Re: fourier transform in higher dimensions
- From: iredshift@xxxxxxxxx
- Date: 2 Apr 2006 14:00:42 -0700
Stephen Montgomery-Smith wrote:
iredshift@xxxxxxxxx wrote:
Hi,
I am studying transform methods for solving pdes and I am having
trouble seeing how transform properties for one variable generalize to
higher dimensions. For instance, i can show that:
F( u'(x) ) = (i*w)*F( u(x) )
F( u''(x) ) = (i*w)^2*F( u(x) )
or something similar, depending on how you define the fourier
transform. I can also see that in multiple dimensions:
F( laplacian(u(x1,...,xn)) ) = (I*w)^2*F( u(x1,..,xn) )
since the laplacian just gives a scalar for u: R^n->R and the transform
is linear
But what about vector functions? For instance,
is F( Grad(u(x1,..,xn)) ) = (i*w)*F( u(x1,..,xn) ) ?
If so, how can I see that this indeed should be the case.
Thank you for your help!
The Fourier transform of u(x1,...,xn) is a function of w1,...,wn. Thus
F(Grad u) = I (w1,...,wn) F(u)
and
F(Lap u) = ((Iw1)^2+...(Iwn)^2) F(u).
Stephen, thanks for your help! I am just slightly confused by the
notation. Let's say I have a function u: R^n-> R of period 1 and the
vectors:
X = (x1,..,xn)
K = (k1,...,kn)
Now,
F[ u(X) ](K) = integral{ exp(-2*pi*i*X*K) * f(X) dX }
So in this notation I write,
Lap u(X) = (d^2/dx1^2)u + ... + (d^2/dxn^2)u which is a scalar and
F[ Lap u(X) ] = (-2*pi*i*K)*(-2*pi*i*K) * integral{ exp(-2*pi*i*X*k) *
f(X) dX } =
= -4*pi^2*|K|^2 * F[ u(X) ](K)
So how would I express,
F[ Grad u(X) ] = F[ ( (d/dx1)u,...,(d/dxn)u ) ]?
Thank you for your help!
.
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