# Visualizing the Laplace Transform (for beginners)

I'm taking a differential equations course and have just been introduced to the concept of the Laplace Transform. I know its usefulness as a tool for solving IVPs, but I'm having trouble visualizing what the transform actually is. How would one explain the physical interpretation of the Laplace transform to a dummy? If possible, please refrain from using too much real/complex analysis (I only have experience in Linear Algebra, Multivariable Calculus, and Diff Eqs with some dabbling in function analysis).

• I very highly recommend to watch this lecture given by Arthur Mattuck - youtube.com/watch?v=sZ2qulI6GEk. It explains the Laplace transform in a beautiful manner, and shows the intuition behind it. Jan 20, 2019 at 18:00
• @Kolja Thank you so much! That video helped a ton with my understanding. Jan 21, 2019 at 16:34

The Laplace transform and the Fourier transform are closely related. One way to view them is to analyse what happens in convolution of a function with anexponential functions. Recall that convolution is defined as $$(f*g)(x) := \int_{-\infty}^\infty f(t)\,g(x-t) dt. \tag{1}$$ If we let $$\,g_s(x) := e^{s\,x},\,$$ a nice test function, then $$(f*g_s)(x) = \int_{-\infty}^\infty f(t)\,e^{s\,(x-t)} dt = e^{s\, x}\, (f*g_s)(0).\tag{2}$$ Now notice that $$\,(f*g_s)(0)\,$$ is just the bilateral Laplace transform of $$\,f(t).\,$$ A similar situation arises for the Fourier transform. One interpretation of this is that the linear transformation $$\, g \to f*g\,$$ has exponential eigenvectors $$\,g_s\,$$ and the associated eigenvalue is the Laplace transform $$\,F(s) = (f*g_s)(0).$$