Week 9: Existence and interior regularity
Existence of weak solutions (continued)
We can actually obtain more information on the values of μ for which the boundary-value problem of finding u∈H01,2(Ω) such that Lμu=f admits a unique solution by making use of an important property of compact linear mappings. We recall the following spectral theorem.
Theorem 9.1 (spectral theorem for compact operators). Suppose K:X→X is a compact operator on a Banach space X and let
σ(K)=C\{ξ∈C:K−ξI is bijective}denote the spectrum of K. We have the following properties:
- σ(K) is compact.
- If ξ∈σ(K)\{0}, then ξ is an eigenvalue of K, i.e. there exists a v∈X\{0} such that Kv=ξv.
- σ(K) is at most countable, and its only accumulation point, if it has one, is 0.
Remark 9.2. Note that if λ∈C is an eigenvalue of K, then it is automatically in σ(K).
Theorem 9.3 (third existence theorem). There exists an at most countable set Σ⊂R such that whenever λ∈/Σ, there exists a unique weak solution u∈H01,2(Ω) to the boundary-value problem
Luu∣∂Ω=λu+f on Ω≡0.(###)If Σ is infinite, then Σ={λk}k=1∞ for a nondecreasing sequence λk→∞.
Proof. Let μ>max{C,0} be as in the first existence theorem. By that theorem, the boundary value problem (###) possesses a unique solution if −λ≥μ, i.e. if λ+μ≤0, so suppose λ+μ>0. By the Fredholm alternative, this boundary-value problem has a unique weak solution if and only if λ is not an eigenvalue of L, i.e.
Lv=λv⇒v=0for v∈H01,2(Ω). Fixing μ>0 satisfying the condition of the first existence theorem, this in turn is equivalent to the condition that
Lμv=(λ+μ)v⇒v=0,or Kv=μ+λμv⇒v=0, i.e. μ+λμ is not an eigenvalue of K. Since K is compact, the spectral theorem tells us that the eigenvalues of K form an at most countable set {νk}k=1∞ with 0 as the only possible cluster point; therefore, λ is not an eigenvalue of L iff μ+λμ∈/{νk}⇔λ∈/{νkμ−μ}. The sequence {νkμ−μ} tends to ∞ if it possesses a cluster point so that we may reörder it to obtain a nondecreasing sequence {λk} with λk→∞.
Remark 9.4. The set Σ of all (real) eigenvalues λ of L is called the spectrum of L.
Remark 9.5. The equation in (###) is sometimes referred to as Helmholtz's equation.
We may also establish the boundedness of the inverse of L+λI (λ∈/Σ) as a mapping into L2(Ω) (cf. remarks following first existence theorem).
Theorem 9.6. Let λ∈/Σ. There exists a constant C depending on λ, Ω and L such that whenever f∈L2(Ω) and u solves (###),
∣∣u∣∣2≤C∣∣f∣∣2.
Proof. Suppose that this is not the case so that there exist sequences {uk}k=1∞⊂H01,2(Ω) and {fk}k=1∞⊂L2(Ω) in L2(Ω) such that Luk=λuk+fk and ∣∣uk∣∣2≥k∣∣f∣∣2 for all k∈N. By dividing through by ∣∣uk∣∣2, we may assume that ∣∣uk∣∣2=1 and ∣∣fk∣∣2≤k1 for all k so that fk→0. Now, by the energy estimates and Hölder's inequality, we actually have that
∣∣uk∣∣1,22≤C(Ω,λ,L)(BL−λI(uk,uk)+∣∣uk∣∣2)≤C(Ω,λ,L)(∣∣fk∣∣22+2∣∣uk∣∣22)so that {uk} is bounded in H01,2(Ω). By the Banach-Alaoǧlu theorem and Rellich-Kondrachov, we may pass to a subsequence, again denoted {uk}, such that uk→u in L2(Ω) for some u∈H01,2(Ω), and uk⇁u in H01,2(Ω). In particular, ∣∣u∣∣2=limk→∞∣∣uk∣∣2=1 so that u=0 and for all v∈H01,2(Ω),
BL−λI(u,v)=k→∞limBL−λI(uk,v)=k→∞lim⟨fk,v⟩=0so that Lu=λu on Ω. But this implies that λ∈Σ!
Interior regularity
We now turn our attention to the regularity of solutions to the equation Lu=f under the assumptions stated in Week 8. We first show that if u∈W1,2(Ω) solves this equation and aij∈C1(Ω), then u is actually twice weakly differentiable.
Theorem 9.7. Suppose u∈W1,2(Ω) solves the equation
Lu=f on Ωweakly, i.e. for all v∈H01,2(Ω)
BL(u,v)=∫Ωf⋅vand L satisfies the hypotheses given in Week 8. Suppose furthermore that for all i,j∈{1,…,n}, aij∈C1(Ω). Then u∈Wloc2,2(Ω) and whenever U⋐Ω, the inequality
∣∣u∣∣2,2,U≤C(∣∣f∣∣2,Ω+∣∣u∣∣2,Ω)holds, where C>0 is a constant depending only on U, Ω and L.
