PDE Notes, Lecture #11

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1 PDE Notes, Lecture # from Professor Jalal Shatah s Lectures Febuary 9th, 2009 Sobolev Spaces Recall that for u L loc we can efine the weak erivative Du by Du, φ := udφ φ C0 If v L loc such that Du, φ = vφ then Du is a L loc function an Du = v. We can verify this by approximating u using convolution with an approximate ientity. Take η ɛ an approximate ientity, an consier D(u η ɛ ) = u(y)dη ɛ (x y)y = (Du) η ɛ = v(y)η ɛ (x y)y Here Du is a istribution an η ɛ C 0, so (Du) η ɛ is C. Observe that (Du) η ɛ = D(u η ɛ ) v as ɛ 0 Definition.. We now efine the Sobolev Spaces: W,p (R ) = {u : u, Du L p, u W,p = ( u p L p + Du p L p)/p < } W,2 = H W k,p (R ) = {u : u, Du,..., D k u L p, u W k,p = u L p D k u L p < } W k,2 = H k By the equivalence of finite-imensional norms, it oes not matter in the efinition above which norm we use to combine the L p norms of the function an its erivatives. (As an asie, H s,p = {u : (( + ξ 2 ) s/2 û ) L p }, an can efine W s,p for non-integer s as an interpolation space ) We can talk about the trace of the Sobolev spaces, like we i for the H s spaces. What follows is a baby version of the trace theorem. Proposition.2. Let u W,p an v = u x =0. Then v L p (R ) C u W,p (R ) Proof. Apply the Funamental Theorem of Calculus to v p. Write v p = 0 D u p (..., s)s

2 an take the L norm to get v L p (R ) = v p R D u p R p u p D u p u p q D u p p u p p D u p p u p W,p (R ) Above we have Höler s inequality, noting that u p L q since q = Taking p-th roots, we conclue that ( u p q = ) p u p p = u p p v L p (R ) C u W,p (R ) p p an Remark. Morally, we can o better. Analogously to the trace theorem for H s, in general on the bounary we shoul only lose /p-th of a erivative (recall for u H s, u x H s /2 ). But the proof is much more complicate. So why o we nee this? Consier the following problem: u = f u H u Ω = g (Dirichlet) Here we can talk about g on the bounary in H /2. However, if we replace the bounary conitions by Neumann bounary conitions: u Ω = g then this is problematic. There exists a solution but as state the problem is wrong. We nee u H 2 to even talk about the erivative on the bounary, since then u H an its trace woul be in H /2.. Sobolev Embeing Theorem (Potential Estimates) Consier the following problem u = ivf x R > How o we solve such a problem? We can take the Fourier transform: Now we can ivie by ξ 2 because ξ ˆf ξ 2 ξ 2 û = ivf = ξ ˆf is locally integrable. Then û = ξ 2 ivf 2

3 From homework, we know that (If = 2, then invert ξ ξ 2 ( ) ξ 2 = C ξ 2 ( > 2) instea an take weak erivatives) Thus u = C ivf x y 2 Applying integration by parts, we get u = C x y x y f(y)y Now by assumption, we have u = iv( u) so that f = u. Thus x y u(x) = C x y uy This implies that Now we use the fractional inequality: u(x) C u y x y u L r C u L p where + r = + p, r an p. This reuces to r = p where < p <. In papers, this r is enote p : p := p p < For p =, the inequality above is still vali, but we obtain this more irectly. Applying the Funamental Theorem on each coorinate, we get that u(x) = i= ( xi Taking absolute values an ( )-th roots, ) i u(x,..., s i,..., x )s i u(x) i= ( xi i u(s i ) s i Now we integrate with respect to each coorinate an apply Höler: u(x) x i= ( x ( x ( xi i u(x, s i ) s i u(s ) s u(s ) s i=2 ( i=2 x ( xi x i u(x, s i ) s i xi x i u(x, s i ) s i x 3

