Re: fourier transform in higher dimensions


Stephen Montgomery-Smith wrote:
iredshift@xxxxxxxxx wrote:
Stephen Montgomery-Smith wrote:

iredshift@xxxxxxxxx wrote:

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

You mean "period 1 in all n co-ordinates" because different coordinates
might have different periods.


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 ) ]?


= -2*pi*i (k1,...,kn) F[u].


Ok thanks, yes I see, if it's not period 1 in all n then the -2*pi*i
can't factor. Although I guess by a u-substitution you can bring the
factors outside the integral. (i.e. multiply by something like
1/(p1*p2*...*pn) )

So I guess it's:

F[ Grad u(X) ](k1,...,kn) = F[ ( (d/dx1)u,...,(d/dxn)u ) ](k1,...,kn) =
-2*pi*i * F[u](k1,...,kn)

i.e. where the last part if the fourier transform of u evaluated at the
vector (k1,..,kn).

But say I have some vector V = (v1,...,vn) and I do:

F[ V*Grad u(X) ] = v1*F[ (d/dx1)u ](K) + ... + v2*F[ (d/dxn)u ](K) by
linearity

So now does that = (-2*pi*i)*F[u](K)*(v1 + ... + vn)?

No, (-2*pi*i)*F(u)(K)*(k1*v1+...+kn*vn).



I guess my question comes down to whether in the n-dim case,

F[ (d/dxA)u ](K) = F[ (d/dxB)u ](K) = (-2*pi*i)*F[u](K) for 1<A<B<n
assuming the periods in the variables are 1.


No. The left hand side is

-2*pi*i kA F(u),

and similarly for the right hand side.


Thanks!



Ah, yes, thanks you are right:

F[ u(X) ](K) = integral{ exp(-2*pi*i*X*K) * f(X) dX }

so that F[ (d/dxA)u ](K) = -2*pi*i * (kA)* F[u](K)

That makes sense, thanks!

.

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