Section 3.1 Ellipticity and uniform ellipticity
Throughout this chapter, \(L\) will denote a second-order linear operator of the form
\begin{equation}
L = a_{ij}(x) \partial_{ij} + b_i(x) \partial_i + c(x)\text{,}\tag{3.1}
\end{equation}
acting on functions \(u \in C^2(\Omega)\) via
\begin{gather*}
Lu(x) = a_{ij}(x) \partial_{ij} u(x) + b_i(x) \partial_i u(x) + c(x) u(x)\text{.}
\end{gather*}
Here as always the summation convention from Notation 2.1 is in force, so that repeated indices are understood to summed from \(1\) through \(N\text{.}\)
Throughout this chapter we assume that the coefficients \(a_{ij}, b_i, c\) are bounded functions on \(\overline \Omega\text{.}\) Moreover, we assume that \(a_{ij} = a_{ji}\) so that the \(N \times N\) matrix \(a\) is symmetric.
Definition 3.1. Ellipticity.
We say that \(L\) is elliptic if for every \(x \in \Omega\) there is a number \(\lambda(x)\gt0\) such that
\begin{gather}
a_{ij}(x)\xi_i \xi_j \ge \lambda(x)|\xi|^2 \tag{3.2}
\end{gather}
for all \(\xi \in \R^N\text{.}\) \(L\) is uniformly elliptic if (3.2) holds with \(\lambda(x)=\lambda_0\) for some constant \(\lambda \gt 0\) independent of \(x \in \Omega\text{.}\) We call \(\lambda_0\) an ellipticity constant for \(L\text{.}\)
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 \(x \in \Omega\text{,}\) and uniformly elliptic if the minimum eigenvalue of \(a(x)\) is \(\ge \lambda_0 \gt 0\) for all \(x \in \Omega\text{.}\)
Exercises Exercises
1. (PS4) Checking uniform ellipticity.
(a)
Show that \(L=\Delta\) is a uniformly elliptic operator for any open \(\Omega \subseteq
\R^N\text{.}\)
Solution.
We have \(a_{ij} = \delta_{ij}\text{,}\) and hence
\begin{gather*}
a_{ij} \xi_i \xi_j = \delta_{ij} \xi_i \xi_j = \abs \xi^2 \ge 1 \cdot \xi^2
\end{gather*}
for any \(\xi \in \R^N\text{.}\) Thus \(\Delta\) is uniformly elliptic with ellipticity constant \(\lambda_0 = 1\text{.}\)
(b)
For any \(\Omega \subset \R^2\text{,}\) show that \(L=\partial_{11}+\partial_{12}+\partial_{21}+\partial_{22}\) 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 \(a_{ij} = 1\text{,}\) and hence
\begin{align*}
a_{ij} \xi_i \xi_j
\amp = \xi_1 \xi_1 + \xi_1 \xi_2 + \xi_2 \xi_1 + \xi_2 \xi_2\\
\amp = \xi_1^2 + 2\xi_1 \xi_2 + \xi_2^2\\
\amp = (\xi_1+\xi_2)^2\text{,}
\end{align*}
which is \(\ge 0\) but vanishes for instance when \(\xi = (1,-1)\text{.}\) Thus \(L\) is not elliptic.
Alternatively, we could check that one of the eigenvalues of the matrix
\begin{equation*}
a =
\begin{pmatrix}
1 \amp 1 \\ 1 \amp 1
\end{pmatrix}
\end{equation*}
is \(0\text{,}\) 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 \(\partial_{12}=\partial_{21}\) when acting on \(C^2\) functions, many authors will abbreviate the operator \(L\) in this problem as
\begin{equation*}
L = \partial_{11} + 2\partial_{12} + \partial_{22}\text{.}
\end{equation*}
This does not mean that they consider the non-symmetric matrix
\begin{equation*}
a =
\begin{pmatrix}
1\amp 2 \\
0 \amp 1
\end{pmatrix},
\end{equation*}
as this would violate our assumption \(a_{ij}=a_{ji}\text{.}\) Perhaps more importantly, calculating the eigenvalues of this non-symmetric matrix will no longer provide you with any information about the ellipticity of \(L\text{.}\)
(c)
Let \(\alpha,\beta\) be bounded functions on \(\Omega \subseteq \R^2\text{.}\) Show that
\begin{equation*}
L = (1+\alpha^2)\partial_{11} - \alpha\beta \partial_{12} - \alpha\beta \partial_{21} + (1+\beta^2)\partial_{22}
\end{equation*}
is uniformly elliptic.
