# Solving ODE using Fourier Transform

1. Apr 11, 2010

### absolute76

i have found the general solution which is,

u(x)= (C1 + C2x)e^ax + (1/2a)$$\int f(x-y) e^\left|y\right| dy is this correct?? now, i just want your help to guide me for justifying f(x)=x^5.... is that wrong if i solve the integration and just substitute the integral which is the range( infinity to negative infinity)?? thank you... #### Attached Files: • ###### Presentation1.jpg File size: 15 KB Views: 221 2. Apr 11, 2010 ### D H Staff Emeritus That doesn't look correct. Could you show the work that led you to this result? Also, since there is no one agreed-upon definition of the Fourier transform, tell us what formalism you are using. I know just enough about Fourier transforms to be dangerous. I've asked other homework helpers who are better versed than am I in Fourier transforms to jump in and help. Until then, show some work and try to think about the second part of the question. A hint for that: Are those functions square integrable? 3. Apr 11, 2010 ### absolute76 I Fourier transform both sides and I get this: u~(k) = 1/(k^2 + a^2) . f~(k) From table, 1/(k^2 + a^2) = √(∏/2) (e^-a|x|/a)----> denotes this as g~(k) u~(k) = g~(k)*f~(k) ----> applied convolution After convolution, I get 1/2a ∫f(x-y) e^-a|y| dy ; range -∞<y<∞ Is it correct up until here?? 4. Apr 11, 2010 ### absolute76 can you explain in detail of what you mean by that?? 5. Apr 11, 2010 ### gabbagabbahey You've correctly found the particular solution, $u_p(x)=\frac{1}{2a}\int_{-\infty}^{\infty}f(x-y)e^{-a|y|}dy$, (assuming $a>0$) via Fourier Transform methods. But $(c_1+c_2x)e^{ax}$ does not satisfy the homogeneous equation $-\frac{d^2 u}{dx^2}+a^2u=0$, and so it is not the correct homogeneous solution. 6. Apr 11, 2010 ### absolute76 -d^2 u/dx^2 + a^u = 0 2nd order homogeneous eq. -λ^2(e^ λx)+a^2e ^λx = 0 - λ^2+a^2=0 a^2= λ^2 a= λ ------- y=(c1+c2x)e^ax is this wrong?? 7. Apr 11, 2010 ### gabbagabbahey [tex]a^2=\lambda^2\implies a=\pm\lambda$$

You have two distinct roots, not one double root.

8. Apr 11, 2010

### absolute76

oh issit that way?? i thought

a=(λ^2)^1/2

and will give us a=λ because we cancel the 2x(1/2)??

9. Apr 11, 2010

### gabbagabbahey

No, $(-\lambda)^2=\lambda^2$. You can't only take the positive root, and then claim that it is a repeated root. It's the exact same concept as solving the equation $x^2=c$; both $x=\sqrt{c}$ and $x=-\sqrt{c}$ are solutions.

10. Apr 11, 2010

### absolute76

oh sorry! my mistake!

anyway Im aware that for the second part f(x)=x^5,

1/2a ∫(x-y)^5 e^-a|y| dy -∞<y<∞

right??

I am aware that i have to use convergence test..isn't it the same?

i have to integrate it and substitue the range?

11. Apr 11, 2010

### absolute76

please correct me if im wrong,

for f(x)=x^5

let say i take -∞<y<0 (first)

then i substitute inside---->∫(x-y)^5 e^-a|y| dy

y=0---> (x)^5 [(e^0)=1]

which give us when y=0 x^5

will this solution conclude that f(x)=x^5 is convergence??

12. Apr 11, 2010

### absolute76

If u don’t mind, can I ask one more question..

2 d²u/dx² - x du/dx + u =0

For this question, I already fourier transform both sides which give me,

u~(k)[-2k² - xik + 1]=0… but it is impossible if just u~(k)=0

do you have any idea of solving it?

13. Apr 11, 2010

### gabbagabbahey

That doesn't look like any convergence test I've ever seen.

14. Apr 11, 2010

### gabbagabbahey

$$F\left[x\frac{du}{dx}\right]\neq ikx\tilde{u}(k)$$

There is a rule that tells you how to take the FT of the product of $x^n$ with any function...use that rule.

15. Apr 11, 2010

### absolute76

u~[-2k² + 1] = i/2∏ [d/dk u~(k)]

I transfer, xu’ to the right side

It this correct??

doesnt [d/dk u~(k)] =iku~(k)

is the same?

Last edited: Apr 11, 2010
16. Apr 11, 2010

### gabbagabbahey

You're missing a factor of $k$ and you don't need the $1/2\pi$:

$$F\left[x\frac{du}{dx}\right]=i\frac{d}{dk}\left(F\left[\frac{du}{dx}\right]\right)=i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]=-k\frac{d\tilde{u}}{dk}$$

17. Apr 11, 2010

### absolute76

ok thank you!, so now i rearrange it,

u~(k)=(-k/-2k²+1)(du~/dk)
is this correct if i diffrentiate towards k on the right side?

that will give me u~(k)=2k³-4k²-k/(-2k²+1)²???

Last edited: Apr 11, 2010
18. Apr 11, 2010

### gabbagabbahey

Actually, I made an error in my previous post: