I have got a multivariable function $f$ defined on an open set $V$. Suppose $f$ attains maximum at some point $(x,y)$ inside of $V$. At this point we also have $f_{xx}=f_{yy}=0$. And finally the Laplacian of $f$ on $V$ is greater than or equal to $0$. What can we say? such a point can really exist?
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
|||||||||
|
|
On $\mathbb{R}^n$ with sufficient smoothness assumptions, the second derivative test implies such a scenario is impossible. If we define $M=\det H$ where $H_{ij}=\frac{\partial^2 f}{\partial x_i~\partial x_i}$ is the Hessian matrix, then the test is conclusive as long as $M\ne0$. If $M$ is positive/negative definite (at an interior point of the domain), then $f$ has a local extremum (minimum/maximum). If there are eigenvalues of $H$ of each sign, then the point is a saddle point (the restriction of $f$ along eigenlines of opposite-sign eigenvalues has a minimum on one line but a maximum on the other). Otherwise ($M=0$ or there are complex eigenvalues), the test is inconclusive. In two dimensions, this means that $M=f_{xx}f_{yy}-f_{xy}^2 > 0$ is a necessary condition for a point to be a local extremum. But if that holds, then $f_{xx}f_{yy} > f_{xy}^2 \ge 0$, so that $f_{xx}$ & $f_{yy}$ share the same sign and so the sign of the Laplacian determines whether the point is a minimum or maximum (if $\nabla^2f$ is positive or negative respectively). But the Laplacian alone does not determine whether a point is a local extremum. You will find a brief discussion of the above here, with more under the article on Morse theory. |
|||
|
|
|
A $C^2$-function $f$ with $\Delta f(x,y)\geq0$ for all $(x,y)$ is subharmonic, see here. A nonconstant subharmonic function cannot have a maximum at an interior point. |
|||
|
|
