Counterexamples to the
comparison principle in the special Lagrangian potential equation

The special Lagrangian potential operator is the mapping $F\colon\mathcal{S}^{n}\to\mathbb{R}$ given by

$$F(X):=\sum_{i=1}^{n}\arctan(\lambda_{i}(X))$$

where $\lambda_{1}(X)\leq\cdots\leq\lambda_{n}(X)$ are the eigenvalues of the symmetric $n\times n$ matrix $X$. The corresponding equation

$$F(\mathcal{H}w)=f(x)$$

(1.1)

in $\Omega\subseteq\mathbb{R}^{n}$, including the autonomous version

$$F(\mathcal{H}w)=\theta,$$

(1.2)

has attained much interest since it was introduced in [HL82]. For a right-hand side constant $\theta\in(-n\pi/2,n\pi/2)$ the solutions of (1.2) have a nice geometrical interpretation. The graph of the gradient $\nabla w$ in $\Omega\times\mathbb{R}^{n}$ is a special Lagrangian manifold. i.e., it is a Lagrangian manifold of minimal area. See [HL21] and the references therein.

Recently, [CP21] were able to prove the comparison principle for (1.1) when $f$ is continuous and avoids the special phase values

$$\theta_{k}:=(n-2k)\frac{\pi}{2},\qquad k=1,\dots,n-1.$$

In Section 3 we show that their proof can be somewhat simplified by applying a result in [Bru22] valid for equations on the generic form (1.1). However, our main purpose of this short note is to demonstrate how the comparison principle may fail when $\theta_{k}\in f(\Omega)$. Interestingly, the comparison principle is valid in (1.2) for all $\theta\in\mathbb{R}$. This follows immediately from the facts that $X\leq Y$ implies $\lambda_{i}(X)\leq\lambda_{i}(Y)$, $\lambda_{i}(X+\tau I)=\lambda_{i}(X)+\tau$, and that $\arctan$ is strictly increasing. Thus, $F$ is elliptic and $F(X+\tau I)>F(X)$ for all $X\in\mathcal{S}^{n}$, $\tau>0$. See for example Proposition 2.6 in [Bru22].

In our construction of the counterexamples, we shall take advantage of a couple of symmetries in $F$. Firstly, as $F(X)$ only depends on the eigenvalues of $X$, we have $F(QXQ^{\top})=F(X)$ for every orthogonal matrix $Q$. Secondly, since $\arctan$ is odd – and since $\lambda_{i}(-X)=-\lambda_{n-i+1}(X)$ and the different eigenvalues are treated equally by $F$ – it follows that $F$ is odd as well. Moreover, we shall make use of the $n\times n$ exchange matrix

$$J:=\begin{bmatrix}0&&1\\ &\reflectbox{$\ddots$}&\\ 1&&0\end{bmatrix}.$$

It is the corresponding matrix to the reverse order permutation on the set $\{1,2,\dots,n\}$. Obviously, $J$ is symmetric and orthogonal.

For $k=0,\dots,n$, define $v_{k},u_{k}\in C(\mathbb{R}^{n})$ as

	$\displaystyle v_{k}(x)$	$\displaystyle:=\frac{1}{4}-\sum_{i=1}^{k}|x_{i}|+\sum_{i=k+1}^{n}\frac{1}{2}|x_{i}|^{3/2},$
	$\displaystyle u_{k}(x)$	$\displaystyle:=-v_{n-k}(Jx),$

and let $f_{k}\in C(\mathbb{R}^{n})$ be the continuous extension of

$$f_{k}(x):=-\sum_{i=1}^{k}\arctan\left(\frac{3}{8}|x_{i}|^{-1/2}\right)+\sum_{i=k+1}^{n}\arctan\left(\frac{3}{8}|x_{i}|^{-1/2}\right),\qquad x_{i}\neq 0.$$

Observe that

$$f_{k}(0)=-k\frac{\pi}{2}+(n-k)\frac{\pi}{2}=\theta_{k}.$$

We obviously have $v_{k}(0)-u_{k}(0)=1/2>0$. In order to create a counterexample to the comparison principle it only remains to observe that

	$\displaystyle-u_{k}(x)$	$\displaystyle=v_{n-k}(Jx)$
		$\displaystyle=\frac{1}{4}-\sum_{i=1}^{n-k}|x_{n+1-i}|+\sum_{i=n-k+1}^{n}\frac{1}{2}|x_{n+1-i}|^{3/2}$
		$\displaystyle=\frac{1}{4}-\sum_{j=k+1}^{n}|x_{j}|+\sum_{j=1}^{k}\frac{1}{2}|x_{j}|^{3/2}$

and

	$\displaystyle v_{k}(x)-u_{k}(x)$	$\displaystyle=\frac{1}{2}+\sum_{i=1}^{n}\frac{1}{2}|x_{i}|^{3/2}-|x_{i}|$
		$\displaystyle=\frac{1}{2}+\sum_{i=1}^{n}\frac{1}{2}|x_{i}|\left(|x_{i}|^{1/2}-2\right)$

independently of $k$. If we take the domain to be the unit ball in the infinity-norm,

$$\Omega:=\left\{x\in\mathbb{R}^{n}\;\colon\;|x_{i}|<1\right\},$$

it follows that $v_{k}(x)-u_{k}(x)\leq 0$ whenever $|x|_{\infty}\leq 1$ and when there is at least one index $j$ such that $|x_{j}|=1$. That is, for $x\in\partial\Omega$.