Hint.
Same as for previous part.
Solution.
In this case we have, for any \(\xi\in \R^2\text{,}\)
\begin{align*}
a_{ij} \xi_i \xi_j
\amp
= (1+\alpha^2) \xi_1^2 - 2 \alpha\beta \xi_1 \xi_2 + (1+\beta^2) \xi_2^2\\
\amp
= \xi_1^2 + \xi_2^2
+ (\alpha \xi_1 - \beta \xi_2)^2\\
\amp
\ge \abs \xi^2\text{.}
\end{align*}
Thus \(L\) is uniformly elliptic with ellipticity constant \(\lambda_0 =
1\text{.}\)
Alternatively, one can simply calculate the eigenvalues of the relevant matrix
\begin{equation*}
a =
\begin{pmatrix}
1+\alpha^2 \amp -\alpha\beta \\
-\alpha\beta \amp 1+\beta^2
\end{pmatrix}
\end{equation*}
by brute force. They are \(1\) and \(1+\alpha^2+\beta^2 \ge 1\text{,}\) which again implies uniform ellipticity with \(\lambda_0 = 1\text{.}\)
(d)
Show that \(L = \partial_{11} + e^{-x_1} \partial_{22} \) is elliptic on \(\Omega =
\{ x \in \R^2 : x_1 \gt 0 \} \subset \R^2\text{,}\) but not uniformly elliptic.
Solution.
The eigenvalues of the matrix
\begin{equation*}
a(x) =
\begin{pmatrix}
1 \amp 0 \\
0 \amp e^{-x_1}
\end{pmatrix}
\end{equation*}
are clearly \(1\) and \(e^{-x_1}\text{,}\) both of which are strictly positive for any fixed \(x \in \Omega\text{.}\) Thus \(L\) is elliptic. On the other hand, as \(x_1 \to \infty\text{,}\) the second eigenvalue \(e^{-x_1} \to 0\text{,}\) 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 \(\ge e^{-x_1}\text{,}\) or equivalently that \(a_{ij} \xi_i \xi_j \ge e^{-x_1} \abs \xi^2\text{.}\) Indeed, these properties also hold for the Laplacian \(\Delta\text{,}\) which is uniformly elliptic on \(\Omega\text{.}\) To disprove uniform ellipticity need the reverse inequality that the smallest eigenvalue of \(a\) is \(\le e^{-x_1}\text{,}\) or alternatively that for \(\xi=e_2\) we get \(a_{ij} \xi_i \xi_j \le e^{-x_1} \abs
\xi^2\text{.}\)
Comment 2.
Note that, because of our choice of \(\Omega\text{,}\) the coefficients of \(L\) are indeed bounded functions.
2. (PS4) Ellipticity after shifting and scaling.
Consider a differential operator
\begin{equation*}
L = a_{ij}(x) \partial_{ij} + b_i(x) \partial_i + c(x)
\end{equation*}
acting on functions in \(C^2(B_r(y))\) for some \(r \gt 0\) and \(y \in \R^N\text{.}\) Given \(u \in C^2(B_r(y))\text{,}\) define \(v \in C^2(B_1(0))\) by
\begin{gather*}
v(x) = u(y+rx)
\end{gather*}
as in Exercise 2.3.1.
