Week 11: Maximum principles and the eigenvalue problem
We now turn our attention to classical solutions to second-order elliptic PDE and employ more classical techniques to glean information on their behaviour. Throughout this lecture, we shall suppose Ω⊂Rn is open and bounded.
Write L for the non-divergence form operator
Lu:=−i,j=1∑naij⋅∂i∂ju+i=1∑nbi⋅∂iu+c⋅u,
where aij, bi and c are continuous functions on Ω, (aij(x))i,j=1n is a symmetric matrix for each x∈Ωand for some constant λ0>0 the bound
i,j=1∑naij(x)vivj≥λ0∣v∣2
holds for all x∈Ω and v∈Rn.
If a function u∈C2(Ω) satisfies the inequality Lu≤0 on Ω, we call u a subsolution to the equation Lu=0, and if Lu≥0, we say that u is a supersolution.
The weak maximum principle
It turns out that in the absence of the zeroth order term of L, the pointwise behaviour of subsolutions and supersolutions are in a sense controlled in terms of their boundary data. We shall require some auxiliary results from calculus and linear algebra to establish this.
Lemma 11.1. If u∈C2(Ω) attains its maximum in Ω, i.e. if there is an x0∈Ω such that supΩu=u(x0), then Du(x0)=0 and the Hessian D2u(x0) is nonpositive definite.
Proof. Fix v∈Rn and consider t↦u(x0+tv) for t sufficiently close to 0. By classical calculus results, we must have that ∑i=1nvi∂iu(x0)=dtd∣∣t=0u(x0+tv)=0 and ∑i,j=1nvi∂i∂ju(x0)vj=dt2d2∣∣∣t=0u(x0+tv)=0. Since v is arbitrary, this establishes the claim.
Lemma 11.2. Suppose A and B are n×n non-negative definite matrices and A is symmetric. Then
tr(ATB)=i,j=1∑naijbij≥0.
Proof. By classical linear algebra, there exists an orthonormal basis {vi}i=1n for Rn and non-negative numbers {λi}i=1n (not all distinct) such that Avi=λivi. By definition of the trace,
tr(ATB)=i=1∑nviTATBvi=i=1∑nλiviTBvi≥0.
Theorem 11.3 (weak maximum principle). Suppose u∈C2(Ω)∩C0(Ω). If u is a subsolution to the equation Lu=0 with c≡0, then it must attain its maximum on ∂Ω, i.e.
Ωmaxu=∂Ωmaxu.
If instead u is a supersolution to the equation Lu=0 with c≡0, then it must attention its minimum on ∂Ω, i.e.
Ωminu=∂Ωminu.
Proof. First suppose that u is a strict subsolution in the sense that Lu<0 on Ω and assume there exists an x0∈Ω such that maxΩu=u(x0). By the preceding lemmas, we compute that
which contradicts the assumption that Lu<0 on Ω so that we must have x0∈∂Ω. More generally, suppose Lu≤0 and define v:Ω→R by v(x)=u(x)+εekx1 for fixed ε>0 and k∈R to be determined. We compute that
On the other hand, we clearly have maxΩu≥max∂Ωu. This establishes the claim for subsolutions. If u is a supersolution, then clearly −u is a subsolution so that
Ωminu=−Ωmax(−u)=−∂Ωmax(−u)=∂Ωminu.
Remark 11.4. We deduce that if Lu=0 on Ω, then u attains its minimum and maximum on ∂Ω.
In the case where c≥0, we can apply the weak maximum principle to extract the following maximum principle.
Corollary 11.5. Suppose u∈C2(Ω)∩C0(Ω). If u is a subsolution to the equation Lu=0 with c≥0, then we have the inequality
Ωmaxu≤∂Ωmaxu+,
where u+(x):=max{u(x),0)}. If instead u is a supersolution to the equation Lu=0 with c≥0, then it satisfies the inequality
Ωminu≥−∂Ωmaxu−,
where u−(x):=−min{u(x),0}.
