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).

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    $\begingroup$ 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. $\endgroup$
    – Kolja
    Jan 20, 2019 at 18:00
  • $\begingroup$ @Kolja Thank you so much! That video helped a ton with my understanding. $\endgroup$
    – Hyperion
    Jan 21, 2019 at 16:34

1 Answer 1


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 $\,F(s) = (f*g_s)(0).$

As for visualization, you can consult the Wikipedia article on convolution for illustrative examples.


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