From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: Meaning of Laplace transform Date: 7 Sep 1999 22:52:52 -0400 Newsgroups: sci.math In article <7r434c$8ij$1@nnrp1.deja.com>, wrote: >I'm sorry I don't know how to edit my post, so I add my another >question about Laplace transform to another posting. > >I know where Laplace transform is useful (solving Diff.equ) ,,But >I can't hardly grasp what intrgral transform (in general) does tell >about the original function.(Any geometrical meaning?) > [...] There is a "geometric", actually linear-algebraic meaning, the one that makes it useful in differential equations: Take the operator of the derivative D, defined on the differentiable functions on [0, inf} that vanish at 0 and are dominated by an exponential (such functions are called Laplace transformable). Call the Laplace transform operator L. Then (integration by parts) (LDf)(s) = s * (Lf)(s) (If f does not vanish at 0, the correction is well-known.) One more piece of notation, plus the fact that L is one-to-one: S will be the operator of multiplication by s; it turns a function s |-> F(s) into a function s |-> s*F(s). Then L*D = S*L, that is, D = L^(-1) * S * L If you imagine S as a "diagonal" operator (acting similarly to diagonal matrices on vectors), then the above equation says that D (a complicated operator) has been "diagonalized" to become S, a simple-looking operator. The eigenvectors of D are the exponentials (of course, Dexp(a*t)=a*exp(a*t)), so that L sort of expands D in its system of eigenvectors. D has no claim to be symmetric - that is, on the set of transformable and differentiable functions on [0,inf), which does not even have a reasonable inner product, so one cannot expect anything near orthogonality of eigenvectors. If you are somehow familiar with Fourier Transform, there is another "apology" for L: Take a (piecewise continuous) function f, vanishing on (-inf, 0), and try to find its Fourier Transform (i.e. determine the frequency image of f). An obstacle could be that |f| does not have a finite integral. In an attempt to extract some frequency information anyway, apply a damping factor to f, i.e. consider a function x |-> exp(-u*x)*f(x) Now if f is not too fast growing, there will be values of u for which the damped function will be Fourier transformable. Then the Fourier Transform will be integral[0 to inf) exp(-i*v*x) * exp(-u*x) * f(x) dx All you have to do now is to wrap up u and v into one complex variable u + i*v = s and Laplace Transform pops out. Summary: it is a coded version of Fourier Transform applied to an exponentially damped function. Hope some of it helps, ZVK(Slavek). ============================================================================== From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: LaPlace Invariant Date: 6 Mar 1999 02:58:57 -0500 Newsgroups: sci.math In article <36E07877.3627@rochester.rr.com>, Jonathan W. Hoyle wrote: >As my Differential Equation knowledge has gotten a bit rusty lately, I >have been recently thinking about Laplace Transforms, particularly, if >f(x) and g(x) are two real valued functions over the positive reals, and >L{f(x)} = L{g(x)}, what can we say about f(x) and g(x)? > >It's certainly not true that f(x) must equal g(x), as they can differ >over a set of measure 0 without affecting the integral. But more than >that, I have the following questions I was wondering if are already >known: > >1. If L{f(x)} = L{g(x)}, and f and g are both continuous, must f = g? > Yes; it boils down (subtract right side from left side) to: If L{h(x)} = 0, and if h is continuous, must h = 0? And this follows, after a change of variable, from Weierstrass's Approximation Theorem. >2. When does L^-1{L{f(x)}} = f(x)? (Same question with L{L^-1{f(x)}}.) > Every time it's defined. >3. If L{f(x)} exists, does there always exist a continuous g(x) such >that L{f(x)} = L{g(x)}? > No. Recall that for the step function b: b(x)=1 in (0,A) and 0 otherwise, L{b(x)}(s) = (1-e^(A*s))/s, and no continuous function can equal b almost everywhere. >4. Finally, given arbitrary f(x) such that L{f(x)} exists, can one >determine the family of functions {fn(x)} such that L{fn(x)} = L{f(x)} >for all n? Too easy as stated: simply put f_n = f for all f. I wonder what the apparently more difficult question could have been. (Provided n was a subscript. If it was something else, description in words will help.) Hope it helps, ZVK(Slavek).