The ideas behind the above constructions are all contained, and therefore best illustrated, by the case $n=2$, $k=1$. Then also $n-k=1$ and, dropping the subscript 1 yields,

	$\displaystyle v(x,y)$	$\displaystyle=\frac{1}{4}-|x|+\frac{1}{2}|y|^{3/2},$
	$\displaystyle u(x,y)$	$\displaystyle=-v(y,x)=-\frac{1}{4}-\frac{1}{2}|x|^{3/2}+|y|,$

with phase

$$f(x,y)=\begin{cases}-\arctan(\frac{3}{8}|x|^{-1/2})+\arctan(\frac{3}{8}|y|^{-1/2}),\quad&x\neq 0,y\neq 0,\\ -\pi/2+\arctan(\frac{3}{8}|y|^{-1/2}),&x=0,y\neq 0,\\ -\arctan(\frac{3}{8}|x|^{-1/2})+\pi/2,&x\neq 0,y=0,\\ 0,&x=0,y=0.\end{cases}$$

In addition to the difference

$$v(x,y)-u(x,y)=\frac{1}{2}+\frac{1}{2}|x|\left(|x|^{1/2}-2\right)+\frac{1}{2}|y|\left(|y|^{1/2}-2\right)$$

the graph of these functions over the square

$$\Omega:=\left\{(x,y)\in\mathbb{R}^{2}\;:\;|x|<1,|y|<1\right\}$$

are shown in Figure 1.

Theorem 6.18 [CP21] establish the comparison principle for the equation

$$\sum_{i=1}^{n}\arctan(\lambda_{i}(\mathcal{H}w))=f(x)$$

in every open and bounded $\Omega\subseteq\mathbb{R}$ whenever $f\in C(\Omega)$, $\theta_{n}<f<\theta_{0}$, and

$$\theta_{k}:=(n-2k)\frac{\pi}{2}\notin f(\Omega),\qquad k=1,\dots,n-1.$$

The main idea in our proof, as conducted in Example 2.2 in [Bru22], is the same as in [CP21]. Namely, to reach a contradiction when, for each $i=1,\dots,n$, $|\lambda_{i}(X_{j})|\to\infty$ as $j\to\infty$ for some sequence $X_{j}\in\mathcal{S}^{n}$. Our contribution is to show how this follows almost immediately from a general result for equations on the form $F(\mathcal{H}w)=f(x)$. For convenience, we reproduce the proof below.

Note that the pathological situation when $f$ takes values outside the interval $[\theta_{n},\theta_{0}]$ comes for free: If, say $f>\theta_{0}$ somewhere in $\Omega$, the comparison principle vacuously holds since the equation will have no subsolutions. On the other hand, the case $\theta_{0}\in f(\Omega)$, which is covered by our counterexample, is not pathological. For example, one can easily confirm that $w(x)=|x|^{3/2}$ is a viscosity solution to the equation in $\mathbb{R}^{n}$ when the right-hand side is the continuous extension of $f(x):=F(\mathcal{H}w(x))$.

Assume that

$$\theta_{k}\notin f(\Omega)\subseteq[\theta_{n},\theta_{0}]$$

(3.1)

for all $k=0,\dots,n$. Proposition 2.7 in [Bru22] states that the comparison principle will hold if whenever $X_{j}\in\mathcal{S}^{n}$ is a sequence such that $\lim_{j\to\infty}F(X_{j})=\theta\in f(\Omega)$, then

$$\liminf_{j\to\infty}F(X_{j}+\tau I)>\theta$$

for every $\tau>0$. Suppose to the contrary that this is not true. Then there are numbers $\theta\in f(\Omega)$ and $\tau>0$, and a sequence $X_{j}\in\mathcal{S}^{n}$ with $F(X_{j})\to\theta$, such that

	$\displaystyle 0$	$\displaystyle=\lim_{j\to\infty}F(X_{j}+\tau I)-F(X_{j})$
		$\displaystyle=\lim_{j\to\infty}\sum_{i=1}^{n}\arctan(\lambda_{i}(X_{j})+\tau)-\arctan(\lambda_{i}(X_{j}))\geq 0,$

which – since $\arctan$ is strictly increasing – is possible only if each $\lambda_{i}(X_{j})$ is unbounded as $j\to\infty$. There is thus a subsequence (still indexed by $j$) such that either $\lambda_{i}(X_{j})\to+\infty$ or $\lambda_{i}(X_{j})\to-\infty$. But this is a contradiction of (3.1) as

$$f(\Omega)\ni\theta=\lim_{j\to\infty}F(X_{j})\\ =\lim_{j\to\infty}\sum_{i=1}^{n}\arctan(\lambda_{i}(X_{j}))\\ =\sum_{i=1}^{n}\pm\frac{\pi}{2}=\theta_{k}$$

for some $k=0,\dots,n$.