperbolic mean?

Elliptic equations (Laplace) describe equilibrium and smooth everything; parabolic equations (heat) describe diffusion forward in time; hyperbolic equations (waves) carry signals at finite speed and keep their sharp edges. The type decides which data make a sensible problem.

Viac v kategórii Partial Differential Equations