4 Here we note quickly that we have applie x f f x f L p f L p where p p = (Höler s inequality) using p =... = p =. Continuing the process with the other coorinates results in ( u(x) x i= i u x An noting i u u we have ( u(x) ) u x Finally taking both sies to the power yiels u C u L L In summary, we now have Theorem.3 (Sobolev Embeing). u L p C u L p p < Note that p = is not allowe, using the following stanar counterexample. On the space H (R 2 ) consier the case p = = 2 with where φ(r) is a cutoff function. Then u r 2 u = ln(ln r)φ(r) = r ln r 2 = r 2 (ln r) 2 an supposing the cutoff function has support in the ball of raius /2, B(/2) u 2 x /2 0 [ r 2 (ln r) 2 rr = ] /2 = ln 2 ln r 0 Thus we note that u L 2 = ln 2 <, but u L = since u blows up near 0, so the inequality oes not hol in this case..2 Boune Domains Now we consier boune omains Ω. Assume the bounary is nice, say C, or even Lipschitz. The notion of weak erivatives an istributions remains the same, where we replace C0 with C0 (Ω). Definition.4. ( /p W,p (Ω) = {u : u W,p(Ω) = u p + Du p < } Ω 4

5 The other spaces are efine in the same way (with Ω). The trace theorem still hols, but in a slightly ifferent form. Consier u efine on Ω, v = u Ω. In orer to eal with the bounary, we consier a partition of unity p α =, which allows us to focus inepenently on ifferent pieces of the omain, each piece corresponing to up α. Note up α = u. Using a change of variable, we can map each piece so that the bounary is mappe into {x = 0} an the part in Ω is mappe to {x > 0}. If ũ α = (p α u) Y α where Y α is the change of variable, then we can express the values on the bounary using the Funamental Theorem of Calculus: ũ α (y,..., y, 0) = 0 ũ y y Here is where we use the assumption that the bounary is nice, so that Y α is sufficiently nice. (Extension) Now our function ũ α which is efine in {y 0} can be extene by reflection across {y = 0}, while increasing the norm by a factor of two. This means that for our function u W,p (Ω), we can exten it to some function ũ W,p ( Ω) where Ω Ω such that u = ũ on Ω an ũ W,p ( Ω) C(, Ω) u W,p (Ω) Now consier the cutoff function η such that η C0 where x Ω η(x) = 0 x Ω ecays smoothly towars Ω Then letting v = ηũ, we have that. v = u on Ω 2. v = 0 on Ω 3. v W,p) ( Ω C(, Ω) u W,p (Ω) (Note that ũ i not satisfy the secon conition) Thus given any u W,p (Ω) we can fin an extension that vanishes outsie a compact set. This allows us to express the problem in terms of the original problem on the unboune omain. Definition.5. W,p 0 (Ω) = {u : u W,p (Ω), u Ω = 0} Again, the other spaces are efine analogously. So for u W,p (Ω), we can fin Ω, ũ W,p 0 ( Ω) such that ũ = u on Ω an ũ W,p 0 ( Ω) C u W,p (Ω) Remark. If u W,p 0 (Ω), can efine v W,p (R ) by { u x Ω v = 0 x Ω Since u vanishes on Ω, the weak erivative of v will exist across the bounary. (Note we coul not o this straight from W,p (Ω), since the function i not vanish on the bounary) 5

6 Note that since we can now exten functions on boune omains to finitely-supporte functions on the whole omain, Sobolev Embeing inequality now works for W,p 0 (Ω) an W,p (Ω), except that W,p (Ω) nees the whole W,p norm (as oppose to just L p ): u L p C ũ L p ( Ω) C ũ L p ( Ω) C u W,p (Ω) Here we use the thir property of the extension ũ an ũ L p ( Ω) ũ W,p ( Ω). We cannot use just the L p norm on the RHS, because consier the constant function u =. The LHS woul then be nonzero, but u L p = 0, so the inequality woul not hol. 6