We now prove a higher regularity theorem which states that if the coefficients of L are sufficiently smooth and f is k-times weakly differentiable, then a solution u∈W1,2(Ω) to Lu=f on Ω is (k+2)-times weakly differentiable.
with aij,bi,c∈Ck+1(Ω) and f∈Hk,2(Ω) for some k∈N∪{0}. Then u∈Wlock+2,2(Ω)and for each U⋐Ω,
∣∣u∣∣k+2,2,U≤Ck⋅(∣∣f∣∣k,2+∣∣u∣∣2),
where Ck>0 depends only on k, U, Ω and L.
Proof. We have already proved the case k=0. Suppose the claim is true for k∈N fixed. If u∈W1,2(Ω) solves Lu=f with aij,bi,c∈Ck+2(Ω) and f∈Wk+1,2(Ω), then the inductive hypothesis implies that u∈Wlock+2,2(Ω). Fix r∈{1,…,n}, V⋐Ω and φ∈C0∞(V). Plugging v=−∂rφ into (*) and using the fact that u∈W2,2(V), we see that the left-hand side may be written as
Therefore, by the inductive hypothesis, ∂ru∈Wlock+2,2(V) so that ∂ru∈Wlock+2,2(Ω) and ultimately u∈Wlock+3,2(Ω). The inductive hypothesis also yields the estimate for U⋐V⋐Ω
However, it follows from Hölder's inequality and the Leibniz rule, as well as the fact that the coefficients of L and their derivatives up to order k+2 are bounded on V that
∣∣f1∣∣k,2,V≤Ck′′(∣∣f∣∣k+1,2+∣∣u∣∣k+2,2,V),
where Ck′′ depends only on k, V and the coefficients of L. Finally, combining these two estimates and the inductive step's estimate yields the desired estimate.
Corollary 10.2 (smoothness). Suppose u∈W1,2(Ω) solves Lu=f on Ω with aij,bi,c,f∈C∞(Ω). Then u∈C∞(Ω).
Proof. By the preceding theorem, for each k∈N, we see that u∈Wlock,2(Ω). By the Sobolev embedding theorem, for each m∈N, we may choose k sufficiently large so as to deduce that u is equal to a Cm(Ω) function almost everywhere. Since m is arbitrary, we have established the claim.
Boundary regularity
We now turn to the question of whether weak solutions to Lu=f lie in Wk,2(Ω) for k>1 and under appropriate additional assumptions on L and Ω, a phenomenon we refer to as boundary regularity. Our approach will be similar to the case of interior regularity. We first note the following lemma, which is an analogue of Lemma 6.3 and Corollary 6.6.
Lemma 10.3. Let 0<r1<r2<1 and p∈]1,∞[.
If u∈Hk,p(B+(0,r2)), then whenever 0<∣h∣<r2−r1 and i∈{1,…,n−1}, the inequality
∣∣δihu∣∣k−1,p,B+(0,r1)≤∣∣u∣∣k,p,B+(0,r2)
holds.
If u∈Hk−1,p(B+(0,r2)) and there exists a constant C such that for all 0<r1<r2, 0<∣h∣<r2−r1 and i∈{1,…,n−1}
∣∣δihu∣∣k−1,p,B+(0,r1)≤C,
then ∂1u,…,∂n−1u∈Hk−1,p(B+(0,r2)) and ∣∣∂iu∣∣k−1,p,B+(0,r2)≤C.
We first consider the case Ω=B+(0,r) for appropriate r>0. The following lemma gets us regularity up to (part of) the boundary lying in the hyperplane {xn=0}.
Lemma 10.4. Let r∈]21,1[ and suppose L is an elliptic operator over B+(0,r) satisfying the additional condition aij∈C1(B+(0,r)). If u∈H01,2(B+(0,r)) is a weak solution to the equation
Lu=f
on B+(0,r), then u∈W2,2(B+(0,21))and
∣∣u∣∣2,2,B+(0,21)≤C(L,r)⋅(∣∣f∣∣2+∣∣u∣∣2).
Proof. Let η∈C0∞(Rn) be such that η∣B(0,21)≡1, 0≤η≤1 and suppη⊂B(0,r). As in the proof of Theorem 9.7, we may write the condition 'Lu=f weakly' as the equation
∫aij⋅∂iu⋅∂jv=∫f⋅v,
which should hold for all v∈H01,2(B+(0,r)), where f=f−bi∂iu−cu. We now let v=−δk−h(η2δkhu) for small h∈R and k∈{1,…,n−1}. Note that v is clearly an element of W1,2(B+(0,r)). To see that it lies in H01,2(B+(0,r)), we write it out more explicitly:
Taking traces of both sides and using the properties of trace operator (and the fact that h is small), we see that v has trace zero so that it lies in H01,2(B+(0,r)). Now, proceeding analogously to Theorem 9.7, we deduce that
Using the preceding lemma and its proof, we deduce that for all k∈{1,…,n−1}, ∂ku∈W1,2(B+(0,21))and
∫B+(0,21)∣D∂ku∣2≤c3∫B+(0,r)∣f∣2+∣u∣2+∣Du∣2.
