# Fourier Transformations: Rewriting a F.T.

1. Oct 10, 2008

### Niles

1. The problem statement, all variables and given/known data
Hi all.

I have the following Fourier transformation:

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

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

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

sincerely Niles.

2. Oct 10, 2008

### marcusl

Hint: look at the convolution theorem...

3. Oct 11, 2008

### Niles

Ok, I have that:

$$\widehat u(\omega ,t) = \widehat f_s (\omega )e^{ - c^2 \omega ^2 t},$$

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.

4. Oct 11, 2008

### 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 $g(x)$?

Last edited: Oct 11, 2008
5. Oct 11, 2008

### 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?

6. Oct 11, 2008

### gabbagabbahey

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

7. Oct 11, 2008

### Niles

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

$$\widehat{u}(\omega ,t) = \mathcal{F}_s[f(x)]\mathcal{F}[g_G(x)]$$

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.?

8. Oct 11, 2008

### 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.

9. Oct 11, 2008

### 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:

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

Last edited: Oct 11, 2008
10. Oct 11, 2008

### Niles

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?

Last edited: Oct 11, 2008
11. Oct 13, 2008

Yes.

Yes again.