Proof. First note that we have the inequality u≤u+ on Ω so that
Ωmaxu≤Ωmaxu+.
Let U:={x∈Ω:u+>0}. If U is empty, then u<0 and we are done. Otherwise, maxΩu+=maxUu+, and since on U,
−i,j=1∑naij⋅∂i∂ju+i=1∑nbi⋅∂iu≤−cu≤0,
the weak maximum principle implies that
Ωmaxu≤Umaxu+=∂Umaxu+.
Since ∂U=(U∩∂Ω)∩{x∈Ω:u(x)=0}, we see that max∂Uu+=max∂Ωu+, establishing the claim for subsolutions. As for supersolutions, we again replace u with −u and apply the above argument again to obtain
In the case where c≡0, the weak maximum principle tells us that a (sub)solution to Lu=0 on Ω attains its maximum on ∂Ω. We now show that if the maximum is actually attained in the interior of Ω, then u must actually be constant. We first prove the following technical lemma.
Lemma 11.6 (Hopf). Suppose u∈C2(Ω)∩C1(Ω) is a subsolution to the equation Lu=0 on Ω and there exists an x0∈∂Ω such that u(x0)>u(x) for all x∈Ω. If Ω satisfies the interior ball condition at x0, i.e. there exists an open ball B⊂Ω with x0∈∂B, and if either c≡0orc≥0 and u(x0)≥0, then ∂ν∂u(x0)>0, where ν is the outer unit normal to B at x0.
Proof. We may assume without loss of generality that we are dealing with the second case (otherwise replace u with u−u(x0)). Moreover, we may suppose that B=B(0,r) for some r>0. Now, define v:B(0,r)→R by
v(x)=e−k∣x∣2−e−kr2.
Note that v∣∂B(0,r)≡0. Moreover, we compute that ∂iv(x)=−2kxie−k∣x∣2 and ∂i∂jv(x)=−2kδije−k∣x∣2+4k2xixje−k∣x∣2 so that
Theorem 11.7 (strong maximum principle). Suppose u∈C2(Ω)∩C0(Ω), c≡0 on Ω and Ω is connected. If u is a subsolution to the equation Lu=0 on Ω and maxΩu=u(x0) for some x0∈Ω, then u is constant. Likewise, if u is a supersolution to the equation Lu=0 on Ω and minΩu=u(x0) for some x0∈Ω, then u is constant.
Proof. Suppose u is a subsolution and let C:={x∈Ω:u(x)=maxΩu} and U:=Ω\C={x∈Ω:u(x)<maxΩu}. Suppose U is nonempty, fix y∈U such that dist(y,C)<dist(y,∂Ω) and let B be the largest ball centred at y such that B⊂U so that ∂B∩C is nonempty. Let y0∈∂B∩C. By construction, U satisfies the interior ball condition at y0 so that we may apply the Hopf lemma to conclude that ∂ν∂u(y0)>0, but y0 is a local maximum so that Du(y0)=0, which is a contradiction.
Similarly, we have the following result in the case where c≥0. The proof is analogous to that of Theorem 11.7, using the latter part of the Hopf lemma.
Theorem 11.8 (strong maximum principle). Suppose u∈C2(Ω)∩C0(Ω), c≥0 on Ω and Ω is connected. If u is a subsolution to the equation Lu=0 on Ω and maxΩu=u(x0)≥0 for some x0∈Ω, then u is constant. Likewise, if u is a supersolution to the equation Lu=0 on Ω and minΩu=u(x0)≤0 for some x0∈Ω, then u is constant.
Harnack inequality
In addition to the maximum principles, we have the following inequality for nonnegative solutions to Lu=0 on Ω, which tells us that the supremum of such a solution is controlled by its infimum away from ∂Ω.
Theorem 11.9. For any connectedU⋐Ω, there exists a constant C>0 depending only on U and L such that whenever u∈C2(Ω) is nonnegative and solves Lu=0 on Ω,