Laplacian intuition

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Recommends Kong · 2018-02 · Mathematics Stack Exchange · on “Intuition for the laplacian equation”

I studied about the Laplacian at [this video] and I understand that it can be thought of as the second order derivative test where the value $ \triangle f(x,y) $ will be high at a local minima and low at a local maxima. But what does it mean to solve for the laplacian i.e. $ \triangle f(x,y) = 0$ ? I have extremely little background in physics so it would be nice if someone, bearing that it mind, explain what it means to solve for the laplacian. I come from an image processing background and have only used the laplacian operator for edge detection so "solving" it sounds completely alien to me.

Recommends Antoni Parellada · 2019-10 · Mathematics Stack Exchange · on “How to reconcile two different intuitions of the Laplacian operator?”

In here [LINK] the Laplacian $u_{xx}+u_{yy}$ is said to: The Laplacian measures the degree by which the value at a point differs from the average of its neighbors. whereas here [this video] the intuition departs from a more stringent set up: there is a scalar-valued function $f(x,y),$ from which the Laplacian is defined as the divergence of the gradient, $\nabla\cdot\nabla f.$ The intuition is: A measure of how much of a minimum a point is in a scalar valued multivariate function. The question is where the average appears in the first interpretation - is it truly the average as understood in statistics? Because the second derivative in different directions seems to directly speak about the curvat

Cites sigma · 2018-09 · Mathematics Stack Exchange · on “Intuition of Laplacian of $f(x,y)$”

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 [this video] they gave an intuition of $\nabla^2 f$ . The main theme of this video (as I've understood) 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 i

Cites Mark S. · 2021-04 · Mathematics Stack Exchange · on “what do holomorphic functions 'look like'? Please provide references.”

Setup A holomorphic function $f$ may be constant (and every constant function is holomorphic), in which case its effect on the input plane would be to send it all to a single point, and its graph in four real dimensions would look like a translation of the input plane. Constant functions are exceptional in many ways, so for the rest of this discussion, suppose that that $f$ is not constant. Local One way to “look” at a function is to consider “local” properties that can be seen when you look at the effect on a small neighborhood of a point in the domain. Complex The derivative $f'$ exists and is also holomorphic, and is not constantly zero since $f$ is not constant, so $f'$ has countably man