Fourier Transformations: Rewriting a F.T.

[Niles]

Homework Statement

Hi all.

I have the following Fourier transformation:

<br /> u(x,t) = \sqrt {\frac{2}{\pi }} \int_0^\infty {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} <br /> \over f} _s (\omega )\,e^{ - c^2 \omega ^2 t} \sin \omega x\,d\omega },<br />
where f_s is the sine-transform of a function f. I am supposed to rewrite this into the form:

<br /> u(x,t) = \frac{1}{{2c\sqrt {\pi t} }}\int_0^\infty {f(s)\left[ {\exp \left( { - \frac{{(x - s)^2 }}{{4c^2 t}}} \right) - \exp \left( { - \frac{{(x + s)^2 }}{{4c^2 t}}} \right)} \right]} \,ds<br />

I can see I have to start working with the sine transform of f. But I don't know what to write this as?

Thanks in advance,

sincerely Niles.

 

[marcusl]

Hint: look at the convolution theorem...

 

[Niles]

Ok, I have that:

<br /> \widehat u(\omega ,t) = \widehat f_s (\omega )e^{ - c^2 \omega ^2 t},<br />

where the exponential function is the Fourier transform of the heat kernel (Gauss' kernel). But I do not know how the convolution theorem works, when it is the sine-transform of a function.

If it works the regular way, then I do not get the expression above.

 

[gabbagabbahey]

If \widehat{f}_s(\omega) is the Foruier sin transform of f(x), then \widehat{f}_s(\omega)=\mathcal{F}_s[f(x)] and

\widehat{u}(\omega ,t) = \widehat{f}_s (\omega )e^{ - c^2 \omega ^2 t}=\mathcal{F}_s[f(x)]e^{ - c^2 \omega ^2 t}

Can you write e^{ - c^2 \omega ^2 t} as the Fourier transform of a function g(x)?

 

[Niles]

Hmm, I can always find the inverse Fourier sine transform of e^{ - c^2 \omega ^2 t}, but that seems tedious.

Also I can't find the function in a table of Fourier sine transforms. Is it the Gauss' kernel?

Thanks for replying so fast.

 

[gabbagabbahey]

It's the Gauss' kernel, but it is the full Fourier transform (or a cosine Fourier transform) of a function.

 

[Niles]

How can I use convolution in this context? Now I have:

<br /> \widehat{u}(\omega ,t) = \mathcal{F}_s[f(x)]\mathcal{F}[g_G(x)]<br />

where g_G is Gauss' kernel. The Fourier transform of a convolution is the product of each Fourier transformed part of the convolution - in the above expression I have both sine and regular F.T.?

 

[Niles]

Ok, so the convolution theorem does not care whether we are looking at Fourier sine, cosine or "ordinary" transforms?

Can you give me a hint of how I can rewrite the convolution to the expression in my very first post?

Thanks again.

 

[gabbagabbahey]

The convolution theorem says

\mathcal{F}[(f*g)(x)]= \sqrt{2\pi} \hat{f}(\omega)\hat{g}(\omega)

\Rightarrow u(x,t)=\frac{1}{\sqrt{2\pi}}(f*g)(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(s)g(x-s)ds

Use the fact that f must be odd for its FS to be a sine series, f(-s)=-f(s) to rewrite this as:

u(x,t)=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty} f(s)[g(x-s)-g(x+s)]ds

Then substitute your g(x).

 

[Niles]

Use the fact that f must be odd for its FS to be a sine series, f(-s)=-f(s) to rewrite this as ...

Ok, I am a little uncertain about this. Here are my thoughts:

The inverse Fourier transform (i.e. the complex form of the Fourier Integral Representation) is the analogous of the Fourier series, correct? So the inverse Fourier sine transform is the analogous of the Fourier sine representation, and hence f(s) must be odd? Am I correct about this?

And if my thoughts are correct, then the inverse Fourier cosine transformation of a function f means that f is even?

EDIT: In my book there is a theorem explaining the relationships between odd and even transforms. Perhaps that has something to do with this? The theorem is written in this PDF: http://www.math.uAlberta.ca/~thillen/math300/assign5.pdf

It is the comments after question 9 and 10 (at page 3). What is meant by restriction?

 

[marcusl]

Ok, I am a little uncertain about this. Here are my thoughts:

The inverse Fourier transform (i.e. the complex form of the Fourier Integral Representation) is the analogous of the Fourier series, correct? So the inverse Fourier sine transform is the analogous of the Fourier sine representation, and hence f(s) must be odd? Am I correct about this?

Yes.

And if my thoughts are correct, then the inverse Fourier cosine transformation of a function f means that f is even?

Yes again.

EDIT: In my book there is a theorem explaining the relationships between odd and even transforms. Perhaps that has something to do with this? The theorem is written in this PDF: http://www.math.uAlberta.ca/~thillen/math300/assign5.pdf

It is the comments after question 9 and 10 (at page 3). What is meant by restriction?

Sorry, I can't access your link.

 

[Niles]

Sorry, I can't access your link.

I think the document is accessible now (it is to me, atleast).

Thanks for taking a look at it.