As for ∂nu, we first note by interior regularity (Theorem 9.7) that u∈Wloc2,2(B+(0,r)) so that the equation Lu=f actually holds almost everywhere. Therefore,
−∂i(aij∂ju)=f
almost everywhere. In particular, the left-hand side is equal to
Since for all v∈Rn we have the condition aijvivj≥λ0∣v∣2, taking vi=δni yields the inequality ann≥λ0. In particular, we deduce that
∣∂n2u∣2≤C(L)(i+j<2n∑∣∂i∂ju∣2+∣Du∣2+∣f∣2).
Since each term on the right-hand side is summable by the preceding computation, we deduce that ∂n2u∈L2(B+(0,21), and estimating ∣f∣2 from above as usual, we deduce that
∣∣u∣∣2,2,B+(0,21)≤c4(∣∣f∣∣2+∣∣u∣∣1,2).
Finally, taking v=u in the equation `Lu=f weakly,' we obtain after carrying out a similar computation to the above that ∣∣u∣∣1,2,B+(0,r)≤c5(∣∣f∣∣2+∣∣u∣∣2). Combining the above two inequalities establishes the claim.
Theorem 10.5 (boundary regularity). Suppose Ω is a bounded domain of class C2 and u∈H01,2(Ω) is a weak solution to the equation
Lu=f,
where L is the usual (divergence-form) elliptic operator satisfying the additional condition aij∈C1(Ω). Then u∈W2,2(Ω) and the estimate
∣∣u∣∣2,2≤C(∣∣f∣∣2+∣∣u∣∣2),
where C depends only on Ω and L.
Proof. Recall that by definition, for each x∈∂Ω, we may find a neighbourhood Nx∋x and a C2 diffeomorphism Φx:Nx→B(0,1) such that Φx(x)=0, Φx(Nx∩Ω)=B+(0,1) and Φx(Nx∩∂Ω)=B(0,1)∩{xn=0}. It suffices to show that for each x∈∂Ω,
∣∣u∣∣2,2,Φx−1(B+(0,21))≤C(∣∣f∣∣2+∣∣u∣∣1,2),
since we may pick a partition of unity subordinate to a (finite) covering {Φxi−1(B(0,21))}i=1N∪{Ωδ} of Ω to deduce an estimate of the form
∣∣u∣∣2,2≤C(∣∣f∣∣2+∣∣u∣∣1,2);(#)
Choosing v=u in the defining equation of a weak solution to Lu=f and applying the usual tricks then yields a bound of the form ∣∣u∣∣1,2≤C′⋅(∣∣u∣∣2+∣∣f∣∣2).
To streamline the proof, we define the following functions on B+(0,3/4):
Note that u∈W1,2(B+(0,3/4)), c,bi∈L∞(B+(0,3/4)), aij=aji∈C1(B+(0,3/4))andf∈L2(B+(0,3/4)). Moreover, aij is positive definite, and for any v∈Rn,
where C depends only on L and Ω (through Φx). Writing u and f in terms of u and f, we may then use this inequality to establish (#).
As with interior regularity, we may argue by induction to establish higher-order boundary regularity subject to additional conditions on Ω and L. In this case, we again first consider the case of a half ball. We state the relevant result without proof.
Theorem 10.6 (higher boundary regularity). Suppose Ω is a bounded domain of class Ck+2 and u∈H01,2(Ω) is a weak solution to the equation
Lu=f,
where L is the usual (divergence-form) elliptic operator satisfying the additional conditions aij,bi,c∈Ck+1(Ω) and f∈Hk(Ω). Then u∈Wk+2,2(Ω) and the estimate
∣∣u∣∣k+2,2≤Ck(∣∣f∣∣k,2+∣∣u∣∣2),
where Ck depends only on k, Ω and L. Moreover, if Ω is a bounded domain of class C∞ and aij,bi,c,f∈C∞(Ω), then u∈C∞(Ω).
Remark 10.7. If a solution u∈H01,2(Ω) to Lu=f is known to be unique, then we may combine Theorem 9.6 and Theorem 10.6 to obtain an estimate of the form ∣∣u∣∣k+2,2≤Ck′∣∣f∣∣k,2.