i came across this particular problem which says
Suppose that $z=f(x,y)$ is defined, has continuous second partial derivatives and satisfies the Laplace equation $\frac{\partial^2 z}{\partial y^2} + \frac{\partial ^2z}{\partial x^2} = 0$. Assume that ($\frac{\partial^2 z}{\partial x^2})(x_0,y_0) \ne0$. Prove that f cannot have a local maximum or minimum at$(x_0,y_0)$.
This is easy to show :
since $z_{xx}=-z_{yy}$, $(z_{xx})(z_{yy})-(z_{xy})^2$ at $(x_0,y_0)$ is equal to -$[(z_{xx})^2+(z_{xy})^2]$which is always $\lt0$ given the conditions. Hence the given point is a saddle point.
Now here is what i am thinking. No matter what the point is if a function satisfies the Laplace equation, unless all the second order derivatives vanish, every point is gonna be a saddle point. Does there exist a function whose every point is a saddle point? How will the function look?
3 Answers
We seem to have a problem with definitions. Is it not the case that, by definition, a saddle point must be a critical point? So, if every point is a critical point of $f$, then $f$ must be constant. At that stage, no point is a saddle point.
A saddle itself, e.g. $f(x, y) = xy$ or $f(x, y) = x^2 - y^2$, has the property that "every point on the graph is a saddle point". This may not be visually apparent because only the origin is a critical point for these functions. To see the "saddle" at an arbitrary point $(x_0, y_0)$, subtract the linear approximation $$ \ell(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0), $$ leaving the (saddle-shaped) quadratic term.
Note that a harmonic function can have "flat" points on its graph; think of $f(x, y) = x^3 - 3xy^3$ at the origin. That is, it's arguably a bit imprecise to assert that "if a function satisfies the Laplace equation, unless all the second order derivatives vanish, every point is a saddle point". :)
Another example is the pseudosphere, or the catenoid, both of which are like long tubular saddles one could ride on at any point.
