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Section 3.1 Ellipticity and uniform ellipticity

Throughout this chapter, L will denote a second-order linear operator of the form
(3.1)L=aij(x)ij+bi(x)i+c(x),
acting on functions uC2(Ω) via
Lu(x)=aij(x)iju(x)+bi(x)iu(x)+c(x)u(x).
Here as always the summation convention from Notation 2.1 is in force, so that repeated indices are understood to summed from 1 through N.
Throughout this chapter we assume that the coefficients aij,bi,c are bounded functions on Ω. Moreover, we assume that aij=aji so that the N×N matrix a is symmetric.

Definition 3.1. Ellipticity.

We say that L is elliptic if for every xΩ there is a number λ(x)>0 such that
(3.2)aij(x)ξiξjλ(x)|ξ|2
for all ξRN. L is uniformly elliptic if (3.2) holds with λ(x)=λ0 for some constant λ>0 independent of xΩ. We call λ0 an ellipticity constant for L.
Comparing with Definition 2.30 and Corollary 2.29, we see that L is elliptic if and only if the symmetric matrix a(x) is positive definite for all xΩ, and uniformly elliptic if the minimum eigenvalue of a(x) is λ0>0 for all xΩ.

Exercises Exercises

1. (PS4) Checking uniform ellipticity.

(a)
Show that L=Δ is a uniformly elliptic operator for any open ΩRN.
Solution.
We have aij=δij, and hence
aijξiξj=δijξiξj=|ξ|21ξ2
for any ξRN. Thus Δ is uniformly elliptic with ellipticity constant λ0=1.
(b)
For any ΩR2, show that L=11+12+21+22 is not elliptic.
Hint.
One option is to work with (3.2) directly. Another option is to calculate the eigenvalues and get (3.2) from Corollary 2.29.
Solution.
We have aij=1, and hence
aijξiξj=ξ1ξ1+ξ1ξ2+ξ2ξ1+ξ2ξ2=ξ12+2ξ1ξ2+ξ22=(ξ1+ξ2)2,
which is 0 but vanishes for instance when ξ=(1,1). Thus L is not elliptic.
Alternatively, we could check that one of the eigenvalues of the matrix
a=(1111)
is 0, which is not positive. A slick way to do this is to just take the determinant and recall how this is related to the eigenvalues.
Comment.
Since 12=21 when acting on C2 functions, many authors will abbreviate the operator L in this problem as
L=11+212+22.
This does not mean that they consider the non-symmetric matrix
a=(1201),
as this would violate our assumption aij=aji. Perhaps more importantly, calculating the eigenvalues of this non-symmetric matrix will no longer provide you with any information about the ellipticity of L.
(c)
Let α,β be bounded functions on ΩR2. Show that
L=(1+α2)11αβ12αβ21+(1+β2)22
is uniformly elliptic.
Hint.
Same as for previous part.
Solution.
In this case we have, for any ξR2,
aijξiξj=(1+α2)ξ122αβξ1ξ2+(1+β2)ξ22=ξ12+ξ22+(αξ1βξ2)2|ξ|2.
Thus L is uniformly elliptic with ellipticity constant λ0=1.
Alternatively, one can simply calculate the eigenvalues of the relevant matrix
a=(1+α2αβαβ1+β2)
by brute force. They are 1 and 1+α2+β21, which again implies uniform ellipticity with λ0=1.
(d)
Show that L=11+ex122 is elliptic on Ω={xR2:x1>0}R2, but not uniformly elliptic.
Solution.
The eigenvalues of the matrix
a(x)=(100ex1)
are clearly 1 and ex1, both of which are strictly positive for any fixed xΩ. Thus L is elliptic. On the other hand, as x1, the second eigenvalue ex10, and so this operator cannot be uniformly elliptic.
Comment 1.
To show that L is not uniformly elliptic it is not enough to show that the smallest eigenvalue of a is ex1, or equivalently that aijξiξjex1|ξ|2. Indeed, these properties also hold for the Laplacian Δ, which is uniformly elliptic on Ω. To disprove uniform ellipticity need the reverse inequality that the smallest eigenvalue of a is ex1, or alternatively that for ξ=e2 we get aijξiξjex1|ξ|2.
Comment 2.
Note that, because of our choice of Ω, the coefficients of L are indeed bounded functions.

2. (PS4) Ellipticity after shifting and scaling.

Consider a differential operator
L=aij(x)ij+bi(x)i+c(x)
acting on functions in C2(Br(y)) for some r>0 and yRN. Given uC2(Br(y)), define vC2(B1(0)) by
v(x)=u(y+rx)
(a)
Find a second-order differential operator L~ so that
L~v(x)=Lu(y+rx).
Hint.
Keep Notation 2.14 in mind! Calculate Lu in terms of derivatives of v using the chain rule and/or Exercise 2.3.1.
Solution.
By Exercise 2.3.1 and keeping in mind Notation 2.14, we have
u(y+rx)=v(x)iu(y+rx)=r1iv(x)iju(y+rx)=r2ijv(x).
Thus
Lu(y+rx)=aij(y+rx)iju(y+rx)+bi(y+rx)iu(y+rx)+c(y+rx)u(y+rx)=r2aij(y+rx)ijv(x)+r1bi(y+rx)iu(x)+c(y+rx)v(x),
and so we take
L~=a~ij(x)ij+b~i(x)i+c~(x)
where
a~ij(x)=r2aij(y+rx),b~i(x)=r1bi(y+rx).c~(x)=c(y+rx).
Comment.
In problems like this it is important to keep track of where the functions u,v and coefficients aij,bi,c are being evaluated. This year, several students took
L~=1r2aijij+1rbii+c,
but this operator has
L~v(x)=1r2aij(x)ijv(x)+1rbi(x)iv(x)+c(x)v(x),
which is not quite what we want.
(b)
Suppose that L is uniformly elliptic with ellipticity constant λ0. Show that L~ is also uniformly elliptic. What is the ellipticity constant?
Solution.
Let xB1(0) and ξRN. Then y+rxBr(y) and so by our above formulas and the ellipticity of L we have
a~ij(x)ξiξj=r2aij(y+rx)ξiξjr2λ0|ξ|2.
Since x and ξ were arbitrary, we conclude that L~ is uniformly elliptic with ellipticity constant λ~0=r2λ0. (The boundedness and symmetry of the coefficients of L~ easily follow from the corresponding properties of the coefficients of L.)

3. Divergence form operators.

Show that the operator Lu=i(aijju) appearing in Exercise 2.9.3 can be written in the form (3.1) for an appropriate choice of bi and c. Recall that the coefficients aij are assumed to be C1.

4. The symmetry assumption.

Consider an operator L~=a~ijij+bii+c where the coefficients a~ij,bi,c are bounded functions on Ω but a~ is not necessarily symmetric. Show that for any uC2(Ω) we have L~u=Lu where L=aijij+bii+c is an operator with symmetric second-order coefficients aij=12(a~ij+a~ji).