Fourier transform of differentials equation

  • Thread starter naspek
  • Start date
  • #1
naspek
181
0
hey there.. i really dont know how to start answering this question.. can someone please guide me to solve this question..

maths.jpg
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Since the question tells you to use Fourier transforms, you start by taking the FT of both sides of the differential equation...
 
  • #3
naspek
181
0
Since the question tells you to use Fourier transforms, you start by taking the FT of both sides of the differential equation...
still blur... :confused:
 
  • #4
gabbagabbahey
Homework Helper
Gold Member
5,002
7
[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)][/itex]


Call the FT of [itex]f(x)[/itex], [itex]F[f(x)]=\tilde{f}(k)[/itex]. Call the FT of [itex]u(x)[/itex], [itex]F[u(x)]=\tilde{u}(k)[/itex]. Your textbook likely tells you how to express the Fourier transform of [itex]-\frac{d^2u}{dx^2}[/itex] and [itex]a^2u(x)[/itex] in terms of [itex]\tilde{u}(k)[/itex]....do exactly that
 
  • #5
naspek
181
0
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)][/tex]

[tex]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)][/tex]

[tex]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)][/tex]
 
  • #6
gabbagabbahey
Homework Helper
Gold Member
5,002
7
When you take the fourier transform of [itex]-\frac{d^2u}{dx^2}[/itex] (That's what the [itex]F[\ldots][/itex] means) you get [itex]k^2F[/itex], or [itex]k^2\tilde{u}(k)[/itex].

This gives you an algebraic (instead of differential) equation you can solve for [itex]\tilde{u}[/itex]. To find [itex]u[/itex], 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
naspek
181
0
i got the general solution already.. howeverr.. how am i going to prove...
mathss.jpg
 
  • #8
naspek
181
0
When you take the fourier transform of [itex]-\frac{d^2u}{dx^2}[/itex] (That's what the [itex]F[\ldots][/itex] means) you get [itex]k^2F[/itex], or [itex]k^2\tilde{u}(k)[/itex].

This gives you an algebraic (instead of differential) equation you can solve for [itex]\tilde{u}[/itex]. To find [itex]u[/itex], 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.

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
gabbagabbahey
Homework Helper
Gold Member
5,002
7
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]

[tex]\tilde{u}(k)\neq\frac{1}{k^2+a^2}[/tex]

What happened to the FT of [itex]f(x)[/itex]?
 
  • #10
naspek
181
0
accidently double post
 
Last edited:
  • #11
naspek
181
0
mathsss.jpg

ok.. this is my full solution...
am i got it right until here?
 
  • #12
gabbagabbahey
Homework Helper
Gold Member
5,002
7
That looks good, now use the composition rule to take the inverse FT of [tex]\hat{u}=\hat{g}\hat{f}[/itex]...
 
  • #13
naspek
181
0
That looks good, now use the composition rule to take the inverse FT of [tex]\hat{u}=\hat{g}\hat{f}[/itex]...

inverse both side...
i get..
mathssss.jpg


is that my general solution?
 
  • #14
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Does [itex]\frac{1}{2a}\int f(x) e^{-a|x|}dx[/itex] really represent the convolution of [itex]f[/itex] and [itex]g[/itex]?
 
  • #15
naspek
181
0
That looks good, now use the composition rule to take the inverse FT of [tex]\hat{u}=\hat{g}\hat{f}[/itex]...

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

by convolution theorem,
F^−1{F*G} = f ·g
correct?
so, i just substitute it..?
kinda confused :confused:
 
  • #16
naspek
181
0
ok.. i revised bout it already..
mathsssss.jpg
 
  • #17
gabbagabbahey
Homework Helper
Gold Member
5,002
7
That's better!:smile:

However, that's just part of the general solution. You can also add any function, [itex]u_h[/itex], which satisfies the corresponding homogeneous ODE, [itex]-\frac{d^2 u_h}{dx^2}+a^2u_h=0[/itex], to the [itex]u(x)[/itex] 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
naspek
181
0
That's better!:smile:

However, that's just part of the general solution. You can also add any function, [itex]u_h[/itex], which satisfies the corresponding homogeneous ODE, [itex]-\frac{d^2 u_h}{dx^2}+a^2u_h=0[/itex], to the [itex]u(x)[/itex] 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.

should i solve the integration u(x) = 1/2a ∫ f(x - y) e^-a|y| dy?
 
  • #19
gabbagabbahey
Homework Helper
Gold Member
5,002
7
First worry about finding the homogeneous part of the general solution. Solve the Homogeneous equation.
 
