TITLE: Solutions of $u_{xx} + 2 \mathrm i u_{xy} + u_{yy} = 0$
QUESTION [6 upvotes]: A problem in A. Friedman, Partial Differential Equations, 1969, chapter 1, section 14, is concerned with the elliptic, but not strongly elliptic PDE on the open unit disk $\mathbb D = \{z= x+ \mathrm iy \mid |z| < 1\}$ given by
\begin{equation*}
u_{xx} + 2\mathrm i u_{xy} + u_{yy} = 0, \qquad u(z) = 0, \forall z\in \partial \mathbb D.
\end{equation*}
Friedman claims that for any complex analytic function $f(z)$ defined on the closed unit disk $\bar{ \mathbb D}$, the function $u(z) = (1-|z|^2) f(z)$ solves the Dirichlet problem.
However, taking $f = 1$ already shows that this is not true. Can you explain that? Can you give a better example?

REPLY [4 votes]: There appears to be a typo in the equation Friedman treats. Whenever $f(z)$ is analytic,
the function $u(z)=(1-|z|^2)f(z)$ does satisfy
$$ u_{xx}+2{\rm i}u_{xy} - u_{yy}=0.$$
The symbol of this equation is
$$
a(\xi)=\xi_1^2+2{\rm i}\xi_1\xi_2-\xi_2^2 = (\xi_1+{\rm i}\xi_2)^2,
$$
which does not vanish for any real vector $\xi=(\xi_1,\xi_2)$ except for $(0,0)$.
This makes the PDE elliptic in the sense that I believe Friedman's book uses (which I take from a 1955 paper of Nirenberg titled Remarks on strongly elliptic partial differential equations).
But it is not ``strongly elliptic,'' since it is false that
$$\Re a(\xi)=\xi_1^2-\xi_2^2 \ge c(\xi_1^2+\xi_2^2)$$
for some postive constant $c$.