fourier transform of DE

**nameck** · April 5th 2010, 04:09 AM

hey there.. i really dont know how to start answering this question.. can someone please guide me to solve this question..

**shawsend** · April 5th 2010, 06:50 AM

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.

**nameck** · April 5th 2010, 05:14 PM

i dont know how to proceed from here...

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

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

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

**nameck** · April 5th 2010, 10:37 PM

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..

**shawsend** · April 6th 2010, 01:20 AM

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

From above, we had:

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

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 $e^{x^2}$ .

**nameck** · April 6th 2010, 06:11 AM

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?

**shawsend** · April 6th 2010, 06:38 AM

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.

**nameck** · April 6th 2010, 07:11 PM

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?

**shawsend** · April 6th 2010, 11:14 PM

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:

$u(x)=c_1 e^{-2x}+c_2 e^{2x}+1/5 \sin(x)$

**nameck** · April 7th 2010, 12:43 AM

ok.. this is my full solution...

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

am i right until there?

**nameck** · April 7th 2010, 12:51 AM

ok.. this is my full solution...
am i got it right until here?

**nameck** · April 7th 2010, 04:14 AM

continue.. i get...

**nameck** · April 7th 2010, 10:00 PM

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)?

**nameck** · April 8th 2010, 08:09 PM

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?

because, i need to do integration by parts 5 times

to solve it..

**nameck** · April 9th 2010, 07:05 AM

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

From above, we had:

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

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?

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