# Homework Help: Fourier transform of differentials equation

1. Apr 5, 2010

### naspek

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

2. Apr 5, 2010

### gabbagabbahey

Since the question tells you to use Fourier transforms, you start by taking the FT of both sides of the differential equation...

3. Apr 5, 2010

### naspek

still blur...

4. Apr 5, 2010

$$F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-\frac{d^2u}{dx^2}\right]+F\left[a^2u\right]=F[f(x)][/itex] Call the FT of $f(x)$, $F[f(x)]=\tilde{f}(k)$. Call the FT of $u(x)$, $F[u(x)]=\tilde{u}(k)$. Your textbook likely tells you how to express the Fourier transform of $-\frac{d^2u}{dx^2}$ and $a^2u(x)$ in terms of $\tilde{u}(k)$....do exactly that 5. Apr 5, 2010 ### naspek i dont know how to proceed from here... [tex]F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-\frac{d^2u}{dx^2}\right]+F\left[a^2u\right]=F[f(x)]$$

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

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

6. Apr 5, 2010

### gabbagabbahey

When you take the fourier transform of $-\frac{d^2u}{dx^2}$ (That's what the $F[\ldots]$ means) you get $k^2F$, or $k^2\tilde{u}(k)$.

This gives you an algebraic (instead of differential) equation you can solve for $\tilde{u}$. To find $u$, you then take the inverse FT.

This is a very basic application of using FTs to solve DEs and I have a hard time believing that your textbook doesn't provide a similar example. You should really be in the habit of studying the relevant section of your text/notes before attempting homework problems that are based off of it.

7. Apr 6, 2010

### naspek

i got the general solution already.. howeverr.. how am i going to prove...

8. Apr 6, 2010

### naspek

i've done the inverse..

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]

9. Apr 7, 2010

### gabbagabbahey

$$\tilde{u}(k)\neq\frac{1}{k^2+a^2}$$

What happened to the FT of $f(x)$?

10. Apr 7, 2010

### naspek

accidently double post

Last edited: Apr 7, 2010
11. Apr 7, 2010

### naspek

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

12. Apr 7, 2010

### gabbagabbahey

That looks good, now use the composition rule to take the inverse FT of [tex]\hat{u}=\hat{g}\hat{f}[/itex]...

13. Apr 7, 2010

### naspek

inverse both side...
i get..

is that my general solution?

14. Apr 7, 2010

### gabbagabbahey

Does $\frac{1}{2a}\int f(x) e^{-a|x|}dx$ really represent the convolution of $f$ and $g$?

15. Apr 7, 2010

### naspek

by convolution theorem,
F^−1{F*G} = f ·g
correct?
so, i just substitute it..?
kinda confused

16. Apr 7, 2010

### naspek

ok.. i revised bout it already..

17. Apr 7, 2010

### gabbagabbahey

That's better!

However, that's just part of the general solution. You can also add any function, $u_h$, which satisfies the corresponding homogeneous ODE, $-\frac{d^2 u_h}{dx^2}+a^2u_h=0$, to the $u(x)$ you've found, and the result will still satisfy your ODE. So, you need to find that homogeneous solution and add it to what you've found above in order to obtain the true general solution.

18. Apr 7, 2010

### naspek

should i solve the integration u(x) = 1/2a ∫ f(x - y) e^-a|y| dy?

19. Apr 7, 2010

### gabbagabbahey

First worry about finding the homogeneous part of the general solution. Solve the Homogeneous equation.

20. Apr 8, 2010

### naspek

can u give me more clue.. i really can't get it..
which homogeneous equation i need to solve?

21. Apr 8, 2010

### gabbagabbahey

22. Apr 8, 2010

### naspek

is this my general solution?

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

23. Apr 8, 2010

### naspek

now.. to answer my question, u'' + a^2u = x^5
after solving everything, i manage to get this equation..

[math] 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..

24. Apr 9, 2010

### gabbagabbahey

$u(x)=\frac{1}{2a}\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)dt$ is part of your general solution. You are missing the homogeneous part.

Once you find the true general solution, you need only determine whether the integrals involved in exist/converge for your two $f(x)$'s to know whether or not a solution exists.

25. Apr 9, 2010

### naspek

this is my general solution?

$u(x)=c_1 e^{-ax}+c_2 e^{ax}+y(x)$