# Laplace transformations for dummies

Is there a simple explanation of what the Laplace transformation do exactly and how they work? Reading my math book has left me in a foggy haze of proofs that I don't completely understand. I'm looking for an explanation on lamans terms so that I understand what it is doing as I make these seemingly "magical" transformations. I searched the site and closest to an answer was this however it doesn't explain things for my simplistic mathematical mind.

Note: please edit tags as I don't really know what to put down for this topic

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I would suggest you to go to books.google.com and read a bit here and a bit there - different presentations suit different people, some people learn by examples, other from theory and a third group of people learn from applications in exercises. –  AD. Oct 13 '10 at 5:00
What book are you using? That might allow people to get a gauge of what alternatives would suit you best. –  user1736 Oct 13 '10 at 5:15
Fundamentals of differential equations 7th Ed by Nagle, Saff, and Snider –  KronoS Oct 13 '10 at 5:22
This is not a bad book. –  a.r. Oct 13 '10 at 5:37

There are beautiful video lessons at MIT Opencourseware. I'm particularly in love with this presentation of the Laplace transform.

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I really liked it. It's definitely very well explained –  Andy Oct 13 '10 at 14:49
Great link! Thank you. –  KronoS Oct 13 '10 at 14:56

I'm going to come at this one from left-field. In quantum mechanics, we deal with infinite dimensional vector spaces (Hilbert spaces), so I tend to think of integral transforms in those terms. For instance,

$$\int_{-\infty}^{\infty} K(x,y) f(y) dy = F(x)$$

can be thought of as

$$\mathbf{K} f = F$$

and $x$ and $y$ from the first equation are the indices of the infinite dimensional vectors and matrix (kernel) $f$, $F$, and $\mathbf{K}$. Using that interpretation, if $\mathbf{K}$ is unitary then the integral is just a changing the bases of the function (Hilbert) space. In other words, the integral can be viewed as the decomposition of original vector, $f$, in terms the new basis. For Fourier transforms the kernel is unitary, and while not true of Laplace transforms, the idea of it being a change of basis still holds. It should be noted that unlike in the finite case, in the infinite dimensional case care must be taken to ensure that the transform actually converges, but that is another problem entirely.

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The Laplace transform certainly is not unitary. –  Robin Chapman Oct 13 '10 at 13:32
You're right, in this case $K^-1 \neq K^\dagger$. Edited my answer to reflect this. –  rcollyer Oct 13 '10 at 13:49

I have been teaching the Laplace Transform to a night degree class (mature) of civil engineers. They are good students but not great mathematicians. They couldn't follow the method of how we use the Laplace Transform to solve differential equations until I told them this story:

Suppose that you come across a poem written in English of whose meaning you don't understand. However suppose that you know a French-speaking gentleman who is a master of interpreting poems. So you translate the poem into French and send it to the French gentleman. The French gentleman writes a perfectly good interpretation of the poem in French and sends this back to you where you translate it back into English and you have the meaning of the poem.

Obviously these are simple difficulties that these students are having but I still think it's a nice story.

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Poem in English = Differential Equation. Interpretation in English = Solution of Differential Equation. Translate to French = Take Laplace Transform. Poem in French (better interpreter) = Algebraic Equation (easier to solve). Interpretation in French = Laplace Transform of Solution of Differential Equation. Translate back into English = Inverse Laplace Transform –  Jp McCarthy Oct 23 '12 at 22:16
+1 pour la métaphore! –  Jean-Sébastien Oct 23 '12 at 22:34

Have also a look here - many great resources for the Laplace transform:
http://mathoverflow.net/questions/383/motivating-the-laplace-transform-definition/2141#2141

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A Laplace transform is useful for turning (constant coefficient) ordinary differential equations into algebraic equations, and partial differential equations into ordinary differential equations (though I rarely see these daisy chained together).

Let's say that you have an ordinary DE of the form

$$a y''(t) + b y'(t) + c y(t) = f(t) \quad t \gt 0$$ $$y(0)=y_0$$ $$y'(0)=p_0$$

Then the above equation becomes

$$(a s^2+b s+c) \hat{y}(s) - [a y_0 s + (a p_0 + b y_0)] = \hat{f}(s)$$

where $\hat{y}$ and $\hat{f}$ are Laplace transforms of $y$ and $f$, respectively. Note that we have converted the ODE into an algebraic equation in which we solve for $\hat{y}(s)$. We find $y(t)$ by inverse Laplace transformation, which is usually accomplished through tables, or contour integration if there is facility with that. Note that the initial conditions are built right into the equation we solve.

There are numerous examples for using Laplace transforms in PDE's. Here is a case I did in which I used LT's to solve the heat equation in two dimensions.

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