The Laplacian and the Laplace Transform are two completely different things.

The laplacian is the dot product of the gradient operator with itself, the Laplace transform is an integral transform with the kernel e^(-s*t).

The relation of the Fourier transform to the Laplacian is that FT diagonalises the Laplacian.

> convert Calculus derivative equations into algebraic ones...

This trick is - AFAIR from my classes - the basis of control theory math. You convert a control system to algebra, do your work there, and in the end, you convert back to differential equations.

I remember taking Control theory classes and thinking 'Why the hell did it take me 4-5 years to get to this level of mathematical mastery.'

It was amazing to be able to just do a little bit of odd math here or there, and suddenly have the solution on how to fix something completely unrelated, like how to tilt an airplane safely or make a robot balance a pen upright or something, by modeling it. Granted, we did very simple models, but the sheer scalability and power of these techniques is what made me feel on top of the world.

You can also draw a parallel to category theory/abstract algebra, where Laplace is a functor between algebras.

It should say Laplacian in OP's post, not Laplacian Transform (if I am not mistaken). The Laplacian is a matrix of (partial) derivitives and is used for the equational conversion.

It should be "Laplace transform". This video has nothing about Laplacians at all.

Are you thinking of the Jacobian matrix? I'm not sure what you mean by equational conversion. The Laplacian matrix is from graph theory and doesn't involve derivatives. The Laplacian operator involves differentiation, but is not a matrix.

Sorry, sorry. You are right, I am thinking of the Jacobian.

What does the comment mean then?

> "Also, apparently the Laplacian Transform can be used to convert Calculus derivative equations into algebraic ones..."

That's the primary reason to use the Laplace transform, as seen in the video. A derivative x'(t) gets transformed into a product (and an initial condition), s X(s) - x(0), and similar for higher derivatives, so a differential equation transforms into an algebraic equation, which can be solved by rearranging. This video assumed the initial conditions like x(0) = 0, and its notation was quite sloppy/confusing in places, as it didn't clearly distinguish the names of the two functions, x(t) and X(s).

One more point to sort this out. Is Laplacian always that modified adjacency matrix in graph theory, or does it mean something else as well?

That's the "Laplacian matrix". "Laplacian" as a noun usually refers to the differential operator, and "Laplacian" as an adjective is attached to quite a few things as well as the natrix (mostly developed or worked on by Laplace, or based on such).

> It should say Laplacian in OP's post, not Laplacian Transform (if I am not mistaken).

Neither. It's "Laplace Transform".

> The Laplacian is a matrix of (partial) derivitives and is used for the equational conversion.

Bar the name, this is called Operational Calculus[0].

[0]: https://en.wikipedia.org/wiki/Operational_calculus

You usually don't convert back to differential equations, you convert back to algebraic equations.

In physics literature it is not uncommon to see Fourier transform with complex omega while Laplace transform is more common in engineering. It's the same thing at that point.