I have some confusion in the physical/geometrical interpretation of the Laplacian of a function of two variables, say, $f(x,y)$.

Recently I have found a video of Khan Academy on Youtube. In this video they gave an intuition of $\nabla^2 f$. The main theme of this video (as I've understand) is:

Since the gradient gives you the slope of steepest ascent so each one of the vectors in gradient field points in the direction that you should walk, such that the graph of $f(x,y)$ is kind of a hill on top of you, it tells you the direction you should go to increase the functional value most rapidly.

Now the divergence of a vector field (treating as a fluid flow) represents how much a point in the field acts like a source.

So now let's think about what it might mean when we take the divergence of the gradient field of $~f$. We can see the divergence of the gradient is very high at points that are kind of like minima, at points where everyone around them tends to be higher.But the divergence of the gradient is low at points that look more like maximum points.

So this Laplacian operator is kind of like a measure of how much of a minimum point is this $(x,y)$.You will be very positive when $f$ evaluated at that point tends to give a smaller value than $f$ evaluated at neighbors of that point. But it'll be very negative if when you evaluate $f$ at that point it tends to be bigger than its neighbors.

Upto now everything is pretty much clear to me and I am so convinced with this intuition of Laplacian. But the confusion pops up in my mind when I read the second derivative test in the Multivaiable Calculus book Vol 2 of Tom M Apostol.

In the second derivative test what I understand is that only $f_{xx}$ and $f_{yy}$ is not sufficient for the conclusion about a local minima or maxima. We have to take care about the mixed partials also.

But in the Khan academy's video I think $\text{div}~ \text{grad} f=f_{xx}+f_{yy}$ should determine a point as local minima or maxima at least .

Now my question is:

if $\nabla^2f(x,y)>0$, can we say that $f$ has a local minima at $(x,y)$? Which I know from second derivative test is not enough for the conclusion whereas according to the first intuitive idea of Khan academy's video it seems pretty reasonable to me.

Can anyone here please help me to figure out what's going on here or at least where am I missunderstood? Thank you.


Well, no. To begin with, you need $f_x=f_y=0$ in order to have a local minimum. And even if that's the case, $f_{xx}+f_{yy}>0$ is not sufficient.

The Laplacian compares the value at the point to the average of the values at surrounding points, which may be greater even if you're not at a local minimum; see the question Intuitive interpretation of the Laplacian, for example.


The laplacian couldn't determine maxima/minima on its own. The reason is simple: knowing the average (or sum) of two or more numbers doesn't tell you if all of them have the same sign, so it couldn't decide whether you have a saddle point.

However, you can rule out one possibility if $\nabla^2 f(x_0,y_0)>0$ and $\nabla f(x_0,y_0)=0$, that is, you know $(x_0,y_0)$ cannot be a local max so is either a local min or a saddle.


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