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        riddles >> putnam exam (pure math) >> Fourier Transforms and DEs (M)
        
(Message started by: william wu on Dec 10^th, 2003, 10:54pm)

Title: Fourier Transforms and DEs (M)
Post by william wu on Dec 10^th, 2003, 10:54pm

Suppose we have an ordinary differential equation u'' - u = f. Using Fourier transforms, we can show that this has solution:

u(t) = (1/2) [int]_{-inf to inf} e^-|t-[tau]| f([tau])d[tau]

However, we also know that the general solution of this equation should include a solution of the homogeneous equation u'' - u = 0, and so be of the form

u(t) = c₁e^t + c₂e^-t + (1/2) [int]_{-inf to inf} e^-|t-[tau]| f([tau])d[tau]

Question: Why will methods based on the Fourier Transform not produce such a solution to this DE?

Source: Osgood, Brad

Title: Re: Fourier Transforms and DEs (M)
Post by James Fingas on Dec 16^th, 2003, 8:24am

I'm not familiar with the method you're using. Could you write specifically the steps you used? I've used the Laplace transform to solve ODEs, but when you're taking the transform, you go like this:

L( u(t) ) = U
L( u'(t) ) = sU - u(0)

This takes care of the boundary conditions (which is likely what the solution is missing).

Title: Re: Fourier Transforms and DEs (M)
Post by william wu on Jan 4^th, 2004, 10:08am

Yes, if you use Laplace Transforms I believe you get the missing exponentials thanks to those initial conditions in the Laplace Transform of the derivative. However, this question is concerned specifically with using Fourier Transforms and why they miss out on those exponentials. Here are the steps; I will still use s as the transform domain variable, and the asterisk symbol as the convolution operator:

u''(t) - u(t) = f(t)
(j2[pi]s)² U(s) - U(s) = F(s)
U(s) = F(s) / ( (j2[pi]s)² - 1)
U(s) = (1/2) F(s) [cdot] ( 2 / ( (j2[pi]s)² - 1) )
U(s) = (1/2) F(s) [cdot] ( 2 / ( - (2[pi]s)² - 1) )
U(s) = - (1/2) F(s) [cdot] ( 2 / ((2[pi]s)² + 1) )
u(t) = - (1/2) f(t) * exp{-|t|}
u(t) = - (1/2) [int]_{-inf to inf} e^-|t-[tau]| f([tau])d[tau]

In the third to last step I used the following transform pair, with [alpha] = 1:

exp(-[alpha]|t|), [alpha] > 0 [bigleftrightarrow] 2[alpha] / ([alpha]² + (2[pi]s)²)

I'm missing a negative sign somewhere and I can't seem to see where. Oh well ... the form is correct.

Incidentally this is a conceptual question; the answer can be a one-liner.