# fourier transform of DE

• Apr 5th 2010, 04:09 AM
nameck
fourier transform of DE
hey there.. i really dont know how to start answering this question.. can someone please guide me to solve this question..

http://i99.photobucket.com/albums/l3...ck_5/maths.jpg
• Apr 5th 2010, 06:50 AM
shawsend
Hi. Start taking Fourier Transforms. In general:

$\mathcal{F}\left\{y^{(n)}\right\}=i^n s^n \mathcal{F}\left\{y\right\}$

I get:

\begin{aligned}
\widehat{y}&=\frac{\widehat{f}}{s^2+a^2}\\
&=\widehat{g}\widehat{f}
\end{aligned}

Now take the inverse transform by expressing the right side as a convolution.
• Apr 5th 2010, 05:14 PM
nameck
i dont know how to proceed from here...

$F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-\frac{d^2u}{dx^2}\right]+F\left[a^2u\right]=F[f(x)]$

$F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-\frac{d^2u}{dx^2}\right]+a^2F\left[u\right]=F[f(x)]$

$F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-(-k^2\tilde{u}(k,y))\right]+a^2F\left[u\right]=F[f(x)]$

• Apr 5th 2010, 10:37 PM
nameck
Quote:

Originally Posted by shawsend
Hi. Start taking Fourier Transforms. In general:

$\mathcal{F}\left\{y^{(n)}\right\}=i^n s^n \mathcal{F}\left\{y\right\}$

I get:

\begin{aligned}
\widehat{y}&=\frac{\widehat{f}}{s^2+a^2}\\
&=\widehat{g}\widehat{f}
\end{aligned}

Now take the inverse transform by expressing the right side as a convolution.

i get the general solution already.. but.. how am i going to prove..
http://i99.photobucket.com/albums/l3...k_5/mathss.jpg
• Apr 6th 2010, 01:20 AM
shawsend
First note (from a table or Mathematica):

$\mathcal{F}\left\{e^{-a|x|}\right\}=a\sqrt{2/\pi} \frac{1}{a^2+s^2}$
then:
$\mathcal{F}\left\{1/a \sqrt{\pi/a}e^{-a|x|}\right\}=\frac{1}{a^2+s^2}$

$\widehat{y}=\widehat{g}\widehat{f}$

so use the convolution theorem (look it up and review it):

$\widehat{g}(s)\widehat{f}(s)=\frac{1}{\sqrt{2\pi}} \widehat{(g\star f)}(s)$

and therefore:

$y(x)=\frac{1}{\sqrt{2\pi}}\frac{1}{a}\sqrt{\frac{\ pi}{a}}\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)dt$

and there are various versions of the transform using different coefficients so you need to check the constants I have up there. Now, look at the integral, for positive a, that's going to converge for $f$ that has an order less than $e^{kx}$. Polynomials do but not $e^{x^2}$.
• Apr 6th 2010, 06:11 AM
nameck
Quote:

Originally Posted by shawsend
Now take the inverse transform by expressing the right side as a convolution.

so.. y^ = g^f^ is not my general solution yet?
• Apr 6th 2010, 06:38 AM
shawsend
You're gonna' have to invert that expression as the y-hats represent the Fourier transforms of the function. Also, I didn't include the complimentary components of the solution arising from the homogeneous equation. Add that part to the solution for the general solution. Also, I may not have done the Fourier transform/inverse transform correctly. I'm expecting you to go over it and make sure it's correct and in the process, learn for you self how to work problems this way. You can verify everything by just solving a particular case numerically and comparing it to the solution obtained symbolically.
• Apr 6th 2010, 07:11 PM
nameck
by inverse g^, i get....

F^-1[g^(k)] = g(x) = 1/ √2∏ ∫g^(k)e^ikx dk
.............................. = 1/ √2∏ ∫ 1/ (k^2 + a^2)e^ikx dk
...............................= √2∏ [(e^-a|x|) / 2a]

correct?
so.. what's the next step?(Bow)
• Apr 6th 2010, 11:14 PM
shawsend
let's look at a particular example:

$u''-4u=-\sin(x)=g(x)$

Taking Fourier Transforms of both sides:

$\mathcal{F}\left\{u''-4u=g(x)\right\}$

$-s^2\widehat{u}-4\widehat{u}=\widehat{g}$

so:

$\widehat{u}=-\widehat{g} \frac{1}{s^2+4}$

but I know that:

$\mathcal{F}\left\{\sqrt{\pi/8} e^{-2|x|}\right\}=\frac{1}{s^2+4}$

so I let:

$\widehat{h}=\frac{1}{s^2+4}$

that means

$\widehat{u}=-\widehat{g}\widehat{h}$

I now take the inverse Fourier Transform of both sides:

$\mathcal{F}^{-1}\left\{ \widehat{u}\right\}=\mathcal{F}^{-1}\left\{-\widehat{g}\widehat{h}\right\}$

and the inversse Fourier Transform of a product of transforms is a convolution integral:

$u(x)=-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}-\sin(x-y)\sqrt{\pi/8} e^{-2|y|}dy$

$u(x)=1/4\left(\int_{-\infty}^0 \sin(x-y)e^{2y}dy+\int_0^{\infty} \sin(x-y)e^{-2y}dy\right)$

simplifying that I get:

$u(x)=1/5\sin(x)$

so that the general solution is:

$u(x)=c_1 e^{-2x}+c_2 e^{2x}+1/5 \sin(x)$
• Apr 7th 2010, 12:43 AM
nameck
ok.. this is my full solution...

[img]http://i99.photobucket.com/albums/l313/nameck_5/mathsss.jpg[\img]

am i right until there?
• Apr 7th 2010, 12:51 AM
nameck
http://i99.photobucket.com/albums/l3..._5/mathsss.jpg
ok.. this is my full solution...
am i got it right until here?
• Apr 7th 2010, 04:14 AM
nameck
• Apr 7th 2010, 10:00 PM
nameck
i your example, your g(x) is given.. but, my question, isn't given..

how am i going to solve integration of f(x - y)?
• Apr 8th 2010, 08:09 PM
nameck
now.. to answer my question, u'' + a^2u = x^5
after solving everything, i manage to get this equation..

[tex] u(x)= ∫- ∞ to 0
1/2a(x-y)^{5}e^{ay}dy+ ∫0 to ∞
(x-y)^{5}e^{-ay}dy

should i solve it to get the general solution? (Bow)
because, i need to do integration by parts 5 times (Headbang) to solve it..

• Apr 9th 2010, 07:05 AM
nameck
Quote:

Originally Posted by shawsend
First note (from a table or Mathematica):

$\mathcal{F}\left\{e^{-a|x|}\right\}=a\sqrt{2/\pi} \frac{1}{a^2+s^2}$
then:
$\mathcal{F}\left\{1/a \sqrt{\pi/a}e^{-a|x|}\right\}=\frac{1}{a^2+s^2}$

$\widehat{y}=\widehat{g}\widehat{f}$

so use the convolution theorem (look it up and review it):

$\widehat{g}(s)\widehat{f}(s)=\frac{1}{\sqrt{2\pi}} \widehat{(g\star f)}(s)$

and therefore:

$y(x)=\frac{1}{\sqrt{2\pi}}\frac{1}{a}\sqrt{\frac{\ pi}{a}}\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)dt$

and there are various versions of the transform using different coefficients so you need to check the constants I have up there. Now, look at the integral, for positive a, that's going to converge for $f$ that has an order less than $e^{kx}$. Polynomials do but not $e^{x^2}$.

should i use convergence test to justify the answer?