(a)
Find a second-order differential operator \(\tilde L\) so that
\begin{gather*}
\tilde Lv(x) = Lu(y+rx)\text{.}
\end{gather*}
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
\begin{align*}
u(y+rx) \amp = v(x) \\
\partial_i u(y+rx) \amp = r^{-1} \partial_i v(x) \\
\partial_{ij} u(y+rx) \amp = r^{-2} \partial_{ij} v(x)\text{.}
\end{align*}
Thus
\begin{align*}
Lu(y+rx)
\amp = a_{ij}(y+rx) \partial_{ij} u(y+rx) \\
\amp \qquad + b_i(y+rx) \partial_i u(y+rx) \\
\amp \qquad + c(y+rx) u(y+rx)\\
\amp = r^{-2} a_{ij}(y+rx) \partial_{ij} v(x) \\
\amp \qquad + r^{-1} b_i(y+rx) \partial_i u(x) \\
\amp \qquad + c(y+rx) v(x)\text{,}
\end{align*}
and so we take
\begin{equation*}
\tilde L = \tilde a_{ij}(x) \partial_{ij} + \tilde b_i(x) \partial_i + \tilde c(x)
\end{equation*}
where
\begin{align*}
\tilde a_{ij} (x) \amp= r^{-2} a_{ij}(y+rx),\\
\tilde b_i (x) \amp= r^{-1} b_i(y+rx).\\
\tilde c (x) \amp= c(y+rx)\text{.}
\end{align*}
Comment.
In problems like this it is important to keep track of where the functions \(u,v\) and coefficients \(a_{ij},b_i,c\) are being evaluated. This year, several students took
\begin{gather*}
\tilde L
= \frac 1{r^2} a_{ij} \partial_{ij} + \frac 1r b_i \partial_i + c\text{,}
\end{gather*}
but this operator has
\begin{gather*}
\tilde L v(x)
= \frac 1{r^2} a_{ij}(x) \partial_{ij}v(x) + \frac 1r b_i(x) \partial_iv(x) + c(x)
v(x)\text{,}
\end{gather*}
which is not quite what we want.
(b)
Suppose that \(L\) is uniformly elliptic with ellipticity constant \(\lambda_0\text{.}\) Show that \(\tilde L\) is also uniformly elliptic. What is the ellipticity constant?
Solution.
Let \(x \in B_1(0)\) and \(\xi \in \R^N\text{.}\) Then \(y+rx \in B_r(y)\) and so by our above formulas and the ellipticity of \(L\) we have
\begin{gather*}
\tilde a_{ij}(x)\xi_i \xi_j
= r^{-2} a_{ij}(y+rx)\xi_i \xi_j
\ge r^{-2} \lambda_0|\xi|^2 \text{.}
\end{gather*}
Since \(x\) and \(\xi\) were arbitrary, we conclude that \(\tilde
L\) is uniformly elliptic with ellipticity constant \(\tilde \lambda_0 =
r^{-2} \lambda_0\text{.}\) (The boundedness and symmetry of the coefficients of \(\tilde L\) easily follow from the corresponding properties of the coefficients of \(L\text{.}\))
3. Divergence form operators.
Show that the operator \(Lu = \partial_i(a_{ij}\partial_j u)\) appearing in Exercise 2.9.3 can be written in the form (3.1) for an appropriate choice of \(b_i\) and \(c\text{.}\) Recall that the coefficients \(a_{ij}\) are assumed to be \(C^1\text{.}\)
4. The symmetry assumption.
Consider an operator \(\tilde L = \tilde a_{ij} \partial_{ij} + b_i \partial_i + c\) where the coefficients \(\tilde a_{ij},b_i,c\) are bounded functions on \(\Omega\) but \(\tilde a\) is not necessarily symmetric. Show that for any \(u \in C^2(\Omega)\) we have \(\tilde L u = L u\) where \(L = a_{ij} \partial_{ij} + b_i \partial_i + c\) is an operator with symmetric second-order coefficients \(a_{ij} = \tfrac 12 (\tilde a_{ij} + \tilde a_{ji})
\text{.}\)