  • #20
naspek
181
0
First worry about finding the homogeneous part of the general solution. Solve the Homogeneous equation.

can u give me more clue.. i really can't get it..
which homogeneous equation i need to solve?
 
  • #21
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Re-read post #17
 
  • #22
naspek
181
0
is this my general solution?

[itex]
y(x)=\frac{1}{\sqrt{2\pi}}\frac{1}{a}\sqrt{\frac{\pi}{a}}\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)dt
[/itex]
 
  • #23
naspek
181
0
now.. to answer my question, u'' + a^2u = x^5
after solving everything, i manage to get this equation..


\(\displaystyle 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
gabbagabbahey
Homework Helper
Gold Member
5,002
7
is this my general solution?

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

[itex]u(x)=\frac{1}{2a}\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)dt[/itex] 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 [itex]f(x)[/itex]'s to know whether or not a solution exists.
 
  • #25
naspek
181
0
this is my general solution?

[itex]
u(x)=c_1 e^{-ax}+c_2 e^{ax}+y(x)
[/itex]
 
  • #26
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Yup!:smile:
 
  • #27
naspek
181
0
Yup!:smile:
fuuhh!! (just for a moment..)
so.. now i know how is my general equation looks like..

still got problem here..
i'm having difficulties to solve y(x) when i substitute f(x) = x^5

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

should i solved it?
 
  • #28
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Are you asked to solve it? Or are you only asked to show that a solution exists? If it's the latter option, just show that the integral converges.
 
  • #29
naspek
181
0
Are you asked to solve it? Or are you only asked to show that a solution exists? If it's the latter option, just show that the integral converges.

i try to solve it.. but.. its kinda to long for the solution..

the question ask me to find the general soltuion first..
then find when f(x) = x^5 & e^(x^2) exist or not,
if yes, i need to justify my answer..

how am i suppose to do right now? i kinda lost here..
right now, my general solution is...
[itex]
u(x)=c_1 e^{-ax}+c_2 e^{ax}+\frac{1}{\sqrt{2\pi}}\frac{1}{a}\sqrt{\frac{\ pi}{a}}\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)dt
[/itex]

how am i going to check whether the u(x) f(x) = x^5 will converges or diverges?
 
  • #30
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Haven't you studied any convergence tests for improper integrals?
 
  • #31
naspek
181
0
should i test for e^-ax or x^5 or e^(x^2) or
[itex]
y(x)=\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)dt
[/itex]
i'm sorry coz keep on asking.. but, i really dont know which one should i use to run the test.
 
  • #32
gabbagabbahey
Homework Helper
Gold Member
5,002
7
You want to test to see if [itex]y(x)[/itex] converges for each of your two [itex]f(x)[/itex]'s. If the integral doesn't converge, then no solution exists.
 
  • #33
naspek
181
0
You want to test to see if [itex]y(x)[/itex] converges for each of your two [itex]f(x)[/itex]'s. If the integral doesn't converge, then no solution exists.
so.. i need to test the equations below?
[itex]
y(x)=\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)^{5}dt
[/itex]

and

[itex]
y(x)=\int_{-\infty}^{\infty} e^{-a|x|}f(e^{(x-t)^{2}})dt
[/itex]

correct?
 
  • #34
naspek
181
0
You want to test to see if [itex]y(x)[/itex] converges for each of your two [itex]f(x)[/itex]'s. If the integral doesn't converge, then no solution exists.

by performing the improper integral test, when f(x) = x^5, the integral will converges
when f(x) = e^(x^2), the integral will diverges..

hence.. when f(x) = x^5, the is a solution, when f(x) = e^(x^2), there is no solution..
correct?
 
  • #35
gabbagabbahey
Homework Helper
Gold Member
5,002
7
I just noticed that you have an error in your expression. Convolution tells you that [itex]y(x)=\frac{1}{2a}\int_{-\infty}^{\infty}f(x-t)e^{-a|t|}dt[/itex], not [itex]\frac{1}{2a}\int_{-\infty}^{\infty}f(x-t)e^{-a|x|}dt[/itex].
 

Suggested for: Fourier transform of differentials equation

Replies
1
Views
66
Replies
5
Views
351
  • Last Post
Replies
6
Views
442
Replies
1
Views
240
Replies
2
Views
102
Replies
6
Views
410
Replies
11
Views
604
Replies
2
Views
213
  • Last Post
Replies
6
Views
601
  • Last Post
Replies
8
Views
417
Top