What is a Laplace transform?

This question arose in this discussion .

By Dan Goodman (Dfmg2) on Wednesday, August 2, 2000 - 12:49 am :

Simon, could you tell me what a Laplace transform is? I'm reasonably well acquainted with Fourier transforms and some of their properties, but not Laplace transforms.

By Simon Judes (P2636) on Wednesday, August 2, 2000 - 11:59 pm :

I can't give you any rigor on this, but essentially, they are just a real version of Fourier transforms, i.e. the Laplace transform of a function f(x) is: F(p)=ň₀^Ą f(x)exp(-p x)dx The differences from the fourier transform are the real exponential, and the lower limit of 0, which is obviously necessary for the integral to converge for f(x)=sin(x) say.
In the same way as in the Fourier case, you can take the Laplace transform of functions of x, to get functions of p. A common notation for the Laplace transform of f(x) is L(f).
An example: the Laplace transform of f(x)=exp(k x) is ň₀^Ą exp(x(k-p)) dx=1/(k-p)
Their usefulness in solving differential equations stems from what happens when you take the laplace transform of the differential of a function.

L(df/dx)=ň₀^Ąexp(-p x) df/dx dx=pL(f)-f(x=0) (after integrating by parts) This is great, because by taking the Laplace transform of both sides of an equation, we turn a DE into an algebraic equation. We solve for L, as a function of p. Note also, that the lower limit of zero means that the initial condition required is exactly the one we usually specify, i.e. the conditions at t=0 say. You can work out similar things for higher derivatives of f.

So now, once we have got L as a function of p, we can take the inverse Laplace transform, which is the solution of the DE. There is a problem here which that there is usually no unique inverse, but fortunately in most situations, the inverse is only arbitrary up to a null function. Since I am a physicist, I am allowed to ignore those.
How to get the inverse function?

1. look it up in a table (I like this one)
2. use a Bromwich integral
3. numerical techniques

1 is best. 2 is quite nasty - it is a contour integral along an infinite vertical line from g-iĄ to g+iĄ where g is a constant chosen so as to put all the singularities of L on the left hand side of the vertical line. 3. you must have had a silly DE to start off with

For normal linear 2nd order DEs, this is a very satisfying method, because you get the complimentary function and the particular integral at the same time rather than separately.
If you want more rigor, I think you can treat the transform as a linear operator on the Hilbert space of Lebesgue square integrable functions, but I know nothing about that.

By Sean Hartnoll (Sah40) on Thursday, August 3, 2000 - 02:04 am :

You can use Fourier transformations in differential equations also. Whether Laplace or Fourier transforms are more useful tends to depend on the equation in question and the boundary/initial conditions. Laplace transforms are more "natural" for exponential processes such as diffusion of heat into a cold material whilst Fourier is more useful in systems that are "wavey".

One advantage of the Laplace transform is that
more functions have a well-defined Laplace transform, because of the negative exponential in the integrand. A disadvantage is that you lose information for t< 0.

Inverting both Fourier and Laplace transformations is not too bad if you are nimble with contour integration. Particularly in the Laplace case it tends to reduce to being a sum of residues, as long as there are no branch points.

Sean

By Simon Judes (P2636) on Thursday, August 3, 2000 - 04:04 pm :

Yes, Fourier transforms can be used too. Which one is more useful depends on what sort of initial conditions you know. If, as in the case with the two armies, you want to specify the conditions at some initial time and see what happens after that, then Laplace transforms are most useful, because of the lower limit of 0.

In the Fourier case, I think you need to work out the limit of f(x)exp(-i p x) as x tends to ±Ą. This might be useful if say you want to calculate the electromagnetic field on an axis in the region of some disturbance, an oscillating dipole perhaps. Then you know that far away from the disturbance, you can take the field to be zero.
Simon