Proof. First note that we may rewrite the weak form of Lu=f as
∫Ωaij∂iu⋅∂jv=∫Ωf~⋅v,(*)where f~:=f−bi∂iu−cu∈L2(Ω). We now choose an appropriate v and exploit ellipticity to 'differentiate' the equation Lu=f. Fix open sets U⋐V⋐Ω and a χ∈C∞(Rn) such that 0≤χ≤1, χ∣U≡1 and supp χ⊂V. For h=0 sufficiently small and k∈{1,…,n} fixed, let
v=−δk−h(χ2⋅δkhu).We now estimate both sides of (*) appropriately, making free use of the properties of the 'difference quotient operator' δkh (product rule and integration by parts specifically). First note that
LHS=−∫aij∂iu⋅δk−h(∂j(χ2⋅δkhu))=∫δkh(aij⋅∂iu)⋅∂j(χ2⋅δkhu)=∫aij⋅∂iδkhu⋅∂jδkhu⋅χ2+2∫δkhaij⋅∂iu⋅δkhu⋅χ⋅∂jχ+2∫aij⋅∂iδkhu⋅χ⋅∂jχ⋅δkhu+∫δkhaij⋅∂iu⋅∂jδkhu⋅χ2.By the ellipticity condition on aij (i.e. aijvivj≥λ0∣v∣2), we may estimate the first term from below:
∫aij⋅∂iδkhu⋅∂jδkhu⋅χ2≥λ0∫∣Dδkhu∣2χ2.Since aij∈C1(Ω) and all of these integrands are supported on the compact set supp χ⊂V, we may find a c0>1 such that
(∣(aij)∣+∣(δkhaij)∣+∣Dχ∣)≤c0on this set so that we may bound the sum of the last two terms from below, using Hölder's inequality, the Peter-Paul inequality with ε=6c02λ0, and (the proof of) Lemma 6.3 to control the L2 norm of the difference quotient of u:
2∫aij⋅∂iδkhu⋅χ⋅∂jχ⋅δkhu+∫δkhaij⋅∂iu⋅∂jδkhu⋅χ2≥−2c02∫∣Dδkhu∣⋅∣δkhu∣⋅χ−c0∫∣Du∣⋅∣Dδkhu∣⋅χ≥−3c02(∫∣Dδkhu∣2χ2)1/2(∫V∣Du∣2)1/2≥−2λ0∫∣Dδkhu∣2χ2−2λ09c04∫V∣Du∣2.Finally, we take care of the second integral:
2∫δkhaij⋅∂iu⋅δkhu⋅χ⋅∂jχ≥−2c02∫∣Du∣⋅∣δkhu∣⋅χ≥−2c02∫V∣Du∣2.Altogether, LHS≥2λ0∫∣Dδkhu∣2χ2−c02(2+2λ09c02)∫V∣Du∣2. We now bound the right-hand side of (*) from above, again using Hölder's inequality and Lemma 6.3:
RHS=−∫f~⋅δk−h(χ2⋅δkhu)≤∫(∣f∣+β∣Du∣+γ∣u∣)⋅∣δk−h(χ2δkhu)∣≤(∣∣f∣∣2,V+β∣∣Du∣∣2,V+γ∣∣u∣∣2,V)⋅∣∣D(χ2δkhu)∣∣2,V.By the product rule, D(χ2δkhu)=χ2⋅Dδkhu+2χ⋅Dχ⋅δkhu so that by the triangle inequality,
∣∣D(χ2δkhu)∣∣2,V≤(∫∣Dδkhu∣2χ2)1/2+2c0∣∣Du∣∣2,V;Using Peter-Paul with ε and using the elementary inequality (∑j=1Nαi)r≤C(N,r)∑j=1Nαir, we arrive at
RHS≤4εC0⋅(∣∣f∣∣2,V2+(β2+γ2)∣∣u∣∣1,2,V2)+C0ε∫∣Dδkhu∣2χ2+4c02C0ε∣∣Du∣∣2,V2.Choosing ε=4C0λ0, we deduce that the inequality LHS≤RHS implies that
4λ0∫∣Dδkhu∣2χ2≤C1⋅(∣∣f∣∣2,V2+∣∣u∣∣1,2,V2),for some constant C1>0 depending only on U, V, Ω and L, and ∫∣Dδkhu∣2χ2≥∫U∣Dδkhu∣2. Thus,
∣∣δkhu∣∣1,2,U2=∫U∣δkhu∣2+∫U∣Dδkhu∣2≤λ04C1∣∣f∣∣2,V2+(1+λ04C1)∣∣u∣∣1,2,V2,which implies a bound on ∣∣δkhu∣∣1,2,U independent of h. By Corollary 6.6, we therefore have that u∈W2,2(U), and we immediately obtain the bound
∣∣u∣∣2,2,U≤C2⋅(∣∣f∣∣2,V+∣∣u∣∣1,2,V),(**)C2 having the same dependencies as C1. To obtain the desired estimate, we now choose an η∈C0∞(Ω) such that 0≤η≤1 and η∣V≡1. Setting v=η2⋅u in (*), we see that the left-hand side may be bounded from below as
∫aij∂iu∂j(η2u)=∫aij⋅∂iu⋅∂ju⋅η2+2∫aij⋅∂iu⋅u⋅∂jη⋅η≥λ0∫∣Du∣2η2−2c02∫∣Du∣⋅∣u∣⋅η≥2λ0∫∣Du∣2η2−λ0c02∫Ω∣u∣2,where we have used the Peter-Paul inequality with ε=4c02λ0. On the other hand, the left-hand side of (*) may be estimated as
∫f~⋅η2u≤∫∣f∣⋅∣u∣+β∣Du∣⋅∣u∣⋅η+γ∣u∣2≤(21+γ+λ0β2)∫Ω∣u∣2+4λ0∫∣Du∣2η2+21∫Ω∣f∣2,where we have again used the Peter-Paul inequality. Altogether, after rearranging terms, we obtain the inequality
∫∣Du∣2η2≤λ04(21∣∣f∣∣22+(21+γ+λ0β2+λ0c02)∣∣u∣∣22).Since ∫∣Du∣2η2≥∫V∣Du∣2, plugging the resulting inequality into (**) yields the desired estimate.