$u_{xx}u_{yy}-u_{xy}^2 = 0$ iff $u_{xx}=u_{yy}=u_{xy} = 0$?

Question

I am trying to follow a proof of Bernstein's Theorem. Within this proof we consider a function $u:\mathbb{R^2} \rightarrow \mathbb{R}$ such that

$u$ is a solution of: $$ a(x,y)u_{xx} + 2b(x,y)u_{xy}+c(x,y)u_{yy} = 0 $$ With $a,b,c:\mathbb{R^2}\rightarrow \mathbb{R}$ such that $$ \begin{bmatrix} a(x,y)&b(x,y) \\ b(x,y)&c(x,y) \end{bmatrix} \textit{is positive definite on all }\mathbb{R^2} $$ And also $$ u(x,y) = o(\sqrt{x^2+y^2}) \: \text{ as } \: \sqrt{x^2+y^2}\rightarrow+\infty $$ Here's where I get stuck:

Working through the proof it is easily proven that \begin{equation} u_{xx}u_{yy}-u_{xy}^2\leq 0 \end{equation} And it is then claimed that the equality (ie, $u_{xx}u_{yy}-u_{xy}^2 = 0$) holds only on the points where $u_{xx} = u_{yy} = u_{xy} = 0$.

It seems clear to me that substituting $0$ for the second partial derivatives obviously makes the equality hold, but to claim that the equality holds if and only if the second partial derivatives are all $0$ seems like too far of a stretch. Any help is much appreciated.

P.D I have included all the hypotheses in case it helps but I believe they shouldn't be necessary.

Answer 1

The positive definiteness of the matrix $\begin{bmatrix}a&b\\b&c\end{bmatrix}$ means that $a>0$ and $ac-b^2>0$ at every point. Since $ac-b^2>0$, we have from $$a\,u_{xx}+2\,b\,u_{xy}+c\,u_{yy}=0$$ that $$(a\,u_{xx}+c\,u_{yy})^2=(-2\,b\,u_{xy})^2=4\,b^2\,u_{xy}^2\leq 4\,ac\,u_{xy}^2\,.$$ If $u_{xx}\,u_{yy}-u_{xy}^2>0$, then we must have $$(a\,u_{xx}+c\,u_{yy})^2=4\,ac\,u_{xy}^2<4\,ac\,u_{xx}u_{yy}\,.$$ This leads to a contradiction: $$(a\,u_{xx}-c\,u_{yy})^2<0\,.$$ Thus, $$u_{xx}\,u_{yy}-u_{xy}^2\leq 0$$ must be true.

If $u_{xx}\,u_{yy}-u_{xy}^2=0$, then $$(a\,u_{xx}+c\,u_{yy})^2\leq 4\,ac\,u_{xx}u_{yy}\,.$$ We then get $$(a\,u_{xx}-c\,u_{yy})^2\leq 0\,.$$ However, this can only happen when $a\,u_{xx}=c\,u_{yy}$. Thus, $$u_{yy}=\frac{a}{c}\,u_{xx}\text{ and }u_{xy}=\pm\sqrt{u_{xx}\,u_{yy}}=\pm\sqrt{\frac{a}{c}}\,u_{xx}\,.$$ Plug this into $a\,u_{xx}+2\,b\,u_{xy}+c\,u_{yy}=0$ to get $$2\,\sqrt{\frac{a}{c}}\,\left(\sqrt{ac}\pm b\right)\,u_{xx}=0\,.$$ However, $ac>b^2$ implies $\sqrt{ac}\pm b>0$. Thus, $u_{xx}=0$, and therefore, $u_{xy}=0$ and $u_{yy}=0$.

The converse is trivial. Note that the assumption that $$u(x,y)\in o\left(\sqrt{x^2+y^2}\right)\text{ for }\sqrt{x^2+y^2}\to\infty$$ is not needed here.